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Let $A B C D$ be a cyclic quadrilateral, with $A B<C D$, whose diagonals intersect at the point $F$ and $A D, B C$ intersect at the point $E$. Let also $K, L$ be the projections of $F$ onto the sides $A D, B C$ respectively, and $M, S, T$ be the midpoints of $E F, C F, D F$. Prove that the second intersection point of the circumcircles of the triangles $M K T, M L S$ lies on the side $C D$.
|
Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T, M L S$ pass through $N$.
We will prove first that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle of the triangle $E F C$, so it passes also through $Q .\left(^{*}\right)$
We will prove that
$$
\angle S L Q=\angle Q N S
$$
Indeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and
$$
\angle S L C=\angle S C L=\angle A B C
$$
In addition, since $N, S$ are the midpoints of $D C, F C$ we have that $S N / / F D$.
And finally, $Q, S$ are the midpoints of $E C, C D$, so $Q N / / E D$.
It follows that the angles $\angle E D B$ and $\angle Q N S$ have parallel sides, and since $A B<C D$, they are acute, and as a result we have that
$$
\angle E D B=\angle Q N S
$$
But, from the cyclic quadrilateral $A B C D$, we get that
$$
\angle E D B=\angle A C B
$$
Now, from (2),(3) and (4) we obtain immediately (1), so $\angle S L Q=\angle Q N S$ and the quadrilateral $L N S Q$ is cyclic. Since from $\left(^{*}\right)$ this circle passes also through $M$, we get that the points $M, L, Q, S, N$ are co-cylic and this means that the circumcircle of $M L S$ passes through $N$.
Similarly, that the circumcircle of $M K T$ passes also through $N$ and we have the desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral, with $A B<C D$, whose diagonals intersect at the point $F$ and $A D, B C$ intersect at the point $E$. Let also $K, L$ be the projections of $F$ onto the sides $A D, B C$ respectively, and $M, S, T$ be the midpoints of $E F, C F, D F$. Prove that the second intersection point of the circumcircles of the triangles $M K T, M L S$ lies on the side $C D$.
|
Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T, M L S$ pass through $N$.
We will prove first that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle of the triangle $E F C$, so it passes also through $Q .\left(^{*}\right)$
We will prove that
$$
\angle S L Q=\angle Q N S
$$
Indeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and
$$
\angle S L C=\angle S C L=\angle A B C
$$
In addition, since $N, S$ are the midpoints of $D C, F C$ we have that $S N / / F D$.
And finally, $Q, S$ are the midpoints of $E C, C D$, so $Q N / / E D$.
It follows that the angles $\angle E D B$ and $\angle Q N S$ have parallel sides, and since $A B<C D$, they are acute, and as a result we have that
$$
\angle E D B=\angle Q N S
$$
But, from the cyclic quadrilateral $A B C D$, we get that
$$
\angle E D B=\angle A C B
$$
Now, from (2),(3) and (4) we obtain immediately (1), so $\angle S L Q=\angle Q N S$ and the quadrilateral $L N S Q$ is cyclic. Since from $\left(^{*}\right)$ this circle passes also through $M$, we get that the points $M, L, Q, S, N$ are co-cylic and this means that the circumcircle of $M L S$ passes through $N$.
Similarly, that the circumcircle of $M K T$ passes also through $N$ and we have the desired.
|
{
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"problem_match": "\nG2.",
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}
|
0a5ea54e-a0db-5f48-a82a-cd54145524d3
| 605,202
|
Given that $A B C$ is a triangle where $A B<A C$. On the half-lines $B A$ and $C A$ we take points $F$ and $E$ respectively such that $B F=C E=B C$. Let $M, N$ and $H$ be the mid-points of the segments $B F, C E$ and $B C$ respectively and $K$ and $O$ be the circumcircles of the triangles $A B C$ and $M N H$ respectively. We assume that $O K$ cuts $B E$ and $H N$ at the points $A_{1}$ and $B_{1}$ respectively and that $C_{1}$ is the point of intersection of $H N$ and $F E$. If the parallel line from $A_{1}$ to $O C_{1}$ cuts the line $F E$ at $D$ and the perpendicular from $A_{1}$ to the line $D B_{1}$ cuts $F E$ at the point $M_{1}$, prove that $E$ is the orthocenter of the triangle $A_{1} O M_{1}$.
|
The circumcenter of the triangle $\triangle M N H$ coincides with the incentre of the triangle $\triangle A B C$ because the triangles $\triangle B M H$ and $\triangle N H C$ are isosceles and therefore the perpendiculars of the $M H, H N$ are also the bisectors of the angles $\angle A B C, \angle A C B$, respectively.
Let $G, I$ be the points of tangents of the incircle $(O, r)$ of the triangle $\triangle A B C$ with the sides $A B$ and $A C$ respectively. Now if $a, b, c$ are the sides of the triangle $\triangle A B C$ and $s$ the semiperimeter of the triangle , we have
$$
O F^{2}=O G^{2}+F G^{2}=r^{2}+(a-s+b)^{2}
$$
and
$$
O E^{2}=O I^{2}+E I^{2}=r^{2}+(a-s+c)^{2}
$$
Then
$$
O F^{2}-O E^{2}=\alpha(b-c)
$$
Applying two times the theorem of Stewarts at the triangles $\triangle K F B$ and $\triangle K A C$ we get
$$
F A \cdot K A^{2}+c \cdot K F^{2}=a \cdot K A^{2}+a c \cdot F A \quad \text { or } \quad K F^{2}=K A^{2}+a(a-c)
$$
and
$$
E A \cdot K E^{2}+\alpha \cdot K A^{2}=b \cdot K E^{2}+a b \cdot E A \quad \text { or } \quad K E^{2}=K A^{2}+a(b-a)
$$
From (2) and (3) we have
$$
K F^{2}-K E^{2}=a(a-c)-a(b-a)=\alpha(b-c)
$$
From (1),(4), because $O F^{2}-O E^{2}=K F^{2}-K E^{2}$ we have that $F E \perp O K$.
Let $J$ the point of intersection of $F E, O K$.
Because $A_{1} D / / O C_{1}$ we have
$$
\frac{J O}{J A_{1}}=\frac{J C_{1}}{J D}
$$
And since $A_{1} E / / H N$ we get
$$
\frac{J E}{J A_{1}}=\frac{J C_{1}}{J B_{1}}
$$
Therefore, we have
$$
\frac{J O}{J E}=\frac{J B_{1}}{J D}
$$
Thus, from the inverse of Thales theorem we have that $E O / / D B_{1}$, So
$$
A M_{1} \perp E O
$$
Consequently, the point $E$ is the orthocenter of the triangle $\triangle A_{1} O M_{1}$
Remark(PSC): Here in the solution the side $B C$ has been referred as $\alpha$ and $a$, which are equivalent.
## Number theory
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given that $A B C$ is a triangle where $A B<A C$. On the half-lines $B A$ and $C A$ we take points $F$ and $E$ respectively such that $B F=C E=B C$. Let $M, N$ and $H$ be the mid-points of the segments $B F, C E$ and $B C$ respectively and $K$ and $O$ be the circumcircles of the triangles $A B C$ and $M N H$ respectively. We assume that $O K$ cuts $B E$ and $H N$ at the points $A_{1}$ and $B_{1}$ respectively and that $C_{1}$ is the point of intersection of $H N$ and $F E$. If the parallel line from $A_{1}$ to $O C_{1}$ cuts the line $F E$ at $D$ and the perpendicular from $A_{1}$ to the line $D B_{1}$ cuts $F E$ at the point $M_{1}$, prove that $E$ is the orthocenter of the triangle $A_{1} O M_{1}$.
|
The circumcenter of the triangle $\triangle M N H$ coincides with the incentre of the triangle $\triangle A B C$ because the triangles $\triangle B M H$ and $\triangle N H C$ are isosceles and therefore the perpendiculars of the $M H, H N$ are also the bisectors of the angles $\angle A B C, \angle A C B$, respectively.
Let $G, I$ be the points of tangents of the incircle $(O, r)$ of the triangle $\triangle A B C$ with the sides $A B$ and $A C$ respectively. Now if $a, b, c$ are the sides of the triangle $\triangle A B C$ and $s$ the semiperimeter of the triangle , we have
$$
O F^{2}=O G^{2}+F G^{2}=r^{2}+(a-s+b)^{2}
$$
and
$$
O E^{2}=O I^{2}+E I^{2}=r^{2}+(a-s+c)^{2}
$$
Then
$$
O F^{2}-O E^{2}=\alpha(b-c)
$$
Applying two times the theorem of Stewarts at the triangles $\triangle K F B$ and $\triangle K A C$ we get
$$
F A \cdot K A^{2}+c \cdot K F^{2}=a \cdot K A^{2}+a c \cdot F A \quad \text { or } \quad K F^{2}=K A^{2}+a(a-c)
$$
and
$$
E A \cdot K E^{2}+\alpha \cdot K A^{2}=b \cdot K E^{2}+a b \cdot E A \quad \text { or } \quad K E^{2}=K A^{2}+a(b-a)
$$
From (2) and (3) we have
$$
K F^{2}-K E^{2}=a(a-c)-a(b-a)=\alpha(b-c)
$$
From (1),(4), because $O F^{2}-O E^{2}=K F^{2}-K E^{2}$ we have that $F E \perp O K$.
Let $J$ the point of intersection of $F E, O K$.
Because $A_{1} D / / O C_{1}$ we have
$$
\frac{J O}{J A_{1}}=\frac{J C_{1}}{J D}
$$
And since $A_{1} E / / H N$ we get
$$
\frac{J E}{J A_{1}}=\frac{J C_{1}}{J B_{1}}
$$
Therefore, we have
$$
\frac{J O}{J E}=\frac{J B_{1}}{J D}
$$
Thus, from the inverse of Thales theorem we have that $E O / / D B_{1}$, So
$$
A M_{1} \perp E O
$$
Consequently, the point $E$ is the orthocenter of the triangle $\triangle A_{1} O M_{1}$
Remark(PSC): Here in the solution the side $B C$ has been referred as $\alpha$ and $a$, which are equivalent.
## Number theory
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nG3.",
"solution_match": "\nSolution."
}
|
7135c64f-2c2f-58ab-a09c-732a84292627
| 605,215
|
Find all natural numbers $n$ for which $1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)}$ is coprime with $n$.
|
Consider the given expression $(\bmod p)$ where $p \mid n$ is a prime number. $p|n \Rightarrow p-1| \phi(n)$, thus for any $k$ that is not divisible by $p$, one has $k^{\phi(n)} \equiv 1(\bmod p)$. There are $n-\frac{n}{p}$ numbers among $1,2, \ldots, n$ that are not divisible by $p$. Therefore
$$
1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)} \equiv-\frac{n}{p} \quad(\bmod p)
$$
If the given expression is coprime with $n$, it is not divisible by $p$, so $p \nmid \frac{n}{p} \Rightarrow p^{2} X n$. This is valid for all prime divisors $p$ of $n$, thus $n$ must be square-free. On the other hand, if $n$ is square-free, one has $p^{2} \nmid n \Rightarrow p \nmid \frac{n}{p}$, hence the given expression is not divisible by $p$. Since this is valid for all prime divisors $p$ of $n$, the given two numbers are indeed coprime.
The answer is square-free integers.
|
square-free integers
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all natural numbers $n$ for which $1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)}$ is coprime with $n$.
|
Consider the given expression $(\bmod p)$ where $p \mid n$ is a prime number. $p|n \Rightarrow p-1| \phi(n)$, thus for any $k$ that is not divisible by $p$, one has $k^{\phi(n)} \equiv 1(\bmod p)$. There are $n-\frac{n}{p}$ numbers among $1,2, \ldots, n$ that are not divisible by $p$. Therefore
$$
1^{\phi(n)}+2^{\phi(n)}+\ldots+n^{\phi(n)} \equiv-\frac{n}{p} \quad(\bmod p)
$$
If the given expression is coprime with $n$, it is not divisible by $p$, so $p \nmid \frac{n}{p} \Rightarrow p^{2} X n$. This is valid for all prime divisors $p$ of $n$, thus $n$ must be square-free. On the other hand, if $n$ is square-free, one has $p^{2} \nmid n \Rightarrow p \nmid \frac{n}{p}$, hence the given expression is not divisible by $p$. Since this is valid for all prime divisors $p$ of $n$, the given two numbers are indeed coprime.
The answer is square-free integers.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN1.",
"solution_match": "\nSolution."
}
|
a8c3841b-f693-56f0-babc-ccf4b763c47d
| 605,227
|
Find all odd natural numbers $n$ such that $d(n)$ is the largest divisor of the number $n$ different from $n$ $(d(n)$ is the number of divisors of the number $n$ including 1 and $n$ ).
|
From $d(n)\left|n, \frac{n}{d(n)}\right| n$ one obtains $\frac{n}{d(n)} \leq d(n)$.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}$ where $p_{i}, 1 \leq i \leq s$ are prime numbers. The number $n$ is odd from where we get $p_{i}>2,1 \leq i \leq s$. The multiplicativity of the function $d(n)$ implies $d(n)=\left(1+\alpha_{1}\right) \ldots\left(1+\alpha_{s}\right)$.
From Bernoulli inequality, for every $p_{i}>3$ from the factorization of $n$, one obtain
$$
p_{i}^{\frac{\alpha_{i}}{2}}=\left(1+\left(p_{i}-1\right)\right)^{\frac{\alpha_{i}}{2}} \geq 1+\frac{\alpha_{i}}{2}\left(p_{i}-1\right)>1+\alpha_{i} \quad \text { and } \quad 3^{\frac{\beta}{2}} \geq 1+\beta
$$
The equality holds when $\beta=0, \beta=2$, and strict inequality for $\beta>2$. If $\beta=1$ and there is no prime number in the factorization of $n$, then $n=3, d(n)=2$, which is not a solution of the problem. If $\beta=1$ and there exist another prime in the factorization of $n$, then $n=3 p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}>4\left(1+\alpha_{2}\right)^{2} \ldots\left(1+\alpha_{s}\right)^{2}$. Indeed, if there is $p_{i} \geq 7$ in the factorization, then $3 p_{i}^{\alpha_{i}}>4\left(1+\alpha_{i}\right)^{2}$ for every natural number $\alpha_{i}$. If the power of 5 in the factorization is bigger than 1 , then $3 \cdot p_{i}^{\alpha_{i}}>4\left(1+\alpha_{i}\right)^{2}$. Hence, if the power of 5 is 1 , then $n=3 \cdot 5=15, d(n)=4$, which again is not a solution. The conclusion is that $\beta \neq 1$.
Finaly, we obtain $\sqrt{n} \geq d(n)$ and equality holds when there is no $p_{i}>3$ in the factorization of $n$ and $n=3^{2}=9$.
|
9
|
Incomplete
|
Yes
|
math-word-problem
|
Number Theory
|
Find all odd natural numbers $n$ such that $d(n)$ is the largest divisor of the number $n$ different from $n$ $(d(n)$ is the number of divisors of the number $n$ including 1 and $n$ ).
|
From $d(n)\left|n, \frac{n}{d(n)}\right| n$ one obtains $\frac{n}{d(n)} \leq d(n)$.
Let $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}$ where $p_{i}, 1 \leq i \leq s$ are prime numbers. The number $n$ is odd from where we get $p_{i}>2,1 \leq i \leq s$. The multiplicativity of the function $d(n)$ implies $d(n)=\left(1+\alpha_{1}\right) \ldots\left(1+\alpha_{s}\right)$.
From Bernoulli inequality, for every $p_{i}>3$ from the factorization of $n$, one obtain
$$
p_{i}^{\frac{\alpha_{i}}{2}}=\left(1+\left(p_{i}-1\right)\right)^{\frac{\alpha_{i}}{2}} \geq 1+\frac{\alpha_{i}}{2}\left(p_{i}-1\right)>1+\alpha_{i} \quad \text { and } \quad 3^{\frac{\beta}{2}} \geq 1+\beta
$$
The equality holds when $\beta=0, \beta=2$, and strict inequality for $\beta>2$. If $\beta=1$ and there is no prime number in the factorization of $n$, then $n=3, d(n)=2$, which is not a solution of the problem. If $\beta=1$ and there exist another prime in the factorization of $n$, then $n=3 p_{2}^{\alpha_{2}} \ldots p_{s}^{\alpha_{s}}>4\left(1+\alpha_{2}\right)^{2} \ldots\left(1+\alpha_{s}\right)^{2}$. Indeed, if there is $p_{i} \geq 7$ in the factorization, then $3 p_{i}^{\alpha_{i}}>4\left(1+\alpha_{i}\right)^{2}$ for every natural number $\alpha_{i}$. If the power of 5 in the factorization is bigger than 1 , then $3 \cdot p_{i}^{\alpha_{i}}>4\left(1+\alpha_{i}\right)^{2}$. Hence, if the power of 5 is 1 , then $n=3 \cdot 5=15, d(n)=4$, which again is not a solution. The conclusion is that $\beta \neq 1$.
Finaly, we obtain $\sqrt{n} \geq d(n)$ and equality holds when there is no $p_{i}>3$ in the factorization of $n$ and $n=3^{2}=9$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN2.",
"solution_match": "\nSolution."
}
|
d5333d14-72fd-576e-a64b-a683e47a862f
| 605,237
|
Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
|
We directly check the identity
$$
(x+y+z)^{5}-(-x+y+z)^{5}-(x-y+z)^{5}-(x+y-z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
Therefore, if integers $x, y$ and $z$ satisfy the equation from the statement, we then have
$$
(-x+y+z)^{5}+(x-y+z)^{5}+(x+y-z)^{5}=0
$$
By Fermat's theorem at least one of the parenthesis equals 0 . Let, w.l.o.g., $x=y+z$. Then the previous equation reduces to $(2 z)^{5}+(2 y)^{5}=0$, which is equivalent to $y=-z$. Therefore, the solution set of the proposed equation is
$$
(x, y, z) \in\{(0, t,-t),(t, 0,-t),(t,-t, 0): t \in \mathbb{Z}\}
$$
Remark(PSC): Due to the volume of calculations that are involved in the solution of this problem, a more simplified version can be:
Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{3}=24 x y z
$$
The solution would be similar to the one proposed in the original problem using the identity
$$
(x+y+z)^{3}-(-x+y+z)^{3}-(x-y+z)^{3}-(x+y-z)^{3}=24 x y z
$$
|
(x, y, z) \in\{(0, t,-t),(t, 0,-t),(t,-t, 0): t \in \mathbb{Z}\}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
|
We directly check the identity
$$
(x+y+z)^{5}-(-x+y+z)^{5}-(x-y+z)^{5}-(x+y-z)^{5}=80 x y z\left(x^{2}+y^{2}+z^{2}\right)
$$
Therefore, if integers $x, y$ and $z$ satisfy the equation from the statement, we then have
$$
(-x+y+z)^{5}+(x-y+z)^{5}+(x+y-z)^{5}=0
$$
By Fermat's theorem at least one of the parenthesis equals 0 . Let, w.l.o.g., $x=y+z$. Then the previous equation reduces to $(2 z)^{5}+(2 y)^{5}=0$, which is equivalent to $y=-z$. Therefore, the solution set of the proposed equation is
$$
(x, y, z) \in\{(0, t,-t),(t, 0,-t),(t,-t, 0): t \in \mathbb{Z}\}
$$
Remark(PSC): Due to the volume of calculations that are involved in the solution of this problem, a more simplified version can be:
Find all the integer solutions $(x, y, z)$ of the equation
$$
(x+y+z)^{3}=24 x y z
$$
The solution would be similar to the one proposed in the original problem using the identity
$$
(x+y+z)^{3}-(-x+y+z)^{3}-(x-y+z)^{3}-(x+y-z)^{3}=24 x y z
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN3.",
"solution_match": "\nSolution."
}
|
37000380-1c7e-5f37-adac-bddf1ad62cda
| 605,248
|
Find all monic polynomials $f$ with integer coefficients satisfying the following condition:
There exists a positive integer $N$ such that for every prime $p>N, p$ divides $2(f(p))!+1$.
|
From the divisibility relation $p \mid 2(f(p))!+1$ we conclude that:
$$
f(p)<p, \text { for all primes } p>N
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd. Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg} Q(x) \leq m-1$ and so $f(p)=p^{m}+Q(p)$. Hence for some large enough prime number $p$ holds that $f(p)>p$, which contradicts $\left({ }^{*}\right)$. Therefore we must have $\operatorname{deg} Q(x)=1$ and $Q(x)=x-a$, for some positive integer $a$. Thus the given condition becomes:
$$
p \mid 2(p-a)!+1
$$
But we have (using Wilsons theorem)
$$
\begin{gathered}
2(p-3)!\equiv-(p-3)!(p-2) \equiv-(p-2)!\equiv-1(\bmod p) \\
\Rightarrow p \mid 2(p-3)!+1
\end{gathered}
$$
From (1) and (2) we get $(p-3)!\equiv(p-a)!(\bmod p)$. Since $p-3<p$ and $p-a<p$, we conclude that $a=3$ and $f(p)=p-3$, for every prime $p>N$. Finally, since the number of all these primes is infinite we conclude that $f(x)=x-3$, for all $x$.
Remark(PSC): There is a typo in the original solution received, the part of the solution $\operatorname{deg} Q(x)=1$ and $Q(x)=x-a$ should be replaced with $\operatorname{deg} f(x)=1$ and $f(x)=x-a$.
|
f(x)=x-3
|
Yes
|
Yes
|
proof
|
Number Theory
|
Find all monic polynomials $f$ with integer coefficients satisfying the following condition:
There exists a positive integer $N$ such that for every prime $p>N, p$ divides $2(f(p))!+1$.
|
From the divisibility relation $p \mid 2(f(p))!+1$ we conclude that:
$$
f(p)<p, \text { for all primes } p>N
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd. Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg} Q(x) \leq m-1$ and so $f(p)=p^{m}+Q(p)$. Hence for some large enough prime number $p$ holds that $f(p)>p$, which contradicts $\left({ }^{*}\right)$. Therefore we must have $\operatorname{deg} Q(x)=1$ and $Q(x)=x-a$, for some positive integer $a$. Thus the given condition becomes:
$$
p \mid 2(p-a)!+1
$$
But we have (using Wilsons theorem)
$$
\begin{gathered}
2(p-3)!\equiv-(p-3)!(p-2) \equiv-(p-2)!\equiv-1(\bmod p) \\
\Rightarrow p \mid 2(p-3)!+1
\end{gathered}
$$
From (1) and (2) we get $(p-3)!\equiv(p-a)!(\bmod p)$. Since $p-3<p$ and $p-a<p$, we conclude that $a=3$ and $f(p)=p-3$, for every prime $p>N$. Finally, since the number of all these primes is infinite we conclude that $f(x)=x-3$, for all $x$.
Remark(PSC): There is a typo in the original solution received, the part of the solution $\operatorname{deg} Q(x)=1$ and $Q(x)=x-a$ should be replaced with $\operatorname{deg} f(x)=1$ and $f(x)=x-a$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN4.",
"solution_match": "\nSolution."
}
|
9faec7c7-39d4-578d-bd95-7af9c017af74
| 605,257
|
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all downhill positive integers $n$. Is it true that every downhill-integer-valued polynomial is also integer-valued?
|
No, it is not.
A downhill number can always be written as $a-b_{1}-b_{2}-\ldots-b_{9}$, where $a$ is of the form $99 \ldots 99$ and each $b_{i}$ either equals 0 or is of the form $\overline{11 \ldots 11}$.
Let $n$ be a positive integer. The numbers of the form $99 \ldots 99$ yield at most $n$ different remainders upon division by $2^{n}$, as do the numbers of the form $11 \ldots 11$. Therefore, downhill numbers yield at most $n(n+1)^{9}$ different remainders upon division by $2^{n}$.
Let $n$ be so large that $n(n+1)^{9}<2^{n}$. ( $n=63$ works: $63 \times 64^{9}<64^{10}=2^{60}<2^{63}$.) Let $0 \leq r<2^{n}$ be such that no downhill number is congruent to $r$ modulo $2^{n}$.
Consider the polynomial
$$
P(x)=\frac{1}{2 \times\left(2^{n}-1\right)!} \prod_{1 \leq i<2^{n}}(x-r+i) .
$$
We have that $P(r)=\frac{1}{2}$ is not an integer.
Let, then, $x$ be a downhill number. The number $(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ is a multiple of $\left(2^{n}-1\right)$ ! (as a product of $2^{n}-1$ consecutive integers); therefore, $2 P(x)$ is an integer. On the other hand, the number $(x-r)(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ ia a multiple of $2^{n}$ ! (as a product of $2^{n}$ consecutive integers); therefore, $2(x-r) P(x)$ is an integer multiple of $2^{n}$. Since $x$ is downhill, $x-r$ is not divisible by $2^{n}$. Therefore, $2 P(x)$ is even and $P(x)$ is an integer.
Alternative version. A positive integer $n$ is uphill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \leq a_{k-1} \leq \ldots \leq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and uphill-integer-valued if $P(n)$ is an integer for all uphill positive integers $n$. Is it true that every uphill-integer-valued polynomial is also integer-valued?
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all downhill positive integers $n$. Is it true that every downhill-integer-valued polynomial is also integer-valued?
|
No, it is not.
A downhill number can always be written as $a-b_{1}-b_{2}-\ldots-b_{9}$, where $a$ is of the form $99 \ldots 99$ and each $b_{i}$ either equals 0 or is of the form $\overline{11 \ldots 11}$.
Let $n$ be a positive integer. The numbers of the form $99 \ldots 99$ yield at most $n$ different remainders upon division by $2^{n}$, as do the numbers of the form $11 \ldots 11$. Therefore, downhill numbers yield at most $n(n+1)^{9}$ different remainders upon division by $2^{n}$.
Let $n$ be so large that $n(n+1)^{9}<2^{n}$. ( $n=63$ works: $63 \times 64^{9}<64^{10}=2^{60}<2^{63}$.) Let $0 \leq r<2^{n}$ be such that no downhill number is congruent to $r$ modulo $2^{n}$.
Consider the polynomial
$$
P(x)=\frac{1}{2 \times\left(2^{n}-1\right)!} \prod_{1 \leq i<2^{n}}(x-r+i) .
$$
We have that $P(r)=\frac{1}{2}$ is not an integer.
Let, then, $x$ be a downhill number. The number $(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ is a multiple of $\left(2^{n}-1\right)$ ! (as a product of $2^{n}-1$ consecutive integers); therefore, $2 P(x)$ is an integer. On the other hand, the number $(x-r)(x-r+1) \ldots\left(x-r+2^{n}-1\right)$ ia a multiple of $2^{n}$ ! (as a product of $2^{n}$ consecutive integers); therefore, $2(x-r) P(x)$ is an integer multiple of $2^{n}$. Since $x$ is downhill, $x-r$ is not divisible by $2^{n}$. Therefore, $2 P(x)$ is even and $P(x)$ is an integer.
Alternative version. A positive integer $n$ is uphill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \leq a_{k-1} \leq \ldots \leq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and uphill-integer-valued if $P(n)$ is an integer for all uphill positive integers $n$. Is it true that every uphill-integer-valued polynomial is also integer-valued?
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN5.",
"solution_match": "\nSolution."
}
|
a8680e2a-02f7-5c7b-a09e-76b02c03c907
| 605,272
|
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all downhill positive integers $n$. Is it true that every downhill-integer-valued polynomial is also integer-valued?
|
First we show that no uphill number is congruent to 10 modulo 11.
To this end, notice that an uphill number can always be written as $b_{1}+b_{2}+\ldots+b_{m}$, where $m \leq 9$, $b_{1} \leq b_{2} \leq \ldots \leq b_{m}$, and each $b_{i}$ is of the form $11 \ldots 11$. Since the remainder of each $b_{i}$ modulo 11 is either 0 or 1 , the remainder of $b_{1}+b_{2}+\ldots+b_{m}$ modulo 11 is at most 9 , as required.
Consider, then, the polynomial
$$
P(x)=\frac{1}{11} x(x-1) \cdots(x-9)
$$
It value is an integer at every uphill number; however, $P(10)$ is not an integer.
Remark(PSC): PSC believes the alternative version of this problem has medium as level of difficulty.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
A positive integer $n$ is downhill if its decimal representation $\overline{a_{k} a_{k-1} \ldots a_{0}}$ satisfies $a_{k} \geq a_{k-1} \geq \ldots \geq a_{0}$. A real-coefficient polynomial $P$ is integer-valued if $P(n)$ is an integer for all integer $n$, and downhill-integervalued if $P(n)$ is an integer for all downhill positive integers $n$. Is it true that every downhill-integer-valued polynomial is also integer-valued?
|
First we show that no uphill number is congruent to 10 modulo 11.
To this end, notice that an uphill number can always be written as $b_{1}+b_{2}+\ldots+b_{m}$, where $m \leq 9$, $b_{1} \leq b_{2} \leq \ldots \leq b_{m}$, and each $b_{i}$ is of the form $11 \ldots 11$. Since the remainder of each $b_{i}$ modulo 11 is either 0 or 1 , the remainder of $b_{1}+b_{2}+\ldots+b_{m}$ modulo 11 is at most 9 , as required.
Consider, then, the polynomial
$$
P(x)=\frac{1}{11} x(x-1) \cdots(x-9)
$$
It value is an integer at every uphill number; however, $P(10)$ is not an integer.
Remark(PSC): PSC believes the alternative version of this problem has medium as level of difficulty.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nN5.",
"solution_match": "\nSolution."
}
|
a8680e2a-02f7-5c7b-a09e-76b02c03c907
| 605,272
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} \leq 1 .
$$
|
First we remark that
$$
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right)
$$
Indeed
$$
\begin{aligned}
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right) & \Leftrightarrow a^{5}-a^{4} b-a b^{4}+b^{5} \geq 0 \\
& \Leftrightarrow a^{4}(a-b)-b^{4}(a-b) \geq 0 \\
(a-b)\left(a^{4}-b^{4}\right) \geq 0 & \Leftrightarrow(a-b)^{2}\left(a^{2}+b^{2}\right)(a+b) \geq 0
\end{aligned}
$$
We rewrite the inequality as
$$
\frac{1}{a^{5}+b^{5}+a b c^{3}}+\frac{1}{b^{5}+c^{5}+b c a^{3}}+\frac{1}{c^{5}+a^{5}+c a b^{3}} \leq 1 .
$$
On the other hand the following inequality is true
$$
a^{5}+b^{5}+a b c^{3} \geq a b\left(a^{3}+b^{3}+c^{3}\right)
$$
and similar for the other two.
Finally, using AM-GM we get:
$$
\begin{aligned}
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} & \leq \frac{1}{a^{3}+b^{3}+c^{3}}\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)=\frac{a+b+c}{a^{3}+b^{3}+c^{3}} \\
& \leq \frac{a+b+c}{\frac{(a+b+c)^{3}}{9}}=\frac{9}{(a+b+c)^{2}} \leq \frac{9}{(3 \sqrt[3]{a b c})^{2}}=1
\end{aligned}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} \leq 1 .
$$
|
First we remark that
$$
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right)
$$
Indeed
$$
\begin{aligned}
a^{5}+b^{5} \geq a b\left(a^{3}+b^{3}\right) & \Leftrightarrow a^{5}-a^{4} b-a b^{4}+b^{5} \geq 0 \\
& \Leftrightarrow a^{4}(a-b)-b^{4}(a-b) \geq 0 \\
(a-b)\left(a^{4}-b^{4}\right) \geq 0 & \Leftrightarrow(a-b)^{2}\left(a^{2}+b^{2}\right)(a+b) \geq 0
\end{aligned}
$$
We rewrite the inequality as
$$
\frac{1}{a^{5}+b^{5}+a b c^{3}}+\frac{1}{b^{5}+c^{5}+b c a^{3}}+\frac{1}{c^{5}+a^{5}+c a b^{3}} \leq 1 .
$$
On the other hand the following inequality is true
$$
a^{5}+b^{5}+a b c^{3} \geq a b\left(a^{3}+b^{3}+c^{3}\right)
$$
and similar for the other two.
Finally, using AM-GM we get:
$$
\begin{aligned}
\frac{1}{a^{5}+b^{5}+c^{2}}+\frac{1}{b^{5}+c^{5}+a^{2}}+\frac{1}{c^{5}+a^{5}+b^{2}} & \leq \frac{1}{a^{3}+b^{3}+c^{3}}\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)=\frac{a+b+c}{a^{3}+b^{3}+c^{3}} \\
& \leq \frac{a+b+c}{\frac{(a+b+c)^{3}}{9}}=\frac{9}{(a+b+c)^{2}} \leq \frac{9}{(3 \sqrt[3]{a b c})^{2}}=1
\end{aligned}
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## A1",
"solution_match": "\n## Solution"
}
|
50339cb5-0842-59e3-9edc-0283a5c3e00c
| 605,291
|
Consider the sequence of rational numbers defned by $x_{1}=\frac{4}{3}$ and $x_{n+1}=\frac{x_{n}^{2}}{x_{n}^{2}-x_{n}+1}, n \geq 1$.
Show that the numerator of the lowest term expression of each sum $\sum_{k=1}^{n} x_{k}$ is a perfect square.
|
It is easily seen that the $x_{n}$ are all rational numbers greater than 1 . Rewrite the recurrence formula in the form $x_{n}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{n}-1}, n \geq 1$, to get
$$
\sum_{k=1}^{n} x_{k}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{1}-1}=\frac{x_{n}^{2}-x_{n}+1}{x_{n}-1}-3=\frac{\left(x_{n}-2\right)^{2}}{x_{n}-1}
$$
Finally, express the positive rational number $x_{n}-1$ in lowest terms, $x_{n}-1=\frac{a}{b}$, to deduce that $\frac{(a-b)^{2}}{a b}$ expresses $\sum_{k=1}^{n} x_{k}$ the lowest terms.
Since $\operatorname{gcd}(a, b)=1 \Rightarrow \operatorname{gcd}(a-b, b)=1 \Rightarrow \operatorname{gcd}\left((a-b)^{2}, b\right)=1$. Similarly we can prove that $\operatorname{gcd}\left((a-b)^{2}, a\right)=1$, which implies that $\operatorname{gcd}\left((a-b)^{2}, a b\right)=1$.
The conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Consider the sequence of rational numbers defned by $x_{1}=\frac{4}{3}$ and $x_{n+1}=\frac{x_{n}^{2}}{x_{n}^{2}-x_{n}+1}, n \geq 1$.
Show that the numerator of the lowest term expression of each sum $\sum_{k=1}^{n} x_{k}$ is a perfect square.
|
It is easily seen that the $x_{n}$ are all rational numbers greater than 1 . Rewrite the recurrence formula in the form $x_{n}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{n}-1}, n \geq 1$, to get
$$
\sum_{k=1}^{n} x_{k}=\frac{1}{x_{n+1}-1}-\frac{1}{x_{1}-1}=\frac{x_{n}^{2}-x_{n}+1}{x_{n}-1}-3=\frac{\left(x_{n}-2\right)^{2}}{x_{n}-1}
$$
Finally, express the positive rational number $x_{n}-1$ in lowest terms, $x_{n}-1=\frac{a}{b}$, to deduce that $\frac{(a-b)^{2}}{a b}$ expresses $\sum_{k=1}^{n} x_{k}$ the lowest terms.
Since $\operatorname{gcd}(a, b)=1 \Rightarrow \operatorname{gcd}(a-b, b)=1 \Rightarrow \operatorname{gcd}\left((a-b)^{2}, b\right)=1$. Similarly we can prove that $\operatorname{gcd}\left((a-b)^{2}, a\right)=1$, which implies that $\operatorname{gcd}\left((a-b)^{2}, a b\right)=1$.
The conclusion follows.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## A2",
"solution_match": "\n## Solution"
}
|
afcd1472-bc80-5fba-9715-09d63df8d2c6
| 605,301
|
Find all the functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \mid f(n)+n f(m)
$$
for any $m, n \in \mathbb{N}$
|
We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function take infinite or finite values.
Case 1. The Function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have
$$
n+f(m)\left|f(n)+n f(m)=f(n)-n^{2}+n(f(m)+n) \quad \Rightarrow \quad n+f(m)\right| f(n)-n^{2}
$$
Since $n$ is a fixed natural number, then $f(n)-n^{2}$ is as well a fixed natural number, and since the above results is true for any $m$ and the function $f$ has an infinite range, we can choose $m$ such that $n+f(m)>\left|f(n)-n^{2}\right|$. This implies that $f(n)=n^{2}$ for any natural number $n$. We now check that it is a solution. Since
$$
n+f(m)=n+m^{2}
$$
and
$$
f(n)+n f(m)=n^{2}+n m^{2}=n\left(n+m^{2}\right)
$$
it is straightforward that $n+f(m) \mid f(n)+n f(m)$
Case 2. The Function has a finite range. Since the function takes finite values, then it exists a natural number $k$ such that $\mathrm{I} \leq f(n) \leq k$ for any natural number $n$. It is clear that it exists at least one natural number $s$ (where $1 \leq s \leq k$ ) such that $f(n)=s$ for infinite natural numbers $n$. Let $m, n$ be any natural numbers such that $f(m)=f(n)=s$. Using (1) we have
$$
n+s\left|s+n s=s-s^{2}+s(n+s) \Rightarrow n+s\right| s^{2}-s
$$
Since this is true for any natural number $n$ such that $f(n)=s$ and since exist infinite natural numbers $n$ such that $f(n)=s$, we can choose the natural number $n$ such that $n+s>s^{2}-s$, which implies that $s^{2}=s \Rightarrow s=1$, or in other words $f(n)=1$ for an infinite natural number n
Let's fix a random natural number $m$ and let $n$ be any natural number $f(n)=1$. Then using (1) we have
$$
n+f(m)\left|1+n f(m)=1-(f(m))^{2}+f(m)(n+f(m)) \Rightarrow n+f(m)\right|(f(m))^{2}-1
$$
Since $m$ is a fixed a random natural number, then $(f(m))^{2}-1$ is a fixed non-negative integer and since $n$ is any natural nummber such that $f(n)=1$ and since exist infinite numbers $n$ such that $f(n)=1$, we can choose the the natural number $n$ such that $n+f(m)>(f(m))^{2}-1$. This implies $f(m)=1$ for any natural number $m$. We now check that it is a solution. Since
$$
n+f(m)=n+1
$$
and
$$
f(n)+n f(m)=1+n
$$
it is straightforward that $n+f(m) \mid f(n)+n f(m)$.
So, all the functions that satisfy the given condition are $f(n)=n^{2}$ for any $n \in \mathbb{N}$ or $f(n)=1$ for any $n \in \mathbb{N}$.
|
f(n)=n^{2} \text{ or } f(n)=1
|
Yes
|
Yes
|
proof
|
Number Theory
|
Find all the functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \mid f(n)+n f(m)
$$
for any $m, n \in \mathbb{N}$
|
We will consider 2 cases, whether the range of the functions is infinite or finite or in other words the function take infinite or finite values.
Case 1. The Function has an infinite range. Let's fix a random natural number $n$ and let $m$ be any natural number. Then using (1) we have
$$
n+f(m)\left|f(n)+n f(m)=f(n)-n^{2}+n(f(m)+n) \quad \Rightarrow \quad n+f(m)\right| f(n)-n^{2}
$$
Since $n$ is a fixed natural number, then $f(n)-n^{2}$ is as well a fixed natural number, and since the above results is true for any $m$ and the function $f$ has an infinite range, we can choose $m$ such that $n+f(m)>\left|f(n)-n^{2}\right|$. This implies that $f(n)=n^{2}$ for any natural number $n$. We now check that it is a solution. Since
$$
n+f(m)=n+m^{2}
$$
and
$$
f(n)+n f(m)=n^{2}+n m^{2}=n\left(n+m^{2}\right)
$$
it is straightforward that $n+f(m) \mid f(n)+n f(m)$
Case 2. The Function has a finite range. Since the function takes finite values, then it exists a natural number $k$ such that $\mathrm{I} \leq f(n) \leq k$ for any natural number $n$. It is clear that it exists at least one natural number $s$ (where $1 \leq s \leq k$ ) such that $f(n)=s$ for infinite natural numbers $n$. Let $m, n$ be any natural numbers such that $f(m)=f(n)=s$. Using (1) we have
$$
n+s\left|s+n s=s-s^{2}+s(n+s) \Rightarrow n+s\right| s^{2}-s
$$
Since this is true for any natural number $n$ such that $f(n)=s$ and since exist infinite natural numbers $n$ such that $f(n)=s$, we can choose the natural number $n$ such that $n+s>s^{2}-s$, which implies that $s^{2}=s \Rightarrow s=1$, or in other words $f(n)=1$ for an infinite natural number n
Let's fix a random natural number $m$ and let $n$ be any natural number $f(n)=1$. Then using (1) we have
$$
n+f(m)\left|1+n f(m)=1-(f(m))^{2}+f(m)(n+f(m)) \Rightarrow n+f(m)\right|(f(m))^{2}-1
$$
Since $m$ is a fixed a random natural number, then $(f(m))^{2}-1$ is a fixed non-negative integer and since $n$ is any natural nummber such that $f(n)=1$ and since exist infinite numbers $n$ such that $f(n)=1$, we can choose the the natural number $n$ such that $n+f(m)>(f(m))^{2}-1$. This implies $f(m)=1$ for any natural number $m$. We now check that it is a solution. Since
$$
n+f(m)=n+1
$$
and
$$
f(n)+n f(m)=1+n
$$
it is straightforward that $n+f(m) \mid f(n)+n f(m)$.
So, all the functions that satisfy the given condition are $f(n)=n^{2}$ for any $n \in \mathbb{N}$ or $f(n)=1$ for any $n \in \mathbb{N}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## A3",
"solution_match": "\n## Solution"
}
|
cd7391c9-4d31-52b1-aa9a-752897c82ff8
| 605,311
|
Let $M=\left\{(a, b, c) \in \mathbb{R}^{3}: 0<a, b, c<\frac{1}{2}\right.$ with $\left.a+b+c=1\right\}$ and $f: M \rightarrow \mathbb{R}$ given as
$$
f(a, b, c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{a b c}
$$
Find the best (real) bounds $\alpha$ and $\beta$ such that
$$
f(M)=\{f(a, b, c):(a, b, c) \in M\} \subseteq[\alpha, \beta]
$$
and determine whether any of them is achievable.
|
Let $\forall(a, b, c) \in M, \alpha \leq f(a, b, c) \leq \beta$ and supose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now,
$$
\begin{aligned}
\alpha \leq f(a, b, c) \leq \beta & \Leftrightarrow \alpha a b c \leq 4(a b+b c+c a)-1 \leq \beta a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 4(a b+b c+c a)-8 a b c-1 \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 1-2(a+b+c)+4(a b+b c+c a)-8 a b c \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c
\end{aligned}
$$
For $\alpha<8$, we have
$$
(1-2 a)(1-2 b)(1-2 c) \geq 0>(\alpha-8) a b c .
$$
So $\alpha \geq 8$. But if we take $\varepsilon>0$ small and $a=b=\frac{1}{4}+\varepsilon, c=\frac{1}{2}-2 \varepsilon$, we'll have:
$$
(\alpha-8)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) \leq\left(\frac{1}{2}-2 \varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) 4 \varepsilon
$$
Taking $\varepsilon \rightarrow 0^{+}$, we get $\alpha-8 \leq 0$. So $\alpha=8$ and it can never be achieved. For the right side, note that there is a triangle whose side-lengths are $a, b, c$. For this triangle, denote $p=\frac{1}{2}$ the half-perimeter, $S$ the area and $r, R$ respectively the radius of incircle,outcircle. Using the relations $R=\frac{a b c}{4 S}$ and $S=p r$, we will have:
$$
\begin{aligned}
(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c & \Leftrightarrow(p-a)(p-b)(p-c) \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow \frac{S^{2}}{p} \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow 2 \frac{S}{p} \leq \frac{(\beta-8) a b c}{4 S} \\
& \Leftrightarrow \frac{R}{r} \geq 2(\beta-8)^{-1}
\end{aligned}
$$
Since the least value of $\frac{R}{r}$ is 2 (this is a well-known classic inequality), and it is achievable at $a=b=c=\frac{1}{3}$, we must have $\beta=9$.
Answer: $\alpha=8$ not achievable and $\beta=9$ achievable.
|
\alpha=8 \text{ not achievable and } \beta=9 \text{ achievable}
|
Yes
|
Yes
|
math-word-problem
|
Inequalities
|
Let $M=\left\{(a, b, c) \in \mathbb{R}^{3}: 0<a, b, c<\frac{1}{2}\right.$ with $\left.a+b+c=1\right\}$ and $f: M \rightarrow \mathbb{R}$ given as
$$
f(a, b, c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{a b c}
$$
Find the best (real) bounds $\alpha$ and $\beta$ such that
$$
f(M)=\{f(a, b, c):(a, b, c) \in M\} \subseteq[\alpha, \beta]
$$
and determine whether any of them is achievable.
|
Let $\forall(a, b, c) \in M, \alpha \leq f(a, b, c) \leq \beta$ and supose that there are no better bounds, i.e. $\alpha$ is the largest possible and $\beta$ is the smallest possible. Now,
$$
\begin{aligned}
\alpha \leq f(a, b, c) \leq \beta & \Leftrightarrow \alpha a b c \leq 4(a b+b c+c a)-1 \leq \beta a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 4(a b+b c+c a)-8 a b c-1 \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq 1-2(a+b+c)+4(a b+b c+c a)-8 a b c \leq(\beta-8) a b c \\
& \Leftrightarrow(\alpha-8) a b c \leq(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c
\end{aligned}
$$
For $\alpha<8$, we have
$$
(1-2 a)(1-2 b)(1-2 c) \geq 0>(\alpha-8) a b c .
$$
So $\alpha \geq 8$. But if we take $\varepsilon>0$ small and $a=b=\frac{1}{4}+\varepsilon, c=\frac{1}{2}-2 \varepsilon$, we'll have:
$$
(\alpha-8)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{4}+\varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) \leq\left(\frac{1}{2}-2 \varepsilon\right)\left(\frac{1}{2}-2 \varepsilon\right) 4 \varepsilon
$$
Taking $\varepsilon \rightarrow 0^{+}$, we get $\alpha-8 \leq 0$. So $\alpha=8$ and it can never be achieved. For the right side, note that there is a triangle whose side-lengths are $a, b, c$. For this triangle, denote $p=\frac{1}{2}$ the half-perimeter, $S$ the area and $r, R$ respectively the radius of incircle,outcircle. Using the relations $R=\frac{a b c}{4 S}$ and $S=p r$, we will have:
$$
\begin{aligned}
(1-2 a)(1-2 b)(1-2 c) \leq(\beta-8) a b c & \Leftrightarrow(p-a)(p-b)(p-c) \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow \frac{S^{2}}{p} \leq \frac{(\beta-8) a b c}{8} \\
& \Leftrightarrow 2 \frac{S}{p} \leq \frac{(\beta-8) a b c}{4 S} \\
& \Leftrightarrow \frac{R}{r} \geq 2(\beta-8)^{-1}
\end{aligned}
$$
Since the least value of $\frac{R}{r}$ is 2 (this is a well-known classic inequality), and it is achievable at $a=b=c=\frac{1}{3}$, we must have $\beta=9$.
Answer: $\alpha=8$ not achievable and $\beta=9$ achievable.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nA4",
"solution_match": "\n## Solution"
}
|
f23c2cab-98bf-5e87-a2a2-27f54f18af05
| 605,323
|
Consider integers $m \geq 2$ and $n \geq 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \ldots, P(n)$ are all perfect powers of $m$.
|
Let $a_{0}, a_{1}, \ldots, a_{n}$ be integers to be chosen later, and consider the polynomial $P(x)=\frac{1}{n!} Q(x)$ where
$$
Q(x)=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} a_{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) .
$$
Observe that for $l \in\{0,1, \ldots, n\}$ we have
$$
\begin{aligned}
P(l) & =\frac{1}{n!}(-1)^{n!}(\stackrel{n}{l}) a_{l} \prod_{\substack{0 \leq i \leq n \\
i \neq 1}}(l-i) \\
& =\frac{1}{n!}(-1)^{n-1}\binom{n}{l} a_{l} l!(-1)^{n-l}(n-l)! \\
& =a_{l}
\end{aligned}
$$
So $P(x)$ is the unique polynomial of degree at most $n$ such that $P(l)=a_{l}$. (Any two polynomials of degree at most $n$ agreeing on $n+1$ distinct values are equal.) Note in particular that
$$
\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i)=n!
$$
If $p$ is a prime dividing $n$ !, we let $r_{p}$ be maximal such that $p^{r_{p}}$ divides $\boldsymbol{n}$ !. If $p$ divides $\boldsymbol{m}$, then there is an integer $d_{p}$ such that $m^{d_{p}} \approx 0\left(\bmod p^{r_{p}}\right)$, for example $d_{p}=r_{p}$ will do. If $p$ does not divide $m$, then there is an integer $d_{p}$ such that $m^{d_{p}} \equiv 1\left(\bmod p^{r_{p}}\right)$, for example, by Euler's theorem, $d_{p}=\varphi\left(p^{r_{p}}\right)$ will do. Let $d=d_{1} d_{2} \ldots d_{p}$ and observe that for every positive integer $t$ we have $m^{\prime d} \equiv 0\left(\bmod p^{r_{p}}\right)$ if $p \mid m$ and $m^{t d} \equiv 1\left(\bmod p^{r_{p}}\right)$ if $p \nmid m$.
Now let $t_{0}, \ldots, t_{n}$ be positive integers to be chosen later and define $a_{k}=m^{t_{k} d}$. We will show that the polynomial $P(x)$ has integer coefficients. We will also show that there is an appropriate choice of $t_{0}, \ldots, t_{n}$ such that $P(x)$ has degree exactly equal to $n$.
To show that $P(x)$ has integer coefficients it is enough to show that for every $\boldsymbol{p}$ dividing $\boldsymbol{n !}$, all coefficients of $Q(x)$ are multiples of $p^{r_{p}}$. This is immediate if $p$ divides $m$ as all $a_{k}$ 's are multiples of $p^{r_{p}}$. If $p$ does not divide $m$ then we have $a_{k} \equiv 1\left(\bmod p^{r_{p}}\right)$ for every $0 \leq k \leq n$ and so by (*)
$$
Q(x) \equiv \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) \equiv n!\equiv 0\left(\bmod p^{r_{p}}\right)
$$
This shows that all coefficients of $Q(x)$ are indeed multiples of $p^{r_{p}}$. It remains to show that there is a choice of $t_{0}, \ldots, t_{n}$ guarantecing that the degree of $P(x)$ is exactly equal to $n$. One such choice is $t_{0}=2$ and $t_{1}=\ldots=t_{n}=1$. This works because if $P(x)$ had degree less than $n$, then looking at the values $P(1), \ldots, P(n)$ we would get that $P(x)$ is constant. But this is impossible as $P(0) \neq P(1)$.
Note. Looking at the coefficient of $x^{n}$ in the definition of $P(x)$ it is not difficult to see that if we fix any $t_{1}, \ldots, t_{n}$ and pick $t_{0}$ large enough we will get that this coefficient is non-zero. In particular, we can additionally guarantee that $P(0), P(1), \ldots, P(n)$ are distinct perfect powers of $m$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Consider integers $m \geq 2$ and $n \geq 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \ldots, P(n)$ are all perfect powers of $m$.
|
Let $a_{0}, a_{1}, \ldots, a_{n}$ be integers to be chosen later, and consider the polynomial $P(x)=\frac{1}{n!} Q(x)$ where
$$
Q(x)=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} a_{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) .
$$
Observe that for $l \in\{0,1, \ldots, n\}$ we have
$$
\begin{aligned}
P(l) & =\frac{1}{n!}(-1)^{n!}(\stackrel{n}{l}) a_{l} \prod_{\substack{0 \leq i \leq n \\
i \neq 1}}(l-i) \\
& =\frac{1}{n!}(-1)^{n-1}\binom{n}{l} a_{l} l!(-1)^{n-l}(n-l)! \\
& =a_{l}
\end{aligned}
$$
So $P(x)$ is the unique polynomial of degree at most $n$ such that $P(l)=a_{l}$. (Any two polynomials of degree at most $n$ agreeing on $n+1$ distinct values are equal.) Note in particular that
$$
\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i)=n!
$$
If $p$ is a prime dividing $n$ !, we let $r_{p}$ be maximal such that $p^{r_{p}}$ divides $\boldsymbol{n}$ !. If $p$ divides $\boldsymbol{m}$, then there is an integer $d_{p}$ such that $m^{d_{p}} \approx 0\left(\bmod p^{r_{p}}\right)$, for example $d_{p}=r_{p}$ will do. If $p$ does not divide $m$, then there is an integer $d_{p}$ such that $m^{d_{p}} \equiv 1\left(\bmod p^{r_{p}}\right)$, for example, by Euler's theorem, $d_{p}=\varphi\left(p^{r_{p}}\right)$ will do. Let $d=d_{1} d_{2} \ldots d_{p}$ and observe that for every positive integer $t$ we have $m^{\prime d} \equiv 0\left(\bmod p^{r_{p}}\right)$ if $p \mid m$ and $m^{t d} \equiv 1\left(\bmod p^{r_{p}}\right)$ if $p \nmid m$.
Now let $t_{0}, \ldots, t_{n}$ be positive integers to be chosen later and define $a_{k}=m^{t_{k} d}$. We will show that the polynomial $P(x)$ has integer coefficients. We will also show that there is an appropriate choice of $t_{0}, \ldots, t_{n}$ such that $P(x)$ has degree exactly equal to $n$.
To show that $P(x)$ has integer coefficients it is enough to show that for every $\boldsymbol{p}$ dividing $\boldsymbol{n !}$, all coefficients of $Q(x)$ are multiples of $p^{r_{p}}$. This is immediate if $p$ divides $m$ as all $a_{k}$ 's are multiples of $p^{r_{p}}$. If $p$ does not divide $m$ then we have $a_{k} \equiv 1\left(\bmod p^{r_{p}}\right)$ for every $0 \leq k \leq n$ and so by (*)
$$
Q(x) \equiv \sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k} \prod_{\substack{0 \leq i \leq n \\ i \neq k}}(x-i) \equiv n!\equiv 0\left(\bmod p^{r_{p}}\right)
$$
This shows that all coefficients of $Q(x)$ are indeed multiples of $p^{r_{p}}$. It remains to show that there is a choice of $t_{0}, \ldots, t_{n}$ guarantecing that the degree of $P(x)$ is exactly equal to $n$. One such choice is $t_{0}=2$ and $t_{1}=\ldots=t_{n}=1$. This works because if $P(x)$ had degree less than $n$, then looking at the values $P(1), \ldots, P(n)$ we would get that $P(x)$ is constant. But this is impossible as $P(0) \neq P(1)$.
Note. Looking at the coefficient of $x^{n}$ in the definition of $P(x)$ it is not difficult to see that if we fix any $t_{1}, \ldots, t_{n}$ and pick $t_{0}$ large enough we will get that this coefficient is non-zero. In particular, we can additionally guarantee that $P(0), P(1), \ldots, P(n)$ are distinct perfect powers of $m$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## A5",
"solution_match": "\n## Solution"
}
|
b5e45747-4398-5c22-bb27-3a5e9d5ae41d
| 605,330
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y)
$$
for all real numbers $x$ and $y$.
|
Let $P(x, y)$ be the assertion $f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y) . P(1,0)$ yields $f(0)=0$. If there exists $x_{0} \neq 0$ satisfying $f\left(x_{0}^{2}\right)=0$, then considering $P\left(x_{0}, y\right)$, we get $f\left(x_{0} y\right)=0$ for all $y \in \mathbb{R}$. In this case, since $x_{0} \neq 0$, we can write any real number $c$ in the form $x_{0} y$ for some $y$ and hence we conclude that $f(c)=0$. It is clear that the zero function satisfies the given equation. Now assume that $f\left(x^{2}\right) \neq 0$ for all $x \neq 0$.
By $P(1, y)$ we have
$$
f(1+y f(1))=f(1)+f(y)
$$
If $f(1) \neq 1$, there exists a real number $y$ satisfying $1+y f(1)=y$ which means that $f(1)=0$ which is a contradiction since $f\left(x^{2}\right) \neq 0$ for all $x \neq 0$. Therefore we get $f(1)=1$. Considering $P\left(x,-x / f\left(x^{2}\right)\right.$ ) for $x \neq 0$, we obtain that
$$
f(x)=-x f\left(-\frac{x^{2}}{f\left(x^{2}\right)}\right) \quad \forall x \in \mathbb{R} \backslash\{0\} .
$$
Replacing $x$ by $-x$ in (1), we obtain $f(x)=-f(-x)$ for all $x \neq 0$. Since $f(0)=0$, we have $f(x)=-f(-x)$ for all $x \in \mathbb{R}$. Since $f$ is odd, $P(x,-y)$ implies
$$
f\left(x-y f\left(x^{2}\right)\right)=f(x)-x f(x y)
$$
and hence by adding $P(x, y)$ and $P(x,-y)$, we get
$$
f\left(x+y f\left(x^{2}\right)\right)+f\left(x-y f\left(x^{2}\right)\right)=2 f(x) \forall x, y \in \mathbb{R}
$$
Putting $y=x / f\left(x^{2}\right)$ for $x \neq 0$ we get
$$
f(2 x)=2 f(x) \quad \forall x \in \mathbb{R}
$$
and hence we have
$$
f\left(x+y f\left(x^{2}\right)\right)+f\left(x-y f\left(x^{2}\right)\right)=f(2 x) \forall x, y \in \mathbb{R}
$$
It is clear that for any two real numbers $u$ and $v$ with $u \neq-v$, we can choose
$$
x=\frac{u+v}{2} \text { and } y=\frac{v-u}{\left.2 f\left(\frac{u+v}{2}\right)^{2}\right)}
$$
yielding
$$
f(u)+f(v)=f(u+v)
$$
Since $f$ is odd, (2) is also true for $u=-v$ and hence we obtain that
$$
f(x)+f(y)=f(x+y) \quad \forall x, y \in \mathbb{R} .
$$
Therefore, $P(x, y)$ implies
$$
f\left(y f\left(x^{2}\right)\right)=x f(x y) \quad \forall x, y \in \mathbb{R}
$$
Hence we have
$$
f\left(f\left(x^{2}\right)\right)=x f(x) \quad \forall x \in \mathbb{R}
$$
and
$$
f\left(x f\left(x^{2}\right)\right)=x f\left(x^{2}\right) \forall x \in \mathbb{R}
$$
Using (2) and (3), we get
$$
\begin{aligned}
x f(x)+y f(y)+x f(y)+y f(x) & =(x+y)(f(x)+f(y)) \\
& =f\left(f\left((x+y)^{2}\right)\right) \\
& =f\left(f\left(x^{2}+2 x y+y^{2}\right)\right) \\
& =f\left(f\left(x^{2}\right)+f\left(y^{2}\right)+f(2 x y)\right) \\
& =f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)+f(f(2 x y)) \\
& =x f(x)+y(f(y))+f(f(2 x y)) \\
& =x f(x)+y(f(y))+2 f(f(x y))
\end{aligned}
$$
and hence
$$
2 f(f(x y))=x f(y)+y f(x) \forall x, y \in \mathbb{R} .
$$
Using (5), we have
$$
2 f(f(x))=x+f(x) \quad \forall x \in \mathbb{R} .
$$
Using (3) and (6), we obtain that
$$
2 x f(x)=2 f\left(f\left(x^{2}\right)\right)=x^{2}+f\left(x^{2}\right) \forall x \in \mathbb{R} .
$$
Putting $y=f\left(x^{2}\right)$ in (5) yields
$$
2 f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(f\left(x^{2}\right)\right)+f\left(x^{2}\right) f(x) \forall x \in \mathbb{R}
$$
Using (4), we get $f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(x^{2}\right)$ and by (3) and (8) we have
$$
2 x f\left(x^{2}\right)=2 f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(f\left(x^{2}\right)\right)+f\left(x^{2}\right) f(x)=x^{2} f(x)+f\left(x^{2}\right) f(x) \forall x \in \mathbb{R}
$$
Now using (7), write $f\left(x^{2}\right)=2 x f(x)-x^{2}$ in (9) to obtain that
$$
2 x\left(2 x f(x)-x^{2}\right)=x^{2} f(x)+\left(2 x f(x)-x^{2}\right) f(x)
$$
which is equivalent to
$$
2 x(x-f(x))^{2}=0 \quad \forall x \in \mathbb{R}
$$
This shows that $f(x)=x \quad \forall x \in \mathbb{R}$ which satisfies the original equation. Therefore all solutions are $f(x)=0 \quad \forall x \in \mathbb{R}$ and $f(x)=x \quad \forall x \in \mathbb{R}$.
## NUMBER THEORY
|
f(x)=0 \quad \forall x \in \mathbb{R} \text{ and } f(x)=x \quad \forall x \in \mathbb{R}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y)
$$
for all real numbers $x$ and $y$.
|
Let $P(x, y)$ be the assertion $f\left(x+y f\left(x^{2}\right)\right)=f(x)+x f(x y) . P(1,0)$ yields $f(0)=0$. If there exists $x_{0} \neq 0$ satisfying $f\left(x_{0}^{2}\right)=0$, then considering $P\left(x_{0}, y\right)$, we get $f\left(x_{0} y\right)=0$ for all $y \in \mathbb{R}$. In this case, since $x_{0} \neq 0$, we can write any real number $c$ in the form $x_{0} y$ for some $y$ and hence we conclude that $f(c)=0$. It is clear that the zero function satisfies the given equation. Now assume that $f\left(x^{2}\right) \neq 0$ for all $x \neq 0$.
By $P(1, y)$ we have
$$
f(1+y f(1))=f(1)+f(y)
$$
If $f(1) \neq 1$, there exists a real number $y$ satisfying $1+y f(1)=y$ which means that $f(1)=0$ which is a contradiction since $f\left(x^{2}\right) \neq 0$ for all $x \neq 0$. Therefore we get $f(1)=1$. Considering $P\left(x,-x / f\left(x^{2}\right)\right.$ ) for $x \neq 0$, we obtain that
$$
f(x)=-x f\left(-\frac{x^{2}}{f\left(x^{2}\right)}\right) \quad \forall x \in \mathbb{R} \backslash\{0\} .
$$
Replacing $x$ by $-x$ in (1), we obtain $f(x)=-f(-x)$ for all $x \neq 0$. Since $f(0)=0$, we have $f(x)=-f(-x)$ for all $x \in \mathbb{R}$. Since $f$ is odd, $P(x,-y)$ implies
$$
f\left(x-y f\left(x^{2}\right)\right)=f(x)-x f(x y)
$$
and hence by adding $P(x, y)$ and $P(x,-y)$, we get
$$
f\left(x+y f\left(x^{2}\right)\right)+f\left(x-y f\left(x^{2}\right)\right)=2 f(x) \forall x, y \in \mathbb{R}
$$
Putting $y=x / f\left(x^{2}\right)$ for $x \neq 0$ we get
$$
f(2 x)=2 f(x) \quad \forall x \in \mathbb{R}
$$
and hence we have
$$
f\left(x+y f\left(x^{2}\right)\right)+f\left(x-y f\left(x^{2}\right)\right)=f(2 x) \forall x, y \in \mathbb{R}
$$
It is clear that for any two real numbers $u$ and $v$ with $u \neq-v$, we can choose
$$
x=\frac{u+v}{2} \text { and } y=\frac{v-u}{\left.2 f\left(\frac{u+v}{2}\right)^{2}\right)}
$$
yielding
$$
f(u)+f(v)=f(u+v)
$$
Since $f$ is odd, (2) is also true for $u=-v$ and hence we obtain that
$$
f(x)+f(y)=f(x+y) \quad \forall x, y \in \mathbb{R} .
$$
Therefore, $P(x, y)$ implies
$$
f\left(y f\left(x^{2}\right)\right)=x f(x y) \quad \forall x, y \in \mathbb{R}
$$
Hence we have
$$
f\left(f\left(x^{2}\right)\right)=x f(x) \quad \forall x \in \mathbb{R}
$$
and
$$
f\left(x f\left(x^{2}\right)\right)=x f\left(x^{2}\right) \forall x \in \mathbb{R}
$$
Using (2) and (3), we get
$$
\begin{aligned}
x f(x)+y f(y)+x f(y)+y f(x) & =(x+y)(f(x)+f(y)) \\
& =f\left(f\left((x+y)^{2}\right)\right) \\
& =f\left(f\left(x^{2}+2 x y+y^{2}\right)\right) \\
& =f\left(f\left(x^{2}\right)+f\left(y^{2}\right)+f(2 x y)\right) \\
& =f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)+f(f(2 x y)) \\
& =x f(x)+y(f(y))+f(f(2 x y)) \\
& =x f(x)+y(f(y))+2 f(f(x y))
\end{aligned}
$$
and hence
$$
2 f(f(x y))=x f(y)+y f(x) \forall x, y \in \mathbb{R} .
$$
Using (5), we have
$$
2 f(f(x))=x+f(x) \quad \forall x \in \mathbb{R} .
$$
Using (3) and (6), we obtain that
$$
2 x f(x)=2 f\left(f\left(x^{2}\right)\right)=x^{2}+f\left(x^{2}\right) \forall x \in \mathbb{R} .
$$
Putting $y=f\left(x^{2}\right)$ in (5) yields
$$
2 f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(f\left(x^{2}\right)\right)+f\left(x^{2}\right) f(x) \forall x \in \mathbb{R}
$$
Using (4), we get $f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(x^{2}\right)$ and by (3) and (8) we have
$$
2 x f\left(x^{2}\right)=2 f\left(f\left(x f\left(x^{2}\right)\right)\right)=x f\left(f\left(x^{2}\right)\right)+f\left(x^{2}\right) f(x)=x^{2} f(x)+f\left(x^{2}\right) f(x) \forall x \in \mathbb{R}
$$
Now using (7), write $f\left(x^{2}\right)=2 x f(x)-x^{2}$ in (9) to obtain that
$$
2 x\left(2 x f(x)-x^{2}\right)=x^{2} f(x)+\left(2 x f(x)-x^{2}\right) f(x)
$$
which is equivalent to
$$
2 x(x-f(x))^{2}=0 \quad \forall x \in \mathbb{R}
$$
This shows that $f(x)=x \quad \forall x \in \mathbb{R}$ which satisfies the original equation. Therefore all solutions are $f(x)=0 \quad \forall x \in \mathbb{R}$ and $f(x)=x \quad \forall x \in \mathbb{R}$.
## NUMBER THEORY
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nA6",
"solution_match": "\n## Solution"
}
|
8963e995-c48c-5048-8c76-1e238c4b5b35
| 605,337
|
Find all pairs $(x, y)$ of positive integers such that
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2}
$$
|
Let $d=(x, y)$ be the greatest common divisor of positive integers $x$ and $y$.
So, $x=a d, y=b d$, where $d \in \mathbb{N},(a, b)=1, a, b \in \mathbb{N}$. We have
$$
\begin{aligned}
x^{3}+y^{3}=x^{2}+42 x y+y^{2} & \Leftrightarrow d^{3}\left(a^{3}+b^{3}\right)=d^{2}\left(a^{2}+42 a b+b^{2}\right) \\
& \Leftrightarrow d(a+b)\left(a^{2}-a b+b^{2}\right)=a^{2}+42 a b+b^{2} \\
& \Leftrightarrow(d a+d b-1)\left(a^{2}-a b+b^{2}\right)=43 a b
\end{aligned}
$$
If we denote $c=d a+d b-1 \in \mathbb{N}$, then the equality $a^{2} c-a b c+b^{2} c=43 a b$ implies the relations
$$
\begin{aligned}
\left.\begin{array}{rl}
b\left|c a^{2} \Rightarrow b\right| c \\
a\left|c b^{2} \Rightarrow a\right| c
\end{array}\right\} & \Rightarrow(a b) \mid c \\
& \Leftrightarrow c=m a b, m \in \mathbb{N}^{+} \\
& \Rightarrow m\left(a^{2}-a b+b^{2}\right)=43 \\
& \Rightarrow\left(a^{2}-a b+b^{2}\right) \mid 43 \\
& \Leftrightarrow a^{2}-a b+b^{2}=1 \text { or } a^{2}-a b+b^{2}=43 .
\end{aligned}
$$
If $a^{2}-a b+b^{2}=1$, then $(a-b)^{2}=1-a b \geq 0 \Rightarrow a=b=1,2 d=44,(x, y)=(22,22)$.
If $a^{2}-a b+b^{2}=43$, then, by virtue of simmetry, we suppose that $x \geq y \Rightarrow a \geq b$. We obtain that $43=a^{2}-a b+b^{2} \geq a b \geq b^{2} \Rightarrow b \in\{1,2,3,4,5,6\}$.
If $b=1$, then $a=7, d=1,(x, y)=(7,1)$ or $(x, y)=(1,7)$.
If $b=6$, then $a=7, d=\frac{43}{13} \notin \mathbb{N}$.
For $b \in\{2,3,4,5\}$ there no positive integer solutions for $a$.
Finally, we have $(x, y) \in\{(1,7),(7,1),(22,22)\}$.
|
(1,7),(7,1),(22,22)
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all pairs $(x, y)$ of positive integers such that
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2}
$$
|
Let $d=(x, y)$ be the greatest common divisor of positive integers $x$ and $y$.
So, $x=a d, y=b d$, where $d \in \mathbb{N},(a, b)=1, a, b \in \mathbb{N}$. We have
$$
\begin{aligned}
x^{3}+y^{3}=x^{2}+42 x y+y^{2} & \Leftrightarrow d^{3}\left(a^{3}+b^{3}\right)=d^{2}\left(a^{2}+42 a b+b^{2}\right) \\
& \Leftrightarrow d(a+b)\left(a^{2}-a b+b^{2}\right)=a^{2}+42 a b+b^{2} \\
& \Leftrightarrow(d a+d b-1)\left(a^{2}-a b+b^{2}\right)=43 a b
\end{aligned}
$$
If we denote $c=d a+d b-1 \in \mathbb{N}$, then the equality $a^{2} c-a b c+b^{2} c=43 a b$ implies the relations
$$
\begin{aligned}
\left.\begin{array}{rl}
b\left|c a^{2} \Rightarrow b\right| c \\
a\left|c b^{2} \Rightarrow a\right| c
\end{array}\right\} & \Rightarrow(a b) \mid c \\
& \Leftrightarrow c=m a b, m \in \mathbb{N}^{+} \\
& \Rightarrow m\left(a^{2}-a b+b^{2}\right)=43 \\
& \Rightarrow\left(a^{2}-a b+b^{2}\right) \mid 43 \\
& \Leftrightarrow a^{2}-a b+b^{2}=1 \text { or } a^{2}-a b+b^{2}=43 .
\end{aligned}
$$
If $a^{2}-a b+b^{2}=1$, then $(a-b)^{2}=1-a b \geq 0 \Rightarrow a=b=1,2 d=44,(x, y)=(22,22)$.
If $a^{2}-a b+b^{2}=43$, then, by virtue of simmetry, we suppose that $x \geq y \Rightarrow a \geq b$. We obtain that $43=a^{2}-a b+b^{2} \geq a b \geq b^{2} \Rightarrow b \in\{1,2,3,4,5,6\}$.
If $b=1$, then $a=7, d=1,(x, y)=(7,1)$ or $(x, y)=(1,7)$.
If $b=6$, then $a=7, d=\frac{43}{13} \notin \mathbb{N}$.
For $b \in\{2,3,4,5\}$ there no positive integer solutions for $a$.
Finally, we have $(x, y) \in\{(1,7),(7,1),(22,22)\}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## NT1",
"solution_match": "\n## Solution"
}
|
a40d1008-c3fe-5e2b-9814-edb2c568220b
| 605,349
|
Find all functions $f: \mathbf{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that the number $x f(x)+f^{2}(y)+2 x f(y)$ is a perfect square for all positive integers $x, y$.
|
Let $p$ be a prime number. Then for $x=y=p$ the given condition gives us that the number $f^{2}(p)+3 p f(p)$ is a perfect square. Then, $f^{2}(p)+3 p f(p)=k^{2}$ for some positive integer $k$. Completing the square gives us that $(2 f(p)+3 p)^{2}-9 p^{2}=4 k^{2}$, or
$$
(2 f(p)+3 p-2 k)(2 f(p)+3 p+2 k)=9 p^{2} .
$$
Since $2 f(p)+3 p+3 k>3 p$, we have the following 4 cases.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 9 p } \\
{ 2 f ( p ) + 3 p - 2 k = p }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=p^{2} \\
2 f(p)+3 p-2 k=9
\end{array}\right.\right. \text { or } \\
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 3 p ^ { 2 } } \\
{ 2 f ( p ) + 3 p - 2 k = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=9 p^{2} \\
2 f(p)+3 p-2 k=1
\end{array}\right.\right.
\end{aligned}
$$
Solving the systems, we have the following cases for $f(p)$.
$$
f(p)=p \text { or } f(p)=\left(\frac{p-3}{2}\right)^{2} \text { or } f(p)=\frac{3 p^{2}-6 p-3}{4} \text { or } f(p)=\left(\frac{3 p-1}{2}\right)^{2} .
$$
In all cases, we see that $f(p)$ can be arbitrary large whenever $p$ grows.
Now fix a positive integer $x$. From the given condition we have that
$$
(f(y)+x)^{2}+x f(x)-x^{2}
$$
is a perfect square. Since for $y$ being a prime, let $y=q, f(q)$ can be arbitrary large and $x f(x)-x^{2}$ is fixed, it means that $x f(x)-x^{2}$ should be zero, since the difference of $(f(q)+x+1)^{2}$ and $(f(q)+x)^{2}$ can be arbitrary large.
After all, we conclude that $x f(x)=x^{2}$, so $f(x)=x$, which clearly satisfies the given condition.
|
f(x)=x
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all functions $f: \mathbf{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that the number $x f(x)+f^{2}(y)+2 x f(y)$ is a perfect square for all positive integers $x, y$.
|
Let $p$ be a prime number. Then for $x=y=p$ the given condition gives us that the number $f^{2}(p)+3 p f(p)$ is a perfect square. Then, $f^{2}(p)+3 p f(p)=k^{2}$ for some positive integer $k$. Completing the square gives us that $(2 f(p)+3 p)^{2}-9 p^{2}=4 k^{2}$, or
$$
(2 f(p)+3 p-2 k)(2 f(p)+3 p+2 k)=9 p^{2} .
$$
Since $2 f(p)+3 p+3 k>3 p$, we have the following 4 cases.
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 9 p } \\
{ 2 f ( p ) + 3 p - 2 k = p }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=p^{2} \\
2 f(p)+3 p-2 k=9
\end{array}\right.\right. \text { or } \\
& \left\{\begin{array} { l }
{ 2 f ( p ) + 3 p + 2 k = 3 p ^ { 2 } } \\
{ 2 f ( p ) + 3 p - 2 k = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
2 f(p)+3 p+2 k=9 p^{2} \\
2 f(p)+3 p-2 k=1
\end{array}\right.\right.
\end{aligned}
$$
Solving the systems, we have the following cases for $f(p)$.
$$
f(p)=p \text { or } f(p)=\left(\frac{p-3}{2}\right)^{2} \text { or } f(p)=\frac{3 p^{2}-6 p-3}{4} \text { or } f(p)=\left(\frac{3 p-1}{2}\right)^{2} .
$$
In all cases, we see that $f(p)$ can be arbitrary large whenever $p$ grows.
Now fix a positive integer $x$. From the given condition we have that
$$
(f(y)+x)^{2}+x f(x)-x^{2}
$$
is a perfect square. Since for $y$ being a prime, let $y=q, f(q)$ can be arbitrary large and $x f(x)-x^{2}$ is fixed, it means that $x f(x)-x^{2}$ should be zero, since the difference of $(f(q)+x+1)^{2}$ and $(f(q)+x)^{2}$ can be arbitrary large.
After all, we conclude that $x f(x)=x^{2}$, so $f(x)=x$, which clearly satisfies the given condition.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## NT2",
"solution_match": "\n## Solution"
}
|
e097b60f-4835-5b50-b223-f08a6c9aa17f
| 605,360
|
Prove that for all positive integer $n$, there is a positive integer $m$, that $7^{n} \mid 3^{m}+5^{m}-1$.
|
We prove this by induction on $\boldsymbol{n}$. The case $\boldsymbol{n}=1$ is indeed trivial for $\boldsymbol{m}=1$. Assume that the statement of the problem holds true for $n$, and we have $3^{m}+5^{m}-1=7^{n} l$ for some positive integer $l$ which is not divisible by 7 (if not we are done). Since $3^{6} \equiv 1(\bmod 7)$ and $5^{6} \equiv 1(\bmod 7)$ we conclude that,
$$
3^{6 \cdot 7^{n-1}} \equiv 1\left(\bmod 7^{n}\right), 5^{6 \cdot 7^{n-1}} \equiv 1\left(\bmod 7^{n}\right) .
$$
Since
$$
v_{7}\left(3^{6.7^{n-1}}-1\right)=v_{7}\left(3^{6}-1\right)+v_{7}\left(7^{n-1}\right)=n \text { and } v_{7}\left(5^{6.7^{n-1}}-1\right)=v_{7}\left(5^{6}-1\right)+v_{7}\left(7^{n-1}\right)=n
$$
Thus we can say that: $3^{6.7^{n-1}}=1+7^{n} r, 5^{6.7^{n-1}}=1+7^{n} s$ for some positive integers $r, s$. We find the reminder of $r, s$ module 7 . Note that:
$$
\frac{y^{7^{k}}-1}{7^{k+1}}=\frac{y-1}{7} \cdot \frac{1+y+\ldots+y^{6}}{7} \cdot \ldots \cdot \frac{1+y^{7^{x-1}}+\ldots+y^{67^{k-1}}}{7}
$$
We use the above identity for $y=3^{6}, 5^{6}$. Note that in both cases $y=1(\bmod 7)$. Now we use the following lemma
Lemma. Let $p$ be an odd prime such that $p \mid a-1$ then $\frac{a^{p}-1}{a-1} \equiv p\left(\bmod p^{2}\right)$.
Proof. Take $a-1=b$, then $p \mid b$ now
$$
\frac{a^{p}-1}{a-1}=\frac{(b+1)^{p}-1}{b}=b^{p-1}+\ldots+\binom{p}{2} b+p \equiv p\left(\bmod p^{2}\right)
$$
since all the binomial coefficients is divisible by $p$. So our proof is complete.
Then by use of the lemma repeatedly we find that all the terms of the above identity ( ${ }^{*}$ ) except the first term is congruent to 1 modulo 7 . Thus we can find that :
$$
\frac{y^{7^{k}}-1}{7^{k+1}} \equiv \frac{y-1}{7}(\bmod 7)
$$
Since
$$
\frac{3^{6}-1}{7}=104 \equiv-1, \frac{5^{6}-1}{7} \equiv 2232 \equiv-1(\bmod 7)
$$
we find that $r \equiv s \equiv-I(\bmod 7)$, and by use of binomial theorem, we can easily find that
$$
3^{6 l \cdot 7^{n-1}}=1+7^{n} r t\left(\bmod 7^{n+1}\right), \quad 5^{6 i \cdot 7^{n-1}}=1+7^{n} s t\left(\bmod 7^{n+1}\right)
$$
for all positive integers $t$.
Now take $\boldsymbol{m}+\boldsymbol{t r} \cdot 7^{\boldsymbol{n - 1}}$ instead of $\boldsymbol{m}$,(while we will specify the number $t$ later) we can find that:
$$
3^{m+6 t \cdot 7^{n-1}}+5^{m+6 r \cdot 7^{n-1}}-1=3^{m} \cdot 3^{6 r} \cdot 7^{n-1}+5^{m} \cdot 5^{61 \cdot 7^{n-1}}-1
$$
Taking modulo $7^{n+1}$ we can find that, the above expression is reduced to
$$
\begin{aligned}
3^{m}\left(1+7^{n} r t\right)+5^{m}\left(1+7^{n} s t\right)-1 & \equiv 3^{m}+5^{m}-1+5^{m} \cdot 7^{n} s t+3^{m} \cdot 7^{n} r t \\
& \equiv 7^{n}\left(l+\left(5^{m} s+3^{m} r\right) t\right)\left(\bmod 7^{n+1}\right)
\end{aligned}
$$
Now, the problem reduced to finding a positive integer $t$ such that
$$
l+\left(5^{m} s+3^{m} \cdot r\right) t \equiv 0(\bmod 7)
$$
since
$$
5^{m} s+3^{m} r \equiv-5^{m}-3^{m} \equiv 1(\bmod 7) .
$$
Since $3^{m}+5^{m}-1 \equiv 0(\bmod 7)$ whence, we find that $\operatorname{gcd}\left(5^{m} s+3^{m} r, 7\right)=1$. Thus such integer $t$ exists, so we are done!
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that for all positive integer $n$, there is a positive integer $m$, that $7^{n} \mid 3^{m}+5^{m}-1$.
|
We prove this by induction on $\boldsymbol{n}$. The case $\boldsymbol{n}=1$ is indeed trivial for $\boldsymbol{m}=1$. Assume that the statement of the problem holds true for $n$, and we have $3^{m}+5^{m}-1=7^{n} l$ for some positive integer $l$ which is not divisible by 7 (if not we are done). Since $3^{6} \equiv 1(\bmod 7)$ and $5^{6} \equiv 1(\bmod 7)$ we conclude that,
$$
3^{6 \cdot 7^{n-1}} \equiv 1\left(\bmod 7^{n}\right), 5^{6 \cdot 7^{n-1}} \equiv 1\left(\bmod 7^{n}\right) .
$$
Since
$$
v_{7}\left(3^{6.7^{n-1}}-1\right)=v_{7}\left(3^{6}-1\right)+v_{7}\left(7^{n-1}\right)=n \text { and } v_{7}\left(5^{6.7^{n-1}}-1\right)=v_{7}\left(5^{6}-1\right)+v_{7}\left(7^{n-1}\right)=n
$$
Thus we can say that: $3^{6.7^{n-1}}=1+7^{n} r, 5^{6.7^{n-1}}=1+7^{n} s$ for some positive integers $r, s$. We find the reminder of $r, s$ module 7 . Note that:
$$
\frac{y^{7^{k}}-1}{7^{k+1}}=\frac{y-1}{7} \cdot \frac{1+y+\ldots+y^{6}}{7} \cdot \ldots \cdot \frac{1+y^{7^{x-1}}+\ldots+y^{67^{k-1}}}{7}
$$
We use the above identity for $y=3^{6}, 5^{6}$. Note that in both cases $y=1(\bmod 7)$. Now we use the following lemma
Lemma. Let $p$ be an odd prime such that $p \mid a-1$ then $\frac{a^{p}-1}{a-1} \equiv p\left(\bmod p^{2}\right)$.
Proof. Take $a-1=b$, then $p \mid b$ now
$$
\frac{a^{p}-1}{a-1}=\frac{(b+1)^{p}-1}{b}=b^{p-1}+\ldots+\binom{p}{2} b+p \equiv p\left(\bmod p^{2}\right)
$$
since all the binomial coefficients is divisible by $p$. So our proof is complete.
Then by use of the lemma repeatedly we find that all the terms of the above identity ( ${ }^{*}$ ) except the first term is congruent to 1 modulo 7 . Thus we can find that :
$$
\frac{y^{7^{k}}-1}{7^{k+1}} \equiv \frac{y-1}{7}(\bmod 7)
$$
Since
$$
\frac{3^{6}-1}{7}=104 \equiv-1, \frac{5^{6}-1}{7} \equiv 2232 \equiv-1(\bmod 7)
$$
we find that $r \equiv s \equiv-I(\bmod 7)$, and by use of binomial theorem, we can easily find that
$$
3^{6 l \cdot 7^{n-1}}=1+7^{n} r t\left(\bmod 7^{n+1}\right), \quad 5^{6 i \cdot 7^{n-1}}=1+7^{n} s t\left(\bmod 7^{n+1}\right)
$$
for all positive integers $t$.
Now take $\boldsymbol{m}+\boldsymbol{t r} \cdot 7^{\boldsymbol{n - 1}}$ instead of $\boldsymbol{m}$,(while we will specify the number $t$ later) we can find that:
$$
3^{m+6 t \cdot 7^{n-1}}+5^{m+6 r \cdot 7^{n-1}}-1=3^{m} \cdot 3^{6 r} \cdot 7^{n-1}+5^{m} \cdot 5^{61 \cdot 7^{n-1}}-1
$$
Taking modulo $7^{n+1}$ we can find that, the above expression is reduced to
$$
\begin{aligned}
3^{m}\left(1+7^{n} r t\right)+5^{m}\left(1+7^{n} s t\right)-1 & \equiv 3^{m}+5^{m}-1+5^{m} \cdot 7^{n} s t+3^{m} \cdot 7^{n} r t \\
& \equiv 7^{n}\left(l+\left(5^{m} s+3^{m} r\right) t\right)\left(\bmod 7^{n+1}\right)
\end{aligned}
$$
Now, the problem reduced to finding a positive integer $t$ such that
$$
l+\left(5^{m} s+3^{m} \cdot r\right) t \equiv 0(\bmod 7)
$$
since
$$
5^{m} s+3^{m} r \equiv-5^{m}-3^{m} \equiv 1(\bmod 7) .
$$
Since $3^{m}+5^{m}-1 \equiv 0(\bmod 7)$ whence, we find that $\operatorname{gcd}\left(5^{m} s+3^{m} r, 7\right)=1$. Thus such integer $t$ exists, so we are done!
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## NT3",
"solution_match": "\n## Solution"
}
|
beb64582-ba15-5e9c-a31f-9e9ec3cf168f
| 605,371
|
Find all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$.
|
If $y=1$, then $2 x \mid x^{2} \Leftrightarrow x=2 n, n \in \mathbb{N}$. So, the pairs $(x, y)=(2 n, 1), n \in \mathbb{N}$ satisfy the required divizibility.
Let $y>1$ such, that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. There exist $m \in \mathbb{N}$ such that
$$
x^{2}=m\left(2 x y^{2}-y^{3}+1\right), \text { e.t. } x^{2}-2 m y^{2} x+\left(m y^{3}-m\right)=0
$$
The discriminant of last quadratic equation is equal to $\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m$. Denote
$$
A=4 m\left(y^{2}-1\right)+(y-1)^{2}, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}
$$
For $y>1, y \in \mathbb{N}$ and $m \in \mathbb{N}$ we have
$$
A>0, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}>2\left(y^{2}+1\right)-(y+1)^{2}=(y-1)^{2} \geq 0 \Rightarrow B>0
$$
We obtain the following estimations for the discriminant $\Delta$ :
$$
\begin{aligned}
& \Delta+A=\left(2 m y^{2}-y+1\right)^{2} \geq 0 \Rightarrow \Delta<\left(2 m y^{2}-y+1\right)^{2} ; \\
& \Delta-B=\left(2 m y^{2}-y-1\right)^{2} \geq 0 \Rightarrow \Delta>\left(2 m y^{2}-y-1\right)^{2} .
\end{aligned}
$$
Because the discriminant $\Delta$ must be a perfect square, we obtain the equalities:
$$
\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m=\left(2 m y^{2}-y\right)^{2} \Leftrightarrow y^{2}=4 m \Rightarrow y=2 k, k \in \mathbb{N}, m=k^{2}, k \in \mathbb{N}
$$
The equation $x^{2}-8 k^{4} x+k\left(8 k^{4}-k\right)=0$ has the solutions $x=k$ and $x=8 k^{4}-k$, where $k \in \mathbb{N}$.
Finaly, we obtain that all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$, are equal to $(x, y) \in\left\{(2 k, 1),(k, 2 k),\left(8 k^{4}-k, 2 k\right) \mid k \in \mathbb{N}\right\}$.
|
(x, y) \in\left\{(2 k, 1),(k, 2 k),\left(8 k^{4}-k, 2 k\right) \mid k \in \mathbb{N}\right\}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$.
|
If $y=1$, then $2 x \mid x^{2} \Leftrightarrow x=2 n, n \in \mathbb{N}$. So, the pairs $(x, y)=(2 n, 1), n \in \mathbb{N}$ satisfy the required divizibility.
Let $y>1$ such, that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$. There exist $m \in \mathbb{N}$ such that
$$
x^{2}=m\left(2 x y^{2}-y^{3}+1\right), \text { e.t. } x^{2}-2 m y^{2} x+\left(m y^{3}-m\right)=0
$$
The discriminant of last quadratic equation is equal to $\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m$. Denote
$$
A=4 m\left(y^{2}-1\right)+(y-1)^{2}, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}
$$
For $y>1, y \in \mathbb{N}$ and $m \in \mathbb{N}$ we have
$$
A>0, \quad B=4 m\left(y^{2}+1\right)-(y+1)^{2}>2\left(y^{2}+1\right)-(y+1)^{2}=(y-1)^{2} \geq 0 \Rightarrow B>0
$$
We obtain the following estimations for the discriminant $\Delta$ :
$$
\begin{aligned}
& \Delta+A=\left(2 m y^{2}-y+1\right)^{2} \geq 0 \Rightarrow \Delta<\left(2 m y^{2}-y+1\right)^{2} ; \\
& \Delta-B=\left(2 m y^{2}-y-1\right)^{2} \geq 0 \Rightarrow \Delta>\left(2 m y^{2}-y-1\right)^{2} .
\end{aligned}
$$
Because the discriminant $\Delta$ must be a perfect square, we obtain the equalities:
$$
\Delta=4 m^{2} y^{4}-4 m y^{3}+4 m=\left(2 m y^{2}-y\right)^{2} \Leftrightarrow y^{2}=4 m \Rightarrow y=2 k, k \in \mathbb{N}, m=k^{2}, k \in \mathbb{N}
$$
The equation $x^{2}-8 k^{4} x+k\left(8 k^{4}-k\right)=0$ has the solutions $x=k$ and $x=8 k^{4}-k$, where $k \in \mathbb{N}$.
Finaly, we obtain that all pairs of positive integers $(x, y)$, such that $x^{2}$ is divisible by $2 x y^{2}-y^{3}+1$, are equal to $(x, y) \in\left\{(2 k, 1),(k, 2 k),\left(8 k^{4}-k, 2 k\right) \mid k \in \mathbb{N}\right\}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## NT4",
"solution_match": "\n## Solution"
}
|
bc479a28-f032-5ee8-884c-6b33422ecf03
| 605,383
|
Given a positive odd integer $n$, show that the arithmetic mean of fractional parts $\left\{\frac{k^{2 n}}{p}\right\}$, $k=1, \ldots, \frac{p-1}{2}$, is the same for infinitely many primes $p$.
|
We show that the arithmetic mean in question is $\frac{1}{2}$ for infinitely many primes congruent to 1 modulo 4.
Notice that $\left\{\frac{k^{2 n}}{p}\right\}=\frac{r_{k}}{p}$, where $r_{k}$ is the remainder $k^{2 n}$ leaves upon division by $p$. Clearly, the $r_{k}$ are quadratic residues modulo $p$.
If $p$ is prime, and $p-1$ and $n$ are relatively prime, then the $r_{k}, k=1, \ldots, \frac{p-1}{2}$, are pairwise distinct, since the $k^{2 n}, k=1, \ldots, \frac{p-1}{2}$, are pairwise distinct modulo $p$, by Fermat's little theorem. In this case, the $r_{k}, k=1, \ldots, \frac{p-1}{2}$, form the set $R$ of all $\frac{p-1}{2}$ quadratic residues modulo $p$ in the range 1 through $p-1$.
If, in addition, $p$ is congruent to 1 modulo 4 , then -1 is a quadratic residue modulo $p$, and the assignment $r \mapsto p-r, r \in R$, defines a permutation of $R$. In this case,
$$
\sum_{r \in R} r=\sum_{r \in R}(p-r)=\frac{p(p-1)}{2}-\sum_{r \in R} r
$$
so
$$
\sum_{r \in R} r=\frac{p(p-1)}{4},
$$
and the arithmetic mean in question is $\frac{1}{2}$.
Finally, since $n$ is odd, infinitely many primes congruent to 1 modulo 4 are also congruent to 2 modulo $n$, by Dirichlet's theorem on arithmetic sequences of integers; for such a prime $p$, the numbers $p-1$ and $n$ are clearly relatively prime. This completes the proof.
## GEOMETRY
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Given a positive odd integer $n$, show that the arithmetic mean of fractional parts $\left\{\frac{k^{2 n}}{p}\right\}$, $k=1, \ldots, \frac{p-1}{2}$, is the same for infinitely many primes $p$.
|
We show that the arithmetic mean in question is $\frac{1}{2}$ for infinitely many primes congruent to 1 modulo 4.
Notice that $\left\{\frac{k^{2 n}}{p}\right\}=\frac{r_{k}}{p}$, where $r_{k}$ is the remainder $k^{2 n}$ leaves upon division by $p$. Clearly, the $r_{k}$ are quadratic residues modulo $p$.
If $p$ is prime, and $p-1$ and $n$ are relatively prime, then the $r_{k}, k=1, \ldots, \frac{p-1}{2}$, are pairwise distinct, since the $k^{2 n}, k=1, \ldots, \frac{p-1}{2}$, are pairwise distinct modulo $p$, by Fermat's little theorem. In this case, the $r_{k}, k=1, \ldots, \frac{p-1}{2}$, form the set $R$ of all $\frac{p-1}{2}$ quadratic residues modulo $p$ in the range 1 through $p-1$.
If, in addition, $p$ is congruent to 1 modulo 4 , then -1 is a quadratic residue modulo $p$, and the assignment $r \mapsto p-r, r \in R$, defines a permutation of $R$. In this case,
$$
\sum_{r \in R} r=\sum_{r \in R}(p-r)=\frac{p(p-1)}{2}-\sum_{r \in R} r
$$
so
$$
\sum_{r \in R} r=\frac{p(p-1)}{4},
$$
and the arithmetic mean in question is $\frac{1}{2}$.
Finally, since $n$ is odd, infinitely many primes congruent to 1 modulo 4 are also congruent to 2 modulo $n$, by Dirichlet's theorem on arithmetic sequences of integers; for such a prime $p$, the numbers $p-1$ and $n$ are clearly relatively prime. This completes the proof.
## GEOMETRY
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## NT5",
"solution_match": "\n## Solution"
}
|
84bded69-459c-51e8-921c-c8c4953b6d5a
| 605,393
|
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$.
|
1
Let $H$ be the ortocenter of $A B C$ and $K, L, M$ be the feet of perpendiculars respectively from $A, B, C$ to their opposite sides of $A B C$. Also let $D$ be the intersection point of lines $B E$ and $C F$. From power of point we have
$$
B A \cdot B M=B C \cdot B K
$$
and
$$
C A \cdot C L=C B \cdot C K
$$
Adding (1) and (2) we have:
$$
C A \cdot C L+B A \cdot B M=B C \cdot B K+C B \cdot C K=B C(B K+C K)=B C^{2}
$$
Combining (3) with the problem statement $B C^{2}=B A \cdot B F+C E \cdot C A$ we have:
$$
\begin{aligned}
& B A \cdot B F-B A \cdot B M=C A \cdot C L-C E \cdot C A \\
& B A(B F-B M)=C A(C L-C E) \\
& B A \cdot F M=C A \cdot L E \\
& \quad \frac{L E}{F M}=\frac{A B}{A C}=\frac{B L}{C M}
\end{aligned}
$$
Where the last equality follows from $\triangle A M C \sim \triangle A L B$. Now since $\frac{L E}{F M}=\frac{B L}{C M}$ and $\angle F M C=\angle E L B=90^{\circ}$ we get that triangles $\triangle F M C \sim \triangle E L B$. From this similarity we get
$$
\measuredangle A E D=\measuredangle A E B=\angle L E B=\measuredangle M F C=180^{\circ}-\angle A F C=180-\angle A F D,
$$
meaning points $A, D, E, F$ are concylic.
Since both pairs $\{E, F\}$ and $\{M, L\}$ satisfy the problem condition, we must have this fixed point we are looking for is the second intersection of the circumcircles around $A F D E$ and AMHL. Let this point be $X$. We now prove that $X$ is fixed on the circumcircle of $A M H L$ (which would imply $X$ is fixed).
From the concylicity we have
$$
\angle X L E=180^{\circ}-\measuredangle X L A=\measuredangle X M A=\angle X M F \text { and } \measuredangle X E L=\measuredangle X E A=180-\measuredangle X F A=\measuredangle X F M
$$
and from here we get $\triangle X L E \sim \triangle X M F$. This similarity gives us
$$
\frac{X L}{X M}=\frac{L E}{M F} .
$$
Now combining (4) and (5) we get $\frac{X L}{X M}=\frac{A B}{A C}$ which is a fixed quantity. Since points $M, L$, the circumcircle of $A M L$, and ratio $\frac{X L}{X M}$ are fixed, this implies that point $X$ is fixed.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$.
|
1
Let $H$ be the ortocenter of $A B C$ and $K, L, M$ be the feet of perpendiculars respectively from $A, B, C$ to their opposite sides of $A B C$. Also let $D$ be the intersection point of lines $B E$ and $C F$. From power of point we have
$$
B A \cdot B M=B C \cdot B K
$$
and
$$
C A \cdot C L=C B \cdot C K
$$
Adding (1) and (2) we have:
$$
C A \cdot C L+B A \cdot B M=B C \cdot B K+C B \cdot C K=B C(B K+C K)=B C^{2}
$$
Combining (3) with the problem statement $B C^{2}=B A \cdot B F+C E \cdot C A$ we have:
$$
\begin{aligned}
& B A \cdot B F-B A \cdot B M=C A \cdot C L-C E \cdot C A \\
& B A(B F-B M)=C A(C L-C E) \\
& B A \cdot F M=C A \cdot L E \\
& \quad \frac{L E}{F M}=\frac{A B}{A C}=\frac{B L}{C M}
\end{aligned}
$$
Where the last equality follows from $\triangle A M C \sim \triangle A L B$. Now since $\frac{L E}{F M}=\frac{B L}{C M}$ and $\angle F M C=\angle E L B=90^{\circ}$ we get that triangles $\triangle F M C \sim \triangle E L B$. From this similarity we get
$$
\measuredangle A E D=\measuredangle A E B=\angle L E B=\measuredangle M F C=180^{\circ}-\angle A F C=180-\angle A F D,
$$
meaning points $A, D, E, F$ are concylic.
Since both pairs $\{E, F\}$ and $\{M, L\}$ satisfy the problem condition, we must have this fixed point we are looking for is the second intersection of the circumcircles around $A F D E$ and AMHL. Let this point be $X$. We now prove that $X$ is fixed on the circumcircle of $A M H L$ (which would imply $X$ is fixed).
From the concylicity we have
$$
\angle X L E=180^{\circ}-\measuredangle X L A=\measuredangle X M A=\angle X M F \text { and } \measuredangle X E L=\measuredangle X E A=180-\measuredangle X F A=\measuredangle X F M
$$
and from here we get $\triangle X L E \sim \triangle X M F$. This similarity gives us
$$
\frac{X L}{X M}=\frac{L E}{M F} .
$$
Now combining (4) and (5) we get $\frac{X L}{X M}=\frac{A B}{A C}$ which is a fixed quantity. Since points $M, L$, the circumcircle of $A M L$, and ratio $\frac{X L}{X M}$ are fixed, this implies that point $X$ is fixed.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G1",
"solution_match": "\n## Solution"
}
|
3aa178da-fdfa-5e63-8f0d-7fe30dd8a643
| 605,400
|
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$.
|
2
Let the $D$ be the intersection of $B E$ and $C F$ and let circumcircle of triangle $C F A$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B F \cdot B A
$$
Combining (6) with the problem statement we get
$$
B C^{2}=B A \cdot B F+C E \cdot C A=B G \cdot B C+C E \cdot C A
$$
and from here we get
$$
C E \cdot C A=B C(B C-B G)=B C \cdot C G
$$
(7) implies that $E, A, B, G$ are concyclic as well.
This gives us
$$
\measuredangle G A C=\measuredangle G A E=\measuredangle G B E=\measuredangle C B D
$$
and
$$
\measuredangle B A C-\measuredangle G A C=\measuredangle G A B=\measuredangle G A F=\measuredangle G C F=\measuredangle B C D .
$$
Adding these two equalities gives us
$$
\measuredangle B A C=\measuredangle C B D+\measuredangle B C D=180^{\circ}-\measuredangle B D C .
$$
This implies that $A, E, D, F$ are concyclic. Now let the second intersection of the circumcircles of $B D C$ and $A F D E$ be $X$. We have
$$
\angle X A B=\angle X A F=\angle X D F=180^{\circ}-\angle X D C=\angle X B C
$$
and
$$
\measuredangle X A C=\angle X A E=180^{\circ}-\measuredangle X D E=\angle X D B=\measuredangle X C B
$$
(8) and (9) imply that $B C$ is tangent to the circumcircles of $\triangle X A B$ and $\triangle X A C$ respectively. Let $A X$, the radical axis of the two circumcircles, intersect $B C$ at $Q$. Now we have by power of point
$$
Q B^{2}=Q X \cdot Q A=Q C^{2}
$$
giving up that $A X$ bisects $B C$. So $X$ is the point on the median from $A$ to side $B C$ such that $\measuredangle B X C=180-\measuredangle B A C$. This point is unique and we have proven that it is always on the circumcircle of $A E D F$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle. Variable points $E$ and $F$ are on sides $A C$ and $A B$ respectively such that $B C^{2}=B A \cdot B F+C E \cdot C A$. As $E$ and $F$ vary prove that the circumcircle of $A E F$ passes through a fixed point other than $A$.
|
2
Let the $D$ be the intersection of $B E$ and $C F$ and let circumcircle of triangle $C F A$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B F \cdot B A
$$
Combining (6) with the problem statement we get
$$
B C^{2}=B A \cdot B F+C E \cdot C A=B G \cdot B C+C E \cdot C A
$$
and from here we get
$$
C E \cdot C A=B C(B C-B G)=B C \cdot C G
$$
(7) implies that $E, A, B, G$ are concyclic as well.
This gives us
$$
\measuredangle G A C=\measuredangle G A E=\measuredangle G B E=\measuredangle C B D
$$
and
$$
\measuredangle B A C-\measuredangle G A C=\measuredangle G A B=\measuredangle G A F=\measuredangle G C F=\measuredangle B C D .
$$
Adding these two equalities gives us
$$
\measuredangle B A C=\measuredangle C B D+\measuredangle B C D=180^{\circ}-\measuredangle B D C .
$$
This implies that $A, E, D, F$ are concyclic. Now let the second intersection of the circumcircles of $B D C$ and $A F D E$ be $X$. We have
$$
\angle X A B=\angle X A F=\angle X D F=180^{\circ}-\angle X D C=\angle X B C
$$
and
$$
\measuredangle X A C=\angle X A E=180^{\circ}-\measuredangle X D E=\angle X D B=\measuredangle X C B
$$
(8) and (9) imply that $B C$ is tangent to the circumcircles of $\triangle X A B$ and $\triangle X A C$ respectively. Let $A X$, the radical axis of the two circumcircles, intersect $B C$ at $Q$. Now we have by power of point
$$
Q B^{2}=Q X \cdot Q A=Q C^{2}
$$
giving up that $A X$ bisects $B C$. So $X$ is the point on the median from $A$ to side $B C$ such that $\measuredangle B X C=180-\measuredangle B A C$. This point is unique and we have proven that it is always on the circumcircle of $A E D F$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G1",
"solution_match": "\n## Solution"
}
|
3aa178da-fdfa-5e63-8f0d-7fe30dd8a643
| 605,400
|
Let $A B C$ be an acute triangle and $D$ a variable point on side $A C$. Point $E$ is on $B D$ such that $B E=\frac{B C^{2}-C D \cdot C A}{B D}$. As $D$ varies on side $A C$ prove that the circumcircle of $A D E$ passes through a fixed point other than $A$.
|
Let the circumcircle of triangle $C E D$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B E \cdot B D .
$$
Combining (1) with the problem statement we get
$$
\frac{B C \cdot B C}{B D}=B E=\frac{B C^{2}-C D \cdot C A}{B D}
$$
and from here we get
$$
C D \cdot C A=B C(B C-B G)=B C \cdot C G .
$$
(2) implies that $D, A, B, G$ are concyclic as well. This gives us
$$
\angle B E C=\measuredangle B G D=180^{\circ}-\measuredangle B A D=180-\measuredangle C A B .
$$
Now let the circumcircle of $A D E$ and $B E C$ intersect again at $X$. Since
$$
\angle X C B=\angle X E B=180^{\circ}-\measuredangle X E D=\angle X A D=\angle X A C
$$
and
$$
\angle B X C=\angle B E C=180^{\circ}-\angle B A C
$$
we have that $X$ is on the unique circle through $A$ and $C$ tangent to side $B C$ at point $C$ and circumcircle of $B H C$ where $H$ is the ortocenter of triangle $A B C$. This intersection is unique and we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle and $D$ a variable point on side $A C$. Point $E$ is on $B D$ such that $B E=\frac{B C^{2}-C D \cdot C A}{B D}$. As $D$ varies on side $A C$ prove that the circumcircle of $A D E$ passes through a fixed point other than $A$.
|
Let the circumcircle of triangle $C E D$ intersect $B C$ at point $G$. From power of point we have
$$
B G \cdot B C=B E \cdot B D .
$$
Combining (1) with the problem statement we get
$$
\frac{B C \cdot B C}{B D}=B E=\frac{B C^{2}-C D \cdot C A}{B D}
$$
and from here we get
$$
C D \cdot C A=B C(B C-B G)=B C \cdot C G .
$$
(2) implies that $D, A, B, G$ are concyclic as well. This gives us
$$
\angle B E C=\measuredangle B G D=180^{\circ}-\measuredangle B A D=180-\measuredangle C A B .
$$
Now let the circumcircle of $A D E$ and $B E C$ intersect again at $X$. Since
$$
\angle X C B=\angle X E B=180^{\circ}-\measuredangle X E D=\angle X A D=\angle X A C
$$
and
$$
\angle B X C=\angle B E C=180^{\circ}-\angle B A C
$$
we have that $X$ is on the unique circle through $A$ and $C$ tangent to side $B C$ at point $C$ and circumcircle of $B H C$ where $H$ is the ortocenter of triangle $A B C$. This intersection is unique and we are done.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nG2",
"solution_match": "\n## Solution"
}
|
d434158f-847e-5995-ba0f-996020e35bf4
| 605,418
|
Let $A B C$ be a triangle with $A B<A C$ inscribed into a circle $c$. The tangent of $c$ at the point $C$ meets the parallel from $B$ to $A C$ at the point $D$. The tangent of $c$ at the point $B$ meets the parallel from $C$ to $A B$ at the point $E$ and the tangent of $c$ at the point $C$ at the point $L$. Suppose that the circumcircle $c_{1}$ of the triangle $B D C$ meets $A C$ at the point $T$ and the circumcircle $c_{2}$ of the triangle $B E C$ meets $A B$ at the point $S$. Prove that the lines $S T, B C, A L$ are concurrent.
|
We will prove first that the circle $c_{1}$ is tangent to $A B$ at the point $B$. In order to prove this, we have to prove that $\measuredangle B D C=\measuredangle A B C$. Indeed, since $B D \| A C$, we have that $\measuredangle D B C=\measuredangle A C B$. Additionally, $\angle B C D=\measuredangle B A C$ (by chord and tangent), which means that the triangles $A B C, B D C$ have two equal angles and so the third ones are also equal. It follows that $\angle B D C=\measuredangle A B C$, so $c_{1}$ is tangent to $A B$ at the point $B$.
Similarly, the circle $c_{2}$ is tangent to $A C$ at the point $C$.
As a consequence, $\measuredangle A B T=\measuredangle A C B$ (by chord and tangent) and also $\measuredangle B S C=\measuredangle A C B$.
By the above, we have that $\measuredangle A B T=\measuredangle B S C$, so the lines $B T, S C$ are parallel.
Now, let $S T$ intersect $B C$ at the point $K$. It suffice to prove that $K$ belongs to $A L$. From the trapezoid $B T C S$ we get that
$$
\frac{B K}{K C}=\frac{B T}{S C}
$$
and from the similar triangles $A B T, A S C$, we have that
$$
\frac{B T}{S C}=\frac{A B}{A S}
$$
By (1), (2) we get that
$$
\frac{B K}{K C}=\frac{A B}{A S}
$$
From the power of point theorem, we have that
$$
A C^{2}=A B \cdot A S \Rightarrow A S=\frac{A C^{2}}{A B}
$$
Going back into (3), it gives that
$$
\frac{B K}{K C}=\frac{A B^{2}}{A C^{2}}
$$
From the last one, it follows that $K$ belongs to the symmedian of the triangle $A B C$.
Finally, recall that the well known fact that since $L B$ and $L C$ are tangents, it follows that $A L$ is the symmedian of the triangle $A B C$, so $K$ belongs to $A L$, as needed.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$ inscribed into a circle $c$. The tangent of $c$ at the point $C$ meets the parallel from $B$ to $A C$ at the point $D$. The tangent of $c$ at the point $B$ meets the parallel from $C$ to $A B$ at the point $E$ and the tangent of $c$ at the point $C$ at the point $L$. Suppose that the circumcircle $c_{1}$ of the triangle $B D C$ meets $A C$ at the point $T$ and the circumcircle $c_{2}$ of the triangle $B E C$ meets $A B$ at the point $S$. Prove that the lines $S T, B C, A L$ are concurrent.
|
We will prove first that the circle $c_{1}$ is tangent to $A B$ at the point $B$. In order to prove this, we have to prove that $\measuredangle B D C=\measuredangle A B C$. Indeed, since $B D \| A C$, we have that $\measuredangle D B C=\measuredangle A C B$. Additionally, $\angle B C D=\measuredangle B A C$ (by chord and tangent), which means that the triangles $A B C, B D C$ have two equal angles and so the third ones are also equal. It follows that $\angle B D C=\measuredangle A B C$, so $c_{1}$ is tangent to $A B$ at the point $B$.
Similarly, the circle $c_{2}$ is tangent to $A C$ at the point $C$.
As a consequence, $\measuredangle A B T=\measuredangle A C B$ (by chord and tangent) and also $\measuredangle B S C=\measuredangle A C B$.
By the above, we have that $\measuredangle A B T=\measuredangle B S C$, so the lines $B T, S C$ are parallel.
Now, let $S T$ intersect $B C$ at the point $K$. It suffice to prove that $K$ belongs to $A L$. From the trapezoid $B T C S$ we get that
$$
\frac{B K}{K C}=\frac{B T}{S C}
$$
and from the similar triangles $A B T, A S C$, we have that
$$
\frac{B T}{S C}=\frac{A B}{A S}
$$
By (1), (2) we get that
$$
\frac{B K}{K C}=\frac{A B}{A S}
$$
From the power of point theorem, we have that
$$
A C^{2}=A B \cdot A S \Rightarrow A S=\frac{A C^{2}}{A B}
$$
Going back into (3), it gives that
$$
\frac{B K}{K C}=\frac{A B^{2}}{A C^{2}}
$$
From the last one, it follows that $K$ belongs to the symmedian of the triangle $A B C$.
Finally, recall that the well known fact that since $L B$ and $L C$ are tangents, it follows that $A L$ is the symmedian of the triangle $A B C$, so $K$ belongs to $A L$, as needed.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G3",
"solution_match": "\n## Solution"
}
|
189f29e7-d484-5e39-9183-d3c1e7c5bde7
| 605,428
|
The acuteangled triangle $A B C$ with circumcenter $O$ is given. The midpoints of the sides $B C$, $C A$ and $A B$ are $D, E$ and $F$ respectivelly. An arbitrary point $M$ on the side $B C$, different of $D$, is choosen. The straight lines $A M$ and $E F$ intersects at the point $N$ and the straight line $O N$ cut again the circumscribed circle of the triangle $O D M$ at the point $P$. Prove that the reflection of the point $M$ with respect to the midpoint of the segment $D P$ belongs on the nine points circle of the triangle $A B C$.
|
The straight lines $D O, E O$ and $F O$ are the perpendicular bisectors of the sides $B C, C A$ and $A B$ respectively. It follows that $[O M]$ is the diameter of the circumscribed circle of the triangle $O D M$ and $M P \perp O N$. The point $O$ is the ortocenter of the triangle $D E F$ (see the picture)
Let $O_{1}$ be the circumcenter of the triangle $D E F$ and $H$ be the diametrically opposite point of $D$. The circumscribed circle of the triangle $D E F$ is the nine points circle of the triangle $A B C$. It follows that $E H \perp D E, F H \perp F D$ and $E D\|A F, D F\| A E$. So, the point $H$ is the ortocenter of the triangle $A E F$.
Let $A D \cap E F=\{I\}$ and $R$ is the reflection of the point N with respect to the point $I$, i.e. $R \in(E F), N I=R I$. The point $I$ is the simmetry center of the parallelogram $A E D F$. It follows that the point $I$ is the midpoint of the segment $[O H]$ and the quadrilaterals $A E D F$, $A N D R, H N O R$ are all parallelograms.
Let $Q$ be the reflection of the point $M$ with respect to the midpoint of the segment $D P$. It follows that the quadrilaterals $P Q D M$ and $M N R D$ are the parallelograms, which imply that the quadrilateral $P Q R N$ is a parallelogram. So, $N O\|H R, N P\| R Q$, which imply that the points $H, R$ and $Q$ are collinear. We obtain that $m(\angle D Q H)=90^{\circ}$, i.e. the point $Q$ belongs on the nine points circle of the triangle $A B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
The acuteangled triangle $A B C$ with circumcenter $O$ is given. The midpoints of the sides $B C$, $C A$ and $A B$ are $D, E$ and $F$ respectivelly. An arbitrary point $M$ on the side $B C$, different of $D$, is choosen. The straight lines $A M$ and $E F$ intersects at the point $N$ and the straight line $O N$ cut again the circumscribed circle of the triangle $O D M$ at the point $P$. Prove that the reflection of the point $M$ with respect to the midpoint of the segment $D P$ belongs on the nine points circle of the triangle $A B C$.
|
The straight lines $D O, E O$ and $F O$ are the perpendicular bisectors of the sides $B C, C A$ and $A B$ respectively. It follows that $[O M]$ is the diameter of the circumscribed circle of the triangle $O D M$ and $M P \perp O N$. The point $O$ is the ortocenter of the triangle $D E F$ (see the picture)
Let $O_{1}$ be the circumcenter of the triangle $D E F$ and $H$ be the diametrically opposite point of $D$. The circumscribed circle of the triangle $D E F$ is the nine points circle of the triangle $A B C$. It follows that $E H \perp D E, F H \perp F D$ and $E D\|A F, D F\| A E$. So, the point $H$ is the ortocenter of the triangle $A E F$.
Let $A D \cap E F=\{I\}$ and $R$ is the reflection of the point N with respect to the point $I$, i.e. $R \in(E F), N I=R I$. The point $I$ is the simmetry center of the parallelogram $A E D F$. It follows that the point $I$ is the midpoint of the segment $[O H]$ and the quadrilaterals $A E D F$, $A N D R, H N O R$ are all parallelograms.
Let $Q$ be the reflection of the point $M$ with respect to the midpoint of the segment $D P$. It follows that the quadrilaterals $P Q D M$ and $M N R D$ are the parallelograms, which imply that the quadrilateral $P Q R N$ is a parallelogram. So, $N O\|H R, N P\| R Q$, which imply that the points $H, R$ and $Q$ are collinear. We obtain that $m(\angle D Q H)=90^{\circ}$, i.e. the point $Q$ belongs on the nine points circle of the triangle $A B C$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G4",
"solution_match": "\n## Solution."
}
|
8b076c43-1180-560c-b234-c4690c2af604
| 605,436
|
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$.
|
1
Note that it is enough to prove that $\angle D P A+\angle M Q A=180^{\circ}$.
Without loss of generality assume that $A B<A C$. Let the reflection of $H$ in point $M$ be $H^{\prime}$. Since $B H C H^{\prime}$ is a paralelogram we get
$$
\measuredangle B H^{\prime} C=\angle B H C=180^{\circ}-\measuredangle B A C
$$
which means $H^{\prime}$ lies on $\omega$. Also we get
$$
\angle A B H^{\prime}=\angle A B C+\angle C B H^{\prime}=\angle A B C+\angle B C H=90^{\circ}
$$
since $C H \perp A B$. This mean $A H^{\prime}$ is the diameter of $\omega$. This means $\measuredangle M P A=\angle H^{\prime} P A=90^{\circ}$. Since $\angle M P A=\angle M D A=90^{\circ}$ we get that $M, D, P, A$ are concyclic.
This gives us $\angle D P A+\angle A M D=180^{\circ}$. So now it is enough to prove that $\measuredangle A M B=\angle M Q A$.
Taking the homothety with center $G$ and factor -2 (the homothety taking the 9 point circle to the circumcircle of $A B C$ ), we get that $Q$ and $A$ are images of $D$ and $M$ respectively.
This means $A Q \| D M$ giving $A Q \| B C$. Since $A Q \| B C$ and $A, B, C, Q$ are concyclic this means $A B C Q$ is an iscoseles trapezoid.
This means $M$ lies on the perpendicular bisector of $A Q$. This gives us $M A=M Q$. Since $A Q \| B C$ this gives us $\angle A Q M=\angle Q A M=\angle A M B$ which concludes our problem.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$.
|
1
Note that it is enough to prove that $\angle D P A+\angle M Q A=180^{\circ}$.
Without loss of generality assume that $A B<A C$. Let the reflection of $H$ in point $M$ be $H^{\prime}$. Since $B H C H^{\prime}$ is a paralelogram we get
$$
\measuredangle B H^{\prime} C=\angle B H C=180^{\circ}-\measuredangle B A C
$$
which means $H^{\prime}$ lies on $\omega$. Also we get
$$
\angle A B H^{\prime}=\angle A B C+\angle C B H^{\prime}=\angle A B C+\angle B C H=90^{\circ}
$$
since $C H \perp A B$. This mean $A H^{\prime}$ is the diameter of $\omega$. This means $\measuredangle M P A=\angle H^{\prime} P A=90^{\circ}$. Since $\angle M P A=\angle M D A=90^{\circ}$ we get that $M, D, P, A$ are concyclic.
This gives us $\angle D P A+\angle A M D=180^{\circ}$. So now it is enough to prove that $\measuredangle A M B=\angle M Q A$.
Taking the homothety with center $G$ and factor -2 (the homothety taking the 9 point circle to the circumcircle of $A B C$ ), we get that $Q$ and $A$ are images of $D$ and $M$ respectively.
This means $A Q \| D M$ giving $A Q \| B C$. Since $A Q \| B C$ and $A, B, C, Q$ are concyclic this means $A B C Q$ is an iscoseles trapezoid.
This means $M$ lies on the perpendicular bisector of $A Q$. This gives us $M A=M Q$. Since $A Q \| B C$ this gives us $\angle A Q M=\angle Q A M=\angle A M B$ which concludes our problem.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G5",
"solution_match": "\n## Solution"
}
|
3647a2cc-2e3c-552b-8ab8-2a96730a3c59
| 605,445
|
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$.
|
2
We prove that $M, D, P, A$ are concyclic same as in solution 1. Let $P D$ intersect $\omega$ again at $S$ We see that this gives us $\measuredangle S A H^{\prime}=\measuredangle S P H^{\prime}=\measuredangle D P M=\measuredangle D A M$. Combining this with $\measuredangle B A H^{\prime}=\measuredangle C A D$ we get:
$$
\measuredangle S A H^{\prime}+\measuredangle S A B=\measuredangle H^{\prime} A B=\measuredangle C A D=\measuredangle D A M+\measuredangle M A C
$$
## Giving us
$$
\measuredangle S A B=\measuredangle M A C
$$
Combining (*) with $\angle A S B=\measuredangle A C B=\measuredangle A C M$ we get triangles $A S B$ and $A C M$ are similar. This gives us
$$
\frac{A B}{A M}=\frac{S B}{C M} .
$$
Analogously we get triangles $A S C$ and $A B M$ are similar. This gives us
$$
\frac{A C}{A M}=\frac{S C}{B M} .
$$
Combining ( ${ }^{* *}$ ) and ( $\left.{ }^{(* *}\right)$ we get
$$
\frac{A B}{A C}=\frac{S B}{S C}
$$
since $B M=C M$. Let $S M$ intersect $\omega$ again at $Q^{\prime}$. Let $Q^{\prime} D$ intersect $A M$ at $G^{\prime}$. We wish to prove that $Q^{\prime} \equiv Q$ and $G^{\prime} \equiv G$. It is enough to prove that $A G^{\prime}=2 G^{\prime} M$.
Since triangles SMB and $C M Q^{\prime}$ are similar we get
$$
\frac{S B}{S M}=\frac{C Q}{C M} .
$$
Analogously SMC and BMQ' are similar and we get
$$
\frac{S M}{S C^{\prime}}=\frac{B M}{B Q^{\prime}} .
$$
Multiplying (2) and (3) we get $\frac{S B}{S C}=\frac{C Q^{\prime}}{B Q^{\prime}}$. Combining that with (1) we get
$$
\frac{A B}{A C}=\frac{C Q^{\prime}}{B Q^{\prime}} .
$$
Since $Q^{\prime}$ and $A$ are on the same side of $B C$ (4) gives us $A Q^{\prime} \| B C$. This means that $A B C Q^{\prime}$ is an isosceles trapezoid.
Let $Q^{\prime} S$ intersect $A D$ at point $T$. Since $M D \| A Q^{\prime}, M A=M Q^{\prime}$ and $\angle T A Q^{\prime}=90^{\circ}$ we get that $M$ is center of the circumcircle of the right triangle TAQ'.
Applying Menelaus' theorem on $D-G^{\prime}-Q^{\prime}$ and triangle $A M T$ we get
$$
\frac{A G^{\prime}}{M G^{\prime}} \cdot \frac{M Q^{\prime}}{T Q^{\prime}} \cdot \frac{T D}{A D}=1
$$
Since $M A=M T$ and $M D \perp A T$ this means
$$
D A=D T .
$$
Also since $M$ is the circumcenter of triangle $T A Q^{\prime}$ we get
$$
2 M Q^{\prime}=T Q^{\prime} .
$$
Combining (5) with (6) and (7) we get $2 M G^{\prime}=A G^{\prime}$. This gives us $G^{\prime} \equiv G$ and $Q^{\prime} \equiv Q$, thus proving the problem statement.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute angled triangle with ortocenter $H$, centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $A H$ and $A G$ with side $B C$. Rays $M H$ and $D G$ interect $\omega$ again at $P$ and $Q$ respectively. Prove that $P D$ and $Q M$ intersect on $\omega$.
|
2
We prove that $M, D, P, A$ are concyclic same as in solution 1. Let $P D$ intersect $\omega$ again at $S$ We see that this gives us $\measuredangle S A H^{\prime}=\measuredangle S P H^{\prime}=\measuredangle D P M=\measuredangle D A M$. Combining this with $\measuredangle B A H^{\prime}=\measuredangle C A D$ we get:
$$
\measuredangle S A H^{\prime}+\measuredangle S A B=\measuredangle H^{\prime} A B=\measuredangle C A D=\measuredangle D A M+\measuredangle M A C
$$
## Giving us
$$
\measuredangle S A B=\measuredangle M A C
$$
Combining (*) with $\angle A S B=\measuredangle A C B=\measuredangle A C M$ we get triangles $A S B$ and $A C M$ are similar. This gives us
$$
\frac{A B}{A M}=\frac{S B}{C M} .
$$
Analogously we get triangles $A S C$ and $A B M$ are similar. This gives us
$$
\frac{A C}{A M}=\frac{S C}{B M} .
$$
Combining ( ${ }^{* *}$ ) and ( $\left.{ }^{(* *}\right)$ we get
$$
\frac{A B}{A C}=\frac{S B}{S C}
$$
since $B M=C M$. Let $S M$ intersect $\omega$ again at $Q^{\prime}$. Let $Q^{\prime} D$ intersect $A M$ at $G^{\prime}$. We wish to prove that $Q^{\prime} \equiv Q$ and $G^{\prime} \equiv G$. It is enough to prove that $A G^{\prime}=2 G^{\prime} M$.
Since triangles SMB and $C M Q^{\prime}$ are similar we get
$$
\frac{S B}{S M}=\frac{C Q}{C M} .
$$
Analogously SMC and BMQ' are similar and we get
$$
\frac{S M}{S C^{\prime}}=\frac{B M}{B Q^{\prime}} .
$$
Multiplying (2) and (3) we get $\frac{S B}{S C}=\frac{C Q^{\prime}}{B Q^{\prime}}$. Combining that with (1) we get
$$
\frac{A B}{A C}=\frac{C Q^{\prime}}{B Q^{\prime}} .
$$
Since $Q^{\prime}$ and $A$ are on the same side of $B C$ (4) gives us $A Q^{\prime} \| B C$. This means that $A B C Q^{\prime}$ is an isosceles trapezoid.
Let $Q^{\prime} S$ intersect $A D$ at point $T$. Since $M D \| A Q^{\prime}, M A=M Q^{\prime}$ and $\angle T A Q^{\prime}=90^{\circ}$ we get that $M$ is center of the circumcircle of the right triangle TAQ'.
Applying Menelaus' theorem on $D-G^{\prime}-Q^{\prime}$ and triangle $A M T$ we get
$$
\frac{A G^{\prime}}{M G^{\prime}} \cdot \frac{M Q^{\prime}}{T Q^{\prime}} \cdot \frac{T D}{A D}=1
$$
Since $M A=M T$ and $M D \perp A T$ this means
$$
D A=D T .
$$
Also since $M$ is the circumcenter of triangle $T A Q^{\prime}$ we get
$$
2 M Q^{\prime}=T Q^{\prime} .
$$
Combining (5) with (6) and (7) we get $2 M G^{\prime}=A G^{\prime}$. This gives us $G^{\prime} \equiv G$ and $Q^{\prime} \equiv Q$, thus proving the problem statement.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G5",
"solution_match": "\n## Solution"
}
|
3647a2cc-2e3c-552b-8ab8-2a96730a3c59
| 605,445
|
Construct outside the acute-angled triangle $A B C$ the isosceles triangles $A B A_{B}, A B B_{A}$, $A C A_{C}, A C C_{A}, B C B_{C}$ and $B C C_{B}$, so that
$$
A B=A B_{A}=B A_{B}, A C=A C_{A}=C A_{C}, B C=B C_{B}=C B_{C}
$$
and
$$
\measuredangle B A B_{A}=\measuredangle A B A_{B}=\measuredangle C A C_{A}=\measuredangle A C A_{C}=\measuredangle B C B_{C}=\measuredangle C B C_{B}=\alpha<90^{\circ} .
$$
Prove that the perpendiculars from $A$ to $B_{A} C_{A}$, from $B$ to $A_{B} C_{B}$ and from $C$ to $A_{C} B_{C}$ are concurrent.
|
Lemma. If $B C D$ is the isoceles triangle which is outside the triangle $A B C$ and has
$$
\measuredangle C B D=\measuredangle B C D=90^{\circ}-\alpha: \stackrel{\text { not }}{=} \beta,
$$
then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that $\measuredangle A B E=\measuredangle A E B=\beta$.
Then $A E=A B=A B_{A}$ and $\measuredangle E A B_{A}=\alpha$, so a rotation of center $A$ and angle $\alpha$ sends $C_{A}$ to $C$ and $B_{A}$ to $E$, hence $\Varangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)=\alpha$ (the angle between vectors is considered oriented). Also triangles $E B A$ and $B C D$ are similar, so a rotation of center $B$ and angle $\beta$, followed by
a dilation of ratio $\frac{E B}{A B}=\frac{B C}{B D}$ sends $E$ to $A$ and $C$ to $D$, hence $\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\beta$ (also oriented angle).
This shows that
$$
\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{A D}\right)=\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)+\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\alpha+\beta=90^{\circ} .
$$
Returning to the solution of the problem, denote $A^{\prime}$ the intersection of $B C$ with the perpendicular from $A$ to $B_{A} C_{A}$. Then $A^{\prime}$ belongs to the segment $B C$ and
$$
\frac{A^{\prime} B}{A^{\prime} C}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)}
$$
Since similar relations are true for the intersections $B^{\prime}, C^{\prime}$ of the other two perpendiculars with the opposite sides, this yields
$$
\frac{A^{\prime} B}{A^{\prime} C} \cdot \frac{B^{\prime} C}{B^{\prime} A} \cdot \frac{C^{\prime} A}{C^{\prime} B}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)} \cdot \frac{B C \sin (C+\beta)}{B A \sin (A+\beta)} \cdot \frac{C A \sin (A+\beta)}{C B \sin (B+\beta)}=1
$$
whence the conclusion.
Remark. The conditions 'acute-angled' and ' $\alpha<90^{\circ}$ ' are not essential, but without them there are cases when $A^{\prime}$ does not belong to the segment $B C$, or the perpendiculars become parallel.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Construct outside the acute-angled triangle $A B C$ the isosceles triangles $A B A_{B}, A B B_{A}$, $A C A_{C}, A C C_{A}, B C B_{C}$ and $B C C_{B}$, so that
$$
A B=A B_{A}=B A_{B}, A C=A C_{A}=C A_{C}, B C=B C_{B}=C B_{C}
$$
and
$$
\measuredangle B A B_{A}=\measuredangle A B A_{B}=\measuredangle C A C_{A}=\measuredangle A C A_{C}=\measuredangle B C B_{C}=\measuredangle C B C_{B}=\alpha<90^{\circ} .
$$
Prove that the perpendiculars from $A$ to $B_{A} C_{A}$, from $B$ to $A_{B} C_{B}$ and from $C$ to $A_{C} B_{C}$ are concurrent.
|
Lemma. If $B C D$ is the isoceles triangle which is outside the triangle $A B C$ and has
$$
\measuredangle C B D=\measuredangle B C D=90^{\circ}-\alpha: \stackrel{\text { not }}{=} \beta,
$$
then $A D \perp B_{A} C_{A}$.
Proof of the lemma. Construct an isosceles triangle $A B E$ outside the triangle $A B C$, so that $\measuredangle A B E=\measuredangle A E B=\beta$.
Then $A E=A B=A B_{A}$ and $\measuredangle E A B_{A}=\alpha$, so a rotation of center $A$ and angle $\alpha$ sends $C_{A}$ to $C$ and $B_{A}$ to $E$, hence $\Varangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)=\alpha$ (the angle between vectors is considered oriented). Also triangles $E B A$ and $B C D$ are similar, so a rotation of center $B$ and angle $\beta$, followed by
a dilation of ratio $\frac{E B}{A B}=\frac{B C}{B D}$ sends $E$ to $A$ and $C$ to $D$, hence $\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\beta$ (also oriented angle).
This shows that
$$
\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{A D}\right)=\measuredangle\left(\overrightarrow{B_{A} C_{A}}, \overrightarrow{E C}\right)+\measuredangle(\overrightarrow{E C}, \overrightarrow{A D})=\alpha+\beta=90^{\circ} .
$$
Returning to the solution of the problem, denote $A^{\prime}$ the intersection of $B C$ with the perpendicular from $A$ to $B_{A} C_{A}$. Then $A^{\prime}$ belongs to the segment $B C$ and
$$
\frac{A^{\prime} B}{A^{\prime} C}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)}
$$
Since similar relations are true for the intersections $B^{\prime}, C^{\prime}$ of the other two perpendiculars with the opposite sides, this yields
$$
\frac{A^{\prime} B}{A^{\prime} C} \cdot \frac{B^{\prime} C}{B^{\prime} A} \cdot \frac{C^{\prime} A}{C^{\prime} B}=\frac{A B \sin (B+\beta)}{A C \sin (C+\beta)} \cdot \frac{B C \sin (C+\beta)}{B A \sin (A+\beta)} \cdot \frac{C A \sin (A+\beta)}{C B \sin (B+\beta)}=1
$$
whence the conclusion.
Remark. The conditions 'acute-angled' and ' $\alpha<90^{\circ}$ ' are not essential, but without them there are cases when $A^{\prime}$ does not belong to the segment $B C$, or the perpendiculars become parallel.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## G6",
"solution_match": "\n## Solution."
}
|
0b8205ab-914d-5836-bd3f-0fd2171e4108
| 605,461
|
Let $A B C$ be an acute triangle with $A B \neq A C$ and circumcirle $\Gamma$. The angle bisector of $B A C$ intersects $B C$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $D E$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $A F$ such that $P B=P C$ and $X$ and $Y$ are feet of perpendiculars from $P$ to $A B$ and $A C$ respectively. Let $H$ and $H^{\prime}$ be the ortocenters of $A B C$ and $A X Y$ respectively. $A H$ meets $\Gamma$ again at $Q$. If $A H^{\prime}$ and $H H^{\prime}$ intersect the circle with diameter $A H$ again at points $S$ and $T$, respectively, prove that the lines $A T, H S$ and $F Q$ are concurrent.
|
WLOG, assume $A B<A C$. Let $M$ be the midpoint of side $B C$ and let the circumcircle of $D F E$ intersect $A F$ again at $K$. Since
$$
90^{\circ}+\measuredangle M E D=180^{\circ}-\measuredangle M D E=\measuredangle A B C+\frac{\measuredangle B A C}{2}=\measuredangle A F E=\angle D F E+\angle A F D=90+\angle A F D
$$
$\measuredangle M^{\prime} C E=\measuredangle E C P$. From the angle bisector theorem we get
$$
\frac{M^{\prime} E}{P E}=\frac{C M^{\prime}}{C P}=\frac{M^{\prime} L}{P L}
$$
Multiplying (1) and (2) we get $\frac{M E}{L M}=\frac{M^{\prime} E}{M^{\prime} L}$ adding 1 on both sides we get $L M=L M^{\prime}$ from which it follows that $M \equiv M^{\prime}$ and thus $C E$ and $C L$ are the bisectors of $\measuredangle M C P$.
Now we have
Since $X$ and $Y$ are perpendicular to $A B$ and $A C$ we have $B X P M$ and $C Y P M$ are concyclic. Here we get
$$
\angle M Y C=\angle M P C=90^{\circ}-\angle B A C
$$
and it follows that $Y M \perp A X$ Similarily we get $X M \perp A Y$ and so $M$ is the ortocentar of $\triangle A X Y$ giving us $M \equiv H^{\prime}$.
Since ATHS and ATQF are both concyclic it is enough to prove that $H S F Q$ is concyclic. Since
$$
\begin{aligned}
\triangle B Q C & =180^{\circ}-\angle B A C \\
& =\angle B H C
\end{aligned}
$$
and $H Q \perp B C$ it follows that $B C$ is the perpendicular bisector of $H Q$. It is enough to prove that $B C$ is the perpendicular bisector of $S F$. Let $A M$ and $T H$ meet $\Gamma$ again at points $A^{\prime}$ and $N$ respec-
tively. Since $H N$ passes through the midpoint of side $B C$ and
$$
\angle B H C=180-\angle B A C=\angle B N C
$$
it follows that $B N C H$ is a paralelogram. From here we get that
$$
\angle N C B=\angle H B C=90^{\circ}-\angle A C B
$$
giving us $\measuredangle N C A=90^{\circ}$ and similarily $\measuredangle N B A=90^{\circ}$. This means $A N$ is the diameter of $\Gamma$, so
$$
\begin{aligned}
\angle N A^{\prime} S & =\angle N A^{\prime} A=90^{\circ} \\
& =\angle H S A=\angle H S A^{\prime}
\end{aligned}
$$
and from here we have $H S \| A^{\prime} N$. Now since $H S \| A^{\prime} N$ and $M$ is the midpoint of $H N$ (because $B H C N$ is a paralelogram) we get
that $H S N A$ is a paralelogram. Since
$$
\measuredangle F A E=\measuredangle E A M=\measuredangle E A A^{\prime}
$$
we get that $F A^{\prime} B C$ is an isocelese trapezoid which means that $M E$ is the perpendicular bisector of $F A^{\prime}$ (since it is the perpendicular bisector of $B C$ ).
This gives us $B F=C A^{\prime}=B S$ and $C F=B A^{\prime}=C S$ giving us that $S B F C$ is a deltoid, meaning that $B C$ is the perpendicular bisector of FS. This means that $H S F Q$ is an isoceles trapezoid. Now from the radical axis theorem of the circumcircles of HSFQ, HSAT and $A T Q F$ we get that $Q F$,
$H S, A T$ are concurrent.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B \neq A C$ and circumcirle $\Gamma$. The angle bisector of $B A C$ intersects $B C$ and $\Gamma$ at $D$ and $E$ respectively. Circle with diameter $D E$ intersects $\Gamma$ again at $F \neq E$. Point $P$ is on $A F$ such that $P B=P C$ and $X$ and $Y$ are feet of perpendiculars from $P$ to $A B$ and $A C$ respectively. Let $H$ and $H^{\prime}$ be the ortocenters of $A B C$ and $A X Y$ respectively. $A H$ meets $\Gamma$ again at $Q$. If $A H^{\prime}$ and $H H^{\prime}$ intersect the circle with diameter $A H$ again at points $S$ and $T$, respectively, prove that the lines $A T, H S$ and $F Q$ are concurrent.
|
WLOG, assume $A B<A C$. Let $M$ be the midpoint of side $B C$ and let the circumcircle of $D F E$ intersect $A F$ again at $K$. Since
$$
90^{\circ}+\measuredangle M E D=180^{\circ}-\measuredangle M D E=\measuredangle A B C+\frac{\measuredangle B A C}{2}=\measuredangle A F E=\angle D F E+\angle A F D=90+\angle A F D
$$
$\measuredangle M^{\prime} C E=\measuredangle E C P$. From the angle bisector theorem we get
$$
\frac{M^{\prime} E}{P E}=\frac{C M^{\prime}}{C P}=\frac{M^{\prime} L}{P L}
$$
Multiplying (1) and (2) we get $\frac{M E}{L M}=\frac{M^{\prime} E}{M^{\prime} L}$ adding 1 on both sides we get $L M=L M^{\prime}$ from which it follows that $M \equiv M^{\prime}$ and thus $C E$ and $C L$ are the bisectors of $\measuredangle M C P$.
Now we have
Since $X$ and $Y$ are perpendicular to $A B$ and $A C$ we have $B X P M$ and $C Y P M$ are concyclic. Here we get
$$
\angle M Y C=\angle M P C=90^{\circ}-\angle B A C
$$
and it follows that $Y M \perp A X$ Similarily we get $X M \perp A Y$ and so $M$ is the ortocentar of $\triangle A X Y$ giving us $M \equiv H^{\prime}$.
Since ATHS and ATQF are both concyclic it is enough to prove that $H S F Q$ is concyclic. Since
$$
\begin{aligned}
\triangle B Q C & =180^{\circ}-\angle B A C \\
& =\angle B H C
\end{aligned}
$$
and $H Q \perp B C$ it follows that $B C$ is the perpendicular bisector of $H Q$. It is enough to prove that $B C$ is the perpendicular bisector of $S F$. Let $A M$ and $T H$ meet $\Gamma$ again at points $A^{\prime}$ and $N$ respec-
tively. Since $H N$ passes through the midpoint of side $B C$ and
$$
\angle B H C=180-\angle B A C=\angle B N C
$$
it follows that $B N C H$ is a paralelogram. From here we get that
$$
\angle N C B=\angle H B C=90^{\circ}-\angle A C B
$$
giving us $\measuredangle N C A=90^{\circ}$ and similarily $\measuredangle N B A=90^{\circ}$. This means $A N$ is the diameter of $\Gamma$, so
$$
\begin{aligned}
\angle N A^{\prime} S & =\angle N A^{\prime} A=90^{\circ} \\
& =\angle H S A=\angle H S A^{\prime}
\end{aligned}
$$
and from here we have $H S \| A^{\prime} N$. Now since $H S \| A^{\prime} N$ and $M$ is the midpoint of $H N$ (because $B H C N$ is a paralelogram) we get
that $H S N A$ is a paralelogram. Since
$$
\measuredangle F A E=\measuredangle E A M=\measuredangle E A A^{\prime}
$$
we get that $F A^{\prime} B C$ is an isocelese trapezoid which means that $M E$ is the perpendicular bisector of $F A^{\prime}$ (since it is the perpendicular bisector of $B C$ ).
This gives us $B F=C A^{\prime}=B S$ and $C F=B A^{\prime}=C S$ giving us that $S B F C$ is a deltoid, meaning that $B C$ is the perpendicular bisector of FS. This means that $H S F Q$ is an isoceles trapezoid. Now from the radical axis theorem of the circumcircles of HSFQ, HSAT and $A T Q F$ we get that $Q F$,
$H S, A T$ are concurrent.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nG7",
"solution_match": "\n## Solution"
}
|
ae7f9da9-4d64-5c04-a20c-2f6cd4284ad7
| 605,472
|
Given an acute triangle $\triangle A B C(A C \neq A B)$ and let $(C)$ be its circumcircle. The excircle $\left(C_{1}\right)$ corresponding to the vertex $A$, of center $I_{a}$, tangents to the side $B C$ at the point $D$ and to the extensions of the sides $A B, A C$ at the points $E, Z$ respectively. Let $I$ and $L$ are the intersection points of the circles $(C)$ and $\left(C_{1}\right), H$ the orthocenter of the triangle $\triangle E D Z$ and $N$ the midpoint of segment $E Z$. The parallel line through the point $I_{a}$ to the line $H L$ meets the line $H I$ at the point $G$. Prove that the perpendicular line ( $e$ ) through the point $N$ to the to the line $B C$ and the parallel line $(\delta)$ through the point $G$ to the line $I L$ meet each other on the line $\mathrm{HI}_{a}$
|
We have $(e) \perp B C$ and $I_{a} D \perp B C$, so $(e) \| I_{a} D$. Let $T$. $S$ be the midpoints of the segments $H I_{a}, H D$ respectively and $Y$ the point of intersection of the lines $H D, E Z$ Then, $T S \| I_{a} D$. $T S \perp B C$ and $S Y \perp E Z$
The Euler circle ( $\omega$ ) of the triangle EDZ passes through the points $N, Y, S$. Therefore, the segment $S N$ is a diameter of the circle $(\omega)$. Thus, the center of $(\omega)$, let $T^{\prime}$, is the midpoint of the segment $S N$
On the other hand, we know that the center of Euler circle ( $\omega$ ) is the midpoint $T$ of $\mathrm{HI}_{a}$. So $T \equiv T^{\prime}$. Therefore, the line ( $e$ ) passes through the points $T, S$.
Therefore, we get that the quadrilateral $H S I_{a} N$ is parallelogram and its diagonals meet each other at the point $T$.
We consider the inversion $I\left(I_{a}, I_{a} Z^{2}\right)$. As $I_{a} Z^{2}=I_{a} A \cdot I_{a} N$ we have $I(N)=A$. Similarly, if $M_{1}, M_{2}$ the midpoints of the segments $D E, D Z$ respectively, we get. $I\left(M_{1}\right)=B$ and $I\left(M_{2}\right)=C$.
Therefore, the circumcircle ( $C$ ) of the triangle $A B C$ is the image of the circle ( $\omega$ ) under the inversion $I$ and the points of the intersection of the circles and ( $\omega$ ) are invariant under this inversion. But it is well known that the circle of inversion passes through the points of the intersection of the circles $(C)$ and ( $\omega$ ). Thus, the Euler circle ( $\omega$ ) passes through the points I,L.
Also, we consider the inversion $J\left(H, r^{2}\right)$ with
$$
r^{2}=H X \cdot H Z=H D \cdot H Y=H W \cdot H E
$$
where $X, W, Y$ the traces of the altitudes of the triangle $E D Z$ on its sides. Then, $J(Z)=X$, $J(D)=Y$ and $J(E)=W$. Therefore, the circumcircle $\left(C_{1}\right)$ of the triangle $A B C$ is the image of the circle $(\omega)$ under the inversion $J$. Thus, the circle of inversion $J$ passes through the points $I, L$.
We conclude that $H I=H L$ and $H I_{a} \perp I L$ and since $(\delta) \| I L$, we have $H I_{a} \perp(\delta)$.
If, $R$ is the point of intersection of the lines $(\delta), H L$, we get that quadrilateral $H R I_{a} G$ is parallelogram and its diagonals meet each other at the point $T$. So, the perpendicular line (e) through the point $N$ to the to the line $B C$ and the parallel line $(\delta)$ through the point $G$ to the line $I L$, meet each other on the line $H I_{a}$.
## COMBINATORICS
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given an acute triangle $\triangle A B C(A C \neq A B)$ and let $(C)$ be its circumcircle. The excircle $\left(C_{1}\right)$ corresponding to the vertex $A$, of center $I_{a}$, tangents to the side $B C$ at the point $D$ and to the extensions of the sides $A B, A C$ at the points $E, Z$ respectively. Let $I$ and $L$ are the intersection points of the circles $(C)$ and $\left(C_{1}\right), H$ the orthocenter of the triangle $\triangle E D Z$ and $N$ the midpoint of segment $E Z$. The parallel line through the point $I_{a}$ to the line $H L$ meets the line $H I$ at the point $G$. Prove that the perpendicular line ( $e$ ) through the point $N$ to the to the line $B C$ and the parallel line $(\delta)$ through the point $G$ to the line $I L$ meet each other on the line $\mathrm{HI}_{a}$
|
We have $(e) \perp B C$ and $I_{a} D \perp B C$, so $(e) \| I_{a} D$. Let $T$. $S$ be the midpoints of the segments $H I_{a}, H D$ respectively and $Y$ the point of intersection of the lines $H D, E Z$ Then, $T S \| I_{a} D$. $T S \perp B C$ and $S Y \perp E Z$
The Euler circle ( $\omega$ ) of the triangle EDZ passes through the points $N, Y, S$. Therefore, the segment $S N$ is a diameter of the circle $(\omega)$. Thus, the center of $(\omega)$, let $T^{\prime}$, is the midpoint of the segment $S N$
On the other hand, we know that the center of Euler circle ( $\omega$ ) is the midpoint $T$ of $\mathrm{HI}_{a}$. So $T \equiv T^{\prime}$. Therefore, the line ( $e$ ) passes through the points $T, S$.
Therefore, we get that the quadrilateral $H S I_{a} N$ is parallelogram and its diagonals meet each other at the point $T$.
We consider the inversion $I\left(I_{a}, I_{a} Z^{2}\right)$. As $I_{a} Z^{2}=I_{a} A \cdot I_{a} N$ we have $I(N)=A$. Similarly, if $M_{1}, M_{2}$ the midpoints of the segments $D E, D Z$ respectively, we get. $I\left(M_{1}\right)=B$ and $I\left(M_{2}\right)=C$.
Therefore, the circumcircle ( $C$ ) of the triangle $A B C$ is the image of the circle ( $\omega$ ) under the inversion $I$ and the points of the intersection of the circles and ( $\omega$ ) are invariant under this inversion. But it is well known that the circle of inversion passes through the points of the intersection of the circles $(C)$ and ( $\omega$ ). Thus, the Euler circle ( $\omega$ ) passes through the points I,L.
Also, we consider the inversion $J\left(H, r^{2}\right)$ with
$$
r^{2}=H X \cdot H Z=H D \cdot H Y=H W \cdot H E
$$
where $X, W, Y$ the traces of the altitudes of the triangle $E D Z$ on its sides. Then, $J(Z)=X$, $J(D)=Y$ and $J(E)=W$. Therefore, the circumcircle $\left(C_{1}\right)$ of the triangle $A B C$ is the image of the circle $(\omega)$ under the inversion $J$. Thus, the circle of inversion $J$ passes through the points $I, L$.
We conclude that $H I=H L$ and $H I_{a} \perp I L$ and since $(\delta) \| I L$, we have $H I_{a} \perp(\delta)$.
If, $R$ is the point of intersection of the lines $(\delta), H L$, we get that quadrilateral $H R I_{a} G$ is parallelogram and its diagonals meet each other at the point $T$. So, the perpendicular line (e) through the point $N$ to the to the line $B C$ and the parallel line $(\delta)$ through the point $G$ to the line $I L$, meet each other on the line $H I_{a}$.
## COMBINATORICS
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nG8",
"solution_match": "\n## Solution"
}
|
9b5bb84d-ba58-54d3-83e9-f6af820c01ed
| 605,480
|
A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to catch the grasshopper. Each second the hunter can choose one integer point in the plane and, if the grasshopper is there, he catches it. Can the hunter always catch the grasshopper in a finite amount of time?
|
The hunter can catch the grasshopper. Here is the strategy for him. Let $f$ be any bijection between the set of positive integers and the set $\{((x, y),(u, v)): x, y, u, v \in \mathbb{Z}\}$, and denote
$$
f(t)=\left(\left(x_{t}, y_{t}\right),\left(u_{t}, v_{t}\right)\right) .
$$
In the second $t$, the hunter should hunt at the point $\left(x_{t}+t u_{t}, y_{t}+t v_{t}\right)$. Let us show that this strategy indeed works.
Assume that the grasshopper starts at the point ( $x^{\prime}, y^{\prime}$ ) and that the jump vector is ( $u^{\prime}, v^{\prime}$ ). Then in the second $t$ the grasshopper is at the point $\left(x^{\prime}+t u^{\prime}, y^{\prime}+t v^{\prime}\right)$. Let
$$
t^{\prime}=f^{-1}\left(\left(x^{\prime}, y^{\prime}\right),\left(u^{\prime}, v^{\prime}\right)\right)
$$
The hunter's strategy dictates that in the second $t^{\prime}$ he searches for the grasshopper at the point $\left(x_{t^{\prime}}+t^{\prime} u_{t^{\prime}}, y_{t^{\prime}}+t^{\prime} v_{t^{\prime}}\right)$, which is actually $\left(x^{\prime}+t^{\prime} u^{\prime}, y^{\prime}+t^{\prime} v^{\prime}\right)$, and this is precisely the point where the grasshopper is in the second $t^{\prime}$. This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
A grasshopper is sitting at an integer point in the Euclidean plane. Each second it jumps to another integer point in such a way that the jump vector is constant. A hunter that knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector for each second is constant) wants to catch the grasshopper. Each second the hunter can choose one integer point in the plane and, if the grasshopper is there, he catches it. Can the hunter always catch the grasshopper in a finite amount of time?
|
The hunter can catch the grasshopper. Here is the strategy for him. Let $f$ be any bijection between the set of positive integers and the set $\{((x, y),(u, v)): x, y, u, v \in \mathbb{Z}\}$, and denote
$$
f(t)=\left(\left(x_{t}, y_{t}\right),\left(u_{t}, v_{t}\right)\right) .
$$
In the second $t$, the hunter should hunt at the point $\left(x_{t}+t u_{t}, y_{t}+t v_{t}\right)$. Let us show that this strategy indeed works.
Assume that the grasshopper starts at the point ( $x^{\prime}, y^{\prime}$ ) and that the jump vector is ( $u^{\prime}, v^{\prime}$ ). Then in the second $t$ the grasshopper is at the point $\left(x^{\prime}+t u^{\prime}, y^{\prime}+t v^{\prime}\right)$. Let
$$
t^{\prime}=f^{-1}\left(\left(x^{\prime}, y^{\prime}\right),\left(u^{\prime}, v^{\prime}\right)\right)
$$
The hunter's strategy dictates that in the second $t^{\prime}$ he searches for the grasshopper at the point $\left(x_{t^{\prime}}+t^{\prime} u_{t^{\prime}}, y_{t^{\prime}}+t^{\prime} v_{t^{\prime}}\right)$, which is actually $\left(x^{\prime}+t^{\prime} u^{\prime}, y^{\prime}+t^{\prime} v^{\prime}\right)$, and this is precisely the point where the grasshopper is in the second $t^{\prime}$. This completes the proof.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## C1",
"solution_match": "\n## Solution"
}
|
7583b59b-ba45-50d1-8e5c-29155163b08f
| 605,493
|
Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose lenght is divisible by $a$ and a side whose lenght is divisible by $b$.
|
Let the colors be $d_{1}, d_{2}, d_{3} \ldots, d_{n}$. Look at the coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
for integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair ( $k, r$ ) we say the color $d_{i}$ is $(k, r)$-good if at least two coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
are colored by color $d_{i}$. Fixing $r$ and taking $k=0, a b, 2 a b, \ldots, n^{2} a b$ get that some color, say $d_{1}$, was $(k, r)$-good for at least $n+1$.
Among the $n+1$ pairs $(x, y)$ there exists two which share the same $x$ cordinate. We call such quadruple $r$-great. In every $r$-great quadruple there are two triangle whose vertecies are all the same color and whose two sides are divisible by $a b$. Taking
$$
r=0.1,2, \ldots, n\left(c\left(\left(^{(n+1)\left(n^{2}+1\right)} 3{ }^{3}\right)+1\right)+1\right)+1
$$
we get that there is one color which is in a $r$-great quadruple for at least
$$
c\left(\left(^{(n+1)\left(n^{2}+1\right)} 3-1\right)+1\right.
$$
different values of $r$. Let this color be $d_{1}$. Since there are less than $\binom{\left.(n+1) n^{2}+1\right)}{3}$ possible triangles in any $r$-great quadruple (among $c\left(\left(^{(n+1)\left(n^{2}+1\right)}\right)+1\right)+1 \quad r$-great quadruples with the color $d_{1}$ ) we get that there are $c+1$ triangles which are the same and the same color $d_{1}$ and with two sides divisible by $a b$. This concludes the problem.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose lenght is divisible by $a$ and a side whose lenght is divisible by $b$.
|
Let the colors be $d_{1}, d_{2}, d_{3} \ldots, d_{n}$. Look at the coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
for integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair ( $k, r$ ) we say the color $d_{i}$ is $(k, r)$-good if at least two coordinates
$$
(k, 0+(n+1) a b r),(k, a b+(n+1) a b r),(k, 2 a b+(n+1) a b r), \ldots,(k, n a b+(n+1) a b r)
$$
are colored by color $d_{i}$. Fixing $r$ and taking $k=0, a b, 2 a b, \ldots, n^{2} a b$ get that some color, say $d_{1}$, was $(k, r)$-good for at least $n+1$.
Among the $n+1$ pairs $(x, y)$ there exists two which share the same $x$ cordinate. We call such quadruple $r$-great. In every $r$-great quadruple there are two triangle whose vertecies are all the same color and whose two sides are divisible by $a b$. Taking
$$
r=0.1,2, \ldots, n\left(c\left(\left(^{(n+1)\left(n^{2}+1\right)} 3{ }^{3}\right)+1\right)+1\right)+1
$$
we get that there is one color which is in a $r$-great quadruple for at least
$$
c\left(\left(^{(n+1)\left(n^{2}+1\right)} 3-1\right)+1\right.
$$
different values of $r$. Let this color be $d_{1}$. Since there are less than $\binom{\left.(n+1) n^{2}+1\right)}{3}$ possible triangles in any $r$-great quadruple (among $c\left(\left(^{(n+1)\left(n^{2}+1\right)}\right)+1\right)+1 \quad r$-great quadruples with the color $d_{1}$ ) we get that there are $c+1$ triangles which are the same and the same color $d_{1}$ and with two sides divisible by $a b$. This concludes the problem.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nC2",
"solution_match": "\n## Solution"
}
|
e285cccc-98e3-58a3-847f-31d6552fd377
| 605,506
|
Let $n \geq 4$ points in the plane, no three of them are collinear. Prove that the number of parallelograms of area 1 , formed by these points, is at most $\frac{n^{2}-3 n}{4}$.
|
Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area 1 formed by these $k$ pairs of points.
Summing over all directions we get that the number of parallelograms of area 1 are at most $\binom{n}{2}-s$ where $s$ is the number of different directions. But in that way we count every parallelogram two times, so the that the number of parallelograms of area 1 is at most $\frac{\binom{n}{2}-s}{2}$. We will prove that $s \geq n$. Indeed, taking the convex hull of the $n$ points, let $x$ be a point on the boundary of the convex hull. Because the convex hull has at least three points on its boundary, we can take two points which are neighbors of $x$ in the convex hull, say $y, z$ these points. Then every segment starting from $x$ has different direction from $y z$. So we have at least $n-1+1=n$ different directions. So the number of parallelograms is at most $\frac{\binom{n}{2}-n}{2}=\frac{n^{2}-3 n}{4}$.
|
\frac{n^{2}-3 n}{4}
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq 4$ points in the plane, no three of them are collinear. Prove that the number of parallelograms of area 1 , formed by these points, is at most $\frac{n^{2}-3 n}{4}$.
|
Fix a direction in the plane. We cannot have three points in the same line parallel to the direction so suppose that in that direction there are $k$ pairs of points, each pair belonging to a parallel line to the fixed direction. Then there are at most $k-1$ parallelograms of area 1 formed by these $k$ pairs of points.
Summing over all directions we get that the number of parallelograms of area 1 are at most $\binom{n}{2}-s$ where $s$ is the number of different directions. But in that way we count every parallelogram two times, so the that the number of parallelograms of area 1 is at most $\frac{\binom{n}{2}-s}{2}$. We will prove that $s \geq n$. Indeed, taking the convex hull of the $n$ points, let $x$ be a point on the boundary of the convex hull. Because the convex hull has at least three points on its boundary, we can take two points which are neighbors of $x$ in the convex hull, say $y, z$ these points. Then every segment starting from $x$ has different direction from $y z$. So we have at least $n-1+1=n$ different directions. So the number of parallelograms is at most $\frac{\binom{n}{2}-n}{2}=\frac{n^{2}-3 n}{4}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nC3",
"solution_match": "\n## Solution"
}
|
4b958447-eb59-52c5-8dee-b06a87032bed
| 605,517
|
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_{,}, A_{k}\right)
$$
|
We start with a lemma.
Lemma. If the triangle $A B C$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse. $r(A, B, C)$ is half the length of its longest side.
## Proof.
Let us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us prove that no smaller circle contains all three vertices. If there is a smaller circle, let its center be $P$. Further, let the circumcenter be $O$
Since $A B C$ is acute, $O$ is in the interior. Consider the line that passes through $O$ and is parallel to $B C$. Let us call it $l_{A}$ and define $l_{B}$ and $l_{C}$. similarly. Now, consider the set of points that are on the opposite side of $I_{A}$ with respect to
$A$. Call this set $S_{A}$ and define $S_{B}$ and
$S_{C}$ similarly. It is easily seen (by geometry) that $S_{A} \cap S_{B} \cap S_{C} \cdot=\varnothing$. As such, assume $P \notin S_{A}$ without loss of generality. That is to say, $P$ is on the same side of $I_{A}$ as $A$. Now. consider the perpendicular bisector of $B C$ and assume that $P$, w.lo.g, is on the same side of this line as $C$. Under these circumstances, $|P B| \geqslant|O B|$. Thus, the smaller circle centered at $P$ must exclude $B$.
In the obtuse case, let $\measuredangle B A C \geq 90^{\circ}$. Then $B C$ is the longest side. The circle with diameter $B C$ contains all three vertices. Therefore. $r(A, B, C)$ is not greater than $\frac{1}{2}|B C|$. But any smaller circle will clearly exclude at least one of $B$ and $C$.
Now, let us return to the original problem. Note that there must be points $A, B, C$ among $A_{1}, A_{2}, \ldots, A_{n}$ such that the circumcircle of $A B C$ contains all $n$ points. One can see this as follows: First start with a large circle that contains all $n$ points. Then shrink it while keeping the center fixed, until one of the $n$ points is on the circle and call this point $A$. Then shrink it keeping the point $A$ in place and moving the center closer to $A$. until another point $B$ is on the circle. Then keep the line $A B$ fixed while moving the center toward it or away from it so
that another $C$ among the $n$ points appears on the circle. It is easy to see that this procedure is doable.
Consider all such triples $A, B, C$ such that the circumcircle of $A B C$ contains all of $A_{1}, A_{2}, \ldots, A_{n}$. Now choose the one among them with the smallest circumradius and let it be $A_{i}, A_{j}, A_{k}$. If $A_{i} A_{j} A_{k}$ is an acute triangle, any smaller circle will exclude one of $A_{1}, A_{j}, A_{k}$ by the lemma above. Therefore,
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=\text { circumradius of } A_{i} A_{j} A_{k}=r\left(A_{i}, A_{j}, A_{k}\right)
$$
If $A_{i} A_{j} A_{k}$ is an obtuse triangle, let $A_{l}$ be its obtuse angle. We wish to prove that the circle with diameter $A_{j} A_{k}$ contains all $n$ points. This will mean that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=\frac{1}{2}\left|A_{j} A_{k}\right|=r\left(A_{i}, A_{j}, A_{k}\right)
$$
and we will be done. If there are no points on the opposite side of $A_{j} A_{k}$ w.r.t. $A_{i}$, then this assertion is clear. If there are some points on that side, choose the one $X$ such that $\measuredangle A_{j} X A_{k}$ is smallest possible. Then the circumcircle of $A_{j} X A_{k}$ contains all $n$ points. However, by the choice of $A_{i}$, the circumradius of $A_{j} X A_{k}$ cannot be less than that of $A_{i} A_{j} A_{k}$. Thus, $\measuredangle A_{j} X A_{k} \geq \measuredangle A_{j} A_{i} A_{k} \geq 90^{\circ}$. As such, the circle with diameter $A_{j} A_{k}$ contains all $n$ points.
Figure 1: The circumcircles of $A_{i} A_{j} A_{k}$ and $A_{j} X A_{k}$ as well as the circle with diameter $A_{j} A_{k}$ are shown.
Remark. The problem selection committee recommended formulation of the task to improve.
## C 5
We have $n$ students sitting at a round table. Initially each student is given one candy. At each step each student having candies either picks one of its candies and gives it to one of its neighbouring students, or distributes all of its candies to its neighbouring students in any way he wishes. A distribution of candies is called legal if it can be reached from the initial distribution via a sequence of steps.
Determine the number of legal distributions. (All the candies are udentical.)
|
proof
|
Yes
|
Incomplete
|
proof
|
Geometry
|
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_{,}, A_{k}\right)
$$
|
We start with a lemma.
Lemma. If the triangle $A B C$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse. $r(A, B, C)$ is half the length of its longest side.
## Proof.
Let us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. Now, let us prove that no smaller circle contains all three vertices. If there is a smaller circle, let its center be $P$. Further, let the circumcenter be $O$
Since $A B C$ is acute, $O$ is in the interior. Consider the line that passes through $O$ and is parallel to $B C$. Let us call it $l_{A}$ and define $l_{B}$ and $l_{C}$. similarly. Now, consider the set of points that are on the opposite side of $I_{A}$ with respect to
$A$. Call this set $S_{A}$ and define $S_{B}$ and
$S_{C}$ similarly. It is easily seen (by geometry) that $S_{A} \cap S_{B} \cap S_{C} \cdot=\varnothing$. As such, assume $P \notin S_{A}$ without loss of generality. That is to say, $P$ is on the same side of $I_{A}$ as $A$. Now. consider the perpendicular bisector of $B C$ and assume that $P$, w.lo.g, is on the same side of this line as $C$. Under these circumstances, $|P B| \geqslant|O B|$. Thus, the smaller circle centered at $P$ must exclude $B$.
In the obtuse case, let $\measuredangle B A C \geq 90^{\circ}$. Then $B C$ is the longest side. The circle with diameter $B C$ contains all three vertices. Therefore. $r(A, B, C)$ is not greater than $\frac{1}{2}|B C|$. But any smaller circle will clearly exclude at least one of $B$ and $C$.
Now, let us return to the original problem. Note that there must be points $A, B, C$ among $A_{1}, A_{2}, \ldots, A_{n}$ such that the circumcircle of $A B C$ contains all $n$ points. One can see this as follows: First start with a large circle that contains all $n$ points. Then shrink it while keeping the center fixed, until one of the $n$ points is on the circle and call this point $A$. Then shrink it keeping the point $A$ in place and moving the center closer to $A$. until another point $B$ is on the circle. Then keep the line $A B$ fixed while moving the center toward it or away from it so
that another $C$ among the $n$ points appears on the circle. It is easy to see that this procedure is doable.
Consider all such triples $A, B, C$ such that the circumcircle of $A B C$ contains all of $A_{1}, A_{2}, \ldots, A_{n}$. Now choose the one among them with the smallest circumradius and let it be $A_{i}, A_{j}, A_{k}$. If $A_{i} A_{j} A_{k}$ is an acute triangle, any smaller circle will exclude one of $A_{1}, A_{j}, A_{k}$ by the lemma above. Therefore,
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=\text { circumradius of } A_{i} A_{j} A_{k}=r\left(A_{i}, A_{j}, A_{k}\right)
$$
If $A_{i} A_{j} A_{k}$ is an obtuse triangle, let $A_{l}$ be its obtuse angle. We wish to prove that the circle with diameter $A_{j} A_{k}$ contains all $n$ points. This will mean that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=\frac{1}{2}\left|A_{j} A_{k}\right|=r\left(A_{i}, A_{j}, A_{k}\right)
$$
and we will be done. If there are no points on the opposite side of $A_{j} A_{k}$ w.r.t. $A_{i}$, then this assertion is clear. If there are some points on that side, choose the one $X$ such that $\measuredangle A_{j} X A_{k}$ is smallest possible. Then the circumcircle of $A_{j} X A_{k}$ contains all $n$ points. However, by the choice of $A_{i}$, the circumradius of $A_{j} X A_{k}$ cannot be less than that of $A_{i} A_{j} A_{k}$. Thus, $\measuredangle A_{j} X A_{k} \geq \measuredangle A_{j} A_{i} A_{k} \geq 90^{\circ}$. As such, the circle with diameter $A_{j} A_{k}$ contains all $n$ points.
Figure 1: The circumcircles of $A_{i} A_{j} A_{k}$ and $A_{j} X A_{k}$ as well as the circle with diameter $A_{j} A_{k}$ are shown.
Remark. The problem selection committee recommended formulation of the task to improve.
## C 5
We have $n$ students sitting at a round table. Initially each student is given one candy. At each step each student having candies either picks one of its candies and gives it to one of its neighbouring students, or distributes all of its candies to its neighbouring students in any way he wishes. A distribution of candies is called legal if it can be reached from the initial distribution via a sequence of steps.
Determine the number of legal distributions. (All the candies are udentical.)
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## C4",
"solution_match": "\n## Solution."
}
|
624f2acf-bb4b-5ede-a6c9-de2d0d4f6ab9
| 605,530
|
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_{,}, A_{k}\right)
$$
|
The answer turns out to be $\binom{2 n-1}{n}$ if $n$ is odd and $\binom{2 n-1}{n}-2\left(\begin{array}{c}\frac{3 n}{2}-1\end{array}\right)$ if $n$ is even.
Case 1. Suppose $n$ is odd, say $n=2 m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed ( $\left.\begin{array}{c}2 n-1 \\ n\end{array}\right)$.
In this case we can achieve the above claim by letting each student to always distribute all of its candies to its two neighbouring students in some way. Thus at each step each candy will move either one position clockwise or one anticlockwise.
We now look at the initial distribution of candies and the required final distribution. We specify arbitrarily for each candy in the initial distribution, the position we wish this candy to end up in the required final distribution. Because $n$ is odd, either the clockwise distance or the anticlockwise distance between the initial position of the candy and the required final position is even and at most $m$.
Thus after an even number of steps (at most $m$ ) we can move each candy to its required final position. (Note that if the candy reaches the required position earlier. we can move it back and forth until all candies reach their required position.) This completes the proof of our claim in this case.
Case 2. Suppose $n$ is even, say $n=2 m$. Let $x_{1} \ldots \ldots x_{2 m}$ be the students in this cyclic order. Observe that initially the students with even indices (even students) have at least one candy in total, and so do the students with odd indices (odd students). This property is preserved after each step.
We will show that every distribution in which the even students have at least one candy in total and the odd students also have at least one candy in total is legal.
Let us suppose that the required final distribution has $a$ candies in odd positions and $b$ candies in even positions. (Where $a, b \geq 1$.) It will be enough to reach any position with $a$ candies in even positions and $b$ candies in odd positions as then we can follow the same approach as in Case 1.
To achieve this we will first move all candies to students $x_{1}$ and $x_{2}$. This is easy by specifying that at each step $x_{1}$ moves all of its candies to $x_{2}$ while for $1 \leq r \leq 2 m-1$ student $x_{r+1}$ moves all of its candies to $x_{r}$.
Suppose that we now have $a+k$ candies at $x_{1}$ and $b-k$ candies at $x_{2}$ where without loss of generality $k \geq 0$. If $k=0$ we have reached our target. If not, in the next step $x_{1}$ moves a candy to $x_{2}$ and $x_{2}$ moves a candy to $x_{3}$. In the next step $x_{1}$ (it still has $a+k-1 \geq a>0$
candies) moves a candy to $x_{2}, x_{2}$ moves a candy to $x_{1}$ and $x_{3}$ moves a candy to $x_{2}$. We now have $a+k-1$ candies in $x_{1}$ and $b+1-k$ in $x_{2}$. Repeating this process another $k-1$ times we end up with $a$ candies in $x_{1}$ and $b$ candies in $x_{2}$ as required.
It remains to count the total number of legal configurations in this case. This is indeed equal to
$$
\binom{2 n-1}{n}-2\binom{\frac{3 n}{2}-1}{n}
$$
as $\binom{2 n-1}{n}$ counts the total number of configurations while $\binom{\frac{3 n}{2}-1}{n}$ counts the number of illegal configurations where either all $n$ candies belong to the $\frac{n}{2}$ odd positions or all $n$ candies belong to the $\frac{n}{2}$ even positions.
|
proof
|
Yes
|
Problem not solved
|
proof
|
Geometry
|
For any set of points $A_{1}, A_{2}, \ldots, A_{n}$ on the plane, one defines $r\left(A_{1}, A_{2}, \ldots, A_{n}\right)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \geq 3$, there are indices $i, j, k$ such that
$$
r\left(A_{1}, A_{2}, \ldots, A_{n}\right)=r\left(A_{1}, A_{,}, A_{k}\right)
$$
|
The answer turns out to be $\binom{2 n-1}{n}$ if $n$ is odd and $\binom{2 n-1}{n}-2\left(\begin{array}{c}\frac{3 n}{2}-1\end{array}\right)$ if $n$ is even.
Case 1. Suppose $n$ is odd, say $n=2 m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed ( $\left.\begin{array}{c}2 n-1 \\ n\end{array}\right)$.
In this case we can achieve the above claim by letting each student to always distribute all of its candies to its two neighbouring students in some way. Thus at each step each candy will move either one position clockwise or one anticlockwise.
We now look at the initial distribution of candies and the required final distribution. We specify arbitrarily for each candy in the initial distribution, the position we wish this candy to end up in the required final distribution. Because $n$ is odd, either the clockwise distance or the anticlockwise distance between the initial position of the candy and the required final position is even and at most $m$.
Thus after an even number of steps (at most $m$ ) we can move each candy to its required final position. (Note that if the candy reaches the required position earlier. we can move it back and forth until all candies reach their required position.) This completes the proof of our claim in this case.
Case 2. Suppose $n$ is even, say $n=2 m$. Let $x_{1} \ldots \ldots x_{2 m}$ be the students in this cyclic order. Observe that initially the students with even indices (even students) have at least one candy in total, and so do the students with odd indices (odd students). This property is preserved after each step.
We will show that every distribution in which the even students have at least one candy in total and the odd students also have at least one candy in total is legal.
Let us suppose that the required final distribution has $a$ candies in odd positions and $b$ candies in even positions. (Where $a, b \geq 1$.) It will be enough to reach any position with $a$ candies in even positions and $b$ candies in odd positions as then we can follow the same approach as in Case 1.
To achieve this we will first move all candies to students $x_{1}$ and $x_{2}$. This is easy by specifying that at each step $x_{1}$ moves all of its candies to $x_{2}$ while for $1 \leq r \leq 2 m-1$ student $x_{r+1}$ moves all of its candies to $x_{r}$.
Suppose that we now have $a+k$ candies at $x_{1}$ and $b-k$ candies at $x_{2}$ where without loss of generality $k \geq 0$. If $k=0$ we have reached our target. If not, in the next step $x_{1}$ moves a candy to $x_{2}$ and $x_{2}$ moves a candy to $x_{3}$. In the next step $x_{1}$ (it still has $a+k-1 \geq a>0$
candies) moves a candy to $x_{2}, x_{2}$ moves a candy to $x_{1}$ and $x_{3}$ moves a candy to $x_{2}$. We now have $a+k-1$ candies in $x_{1}$ and $b+1-k$ in $x_{2}$. Repeating this process another $k-1$ times we end up with $a$ candies in $x_{1}$ and $b$ candies in $x_{2}$ as required.
It remains to count the total number of legal configurations in this case. This is indeed equal to
$$
\binom{2 n-1}{n}-2\binom{\frac{3 n}{2}-1}{n}
$$
as $\binom{2 n-1}{n}$ counts the total number of configurations while $\binom{\frac{3 n}{2}-1}{n}$ counts the number of illegal configurations where either all $n$ candies belong to the $\frac{n}{2}$ odd positions or all $n$ candies belong to the $\frac{n}{2}$ even positions.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\n## C4",
"solution_match": "\n## Solution"
}
|
624f2acf-bb4b-5ede-a6c9-de2d0d4f6ab9
| 605,530
|
What is the least positive integer $k$ such that, in every convex 101 -gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals?
|
Let $P Q=1$. Consider a convex 101 -gon such that one of its vertices is at $P$ and the remaining 100 vertices are within $\varepsilon$ of $Q$ where $\varepsilon$ is an arbitrarily small positive real. Let $k+l$ equal the total number $\frac{101.98}{2}=4949$ of diagonals. When $k \leq 4851$. the sum of the $k$ shortest diagonals is arbitrarily small. When $k \geq 4851$, the sum of the $k$ shortest diagonals is arbitrarily close to $k-4851=98-l$ and the sum of the remaining diagonals is arbitrarily close to $l$. Therefore, we need to have $l \leq 49$ and $k \geq 4900$.
We proceed to show that $k=4900$ works. To this end, colour all $l=49$ remaining diagonals green. To each green diagonal $A B$, apart from, possibly, the last one, we will assign two red diagonals $A C$ and $C B$ so that no green diagonal is ever coloured red and no diagonal is coloured red twice.
Suppose that we have already done this for $0 \leq i \leq 48$ green diagonals (thus forming $i$ red-red-green triangles) and let $A B$ be up next. Let $D$ be the set of all diagonals emanating from $A$ or $B$ and distinct from $A B$ : we have $|D|=2 \cdot 97=194$. Every red-red-green triangle formed thus far has at most two sides in $D$. Therefore, the subset $E$ of all as-of-yetuncoloured diagonals in $D$ contains at least 194-2i elements.
When $i \leq 47,194-2 i \geq 100$. The total number of endpoints distinct from $A$ and $B$ of diagonals in $D$. however, is 99 . Therefore, two diagonals in $E$ have a common endpoint $C$ and we can assign $A C$ and $C B$ to $A B$. as needed.
The case $i=48$ is slightly more tricky: this time, it is possible that no two diagonals in $E$ have a common endpoint other than $A$ and $B$. but, if so, then there are two diagonals in $E$ that intersect in a point interior to both. Otherwise, at least one (say, $a$ ) of the two vertices adjacent to $A$ is cut off from $B$ by the diagonals emanating from $A$ and at least one (say, $b$ ) of the two vertices adjacent to $B$ is cut off from $A$ by the diagonals emanating from $B$ (and $a \neq b$ ). This leaves us with at most 97 suitable endpoints and at least 98 diagonals in $E$. a contradiction.
By the triangle inequality, this completes the solution.
## $\mathbb{O}_{1}^{2}$ $R_{1}$
## Union of Mathematicians of Macedonia
Sponsored by
Homisinan
|
4900
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
What is the least positive integer $k$ such that, in every convex 101 -gon, the sum of any $k$ diagonals is greater than or equal to the sum of the remaining diagonals?
|
Let $P Q=1$. Consider a convex 101 -gon such that one of its vertices is at $P$ and the remaining 100 vertices are within $\varepsilon$ of $Q$ where $\varepsilon$ is an arbitrarily small positive real. Let $k+l$ equal the total number $\frac{101.98}{2}=4949$ of diagonals. When $k \leq 4851$. the sum of the $k$ shortest diagonals is arbitrarily small. When $k \geq 4851$, the sum of the $k$ shortest diagonals is arbitrarily close to $k-4851=98-l$ and the sum of the remaining diagonals is arbitrarily close to $l$. Therefore, we need to have $l \leq 49$ and $k \geq 4900$.
We proceed to show that $k=4900$ works. To this end, colour all $l=49$ remaining diagonals green. To each green diagonal $A B$, apart from, possibly, the last one, we will assign two red diagonals $A C$ and $C B$ so that no green diagonal is ever coloured red and no diagonal is coloured red twice.
Suppose that we have already done this for $0 \leq i \leq 48$ green diagonals (thus forming $i$ red-red-green triangles) and let $A B$ be up next. Let $D$ be the set of all diagonals emanating from $A$ or $B$ and distinct from $A B$ : we have $|D|=2 \cdot 97=194$. Every red-red-green triangle formed thus far has at most two sides in $D$. Therefore, the subset $E$ of all as-of-yetuncoloured diagonals in $D$ contains at least 194-2i elements.
When $i \leq 47,194-2 i \geq 100$. The total number of endpoints distinct from $A$ and $B$ of diagonals in $D$. however, is 99 . Therefore, two diagonals in $E$ have a common endpoint $C$ and we can assign $A C$ and $C B$ to $A B$. as needed.
The case $i=48$ is slightly more tricky: this time, it is possible that no two diagonals in $E$ have a common endpoint other than $A$ and $B$. but, if so, then there are two diagonals in $E$ that intersect in a point interior to both. Otherwise, at least one (say, $a$ ) of the two vertices adjacent to $A$ is cut off from $B$ by the diagonals emanating from $A$ and at least one (say, $b$ ) of the two vertices adjacent to $B$ is cut off from $A$ by the diagonals emanating from $B$ (and $a \neq b$ ). This leaves us with at most 97 suitable endpoints and at least 98 diagonals in $E$. a contradiction.
By the triangle inequality, this completes the solution.
## $\mathbb{O}_{1}^{2}$ $R_{1}$
## Union of Mathematicians of Macedonia
Sponsored by
Homisinan
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2017_bmo_shortlist.jsonl",
"problem_match": "\nC6",
"solution_match": "\n## Solution"
}
|
96677517-ae54-57e1-a5b2-abb6ff75e71f
| 605,552
|
Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
(FYR Macedonia)
|
By the AH mean inequality, we have
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{2}{3(a c+b c)}+\frac{2}{3(a b+a c)}+\frac{2}{3(a b+a c)} \geqslant \frac{3}{a b+a c+b c}
$$
so it only remains to prove that $\frac{3}{a b+a c+b c} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}$, or equivalently
$$
3\left(a^{3}+b^{3}+c^{3}\right) \geqslant(a+b+c)(a b+a c+b c)
$$
The last inequality easily follows by summing $a^{3}+b^{3} \geqslant a b(a+b), a^{3}+c^{3} \geqslant a c(a+c)$, $b^{3}+c^{3} \geqslant b c(b+c)$ and $a^{3}+b^{3}+c^{3} \geqslant 3 a b c$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
(FYR Macedonia)
|
By the AH mean inequality, we have
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{2}{3(a c+b c)}+\frac{2}{3(a b+a c)}+\frac{2}{3(a b+a c)} \geqslant \frac{3}{a b+a c+b c}
$$
so it only remains to prove that $\frac{3}{a b+a c+b c} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}$, or equivalently
$$
3\left(a^{3}+b^{3}+c^{3}\right) \geqslant(a+b+c)(a b+a c+b c)
$$
The last inequality easily follows by summing $a^{3}+b^{3} \geqslant a b(a+b), a^{3}+c^{3} \geqslant a c(a+c)$, $b^{3}+c^{3} \geqslant b c(b+c)$ and $a^{3}+b^{3}+c^{3} \geqslant 3 a b c$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA1.",
"solution_match": "\n## Solution."
}
|
cc9e73fe-4657-5325-a44e-bb59502287d8
| 605,563
|
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exactly the same route within that time. Determine all possible values of $q$.
(United Kingdom)
|
Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ants meet after $n$ minutes, then
$$
0=x_{A}^{(n)}-x_{B}^{(n)}=P(q)
$$
where $P$ is a polynomial with degree at most $n$ and coefficients in $\{-2,-, 1,0,1,2\}$. Thus if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid 2$ and $b \mid 2$, i.e. $q \in\left\{\frac{1}{2}, 1,2\right\}$.
It is clearly possible when $q=1$.
We argue that $q=\frac{1}{2}$ is not possible. Assume that the ants diverge for the first time after the $k$ th minute, for $k \geqslant 0$. Then
$$
\left|x_{B}^{(k+1)}-x_{A}^{(k+1)}\right|+\left|y_{B}^{(k+1)}-y_{A}^{(k+1)}\right|=2 q^{k}
$$
But also $\left|x_{A}^{(\ell+1)}-x_{A}^{(\ell)}\right|+\left|y_{A}^{(\ell+1)}-y_{A}^{(\ell)}\right|=q^{\ell}$ for each $l \geqslant k+1$, and so
$$
\left|x_{A}^{(n)}-x_{A}^{(k+1)}\right|+\left|y_{A}^{(n)}-y_{A}^{(k+1)}\right| \leqslant q^{k+1}+q^{k+2}+\ldots+q^{n-1}
$$
and similarly for the second ant. Combining (1) and (2) with the triangle inequality, we obtain for any $n \geqslant k+1$
$$
\left|x_{B}^{(n)}-x_{A}^{(n)}\right|+\left|y_{B}^{(n)}-y_{A}^{(n)}\right| \geqslant 2 q^{k}-2\left(q^{k+1}+q^{k+2}+\ldots+q^{n-1}\right),
$$
which is strictly positive for $q=\frac{1}{2}$. So for any $n \geqslant k+1$, the ants cannot meet after $n$ minutes. Thus $q \neq \frac{1}{2}$.
Finally, we show that $q=2$ is also not possible. Suppose to the contrary that there is a pair of routes for $q=2$, meeting after $n$ minutes. Now consider rescaling the plane by a factor $2^{-n}$, and looking at the routes in the opposite direction. This would then be an example for $q=1 / 2$ and we have just shown that this is not possible.
|
q=1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exactly the same route within that time. Determine all possible values of $q$.
(United Kingdom)
|
Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ants meet after $n$ minutes, then
$$
0=x_{A}^{(n)}-x_{B}^{(n)}=P(q)
$$
where $P$ is a polynomial with degree at most $n$ and coefficients in $\{-2,-, 1,0,1,2\}$. Thus if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid 2$ and $b \mid 2$, i.e. $q \in\left\{\frac{1}{2}, 1,2\right\}$.
It is clearly possible when $q=1$.
We argue that $q=\frac{1}{2}$ is not possible. Assume that the ants diverge for the first time after the $k$ th minute, for $k \geqslant 0$. Then
$$
\left|x_{B}^{(k+1)}-x_{A}^{(k+1)}\right|+\left|y_{B}^{(k+1)}-y_{A}^{(k+1)}\right|=2 q^{k}
$$
But also $\left|x_{A}^{(\ell+1)}-x_{A}^{(\ell)}\right|+\left|y_{A}^{(\ell+1)}-y_{A}^{(\ell)}\right|=q^{\ell}$ for each $l \geqslant k+1$, and so
$$
\left|x_{A}^{(n)}-x_{A}^{(k+1)}\right|+\left|y_{A}^{(n)}-y_{A}^{(k+1)}\right| \leqslant q^{k+1}+q^{k+2}+\ldots+q^{n-1}
$$
and similarly for the second ant. Combining (1) and (2) with the triangle inequality, we obtain for any $n \geqslant k+1$
$$
\left|x_{B}^{(n)}-x_{A}^{(n)}\right|+\left|y_{B}^{(n)}-y_{A}^{(n)}\right| \geqslant 2 q^{k}-2\left(q^{k+1}+q^{k+2}+\ldots+q^{n-1}\right),
$$
which is strictly positive for $q=\frac{1}{2}$. So for any $n \geqslant k+1$, the ants cannot meet after $n$ minutes. Thus $q \neq \frac{1}{2}$.
Finally, we show that $q=2$ is also not possible. Suppose to the contrary that there is a pair of routes for $q=2$, meeting after $n$ minutes. Now consider rescaling the plane by a factor $2^{-n}$, and looking at the routes in the opposite direction. This would then be an example for $q=1 / 2$ and we have just shown that this is not possible.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA2.",
"solution_match": "\n## Solution."
}
|
5d53ea16-3fe5-5301-ba9c-b388c2b01ce7
| 605,579
|
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exactly the same route within that time. Determine all possible values of $q$.
(United Kingdom)
|
2.
Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, i,-i\}$. If $\alpha_{n}=\beta_{n}$ for some $n>0$, then
$$
\sum_{k=0}^{n-1}\left(a_{k}-b_{k}\right) q^{k}=0, \quad \text { where } \quad a_{k}-b_{k} \in\{0, \pm 1 \pm i, \pm 2, \pm 2 i\}
$$
Note that the coefficient $a_{k}-b_{k}$ is always divisible by $1+i$ in Gaussian integers: indeed,
$$
c_{k}=\frac{a_{k}-b_{k}}{1+i} \in\{0, \pm 1, \pm i, \pm 1 \pm i\}
$$
Canceling $1+i$, we obtain $c_{0}+c_{1} q+\cdots+c_{n-1} q^{n-1}=0$. Therefore if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid c_{0}$ and $b \mid c_{n-1}$ in Gaussian integers, which is only possible if $a=b=1$.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Two ants start at the same point in the plane. Each minute they choose whether to walk due north, east, south or west. They each walk 1 meter in the first minute. In each subsequent minute the distance they walk is multiplied by a rational number $q>0$. They meet after a whole number of minutes, but have not taken exactly the same route within that time. Determine all possible values of $q$.
(United Kingdom)
|
2.
Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, i,-i\}$. If $\alpha_{n}=\beta_{n}$ for some $n>0$, then
$$
\sum_{k=0}^{n-1}\left(a_{k}-b_{k}\right) q^{k}=0, \quad \text { where } \quad a_{k}-b_{k} \in\{0, \pm 1 \pm i, \pm 2, \pm 2 i\}
$$
Note that the coefficient $a_{k}-b_{k}$ is always divisible by $1+i$ in Gaussian integers: indeed,
$$
c_{k}=\frac{a_{k}-b_{k}}{1+i} \in\{0, \pm 1, \pm i, \pm 1 \pm i\}
$$
Canceling $1+i$, we obtain $c_{0}+c_{1} q+\cdots+c_{n-1} q^{n-1}=0$. Therefore if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid c_{0}$ and $b \mid c_{n-1}$ in Gaussian integers, which is only possible if $a=b=1$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA2.",
"solution_match": "\n## Solution"
}
|
5d53ea16-3fe5-5301-ba9c-b388c2b01ce7
| 605,579
|
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$
|
We shall prove that
$$
\binom{n}{k} \geqslant\left(\frac{2 n+1-k}{k+1}\right)^{k} \quad \text { for all } \quad k=0,1, \ldots, n
$$
The result will follow immediately, as $\sum_{k=0}^{n}\binom{n}{k}=2^{n}$.
Note that $(*)$ is trivial for $k=0$ and $k=n$. For $0<k<n$, by Hölder's inequality we have
$$
\binom{n}{k}=\left(1+\frac{n-k}{k}\right) \cdot\left(1+\frac{n-k}{k-1}\right) \cdots\left(1+\frac{n-k}{1}\right) \geqslant\left(1+\frac{n-k}{\sqrt[k]{k!}}\right)^{k} .
$$
Hence, it is enough to prove that
$$
1+\frac{n-k}{\sqrt[k]{k!}} \geqslant \frac{2 n+1-k}{k+1}
$$
This is equivalent to $\sqrt[k]{k!} \leqslant \frac{k+1}{2}$, which follows from $\sqrt[k]{k!} \leqslant \frac{1+2+\cdots+k}{k}=\frac{k+1}{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$
|
We shall prove that
$$
\binom{n}{k} \geqslant\left(\frac{2 n+1-k}{k+1}\right)^{k} \quad \text { for all } \quad k=0,1, \ldots, n
$$
The result will follow immediately, as $\sum_{k=0}^{n}\binom{n}{k}=2^{n}$.
Note that $(*)$ is trivial for $k=0$ and $k=n$. For $0<k<n$, by Hölder's inequality we have
$$
\binom{n}{k}=\left(1+\frac{n-k}{k}\right) \cdot\left(1+\frac{n-k}{k-1}\right) \cdots\left(1+\frac{n-k}{1}\right) \geqslant\left(1+\frac{n-k}{\sqrt[k]{k!}}\right)^{k} .
$$
Hence, it is enough to prove that
$$
1+\frac{n-k}{\sqrt[k]{k!}} \geqslant \frac{2 n+1-k}{k+1}
$$
This is equivalent to $\sqrt[k]{k!} \leqslant \frac{k+1}{2}$, which follows from $\sqrt[k]{k!} \leqslant \frac{1+2+\cdots+k}{k}=\frac{k+1}{2}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA3.",
"solution_match": "\n## Solution."
}
|
2e3db6c1-8421-5b49-8260-aca4298e9aaf
| 605,606
|
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$
|
2.
As in the previous solution, it is enough to prove (*).
First, we prove that
$$
(n-i+1)(n-k+i)(k+1)^{2} \geqslant i(k-i+1)(2 n+1-k)^{2} \quad \text { for all } i=1,2, \ldots, k
$$
Let us denote the left hand side of the previous inequality with $L$ and the left hand side with $R$. Then
$$
\begin{aligned}
& L=(n+1)^{2}(k+1)^{2}-(n+1)(k+1)^{3}+i(k-i+1)(k+1)^{2} \\
& R=4 i(k-i+1)(n+1)^{2}-4 i(k-i+1)(n+1)(k+1)+i(k-i+1)(k+1)^{2}
\end{aligned}
$$
So, it is enough to prove that
$$
(n-k)(k+1)^{2} \geqslant 4 i(k-i+1)(n-k),
$$
which follows from
$$
(k+1)^{2}-4 i(k-i+1)=(k+1-2 i)^{2} \geqslant 0 .
$$
Now, by ( $\#$ ) we have
$$
\binom{n}{k}^{2}=\prod_{i=1}^{k} \frac{(n-i+1)(n-k+i)}{i(k-i+1)} \geqslant \prod_{i=1}^{k}\left(\frac{2 n+1-k}{k+1}\right)^{2}=\left(\frac{2 n+1-k}{k+1}\right)^{2 k}
$$
which completes our proof.
|
proof
|
Yes
|
Incomplete
|
proof
|
Inequalities
|
Show that for every positive integer $n$ we have:
$$
\sum_{k=0}^{n}\left(\frac{2 n+1-k}{k+1}\right)^{k}=\left(\frac{2 n+1}{1}\right)^{0}+\left(\frac{2 n}{2}\right)^{1}+\cdots+\left(\frac{n+1}{n+1}\right)^{n} \leqslant 2^{n}
$$
|
2.
As in the previous solution, it is enough to prove (*).
First, we prove that
$$
(n-i+1)(n-k+i)(k+1)^{2} \geqslant i(k-i+1)(2 n+1-k)^{2} \quad \text { for all } i=1,2, \ldots, k
$$
Let us denote the left hand side of the previous inequality with $L$ and the left hand side with $R$. Then
$$
\begin{aligned}
& L=(n+1)^{2}(k+1)^{2}-(n+1)(k+1)^{3}+i(k-i+1)(k+1)^{2} \\
& R=4 i(k-i+1)(n+1)^{2}-4 i(k-i+1)(n+1)(k+1)+i(k-i+1)(k+1)^{2}
\end{aligned}
$$
So, it is enough to prove that
$$
(n-k)(k+1)^{2} \geqslant 4 i(k-i+1)(n-k),
$$
which follows from
$$
(k+1)^{2}-4 i(k-i+1)=(k+1-2 i)^{2} \geqslant 0 .
$$
Now, by ( $\#$ ) we have
$$
\binom{n}{k}^{2}=\prod_{i=1}^{k} \frac{(n-i+1)(n-k+i)}{i(k-i+1)} \geqslant \prod_{i=1}^{k}\left(\frac{2 n+1-k}{k+1}\right)^{2}=\left(\frac{2 n+1-k}{k+1}\right)^{2 k}
$$
which completes our proof.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA3.",
"solution_match": "\n## Solution"
}
|
2e3db6c1-8421-5b49-8260-aca4298e9aaf
| 605,606
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania)
|
First, we show that
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a b+b c+c a \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c
$$
By AG inequality, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{c}{a}\right)+\frac{1}{3}\left(\frac{b}{c}+\frac{b}{c}+\frac{a}{b}\right)+\frac{1}{3}\left(\frac{c}{a}+\frac{c}{a}+\frac{b}{c}\right) \\
& \geqslant \frac{\sqrt[3]{a c}}{\sqrt[3]{b^{2}}}+\frac{\sqrt[3]{b a}}{\sqrt[3]{c^{2}}}+\frac{\sqrt[3]{c b}}{\sqrt[3]{a^{2}}}=\frac{\sqrt[3]{a b c}}{b}+\frac{\sqrt[3]{a b c}}{c}+\frac{\sqrt[3]{a b c}}{a} \\
& =a b+b c+c a
\end{aligned}
$$
Similarly, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\right)+\frac{1}{3}\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\right)+\frac{1}{3}\left(\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\right) \\
& \geqslant \frac{\sqrt[3]{a^{2}}}{\sqrt[3]{b c}}+\frac{\sqrt[3]{b^{2}}}{\sqrt[3]{c a}}+\frac{\sqrt[3]{c^{2}}}{\sqrt[3]{a b}}=\frac{a}{\sqrt[3]{a b c}}+\frac{b}{\sqrt[3]{a b c}}+\frac{c}{\sqrt[3]{a b c}} \\
& =a+b+c
\end{aligned}
$$
which completes our proof of ( $\dagger$ ).
By Cauchy-Schwarz inequality we have
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}}\right) \geqslant\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}
$$
which together with $\left(a^{2}+b^{2}+c^{2}\right)\left(1 / a^{2}+1 / b^{2}+1 / c^{2}\right) \geqslant 9$ leads to
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}+9 \geqslant 6\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)
$$
Now, the desired inequality follows from ( $\dagger$ ).
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania)
|
First, we show that
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a b+b c+c a \quad \text { and } \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c
$$
By AG inequality, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{c}{a}\right)+\frac{1}{3}\left(\frac{b}{c}+\frac{b}{c}+\frac{a}{b}\right)+\frac{1}{3}\left(\frac{c}{a}+\frac{c}{a}+\frac{b}{c}\right) \\
& \geqslant \frac{\sqrt[3]{a c}}{\sqrt[3]{b^{2}}}+\frac{\sqrt[3]{b a}}{\sqrt[3]{c^{2}}}+\frac{\sqrt[3]{c b}}{\sqrt[3]{a^{2}}}=\frac{\sqrt[3]{a b c}}{b}+\frac{\sqrt[3]{a b c}}{c}+\frac{\sqrt[3]{a b c}}{a} \\
& =a b+b c+c a
\end{aligned}
$$
Similarly, we have
$$
\begin{aligned}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} & =\frac{1}{3}\left(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\right)+\frac{1}{3}\left(\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\right)+\frac{1}{3}\left(\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\right) \\
& \geqslant \frac{\sqrt[3]{a^{2}}}{\sqrt[3]{b c}}+\frac{\sqrt[3]{b^{2}}}{\sqrt[3]{c a}}+\frac{\sqrt[3]{c^{2}}}{\sqrt[3]{a b}}=\frac{a}{\sqrt[3]{a b c}}+\frac{b}{\sqrt[3]{a b c}}+\frac{c}{\sqrt[3]{a b c}} \\
& =a+b+c
\end{aligned}
$$
which completes our proof of ( $\dagger$ ).
By Cauchy-Schwarz inequality we have
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a^{2}}\right) \geqslant\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}
$$
which together with $\left(a^{2}+b^{2}+c^{2}\right)\left(1 / a^{2}+1 / b^{2}+1 / c^{2}\right) \geqslant 9$ leads to
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}+9 \geqslant 6\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)
$$
Now, the desired inequality follows from ( $\dagger$ ).
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA4.",
"solution_match": "\n## Solution."
}
|
63662bcd-26c3-51ba-9f30-203283c4d5ba
| 605,625
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania)
|
2.
Set $a=x^{3}, b=y^{3}, c=z^{3}$ and denote $T_{p, q, r}=\sum_{s y m} x^{p} y^{q} z^{r}=x^{p} y^{q} z^{r}+y^{p} x^{q} z^{r}+\cdots$. The given inequality is expanded into
$$
4 T_{12,6,0}+2 T_{6,6,6} \geqslant 3 T_{8,5,5}+3 T_{7,7,4}
$$
Applying the Schur inequality on triples $\left(x^{4} y^{2}, y^{4} z^{2}, z^{4} x^{2}\right)$ and $\left(x^{2} y^{4}, y^{2} z^{4}, z^{2} x^{4}\right)$ and summing them up yields
$$
T_{12,6,0}+T_{6,6,6} \geqslant T_{10,4,4}+T_{8,8,2}
$$
On the other hand, by the Muirhead inequality we have
$$
T_{12,6,0} \geqslant T_{6,6,6}, \quad T_{10,4,4} \geqslant T_{8,5,5}, \quad T_{8,8,2} \geqslant T_{7,7,4} .
$$
The four inequalities in (1) and (2) imply the desired inequality.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that the following inequality holds:
$$
2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3(a+b+c+a b+b c+c a)
$$
(Romania)
|
2.
Set $a=x^{3}, b=y^{3}, c=z^{3}$ and denote $T_{p, q, r}=\sum_{s y m} x^{p} y^{q} z^{r}=x^{p} y^{q} z^{r}+y^{p} x^{q} z^{r}+\cdots$. The given inequality is expanded into
$$
4 T_{12,6,0}+2 T_{6,6,6} \geqslant 3 T_{8,5,5}+3 T_{7,7,4}
$$
Applying the Schur inequality on triples $\left(x^{4} y^{2}, y^{4} z^{2}, z^{4} x^{2}\right)$ and $\left(x^{2} y^{4}, y^{2} z^{4}, z^{2} x^{4}\right)$ and summing them up yields
$$
T_{12,6,0}+T_{6,6,6} \geqslant T_{10,4,4}+T_{8,8,2}
$$
On the other hand, by the Muirhead inequality we have
$$
T_{12,6,0} \geqslant T_{6,6,6}, \quad T_{10,4,4} \geqslant T_{8,5,5}, \quad T_{8,8,2} \geqslant T_{7,7,4} .
$$
The four inequalities in (1) and (2) imply the desired inequality.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA4.",
"solution_match": "\n## Solution"
}
|
63662bcd-26c3-51ba-9f30-203283c4d5ba
| 605,625
|
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a concave function and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Given that
$$
f(x+y)+f(x-y)-2 f(x)=g(x) y^{2}
$$
for all $x, y \in \mathbb{R}$, prove that $f$ is a quadratic function.
(Bulgaria)
|
We plug in the pairs $(a, x),(a, 2 x),(a+x, x)$ and $(a-x, x)$ to get
$$
\begin{aligned}
f(a+x)+f(a-x)-2 f(a) & =g(a) x^{2} \\
f(a+2 x)+f(a-2 x)-2 f(a) & =4 g(a) x^{2} \\
f(a+2 x)+f(a)-2 f(a+x) & =g(a+x) x^{2} \\
f(a-2 x)+f(a)-2 f(a-x) & =g(a-x) x^{2}
\end{aligned}
$$
respectively. Combining these equations in the form $2 E_{1}-E_{2}+E_{3}+E_{4}$ the left hand side vanishes, yielding an equation in $g:(g(a+x)+g(a-x)-2 g(a)) x^{2}=0$, i.e.
$$
g(a)=\frac{g(a+x)+g(a-x)}{2}
$$
Since $g$ is continuous, it must be linear, i.e. $g(x)=c_{1} x+c_{0}$. However, the original equation for $x=y$ together with the concavity condition now gives us
$$
0 \geqslant f(2 x)+f(0)-2 f(x)=\left(x c_{1}+c_{0}\right) x^{2}
$$
for all $x$, which is only possible if $c_{1}=0$. Thus $g(x) \equiv c_{0}=2 A$ is constant and
$$
f(x+y)+f(x-y)-2 f(x)=2 A y^{2}
$$
This suggests that $f$ is a quadratic function, so we can set $f(x)=A x^{2}+f_{1}(x)$. Then $(*)$ becomes $f_{1}(x+y)+f_{1}(x-y)-2 f_{1}(x)=0$, so an easy induction gives us
$$
f_{1}(n x)-f_{1}(0)=n\left(f_{1}(x)-f_{1}(0)\right) \quad \text { for all } \quad n \in \mathbb{Z}
$$
By setting $f_{1}(0)=C$ and $f_{1}(1)=B+C$ we obtain $f_{1}(x)=B x+C$ and $f(x)=$ $A x^{2}+B x+C$ for all $x \in \mathbb{Q}$. By concavity of $f$ we conclude that $f(x)=A x^{2}+B x+C$ for all real $x$.
## Remark.
In fact, $(*)$ implies that the second derivative of $f$ is constant by taking $y \rightarrow 0$ and the problem is solved. The solution presented here avoids use of derivatives.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a concave function and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Given that
$$
f(x+y)+f(x-y)-2 f(x)=g(x) y^{2}
$$
for all $x, y \in \mathbb{R}$, prove that $f$ is a quadratic function.
(Bulgaria)
|
We plug in the pairs $(a, x),(a, 2 x),(a+x, x)$ and $(a-x, x)$ to get
$$
\begin{aligned}
f(a+x)+f(a-x)-2 f(a) & =g(a) x^{2} \\
f(a+2 x)+f(a-2 x)-2 f(a) & =4 g(a) x^{2} \\
f(a+2 x)+f(a)-2 f(a+x) & =g(a+x) x^{2} \\
f(a-2 x)+f(a)-2 f(a-x) & =g(a-x) x^{2}
\end{aligned}
$$
respectively. Combining these equations in the form $2 E_{1}-E_{2}+E_{3}+E_{4}$ the left hand side vanishes, yielding an equation in $g:(g(a+x)+g(a-x)-2 g(a)) x^{2}=0$, i.e.
$$
g(a)=\frac{g(a+x)+g(a-x)}{2}
$$
Since $g$ is continuous, it must be linear, i.e. $g(x)=c_{1} x+c_{0}$. However, the original equation for $x=y$ together with the concavity condition now gives us
$$
0 \geqslant f(2 x)+f(0)-2 f(x)=\left(x c_{1}+c_{0}\right) x^{2}
$$
for all $x$, which is only possible if $c_{1}=0$. Thus $g(x) \equiv c_{0}=2 A$ is constant and
$$
f(x+y)+f(x-y)-2 f(x)=2 A y^{2}
$$
This suggests that $f$ is a quadratic function, so we can set $f(x)=A x^{2}+f_{1}(x)$. Then $(*)$ becomes $f_{1}(x+y)+f_{1}(x-y)-2 f_{1}(x)=0$, so an easy induction gives us
$$
f_{1}(n x)-f_{1}(0)=n\left(f_{1}(x)-f_{1}(0)\right) \quad \text { for all } \quad n \in \mathbb{Z}
$$
By setting $f_{1}(0)=C$ and $f_{1}(1)=B+C$ we obtain $f_{1}(x)=B x+C$ and $f(x)=$ $A x^{2}+B x+C$ for all $x \in \mathbb{Q}$. By concavity of $f$ we conclude that $f(x)=A x^{2}+B x+C$ for all real $x$.
## Remark.
In fact, $(*)$ implies that the second derivative of $f$ is constant by taking $y \rightarrow 0$ and the problem is solved. The solution presented here avoids use of derivatives.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA5.",
"solution_match": "\n## Solution."
}
|
2db57d65-4f42-5b54-902e-d3cce666732f
| 605,646
|
Let $n$ be a positive integer and let $x_{1}, \ldots, x_{n}$ be real numbers. Show that
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\frac{12\left(\sum_{i=1}^{n} i x_{i}\right)^{2}}{n(n+1)(n+2)(3 n+1)}
$$
|
Let $S=\frac{1}{n+1} \sum_{i=1}^{n} x_{i}$, and $y_{i}=x_{i}-S$ for $1 \leqslant i \leqslant n$. Then we have
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S
$$
and
$$
\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} x_{i}^{2}-2 S \sum_{i=1}^{n} x_{i}+n S^{2}=\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-S^{2}
$$
Now, by the Cauchy-Schwarz inequality
$$
\begin{aligned}
\left(\sum_{i=1}^{n} i x_{i}\right)^{2} & =\left(\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S\right)^{2} \\
& \leqslant\left(\sum_{i=1}^{n} i^{2}+\frac{n^{2}(n+1)^{2}}{4}\right)\left(\sum_{i=1}^{n} y_{i}^{2}+S^{2}\right) \\
& =\frac{n(n+1)(n+2)(3 n+1)}{12} \cdot\left(\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}\right)
\end{aligned}
$$
which completes our proof.
## Remark.
It can be checked that equality holds if and only if $x_{i}=c(n(n+1)+2 i)$ for $1 \leqslant i \leqslant n$ and some $c \in \mathbb{R}$.
## Combinatorics
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $n$ be a positive integer and let $x_{1}, \ldots, x_{n}$ be real numbers. Show that
$$
\sum_{i=1}^{n} x_{i}^{2} \geqslant \frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}+\frac{12\left(\sum_{i=1}^{n} i x_{i}\right)^{2}}{n(n+1)(n+2)(3 n+1)}
$$
|
Let $S=\frac{1}{n+1} \sum_{i=1}^{n} x_{i}$, and $y_{i}=x_{i}-S$ for $1 \leqslant i \leqslant n$. Then we have
$$
\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S
$$
and
$$
\sum_{i=1}^{n} y_{i}^{2}=\sum_{i=1}^{n} x_{i}^{2}-2 S \sum_{i=1}^{n} x_{i}+n S^{2}=\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}-S^{2}
$$
Now, by the Cauchy-Schwarz inequality
$$
\begin{aligned}
\left(\sum_{i=1}^{n} i x_{i}\right)^{2} & =\left(\sum_{i=1}^{n} i y_{i}+\frac{n(n+1)}{2} S\right)^{2} \\
& \leqslant\left(\sum_{i=1}^{n} i^{2}+\frac{n^{2}(n+1)^{2}}{4}\right)\left(\sum_{i=1}^{n} y_{i}^{2}+S^{2}\right) \\
& =\frac{n(n+1)(n+2)(3 n+1)}{12} \cdot\left(\sum_{i=1}^{n} x_{i}^{2}-\frac{1}{n+1}\left(\sum_{i=1}^{n} x_{i}\right)^{2}\right)
\end{aligned}
$$
which completes our proof.
## Remark.
It can be checked that equality holds if and only if $x_{i}=c(n(n+1)+2 i)$ for $1 \leqslant i \leqslant n$ and some $c \in \mathbb{R}$.
## Combinatorics
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nA6.",
"solution_match": "\n## Solution."
}
|
8b7649ad-b004-55f8-bce9-d0a381dae0f2
| 605,656
|
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner had a lower initial ranking than the loser. At the end of the league, the players are ranked according to number of wins, with the initial ranking used to rank players with the same number of wins. It turns out that the final ranking is the same as the initial ranking. What is the largest possible number of upsets?
(United Kingdom)
|
Answer: $\frac{(N-1)(3 N-1)}{8}$.
Suppose the players are ranked $1,2, \ldots, N=2 n+1$, where 1 is the highest ranking.
For $k \leqslant n$, the player ranked $k$ could have beaten at most $k-1$ players with a higher ranking. Thus the top $n$ players could have made at most $\sum_{k=1}^{n}(k-1)=\frac{n(n-1)}{2}$ upsets. On the other hand, the average score of all $2 n+1$ players is $n$, so the average score of the bottom $n+1$ players is not more than $n$, which implies that these $n+1$ players have at most $n(n+1)$ wins in total. Hence the total number of upsets is at most
$$
\frac{n(n-1)}{2}+n(n+1)=\frac{n(3 n+1)}{2}=\frac{(N-1)(3 N-1)}{8} .
$$
An example can be constructed as follows. Suppose that, for $1 \leqslant i \leqslant 2 n+1$, the player ranked $i$ beats the players ranked $i-1, i-2, \ldots, i-n$ (the rankings are counted modulo $N)$ and loses to the rest of the players. Thus each player has exactly $n$ wins. The player ranked $i$ for $i \leqslant n$ made $i-1$ upsets, whereas the player ranked $i$ for $i>n$ made $n$ upsets, so the total number of upsets is exactly $\sum_{i=1}^{n}(i-1)+(n+1) n=\frac{n(3 n+1)}{2}$.
|
\frac{(N-1)(3 N-1)}{8}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner had a lower initial ranking than the loser. At the end of the league, the players are ranked according to number of wins, with the initial ranking used to rank players with the same number of wins. It turns out that the final ranking is the same as the initial ranking. What is the largest possible number of upsets?
(United Kingdom)
|
Answer: $\frac{(N-1)(3 N-1)}{8}$.
Suppose the players are ranked $1,2, \ldots, N=2 n+1$, where 1 is the highest ranking.
For $k \leqslant n$, the player ranked $k$ could have beaten at most $k-1$ players with a higher ranking. Thus the top $n$ players could have made at most $\sum_{k=1}^{n}(k-1)=\frac{n(n-1)}{2}$ upsets. On the other hand, the average score of all $2 n+1$ players is $n$, so the average score of the bottom $n+1$ players is not more than $n$, which implies that these $n+1$ players have at most $n(n+1)$ wins in total. Hence the total number of upsets is at most
$$
\frac{n(n-1)}{2}+n(n+1)=\frac{n(3 n+1)}{2}=\frac{(N-1)(3 N-1)}{8} .
$$
An example can be constructed as follows. Suppose that, for $1 \leqslant i \leqslant 2 n+1$, the player ranked $i$ beats the players ranked $i-1, i-2, \ldots, i-n$ (the rankings are counted modulo $N)$ and loses to the rest of the players. Thus each player has exactly $n$ wins. The player ranked $i$ for $i \leqslant n$ made $i-1$ upsets, whereas the player ranked $i$ for $i>n$ made $n$ upsets, so the total number of upsets is exactly $\sum_{i=1}^{n}(i-1)+(n+1) n=\frac{n(3 n+1)}{2}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nC1.",
"solution_match": "\n## Solution."
}
|
6d36277a-8326-5c75-85e9-10849e17de79
| 605,666
|
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner had a lower initial ranking than the loser. At the end of the league, the players are ranked according to number of wins, with the initial ranking used to rank players with the same number of wins. It turns out that the final ranking is the same as the initial ranking. What is the largest possible number of upsets?
(United Kingdom)
|
2.
Write $N=2 n+1$. We only prove the upper bound.
Consider a tournament $\mathbb{T}$ with correct final ranking, but where not everyone won $n$ matches. Let $A$ be the worst-ranked player with the maximal number of wins, and let $B$ be the best-ranked player with minimal wins. Clearly, $A$ was ranked above $B$.
Assume $A$ beat $B$. Consider the tournament $\mathbb{T}^{\prime}$ obtained from $\mathbb{T}$ by reversing this result, and keeping all others the same. So $B$ beat $A$, which is an upset. $A$ is now the bestranked player with the second-most number of wins; $B$ is now the worst-ranked player with the second-least number of wins, and so the final ranking of $\mathbb{T}^{\prime}$ is still correct, but with one more upset than in $\mathbb{T}$.
Alternatively, assume $B$ beat $A$. Then there must have been a player $C$ such that $A$ beat $C$ and $C$ beat $B$. These are upsets if, respectively, $C$ was ranked above $A$, or
below $B$. It therefore cannot be the case that both of the matches involving $C$ and $\{A, B\}$ were upsets. Consider the tournament $\mathbb{T}^{\prime}$ obtained from $\mathbb{T}$ by reversing these two matches. $C$ 's number of wins stays fixed, while as before $A$ is now the best-ranked player with the second-most wins, and similar for $B$. Thus in $\mathbb{T}^{\prime}$ the final ranking is still correct, with either the same number of upsets as $\mathbb{T}$, or two more upsets than $\mathbb{T}$.
If we iterate this procedure, we eventually obtain a tournament $\overline{\mathbb{T}}$ where everyone won exactly $n$ matches, and with at least as many upsets as in the original tournament $\mathbb{T}$. We now bound the number of upsets in such a tournament $\overline{\mathbb{T}}$. Suppose the player ranked $i \leqslant \frac{N+1}{2}$ beat $K$ higher ranked players. Obviously $K \leqslant i-1$. Then the number of upsets involving $i$ is
$$
2 K+\frac{N+1}{2}-i \leqslant 2(i-1)+\frac{N+1}{2}-i=\frac{N-1}{2}+i-1 .
$$
Similarly, for $i \geqslant \frac{N+1}{2}$ one proves that the number of upsets involving $i$ is at most $\frac{N-1}{2}+i-1$.
Finally, summing over all values of $i$ and dividing by 2 we obtain the desired result.
## Remark.
We demand $N$ odd to avoid candidates providing a case distinction, rather than because the construction or the bounding argument is significantly different.
|
2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner had a lower initial ranking than the loser. At the end of the league, the players are ranked according to number of wins, with the initial ranking used to rank players with the same number of wins. It turns out that the final ranking is the same as the initial ranking. What is the largest possible number of upsets?
(United Kingdom)
|
2.
Write $N=2 n+1$. We only prove the upper bound.
Consider a tournament $\mathbb{T}$ with correct final ranking, but where not everyone won $n$ matches. Let $A$ be the worst-ranked player with the maximal number of wins, and let $B$ be the best-ranked player with minimal wins. Clearly, $A$ was ranked above $B$.
Assume $A$ beat $B$. Consider the tournament $\mathbb{T}^{\prime}$ obtained from $\mathbb{T}$ by reversing this result, and keeping all others the same. So $B$ beat $A$, which is an upset. $A$ is now the bestranked player with the second-most number of wins; $B$ is now the worst-ranked player with the second-least number of wins, and so the final ranking of $\mathbb{T}^{\prime}$ is still correct, but with one more upset than in $\mathbb{T}$.
Alternatively, assume $B$ beat $A$. Then there must have been a player $C$ such that $A$ beat $C$ and $C$ beat $B$. These are upsets if, respectively, $C$ was ranked above $A$, or
below $B$. It therefore cannot be the case that both of the matches involving $C$ and $\{A, B\}$ were upsets. Consider the tournament $\mathbb{T}^{\prime}$ obtained from $\mathbb{T}$ by reversing these two matches. $C$ 's number of wins stays fixed, while as before $A$ is now the best-ranked player with the second-most wins, and similar for $B$. Thus in $\mathbb{T}^{\prime}$ the final ranking is still correct, with either the same number of upsets as $\mathbb{T}$, or two more upsets than $\mathbb{T}$.
If we iterate this procedure, we eventually obtain a tournament $\overline{\mathbb{T}}$ where everyone won exactly $n$ matches, and with at least as many upsets as in the original tournament $\mathbb{T}$. We now bound the number of upsets in such a tournament $\overline{\mathbb{T}}$. Suppose the player ranked $i \leqslant \frac{N+1}{2}$ beat $K$ higher ranked players. Obviously $K \leqslant i-1$. Then the number of upsets involving $i$ is
$$
2 K+\frac{N+1}{2}-i \leqslant 2(i-1)+\frac{N+1}{2}-i=\frac{N-1}{2}+i-1 .
$$
Similarly, for $i \geqslant \frac{N+1}{2}$ one proves that the number of upsets involving $i$ is at most $\frac{N-1}{2}+i-1$.
Finally, summing over all values of $i$ and dividing by 2 we obtain the desired result.
## Remark.
We demand $N$ odd to avoid candidates providing a case distinction, rather than because the construction or the bounding argument is significantly different.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nC1.",
"solution_match": "\n## Solution"
}
|
6d36277a-8326-5c75-85e9-10849e17de79
| 605,666
|
Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, they choose a pile with an even number of coins and move half of the coins of this pile to the other pile. The game ends if a player cannot move, or if we reach a previously reached position. In the first case, the player who cannot move loses. In the second case, the game is declared a draw.
Determine all pairs $(a, b)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy.
(Cyprus)
|
By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\right\}$. We shall prove that Bob has a winning strategy if and only if the initial position is $k$-happy for some even $k$.
- Given a 0-happy position, the player in turn is unable to play and loses.
- Given a $k$-happy position $(a, b)$ with $k \geqslant 1$, the player in turn will transform it into one of the positions $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$ and $\left(b+\frac{1}{2} a, \frac{1}{2} a\right)$, both of which are ( $k-1$ )-happy because $v_{2}\left(a+\frac{1}{2} b\right)=v_{2}\left(\frac{1}{2} b\right)=v_{2}\left(b+\frac{1}{2} a\right)=v_{2}\left(\frac{1}{2} a\right)=k-1$.
Therefore, if the starting position is $k$-happy, after $k$ moves they will get stuck at a 0 -happy position, so Bob will win if and only if $k$ is even.
- Given a $k$-unhappy position $(a, b)$ with $k$ odd and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice can move to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$. Since $v_{2}\left(\frac{1}{2} a\right)=v_{2}\left(b+\frac{1}{2} a\right)=k-1$, this position is ( $k-1$ )-happy with $2 \mid k-1$, so Alice will win.
- Given a $k$-unhappy position $(a, b)$ with $k$ even and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice must not play to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$, because the new position is $(k-1)$-happy and will lead to Bob's victory. Thus she must play to position $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$. We claim that this position is also $k$-unhappy. Indeed, if $\ell>k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)=$ $k<v_{2}\left(\frac{1}{2} b\right)=\ell-1$, whereas if $\ell=k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)>v_{2}\left(\frac{1}{2} b\right)=k$.
Hence a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, they choose a pile with an even number of coins and move half of the coins of this pile to the other pile. The game ends if a player cannot move, or if we reach a previously reached position. In the first case, the player who cannot move loses. In the second case, the game is declared a draw.
Determine all pairs $(a, b)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy.
(Cyprus)
|
By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\right\}$. We shall prove that Bob has a winning strategy if and only if the initial position is $k$-happy for some even $k$.
- Given a 0-happy position, the player in turn is unable to play and loses.
- Given a $k$-happy position $(a, b)$ with $k \geqslant 1$, the player in turn will transform it into one of the positions $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$ and $\left(b+\frac{1}{2} a, \frac{1}{2} a\right)$, both of which are ( $k-1$ )-happy because $v_{2}\left(a+\frac{1}{2} b\right)=v_{2}\left(\frac{1}{2} b\right)=v_{2}\left(b+\frac{1}{2} a\right)=v_{2}\left(\frac{1}{2} a\right)=k-1$.
Therefore, if the starting position is $k$-happy, after $k$ moves they will get stuck at a 0 -happy position, so Bob will win if and only if $k$ is even.
- Given a $k$-unhappy position $(a, b)$ with $k$ odd and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice can move to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$. Since $v_{2}\left(\frac{1}{2} a\right)=v_{2}\left(b+\frac{1}{2} a\right)=k-1$, this position is ( $k-1$ )-happy with $2 \mid k-1$, so Alice will win.
- Given a $k$-unhappy position $(a, b)$ with $k$ even and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice must not play to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$, because the new position is $(k-1)$-happy and will lead to Bob's victory. Thus she must play to position $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$. We claim that this position is also $k$-unhappy. Indeed, if $\ell>k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)=$ $k<v_{2}\left(\frac{1}{2} b\right)=\ell-1$, whereas if $\ell=k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)>v_{2}\left(\frac{1}{2} b\right)=k$.
Hence a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nC2.",
"solution_match": "\n## Solution."
}
|
072f0eb4-10e3-5014-899f-62f4b2fe2937
| 605,687
|
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consecutive emeralds with a sapphire and a ruby, where the sapphire is on the left of the ruby.
(3 ) If we find two consecutive sapphires then we can remove them.
(4 ) If we find consecutively and in this order a ruby, an emerald, and a sapphire, then we can remove them.
Furthermore we can also reverse all of the above operations. For example, by reversing $\left(3^{\circ}\right)$ we can put two consecutive sapphires on any position we wish.
Initially the necklace has one sapphire (and no other precious stones). Decide, with proof, whether there is a finite sequence of steps such that at the end of this sequence the necklace contains one emerald (and no other precious stones).
Remark. A necklace is open if its precious stones are on a line from left to right. We are not allowed to move a precious stone from the rightmost position to the leftmost as we would be able to do if the necklace was closed.
(Cyprus)
|
For each precious stone on the necklace, we define its value as $(-1)^{r} \cdot s$, where $r$ denotes the number of emeralds and sapphires preceding it, and $s$ equals $-2,1$ or -1 for a ruby, emerald or sapphire, respectively.
The value of the necklace is equal to the sum of the values of its precious stones. We claim that the value of the necklace is invariant modulo 6 .
Suppose for example that we remove two consecutive rubies, and suppose there is an even number of emeralds and sapphires preceding them. The value of each ruby is -2 so by removing them we increase the value of the necklace by 4 . The emerald that we add had an even number of emeralds and sapphires preceding it, so its value is 1 . The sapphire that we add has an odd number of emeralds and sapphires preceding it (accounting for the added emerald), so its value is 1 . No other precious stone changes value, so the total increase of the value of the necklace is 6 .
Similarly we can check that all of the other operations and their inverses also leave the value of the necklace invariant modulo 6 .
Since the necklace containing just one sapphire has value -1 , whereas the necklace containing just one emerald has value 1 , there is no desired sequence of steps.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consecutive emeralds with a sapphire and a ruby, where the sapphire is on the left of the ruby.
(3 ) If we find two consecutive sapphires then we can remove them.
(4 ) If we find consecutively and in this order a ruby, an emerald, and a sapphire, then we can remove them.
Furthermore we can also reverse all of the above operations. For example, by reversing $\left(3^{\circ}\right)$ we can put two consecutive sapphires on any position we wish.
Initially the necklace has one sapphire (and no other precious stones). Decide, with proof, whether there is a finite sequence of steps such that at the end of this sequence the necklace contains one emerald (and no other precious stones).
Remark. A necklace is open if its precious stones are on a line from left to right. We are not allowed to move a precious stone from the rightmost position to the leftmost as we would be able to do if the necklace was closed.
(Cyprus)
|
For each precious stone on the necklace, we define its value as $(-1)^{r} \cdot s$, where $r$ denotes the number of emeralds and sapphires preceding it, and $s$ equals $-2,1$ or -1 for a ruby, emerald or sapphire, respectively.
The value of the necklace is equal to the sum of the values of its precious stones. We claim that the value of the necklace is invariant modulo 6 .
Suppose for example that we remove two consecutive rubies, and suppose there is an even number of emeralds and sapphires preceding them. The value of each ruby is -2 so by removing them we increase the value of the necklace by 4 . The emerald that we add had an even number of emeralds and sapphires preceding it, so its value is 1 . The sapphire that we add has an odd number of emeralds and sapphires preceding it (accounting for the added emerald), so its value is 1 . No other precious stone changes value, so the total increase of the value of the necklace is 6 .
Similarly we can check that all of the other operations and their inverses also leave the value of the necklace invariant modulo 6 .
Since the necklace containing just one sapphire has value -1 , whereas the necklace containing just one emerald has value 1 , there is no desired sequence of steps.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nC3.",
"solution_match": "\n## Solution."
}
|
b41b348c-30ca-58db-a5d1-fe1393f107ed
| 605,699
|
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consecutive emeralds with a sapphire and a ruby, where the sapphire is on the left of the ruby.
(3 ) If we find two consecutive sapphires then we can remove them.
(4 ) If we find consecutively and in this order a ruby, an emerald, and a sapphire, then we can remove them.
Furthermore we can also reverse all of the above operations. For example, by reversing $\left(3^{\circ}\right)$ we can put two consecutive sapphires on any position we wish.
Initially the necklace has one sapphire (and no other precious stones). Decide, with proof, whether there is a finite sequence of steps such that at the end of this sequence the necklace contains one emerald (and no other precious stones).
Remark. A necklace is open if its precious stones are on a line from left to right. We are not allowed to move a precious stone from the rightmost position to the leftmost as we would be able to do if the necklace was closed.
(Cyprus)
|
2.
Write $a, b$ and $c$ respectively for a ruby, emerald and sapphire. Each necklace corresponds to an element of a group $G$ containing elements $a, b, c$. If we impose the conditions $a^{2}=b c, b^{3}=c a, c^{2}=1$ and $a b c=1$, the allowed operations will preserve this element. In this group we have $c=a b$ (since $c^{2}=a b c$ ), i.e. $b=a^{-1} c$ and using this relation we obtain $a^{3}=c^{2}=\left(a^{-1} c\right)^{4}=1$. Thus we can take $G=\mathbb{S}_{4}, a=(1,2,3)$, $c=(1,4)$ and $b=a^{-1} c=(1,4,3,2)$. The initial and final necklaces should correspond to elements $c$ and $b$, respectively, so the desired sequence of operations does not exist.
## Geometry
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
An open necklace can contain rubies, emeralds and sapphires. At every step we can perform any of the following operations:
$\left(1^{\circ}\right)$ We can replace two consecutive rubies with an emerald and a sapphire, where the emerald is on the left of the sapphire.
$\left(2^{\circ}\right)$ We can replace three consecutive emeralds with a sapphire and a ruby, where the sapphire is on the left of the ruby.
(3 ) If we find two consecutive sapphires then we can remove them.
(4 ) If we find consecutively and in this order a ruby, an emerald, and a sapphire, then we can remove them.
Furthermore we can also reverse all of the above operations. For example, by reversing $\left(3^{\circ}\right)$ we can put two consecutive sapphires on any position we wish.
Initially the necklace has one sapphire (and no other precious stones). Decide, with proof, whether there is a finite sequence of steps such that at the end of this sequence the necklace contains one emerald (and no other precious stones).
Remark. A necklace is open if its precious stones are on a line from left to right. We are not allowed to move a precious stone from the rightmost position to the leftmost as we would be able to do if the necklace was closed.
(Cyprus)
|
2.
Write $a, b$ and $c$ respectively for a ruby, emerald and sapphire. Each necklace corresponds to an element of a group $G$ containing elements $a, b, c$. If we impose the conditions $a^{2}=b c, b^{3}=c a, c^{2}=1$ and $a b c=1$, the allowed operations will preserve this element. In this group we have $c=a b$ (since $c^{2}=a b c$ ), i.e. $b=a^{-1} c$ and using this relation we obtain $a^{3}=c^{2}=\left(a^{-1} c\right)^{4}=1$. Thus we can take $G=\mathbb{S}_{4}, a=(1,2,3)$, $c=(1,4)$ and $b=a^{-1} c=(1,4,3,2)$. The initial and final necklaces should correspond to elements $c$ and $b$, respectively, so the desired sequence of operations does not exist.
## Geometry
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nC3.",
"solution_match": "\n## Solution"
}
|
b41b348c-30ca-58db-a5d1-fe1393f107ed
| 605,699
|
In an acute triangle $A B C$, the midpoint of the side $B C$ is $M$ and the centers of the excircles relative to $M$ of the triangles $A M B$ and $A M C$ are $D$ and $E$ respectively. The circumcircle of the triangle $A B D$ meets line $B C$ at $B$ and $F$. The circumcircle of the triangle $A C E$ meets line $B C$ at $C$ and $G$. Prove that $B F=C G$.
(Romania)
|
We have $\varangle A D B=90^{\circ}-\frac{1}{2} \varangle A M B$ and $\varangle A E C=90^{\circ}-\frac{1}{2} \varangle A M C$.
Let the circles $A D B$ and $A E C$ respectively meet the line $A M$ again at points $P$ and $P^{\prime}$. Note that $M$ lies outside the circles $A B D$ and $A C E$ because $\varangle A D B+\varangle A M B<180^{\circ}$ and $\varangle A E C+\varangle A M C<180^{\circ}$, so $P$ and $P^{\prime}$ lie on the ray $M A$. Moreover, $\varangle B P M=$ $\varangle B D A=90^{\circ}-\frac{1}{2} \varangle P M B$, implying that $\triangle B P M$ is isosceles with $M P=M B$. Similarly, $M P^{\prime}=M C=M B$, so $P^{\prime} \equiv P$.
Now it follows from the power of point $P$ that $M B \cdot M F=M P \cdot M A=M C \cdot M G$, i.e. $M F=M G=M A$ and hence $B F=C G$.
|
B F=C G
|
Yes
|
Yes
|
proof
|
Geometry
|
In an acute triangle $A B C$, the midpoint of the side $B C$ is $M$ and the centers of the excircles relative to $M$ of the triangles $A M B$ and $A M C$ are $D$ and $E$ respectively. The circumcircle of the triangle $A B D$ meets line $B C$ at $B$ and $F$. The circumcircle of the triangle $A C E$ meets line $B C$ at $C$ and $G$. Prove that $B F=C G$.
(Romania)
|
We have $\varangle A D B=90^{\circ}-\frac{1}{2} \varangle A M B$ and $\varangle A E C=90^{\circ}-\frac{1}{2} \varangle A M C$.
Let the circles $A D B$ and $A E C$ respectively meet the line $A M$ again at points $P$ and $P^{\prime}$. Note that $M$ lies outside the circles $A B D$ and $A C E$ because $\varangle A D B+\varangle A M B<180^{\circ}$ and $\varangle A E C+\varangle A M C<180^{\circ}$, so $P$ and $P^{\prime}$ lie on the ray $M A$. Moreover, $\varangle B P M=$ $\varangle B D A=90^{\circ}-\frac{1}{2} \varangle P M B$, implying that $\triangle B P M$ is isosceles with $M P=M B$. Similarly, $M P^{\prime}=M C=M B$, so $P^{\prime} \equiv P$.
Now it follows from the power of point $P$ that $M B \cdot M F=M P \cdot M A=M C \cdot M G$, i.e. $M F=M G=M A$ and hence $B F=C G$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG1.",
"solution_match": "\n## Solution."
}
|
89661e83-af7d-5ce8-9ce6-6ca125a292aa
| 605,717
|
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K \perp L M$.
(Greece)
|
The polar of $L$ with respect to $\Gamma$ is the line $\ell_{B}$ through $B$ parallel to $A C$, and the polar of $M$ with respect to $\Gamma$ is the line $\ell_{C}$ through $C$ parallel to $A B$. Therefore the pole of the line $L M$ is the intersection $A^{\prime}$ of $\ell_{B}$ and $\ell_{C}$. It follows that $O A^{\prime} \perp L M$, so it remains to show that $O D \| A K$.
Consider the reflection $O^{\prime}$ of $O$ in the midpoint $D$ of $B C$. Since $A^{\prime}$ is the reflection of $A$ in $D, A O A^{\prime} O^{\prime}$ is a parallelogram. Moreover, $A H O^{\prime} O$ is a parallelogram because $\overrightarrow{O O^{\prime}}=2 \overrightarrow{O D}=\overrightarrow{A H}$. It follows that $\overrightarrow{O A^{\prime}}=\overrightarrow{A O^{\prime}}=2 \overrightarrow{A K}$, so $O A^{\prime} \| A K$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K \perp L M$.
(Greece)
|
The polar of $L$ with respect to $\Gamma$ is the line $\ell_{B}$ through $B$ parallel to $A C$, and the polar of $M$ with respect to $\Gamma$ is the line $\ell_{C}$ through $C$ parallel to $A B$. Therefore the pole of the line $L M$ is the intersection $A^{\prime}$ of $\ell_{B}$ and $\ell_{C}$. It follows that $O A^{\prime} \perp L M$, so it remains to show that $O D \| A K$.
Consider the reflection $O^{\prime}$ of $O$ in the midpoint $D$ of $B C$. Since $A^{\prime}$ is the reflection of $A$ in $D, A O A^{\prime} O^{\prime}$ is a parallelogram. Moreover, $A H O^{\prime} O$ is a parallelogram because $\overrightarrow{O O^{\prime}}=2 \overrightarrow{O D}=\overrightarrow{A H}$. It follows that $\overrightarrow{O A^{\prime}}=\overrightarrow{A O^{\prime}}=2 \overrightarrow{A K}$, so $O A^{\prime} \| A K$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG2.",
"solution_match": "\n## Solution."
}
|
53001fd8-1b4c-53b8-b5f9-92ff6dd3b971
| 605,724
|
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K \perp L M$.
(Greece)
|
2.
We introduce the complex plane such that $\Gamma$ is the unit cycle. Also, let the lower-case letters denote complex numbers corresponding to the points denoted by capital letters. First, note that $o=0, \bar{a}=1 / a, \bar{b}=1 / b$ and $\bar{c}=1 / c$.
Since $B L \perp B O$, we have
$$
\frac{b-l}{\bar{b}-\bar{l}}=-\frac{b-o}{\bar{b}-\bar{o}}=-\frac{b}{\bar{b}}=-b^{2}, \quad \text { and hence } \quad \bar{l}=\frac{2 b-l}{b^{2}}
$$
Since $L O \perp A C$, we have
$$
\frac{l}{\bar{l}}=\frac{l-o}{\bar{l}-\bar{o}}=-\frac{a-c}{\bar{a}-\bar{c}}=a c, \quad \text { and hence } \quad \bar{l}=\frac{l}{a c} .
$$
Combining $(\dagger)$ and $(\ddagger)$ we get $l=\frac{2 a b c}{b^{2}+a c}$. By symmetry, $m=\frac{2 a b c}{c^{2}+a b}$ and hence
$$
l-m=\frac{2 a b c(c-b)(b+c-a)}{\left(b^{2}+a c\right)\left(c^{2}+a b\right)} \quad \text { and } \quad \bar{l}-\bar{m}=\frac{2(b-c)(a b+a c-b c)}{\left(b^{2}+a c\right)\left(c^{2}+a b\right)}
$$
By Hamilton's formula $a+b+c=h-o=h$, and hence $k=\frac{h+o}{2}=\frac{a+b+c}{2}$. So,
$$
a-k=\frac{b+c-a}{2} \quad \text { and } \quad \bar{a}-\bar{k}=\frac{a b+a c-b c}{2 a b c}
$$
and hence
$$
\frac{l-m}{\bar{l}-\bar{m}}=-\frac{a-k}{\bar{a}-\bar{k}}
$$
which implies $L M \perp A K$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle inscribed in circle $\Gamma$ with center $O$ and let $H$ its orthocenter and $K$ be the midpoint of $O H$. The tangent of $\Gamma$ at $B$ meets the perpendicular bisector of $A C$ meets at $L$ and the tangent of $\Gamma$ at $C$ meets the perpendicular bisector of $A B$ at $M$. Prove that $A K \perp L M$.
(Greece)
|
2.
We introduce the complex plane such that $\Gamma$ is the unit cycle. Also, let the lower-case letters denote complex numbers corresponding to the points denoted by capital letters. First, note that $o=0, \bar{a}=1 / a, \bar{b}=1 / b$ and $\bar{c}=1 / c$.
Since $B L \perp B O$, we have
$$
\frac{b-l}{\bar{b}-\bar{l}}=-\frac{b-o}{\bar{b}-\bar{o}}=-\frac{b}{\bar{b}}=-b^{2}, \quad \text { and hence } \quad \bar{l}=\frac{2 b-l}{b^{2}}
$$
Since $L O \perp A C$, we have
$$
\frac{l}{\bar{l}}=\frac{l-o}{\bar{l}-\bar{o}}=-\frac{a-c}{\bar{a}-\bar{c}}=a c, \quad \text { and hence } \quad \bar{l}=\frac{l}{a c} .
$$
Combining $(\dagger)$ and $(\ddagger)$ we get $l=\frac{2 a b c}{b^{2}+a c}$. By symmetry, $m=\frac{2 a b c}{c^{2}+a b}$ and hence
$$
l-m=\frac{2 a b c(c-b)(b+c-a)}{\left(b^{2}+a c\right)\left(c^{2}+a b\right)} \quad \text { and } \quad \bar{l}-\bar{m}=\frac{2(b-c)(a b+a c-b c)}{\left(b^{2}+a c\right)\left(c^{2}+a b\right)}
$$
By Hamilton's formula $a+b+c=h-o=h$, and hence $k=\frac{h+o}{2}=\frac{a+b+c}{2}$. So,
$$
a-k=\frac{b+c-a}{2} \quad \text { and } \quad \bar{a}-\bar{k}=\frac{a b+a c-b c}{2 a b c}
$$
and hence
$$
\frac{l-m}{\bar{l}-\bar{m}}=-\frac{a-k}{\bar{a}-\bar{k}}
$$
which implies $L M \perp A K$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG2.",
"solution_match": "\n## Solution"
}
|
53001fd8-1b4c-53b8-b5f9-92ff6dd3b971
| 605,724
|
Let $P$ be a point inside a triangle $A B C$ and let $a, b, c$ be the side lengths and $p$ the semi-perimeter of the triangle. Find the maximum value of
$$
\min \left(\frac{P A}{p-a}, \frac{P B}{p-b}, \frac{P C}{p-c}\right)
$$
over all possible choices of triangle $A B C$ and point $P$.
|
If $A B C$ is an equilateral triangle and $P$ its center, then $\frac{P A}{p-a}=\frac{P B}{p-b}=\frac{P C}{p-c}=\frac{2}{\sqrt{3}}$.
We shall prove that $\frac{2}{\sqrt{3}}$ is the required value. Suppose without loss of generality that $\varangle A P B \geqslant 120^{\circ}$. Then
$$
A B^{2} \geqslant P A^{2}+P B^{2}+P A \cdot P B \geqslant \frac{3}{4}(P A+P B)^{2}
$$
i.e. $P A+P B \leqslant \frac{2}{\sqrt{3}} A B=\frac{2}{\sqrt{3}}((p-a)+(p-b))$, so at least one of the ratios $\frac{P A}{p-a}$ and $\frac{P B}{p-b}$ does not exceed $\frac{2}{\sqrt{3}}$.
|
\frac{2}{\sqrt{3}}
|
Yes
|
Yes
|
math-word-problem
|
Geometry
|
Let $P$ be a point inside a triangle $A B C$ and let $a, b, c$ be the side lengths and $p$ the semi-perimeter of the triangle. Find the maximum value of
$$
\min \left(\frac{P A}{p-a}, \frac{P B}{p-b}, \frac{P C}{p-c}\right)
$$
over all possible choices of triangle $A B C$ and point $P$.
|
If $A B C$ is an equilateral triangle and $P$ its center, then $\frac{P A}{p-a}=\frac{P B}{p-b}=\frac{P C}{p-c}=\frac{2}{\sqrt{3}}$.
We shall prove that $\frac{2}{\sqrt{3}}$ is the required value. Suppose without loss of generality that $\varangle A P B \geqslant 120^{\circ}$. Then
$$
A B^{2} \geqslant P A^{2}+P B^{2}+P A \cdot P B \geqslant \frac{3}{4}(P A+P B)^{2}
$$
i.e. $P A+P B \leqslant \frac{2}{\sqrt{3}} A B=\frac{2}{\sqrt{3}}((p-a)+(p-b))$, so at least one of the ratios $\frac{P A}{p-a}$ and $\frac{P B}{p-b}$ does not exceed $\frac{2}{\sqrt{3}}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG3.",
"solution_match": "\n## Solution."
}
|
53c882ed-44a8-572a-bc80-fe722def841c
| 605,739
|
A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and point $H$ is the foot of the perpendicular from $M$ to $A B$. Given that $\varangle M H C=\varangle M H D$, prove that $A B$ is a diameter of $k$.
|
Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H / H B=K M / M L$.
Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B$ at $H^{\prime}$. Then $A H^{\prime} / H^{\prime} B=K M / M L=A H / H B$, so $H^{\prime} \equiv H$, i.e. $S$ lies on the line $M H$.
The quadrilateral $A B C D$ is not a trapezoid, so $A H \neq B H$. Consider the point $A^{\prime}$ on the ray $H B$ such that $H A^{\prime}=H A$. Since $\varangle S A^{\prime} M=\varangle S A M=\varangle S B M$, quadrilateral $A^{\prime} B S M$ is cyclic and therefore $\varangle A B C=\varangle A^{\prime} B S=\varangle A^{\prime} M H=\varangle A M H=90^{\circ}-\varangle B A C$, which implies that $\varangle A C B=90^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and point $H$ is the foot of the perpendicular from $M$ to $A B$. Given that $\varangle M H C=\varangle M H D$, prove that $A B$ is a diameter of $k$.
|
Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H / H B=K M / M L$.
Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B$ at $H^{\prime}$. Then $A H^{\prime} / H^{\prime} B=K M / M L=A H / H B$, so $H^{\prime} \equiv H$, i.e. $S$ lies on the line $M H$.
The quadrilateral $A B C D$ is not a trapezoid, so $A H \neq B H$. Consider the point $A^{\prime}$ on the ray $H B$ such that $H A^{\prime}=H A$. Since $\varangle S A^{\prime} M=\varangle S A M=\varangle S B M$, quadrilateral $A^{\prime} B S M$ is cyclic and therefore $\varangle A B C=\varangle A^{\prime} B S=\varangle A^{\prime} M H=\varangle A M H=90^{\circ}-\varangle B A C$, which implies that $\varangle A C B=90^{\circ}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG4.",
"solution_match": "\n## Solution."
}
|
25eacde4-78b4-5ab0-821a-7001f180a3c4
| 605,747
|
Let $A B C$ be an acute-angled triangle with $A B<A C<B C$ and let $D$ be an arbitrary point on the extension of $B C$ beyond $C$. The circle $\gamma(A, A D)$ intersects the rays $A C$, $A B, C B$ at points $E, F, G$, respectively. The circumcircle $\omega_{1}$ of triangle $A F G$ intersects the lines $F E, B C, G E, D F$ again at points $J, H, H^{\prime}, J^{\prime}$. The circumcircle $\omega_{2}$ of triangle $A D E$ intersects the lines $F E, B C, G E, D F$ again at points $I, K, K^{\prime}, I^{\prime}$. Prove that the quadrilaterals $H I J K$ and $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ are cyclic and that their circumcenters coincide.
(Greece)
|
From $\varangle F A H=\varangle F G H=\varangle F G D=\frac{1}{2} \varangle F A D=90^{\circ}-\varangle A F D$ we deduce that $A H \perp$ $D F$. Similarly, $\varangle D A I=180^{\circ}-\varangle D E I=180^{\circ}-\varangle D E F=\varangle D G F=\frac{1}{2} \varangle D A F$, so we also have $A I \perp D F$. Therefore, points $A, H, I$ are collinear. Analogously, we find that the triples of points $(A, K, J),\left(A, H^{\prime}, I^{\prime}\right)$ and $\left(A, K^{\prime}, J^{\prime}\right)$ are collinear.
Quadrilateral $H I J K$ is cyclic because $\varangle A I K=\varangle A D K=\varangle A G H=\varangle A J H$. Analogously, quadrilateral $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ is cyclic.
Finally, since $\varangle H^{\prime} J H=\varangle H^{\prime} G H=\varangle E G D=\varangle E F D=\varangle J F J^{\prime}=\varangle J H J^{\prime}$, quadrilateral $H J J^{\prime} H^{\prime}$ is an isosceles trapezoid with $H J \| H^{\prime} J^{\prime}$, so the perpendicular bisectors of $H J$ and $H^{\prime} J^{\prime}$ coincide. Analogously, the perpendicular bisectors of $I K$ and $I^{\prime} K^{\prime}$ coincide. Therefore the circumcenters of $H I J K$ and $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ coincide.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with $A B<A C<B C$ and let $D$ be an arbitrary point on the extension of $B C$ beyond $C$. The circle $\gamma(A, A D)$ intersects the rays $A C$, $A B, C B$ at points $E, F, G$, respectively. The circumcircle $\omega_{1}$ of triangle $A F G$ intersects the lines $F E, B C, G E, D F$ again at points $J, H, H^{\prime}, J^{\prime}$. The circumcircle $\omega_{2}$ of triangle $A D E$ intersects the lines $F E, B C, G E, D F$ again at points $I, K, K^{\prime}, I^{\prime}$. Prove that the quadrilaterals $H I J K$ and $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ are cyclic and that their circumcenters coincide.
(Greece)
|
From $\varangle F A H=\varangle F G H=\varangle F G D=\frac{1}{2} \varangle F A D=90^{\circ}-\varangle A F D$ we deduce that $A H \perp$ $D F$. Similarly, $\varangle D A I=180^{\circ}-\varangle D E I=180^{\circ}-\varangle D E F=\varangle D G F=\frac{1}{2} \varangle D A F$, so we also have $A I \perp D F$. Therefore, points $A, H, I$ are collinear. Analogously, we find that the triples of points $(A, K, J),\left(A, H^{\prime}, I^{\prime}\right)$ and $\left(A, K^{\prime}, J^{\prime}\right)$ are collinear.
Quadrilateral $H I J K$ is cyclic because $\varangle A I K=\varangle A D K=\varangle A G H=\varangle A J H$. Analogously, quadrilateral $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ is cyclic.
Finally, since $\varangle H^{\prime} J H=\varangle H^{\prime} G H=\varangle E G D=\varangle E F D=\varangle J F J^{\prime}=\varangle J H J^{\prime}$, quadrilateral $H J J^{\prime} H^{\prime}$ is an isosceles trapezoid with $H J \| H^{\prime} J^{\prime}$, so the perpendicular bisectors of $H J$ and $H^{\prime} J^{\prime}$ coincide. Analogously, the perpendicular bisectors of $I K$ and $I^{\prime} K^{\prime}$ coincide. Therefore the circumcenters of $H I J K$ and $H^{\prime} I^{\prime} J^{\prime} K^{\prime}$ coincide.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG5.",
"solution_match": "\n## Solution."
}
|
89d0f9c0-819b-5bdc-9d92-a17b70eaaa4f
| 605,757
|
In a triangle $A B C$ with $A B=A C, \omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $B A$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $O A D$ intersects the line $A C$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $D A E H$ is a parallelogram. Line $E H$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $G B$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $G J$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.
(Cyprus)
|
We first observe that $\varangle D O E=\varangle D A E=2 \varangle A B C=\varangle B O A$ and hence $\varangle D O B=$ $\varangle E O A$, which together with $O B=O A$ and $\varangle O B D=\varangle B A O=\varangle O A E$ gives us $\triangle O B D \cong \triangle O A E$. Therefore $B D=A E$.
Next, $O G=O A$ implies $\varangle O D G=\varangle O D A=\varangle O D B$ and hence $\triangle O G D \cong \triangle O B D$. It follows that $D G=D B=A E=D H$. Moreover, since $A D \| E J$, we have $D J=A E=$ $D G$. Thus, the points $B, G, H, J$ lie on a circle $\omega_{2}$ with center $D$.
We deduce that $\varangle A G H=\varangle B G H-\varangle B G A=180^{\circ}-\frac{1}{2} \varangle H D B-\varangle B C A=180^{\circ}-$ $\frac{1}{2} \varangle C A B-\varangle B C A=90^{\circ}$.
We will now invert the diagram through $G$. By $\hat{X}$ we denote the image of any point $X$. The points $\hat{H}, \hat{L}, \hat{N}$ then lie on the lines $\hat{B} \hat{J}, \hat{A} \hat{B}$ and $\hat{A} \hat{J}$, respectively, such that $\varangle \hat{A} G \hat{H}=\hat{B} G \hat{N}=\varangle \hat{J} G \hat{L}=90^{\circ}$. It remains to prove that $\hat{H}, \hat{L}$ and $\hat{N}$ are collinear, which follows from the following statement:
Lemma. Let $X Y Z$ be a triangle and let $U$ be a point in the plane. If the lines through $U$ perpendicular to $U X, U Y, U Z$ meet the lines $Y Z, Z X, X Y$ respectively at points $P, Q, R$, then the points $P, Q$ and $R$ are collinear.
$\underline{\text { Proof. Here we assume that } U \text { is inside } \triangle X Y Z \text { and the angles } X U Y, Y U Z \text { and } Z U X . . . ~}$ are all obtuse - the other cases are similar. We have
$$
\frac{\overrightarrow{Y P}}{\overrightarrow{P Z}}=-\frac{P_{Y U P}}{P_{P U Z}}, \quad \frac{\overrightarrow{Z Q}}{\overrightarrow{Q X}}=-\frac{P_{Z U Q}}{P_{Q U X}}, \quad \frac{\overrightarrow{X R}}{\overrightarrow{R Y}}=-\frac{P_{X U R}}{P_{R U Y}}
$$
On the other hand, since $\varangle Q U X=\varangle Y U P$ are equal and equally directed, we have $\frac{P_{Y U P}}{P_{Q U X}}=\frac{U P \cdot U Y}{U Q \cdot U X}$. Writing the analogous expressions for $\frac{P_{Z U Q}}{P_{R U Y}}$ and $\frac{P_{X U R}}{P_{P U Z}}$
and multiplying them out we obtain $\frac{\overrightarrow{Y P}}{\overrightarrow{P Z}} \cdot \frac{\overrightarrow{Z Q}}{\overrightarrow{Q X}} \cdot \frac{\overrightarrow{X R}}{\overrightarrow{R Y}}=-1$, and the result follows by Menelaus' theorem.
## Remark.
The result remains valid if $D$ is any point on the line $A B$.
Point $L$ does not depend on the choice of $D$. Indeed, $\varangle L C B=\varangle L G B=\varangle J G B-90^{\circ}=$ $\varangle J E A+\varangle A G B-90^{\circ}=\varangle B A C+\varangle A C B-90^{\circ}=90^{\circ}-\varangle A B C$, so $C L \perp A B$.
Also, since $\varangle A O N=\varangle A G N=90^{\circ}-\varangle B G A=90^{\circ}-\varangle B C A=\varangle O A B=\varangle O N D$, $O N D A$ is an isosceles trapezoid, i.e. $O N \| A B$.
## Alternative formulation.
Based on the Remark, the PSC proposes the following modification which hides point $J$ and defines the points in a more natural way:
A triangle $A B C$ with $A B=A C$ is inscribed in a circle $\omega$ with center $O$. Its altitude from $C$ meets $\omega$ again at point $L$. Line $\ell$ through $O$ is parallel to $A B$. A circle $\omega_{1}$ passes through points $A$ and $O$ and meets the lines $A B, A C, \ell$ and circle $\omega$ again at points $D, E, N$ and $G$, respectively. Point $H$ is such that $A D H E$ is a parallelogram.
Prove that $H$ lies on the circumcircle of triangle $G L N$.
## Number Theory
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a triangle $A B C$ with $A B=A C, \omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $B A$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $O A D$ intersects the line $A C$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $D A E H$ is a parallelogram. Line $E H$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $G B$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $G J$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.
(Cyprus)
|
We first observe that $\varangle D O E=\varangle D A E=2 \varangle A B C=\varangle B O A$ and hence $\varangle D O B=$ $\varangle E O A$, which together with $O B=O A$ and $\varangle O B D=\varangle B A O=\varangle O A E$ gives us $\triangle O B D \cong \triangle O A E$. Therefore $B D=A E$.
Next, $O G=O A$ implies $\varangle O D G=\varangle O D A=\varangle O D B$ and hence $\triangle O G D \cong \triangle O B D$. It follows that $D G=D B=A E=D H$. Moreover, since $A D \| E J$, we have $D J=A E=$ $D G$. Thus, the points $B, G, H, J$ lie on a circle $\omega_{2}$ with center $D$.
We deduce that $\varangle A G H=\varangle B G H-\varangle B G A=180^{\circ}-\frac{1}{2} \varangle H D B-\varangle B C A=180^{\circ}-$ $\frac{1}{2} \varangle C A B-\varangle B C A=90^{\circ}$.
We will now invert the diagram through $G$. By $\hat{X}$ we denote the image of any point $X$. The points $\hat{H}, \hat{L}, \hat{N}$ then lie on the lines $\hat{B} \hat{J}, \hat{A} \hat{B}$ and $\hat{A} \hat{J}$, respectively, such that $\varangle \hat{A} G \hat{H}=\hat{B} G \hat{N}=\varangle \hat{J} G \hat{L}=90^{\circ}$. It remains to prove that $\hat{H}, \hat{L}$ and $\hat{N}$ are collinear, which follows from the following statement:
Lemma. Let $X Y Z$ be a triangle and let $U$ be a point in the plane. If the lines through $U$ perpendicular to $U X, U Y, U Z$ meet the lines $Y Z, Z X, X Y$ respectively at points $P, Q, R$, then the points $P, Q$ and $R$ are collinear.
$\underline{\text { Proof. Here we assume that } U \text { is inside } \triangle X Y Z \text { and the angles } X U Y, Y U Z \text { and } Z U X . . . ~}$ are all obtuse - the other cases are similar. We have
$$
\frac{\overrightarrow{Y P}}{\overrightarrow{P Z}}=-\frac{P_{Y U P}}{P_{P U Z}}, \quad \frac{\overrightarrow{Z Q}}{\overrightarrow{Q X}}=-\frac{P_{Z U Q}}{P_{Q U X}}, \quad \frac{\overrightarrow{X R}}{\overrightarrow{R Y}}=-\frac{P_{X U R}}{P_{R U Y}}
$$
On the other hand, since $\varangle Q U X=\varangle Y U P$ are equal and equally directed, we have $\frac{P_{Y U P}}{P_{Q U X}}=\frac{U P \cdot U Y}{U Q \cdot U X}$. Writing the analogous expressions for $\frac{P_{Z U Q}}{P_{R U Y}}$ and $\frac{P_{X U R}}{P_{P U Z}}$
and multiplying them out we obtain $\frac{\overrightarrow{Y P}}{\overrightarrow{P Z}} \cdot \frac{\overrightarrow{Z Q}}{\overrightarrow{Q X}} \cdot \frac{\overrightarrow{X R}}{\overrightarrow{R Y}}=-1$, and the result follows by Menelaus' theorem.
## Remark.
The result remains valid if $D$ is any point on the line $A B$.
Point $L$ does not depend on the choice of $D$. Indeed, $\varangle L C B=\varangle L G B=\varangle J G B-90^{\circ}=$ $\varangle J E A+\varangle A G B-90^{\circ}=\varangle B A C+\varangle A C B-90^{\circ}=90^{\circ}-\varangle A B C$, so $C L \perp A B$.
Also, since $\varangle A O N=\varangle A G N=90^{\circ}-\varangle B G A=90^{\circ}-\varangle B C A=\varangle O A B=\varangle O N D$, $O N D A$ is an isosceles trapezoid, i.e. $O N \| A B$.
## Alternative formulation.
Based on the Remark, the PSC proposes the following modification which hides point $J$ and defines the points in a more natural way:
A triangle $A B C$ with $A B=A C$ is inscribed in a circle $\omega$ with center $O$. Its altitude from $C$ meets $\omega$ again at point $L$. Line $\ell$ through $O$ is parallel to $A B$. A circle $\omega_{1}$ passes through points $A$ and $O$ and meets the lines $A B, A C, \ell$ and circle $\omega$ again at points $D, E, N$ and $G$, respectively. Point $H$ is such that $A D H E$ is a parallelogram.
Prove that $H$ lies on the circumcircle of triangle $G L N$.
## Number Theory
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nG6.",
"solution_match": "\n## Solution."
}
|
d9326439-aa3c-58e5-b92d-bd8e8f7c6c53
| 605,766
|
For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60,126)=d\left(2^{2} \times 3 \times 5,2 \times 3^{2} \times 7\right)=2$. Does there exist a sequence $\left(a_{n}\right)$ of positive integers such that:
(i) $a_{1} \geqslant 2018^{2018}$;
(ii) $a_{m} \leqslant a_{n}$ whenever $m \leqslant n$;
(iii) $d(m, n)=d\left(a_{m}, a_{n}\right)$ for all positive integers $m \neq n$ ?
(United Kingdom)
|
Such a sequence does exist.
Let $p_{1}<p_{2}<p_{3}<\ldots$ be the usual list of primes, and $q_{1}<q_{2}<\ldots, r_{1}<r_{2}<\ldots$ be disjoint sequences of primes greater than $2018^{2018}$. For example, let $q_{i} \equiv 1$ and $r_{i} \equiv 3$ modulo 4. Then, if $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots$, where all but finitely many of the $\alpha_{i}$ will be zero, set
$$
b_{n}:=q_{1}^{\alpha_{1}} q_{2}^{\alpha_{2}} \cdots, \quad \text { for all } n \geqslant 2
$$
This sequence satisfies requirement (iii), but not the ordering conditions (i) and (ii). Iteratively, take $a_{1}=r_{1}$, then given $a_{1}, \ldots, a_{n-1}$, define $a_{n}$ by multiplying $b_{n}$ by as large a power of $r_{n}$ as necessary in order to ensure $a_{n}>a_{n-1}$. Thus $d\left(a_{m}, a_{n}\right)=d\left(b_{m}, b_{n}\right)=$ $d(m, n)$, and so all three requirements are satisfied.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60,126)=d\left(2^{2} \times 3 \times 5,2 \times 3^{2} \times 7\right)=2$. Does there exist a sequence $\left(a_{n}\right)$ of positive integers such that:
(i) $a_{1} \geqslant 2018^{2018}$;
(ii) $a_{m} \leqslant a_{n}$ whenever $m \leqslant n$;
(iii) $d(m, n)=d\left(a_{m}, a_{n}\right)$ for all positive integers $m \neq n$ ?
(United Kingdom)
|
Such a sequence does exist.
Let $p_{1}<p_{2}<p_{3}<\ldots$ be the usual list of primes, and $q_{1}<q_{2}<\ldots, r_{1}<r_{2}<\ldots$ be disjoint sequences of primes greater than $2018^{2018}$. For example, let $q_{i} \equiv 1$ and $r_{i} \equiv 3$ modulo 4. Then, if $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots$, where all but finitely many of the $\alpha_{i}$ will be zero, set
$$
b_{n}:=q_{1}^{\alpha_{1}} q_{2}^{\alpha_{2}} \cdots, \quad \text { for all } n \geqslant 2
$$
This sequence satisfies requirement (iii), but not the ordering conditions (i) and (ii). Iteratively, take $a_{1}=r_{1}$, then given $a_{1}, \ldots, a_{n-1}$, define $a_{n}$ by multiplying $b_{n}$ by as large a power of $r_{n}$ as necessary in order to ensure $a_{n}>a_{n-1}$. Thus $d\left(a_{m}, a_{n}\right)=d\left(b_{m}, b_{n}\right)=$ $d(m, n)$, and so all three requirements are satisfied.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nN1.",
"solution_match": "\n## Solution."
}
|
3c63d421-1cdb-5ea3-9b50-e755b3df3a90
| 605,775
|
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
n!+f(m)!\mid f(n)!+f(m!)
$$
for all $m, n \in \mathbb{N}$.
|
Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$ !, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m, n)=(1, p-1)$ for any prime number $p$ and using Wilson's theorem we obtain $p|(p-1)!+1| f(p-1)!+1$, implying $f(p-1)<p$. Therefore
$$
f(p-1)=p-1
$$
Next, fix a positive integer $m$. For any prime number $p$, setting $n=p-1$ in (*) yields $(p-1)!+f(m)!\mid(p-1)!+f(m!)$, and hence
$$
(p-1)!+f(m)!\mid f(m!)-f(m)!\text { for all prime numbers } p
$$
This implies $f(m!)=f(m)$ ! for all $m \in \mathbb{N}$, so $(*)$ can be rewritten as $n!+f(m)!\mid$ $f(n)!+f(m)$ !. This implies
$$
n!+f(m)!\mid f(n)!-n!\text { for all } n, m \in \mathbb{N}
$$
Fixing $n \in \mathbb{N}$ and taking $m \in \mathbb{N}$ large enough, we conclude that $f(n)$ ! $=n$ !, i.e. $f(n)=n$, for all $n \in \mathbb{N}$.
One readily checks that the identity function satisfies the conditions of the problem.
|
f(n)=n
|
Yes
|
Yes
|
proof
|
Number Theory
|
Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
n!+f(m)!\mid f(n)!+f(m!)
$$
for all $m, n \in \mathbb{N}$.
|
Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$ !, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m, n)=(1, p-1)$ for any prime number $p$ and using Wilson's theorem we obtain $p|(p-1)!+1| f(p-1)!+1$, implying $f(p-1)<p$. Therefore
$$
f(p-1)=p-1
$$
Next, fix a positive integer $m$. For any prime number $p$, setting $n=p-1$ in (*) yields $(p-1)!+f(m)!\mid(p-1)!+f(m!)$, and hence
$$
(p-1)!+f(m)!\mid f(m!)-f(m)!\text { for all prime numbers } p
$$
This implies $f(m!)=f(m)$ ! for all $m \in \mathbb{N}$, so $(*)$ can be rewritten as $n!+f(m)!\mid$ $f(n)!+f(m)$ !. This implies
$$
n!+f(m)!\mid f(n)!-n!\text { for all } n, m \in \mathbb{N}
$$
Fixing $n \in \mathbb{N}$ and taking $m \in \mathbb{N}$ large enough, we conclude that $f(n)$ ! $=n$ !, i.e. $f(n)=n$, for all $n \in \mathbb{N}$.
One readily checks that the identity function satisfies the conditions of the problem.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nN2.",
"solution_match": "\n## Solution."
}
|
9fde6804-f7bf-5199-b89f-c67d414c92e2
| 605,785
|
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
|
Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod r)$. Then $r \mid b^{p} N \equiv a^{p}+1(\bmod r)$, where $a=11 b$. Thus $r \mid a^{2 p}-1$, but $r \nmid a^{p}-1$, which means that $\operatorname{ord}_{r}(a) \mid 2 p$ and $\operatorname{ord}_{r}(a) \nmid p$, i.e. $\operatorname{ord}_{r}(a) \in\{2,2 p\}$.
Note that if $\operatorname{ord}_{r}(a)=2$, then $r \mid a^{2}-1 \equiv\left(11^{2}-17^{2}\right) b^{2}(\bmod r)$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_{r}(a)=2 p$ implies $2 p \mid r-1$. Thus, all prime divisors of $3 p^{q-1}+1$ other than 2 or 7 are congruent to 1 modulo $2 p$, i.e.
$$
3 p^{q-1}+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}
$$
where $p_{i} \notin\{2,7\}$ are prime divisors with $p_{i} \equiv 1(\bmod 2 p)$.
We already know that $\alpha \leqslant 2$. Also, note that
$$
\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\cdots+17^{p-1} \equiv p \cdot 4^{p-1} \quad(\bmod 7)
$$
so $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\beta \leqslant 1$.
If $q=2$, then $(*)$ becomes $3 p+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}$, but $p_{i} \geqslant 2 p+1$, which is only possible if $\gamma_{i}=0$ for all $i$, i.e. $3 p+1=2^{\alpha} 7^{\beta} \in\{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4 \mid 3 p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to 4 or 28 modulo $p$, which gives us $p=3$. Consequently $3^{q}+1 \mid 6244$, which is only possible for $q=3$. The pair $(p, q)=(3,3)$ is indeed a solution.
|
(p, q)=(3,3)
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
|
Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod r)$. Then $r \mid b^{p} N \equiv a^{p}+1(\bmod r)$, where $a=11 b$. Thus $r \mid a^{2 p}-1$, but $r \nmid a^{p}-1$, which means that $\operatorname{ord}_{r}(a) \mid 2 p$ and $\operatorname{ord}_{r}(a) \nmid p$, i.e. $\operatorname{ord}_{r}(a) \in\{2,2 p\}$.
Note that if $\operatorname{ord}_{r}(a)=2$, then $r \mid a^{2}-1 \equiv\left(11^{2}-17^{2}\right) b^{2}(\bmod r)$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_{r}(a)=2 p$ implies $2 p \mid r-1$. Thus, all prime divisors of $3 p^{q-1}+1$ other than 2 or 7 are congruent to 1 modulo $2 p$, i.e.
$$
3 p^{q-1}+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}
$$
where $p_{i} \notin\{2,7\}$ are prime divisors with $p_{i} \equiv 1(\bmod 2 p)$.
We already know that $\alpha \leqslant 2$. Also, note that
$$
\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\cdots+17^{p-1} \equiv p \cdot 4^{p-1} \quad(\bmod 7)
$$
so $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\beta \leqslant 1$.
If $q=2$, then $(*)$ becomes $3 p+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}$, but $p_{i} \geqslant 2 p+1$, which is only possible if $\gamma_{i}=0$ for all $i$, i.e. $3 p+1=2^{\alpha} 7^{\beta} \in\{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4 \mid 3 p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to 4 or 28 modulo $p$, which gives us $p=3$. Consequently $3^{q}+1 \mid 6244$, which is only possible for $q=3$. The pair $(p, q)=(3,3)$ is indeed a solution.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nN3.",
"solution_match": "\n## Solution."
}
|
e1a95f9e-b720-57a7-92f5-e487b7a1587b
| 605,795
|
Let $P(x)=a_{d} x^{d}+\cdots+a_{1} x+a_{0}$ be a non-constant polynomial with nonnegative integer coefficients having $d$ rational roots. Prove that
$$
\operatorname{lcm}(P(m), P(m+1), \ldots, P(n)) \geqslant m\binom{n}{m}
$$
for all positive integers $n>m$.
|
Let $x_{i}=-\frac{p_{i}}{q_{i}}(1 \leqslant i \leqslant d)$ be the roots of $P(x)$, where $p_{i}, q_{i} \in \mathbb{N}$ and $\operatorname{gcd}\left(p_{i}, q_{i}\right)=1$. By Gauss' lemma, we have $P(x)=c\left(q_{1} x+p_{1}\right)\left(q_{2} x+p_{2}\right) \cdots\left(q_{d} x+p_{d}\right)$ for some $c \in \mathbb{N}$, so $q_{1} x+p_{1} \mid P(x)$. Thus it suffices to prove the statement for $P(x)=q_{1} x+p_{1}=q x+p$. Let
$$
\begin{array}{ll}
A=\operatorname{lcm}(q m+p, q(m+1)+p, \ldots, q n+p) & =\prod_{i=1}^{s} p_{i}^{\alpha_{i}}, \\
B=(q m+p)(q(m+1)+p) \cdots(q n+p) & =\prod_{i=1}^{s} p_{i}^{\beta_{i}}
\end{array}
$$
be the prime factorizations of $A$ and $B$.
Consider a prime divisor $p_{i}$. We have $p_{i}^{\alpha_{i}} \mid q x+p$ for some $m \leqslant x \leqslant n$. On the other hand, if $p_{i}^{r} \mid q y+p\left(r \leqslant \alpha_{i}\right)$ for some $m \leqslant y \leqslant n$ with $y \neq x$, then $p_{i}^{r} \mid q(x-y)$, i.e. $p_{i}^{r} \mid x-y$. Taking the product over all $y \neq x$ we obtain that
$$
p_{i}^{\beta_{i}} \quad \text { divides } \quad p_{i}^{\alpha_{i}} \cdot \prod_{\substack{y=m \\ y \neq x}}^{n}|x-y|, \quad \text { which divides } \quad p_{i}^{\alpha_{i}}(n-m)!
$$
It follows that $B \mid A \cdot(n-m)$ !, but $B \geqslant m(m+1) \cdots n$, so the result immediately follows.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $P(x)=a_{d} x^{d}+\cdots+a_{1} x+a_{0}$ be a non-constant polynomial with nonnegative integer coefficients having $d$ rational roots. Prove that
$$
\operatorname{lcm}(P(m), P(m+1), \ldots, P(n)) \geqslant m\binom{n}{m}
$$
for all positive integers $n>m$.
|
Let $x_{i}=-\frac{p_{i}}{q_{i}}(1 \leqslant i \leqslant d)$ be the roots of $P(x)$, where $p_{i}, q_{i} \in \mathbb{N}$ and $\operatorname{gcd}\left(p_{i}, q_{i}\right)=1$. By Gauss' lemma, we have $P(x)=c\left(q_{1} x+p_{1}\right)\left(q_{2} x+p_{2}\right) \cdots\left(q_{d} x+p_{d}\right)$ for some $c \in \mathbb{N}$, so $q_{1} x+p_{1} \mid P(x)$. Thus it suffices to prove the statement for $P(x)=q_{1} x+p_{1}=q x+p$. Let
$$
\begin{array}{ll}
A=\operatorname{lcm}(q m+p, q(m+1)+p, \ldots, q n+p) & =\prod_{i=1}^{s} p_{i}^{\alpha_{i}}, \\
B=(q m+p)(q(m+1)+p) \cdots(q n+p) & =\prod_{i=1}^{s} p_{i}^{\beta_{i}}
\end{array}
$$
be the prime factorizations of $A$ and $B$.
Consider a prime divisor $p_{i}$. We have $p_{i}^{\alpha_{i}} \mid q x+p$ for some $m \leqslant x \leqslant n$. On the other hand, if $p_{i}^{r} \mid q y+p\left(r \leqslant \alpha_{i}\right)$ for some $m \leqslant y \leqslant n$ with $y \neq x$, then $p_{i}^{r} \mid q(x-y)$, i.e. $p_{i}^{r} \mid x-y$. Taking the product over all $y \neq x$ we obtain that
$$
p_{i}^{\beta_{i}} \quad \text { divides } \quad p_{i}^{\alpha_{i}} \cdot \prod_{\substack{y=m \\ y \neq x}}^{n}|x-y|, \quad \text { which divides } \quad p_{i}^{\alpha_{i}}(n-m)!
$$
It follows that $B \mid A \cdot(n-m)$ !, but $B \geqslant m(m+1) \cdots n$, so the result immediately follows.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nN4.",
"solution_match": "\n## Solution."
}
|
de383b2b-aaed-5645-a8f1-c0d49f6a5f14
| 605,804
|
Let $x$ and $y$ be positive integers. If for each positive integer $n$ we have that
$$
(n y)^{2}+1 \mid x^{\varphi(n)}-1
$$
prove that $x=1$.
|
Let us take $n=3^{k}$ and suppose that $p$ is a prime divisor of $\left(3^{k} y\right)^{2}+1$ such that $p \equiv 2$ $(\bmod 3)$.
Since $p$ divides $x^{\varphi(n)}-1=x^{2 \cdot 3^{k-1}}-1$, the order of $x$ modulo $p$ divides both $p-1$ and $2 \cdot 3^{k-1}$, but $\operatorname{gcd}\left(p-1,2 \cdot 3^{k-1}\right) \mid 2$, which implies that $p \mid x^{2}-1$. The result will follow if we prove that the prime $p$ can take infinitely many values.
Suppose, to the contrary, that there are only finitely many primes $p$ with $p \equiv 2(\bmod 3)$ that divide a term of the sequence
$$
a_{k}=3^{2 k} y^{2}+1 \quad(k \geqslant 0)
$$
Let $p_{1}, p_{2}, \ldots, p_{m}$ be these primes. Clearly, we may assume without loss of generality that $3 \nmid y$. Then $a_{0}=y^{2}+1 \equiv 2(\bmod 3)$, so it has a prime divisor of the form $3 s+2$ $\left(s \in \mathbb{N}_{0}\right)$.
For $N=\left(y^{2}+1\right) p_{1} \cdots p_{m}$ we have $a_{\varphi(N)}=3^{2 \varphi(N)} y^{2}+1 \equiv y^{2}+1(\bmod N)$, which means that
$$
a_{\varphi(N)}=\left(y^{2}+1\right)\left(t p_{1} \cdots p_{m}+1\right)
$$
for some positive integer $t$. Since $y^{2}+1 \equiv 2(\bmod 3)$ and $3^{2 \varphi(N)} y^{2}+1 \equiv 1(\bmod 3)$, the number $t p_{1} \cdots p_{m}+1$ must have a prime divisor of the form $3 s+2$, but it cannot be any of the primes $p_{1}, \ldots, p_{m}$, so we have a contradiction as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $x$ and $y$ be positive integers. If for each positive integer $n$ we have that
$$
(n y)^{2}+1 \mid x^{\varphi(n)}-1
$$
prove that $x=1$.
|
Let us take $n=3^{k}$ and suppose that $p$ is a prime divisor of $\left(3^{k} y\right)^{2}+1$ such that $p \equiv 2$ $(\bmod 3)$.
Since $p$ divides $x^{\varphi(n)}-1=x^{2 \cdot 3^{k-1}}-1$, the order of $x$ modulo $p$ divides both $p-1$ and $2 \cdot 3^{k-1}$, but $\operatorname{gcd}\left(p-1,2 \cdot 3^{k-1}\right) \mid 2$, which implies that $p \mid x^{2}-1$. The result will follow if we prove that the prime $p$ can take infinitely many values.
Suppose, to the contrary, that there are only finitely many primes $p$ with $p \equiv 2(\bmod 3)$ that divide a term of the sequence
$$
a_{k}=3^{2 k} y^{2}+1 \quad(k \geqslant 0)
$$
Let $p_{1}, p_{2}, \ldots, p_{m}$ be these primes. Clearly, we may assume without loss of generality that $3 \nmid y$. Then $a_{0}=y^{2}+1 \equiv 2(\bmod 3)$, so it has a prime divisor of the form $3 s+2$ $\left(s \in \mathbb{N}_{0}\right)$.
For $N=\left(y^{2}+1\right) p_{1} \cdots p_{m}$ we have $a_{\varphi(N)}=3^{2 \varphi(N)} y^{2}+1 \equiv y^{2}+1(\bmod N)$, which means that
$$
a_{\varphi(N)}=\left(y^{2}+1\right)\left(t p_{1} \cdots p_{m}+1\right)
$$
for some positive integer $t$. Since $y^{2}+1 \equiv 2(\bmod 3)$ and $3^{2 \varphi(N)} y^{2}+1 \equiv 1(\bmod 3)$, the number $t p_{1} \cdots p_{m}+1$ must have a prime divisor of the form $3 s+2$, but it cannot be any of the primes $p_{1}, \ldots, p_{m}$, so we have a contradiction as desired.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2018_bmo_shortlist.jsonl",
"problem_match": "\nN5.",
"solution_match": "\n## Solution."
}
|
f8ce7770-dd7b-5761-8c2e-0aef4e0c60ed
| 605,813
|
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1}=a_{n}$ for all $n \geq M$.
|
Define $b_{n}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}$ for every positive integer $n$. According to condition, $b_{n}$ is a positive integer for every positive integer $n$.
Since $a_{n+1}$ is the smallest positive integer such that $\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}$ is a positive integer and
$$
\frac{a_{0}+a_{1}+\ldots+a_{n}+b_{n}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}+\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}=b_{n}
$$
which is a positive integer, we get $a_{n+1} \leq b_{n}$ for every positive integer $n$.
Now from last result we have
$$
b_{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1} \leq \frac{a_{0}+a_{1}+\ldots+a_{n}+b_{n}}{n+1}=b_{n} .
$$
Hence the infinite sequence of positive integers $b_{1}, b_{2}, \ldots$ is non-increasing. So there exists a positive integer $T$ such that for all $n \geq T$ we have
$$
\begin{gathered}
b_{n+1}=b_{n} \Rightarrow \frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n} \Rightarrow \\
n\left(a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}\right)=(n+1)\left(a_{0}+a_{1}+\ldots+a_{n}\right) \Rightarrow \\
n a_{n+1}=a_{0}+a_{1}+\ldots+a_{n} \Rightarrow a_{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}=b_{n} .
\end{gathered}
$$
Similarly we get $a_{n+2}=b_{n+1}$, which follows that $a_{n+2}=b_{n+1}=b_{n}=a_{n+1}$. Hence, taking $M=T+1$, we can state that $a_{n+1}=a_{n}$ for every $n \geq M$.
[^3]A2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that:
$$
f(x y)=y f(x)+x+f(f(y)-f(x))
$$
for all $x, y \in \mathbb{R}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1}=a_{n}$ for all $n \geq M$.
|
Define $b_{n}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}$ for every positive integer $n$. According to condition, $b_{n}$ is a positive integer for every positive integer $n$.
Since $a_{n+1}$ is the smallest positive integer such that $\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}$ is a positive integer and
$$
\frac{a_{0}+a_{1}+\ldots+a_{n}+b_{n}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}+\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}=b_{n}
$$
which is a positive integer, we get $a_{n+1} \leq b_{n}$ for every positive integer $n$.
Now from last result we have
$$
b_{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1} \leq \frac{a_{0}+a_{1}+\ldots+a_{n}+b_{n}}{n+1}=b_{n} .
$$
Hence the infinite sequence of positive integers $b_{1}, b_{2}, \ldots$ is non-increasing. So there exists a positive integer $T$ such that for all $n \geq T$ we have
$$
\begin{gathered}
b_{n+1}=b_{n} \Rightarrow \frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n} \Rightarrow \\
n\left(a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}\right)=(n+1)\left(a_{0}+a_{1}+\ldots+a_{n}\right) \Rightarrow \\
n a_{n+1}=a_{0}+a_{1}+\ldots+a_{n} \Rightarrow a_{n+1}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}=b_{n} .
\end{gathered}
$$
Similarly we get $a_{n+2}=b_{n+1}$, which follows that $a_{n+2}=b_{n+1}=b_{n}=a_{n+1}$. Hence, taking $M=T+1$, we can state that $a_{n+1}=a_{n}$ for every $n \geq M$.
[^3]A2. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that:
$$
f(x y)=y f(x)+x+f(f(y)-f(x))
$$
for all $x, y \in \mathbb{R}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nA1b. ${ }^{4}$",
"solution_match": "\nSolution."
}
|
53e6e720-4e7b-5d9d-b994-6b4bcb391e16
| 605,821
|
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1}=a_{n}$ for all $n \geq M$.
|
Firstly, considering $(x, y)=(1,1)$ we get $f(0)=-1$.
Then, setting $y=1$, we see that $-x=f(f(1)-f(x))$, so $f$ must be surjective.
Now let $(x, y)=(a, 0)$ and $(0, a)$ to get
$$
-1=a+f(-1-f(a)) \quad \text { and } \quad-1=-a+f(f(a)+1)
$$
Since $f$ is surjective, for any real $z$ we may write $z=f(a)+1$ and then adding these two results gives $f(z)+f(-z)=-2$.
Letting $(x, y)=(a, 1)$ and $(1, a)$ we get
$$
-a=f(f(1)-f(a)) \quad \text { and } \quad f(a)=a f(1)+1+f(f(a)-f(1)) .
$$
Adding these, and using the previous result with $z=f(a)-f(1)$ gives
$$
f(a)=a f(1)+a-1 .
$$
So $f(x)=k x-1$ for all $x$, for some fixed $k$. Substituting back into the original equation we see that 1 and -1 are the only possibilities for $k$ and that both of these values do give a function that works.
Alternative solution. We prove that $f(x)=x-1$ and $f(x)=-x-1$ are the only solutions. Let $x=y=1$; this gives $f(1)=f(1)+1+f(0)$, so $f(0)=-1$. Then let $(x, y)=(0, a+1),(-a-1,0)$, and $(-a, 1)$ to give the three equalities
$$
\begin{aligned}
f(0)=(a+1) f(0)+f(f(a+1)-f(0)) & \Rightarrow a=f(f(a+1)+1) \\
f(0)=-a-1+f(f(0)-f(-a-1)) & \Rightarrow a=f(-f(-a-1)-1) \\
f(-a)=f(-a)-a+f(f(1)-f(a)) & \Rightarrow a=f(f(1)-f(-a)) .
\end{aligned}
$$
The last of these three implies $f$ is bijective, hence we have
$$
f(a+1)+1=-f(-a-1)-1=f(1)-f(-a)
$$
From the second of these equalities we can deduce the recurrence relation $f(x)=$ $f(x-1)+f(1)+1$, so if $c=f(1)+1$, we have $f(x)=c x-1$ for all $x \in \mathbb{Z}$. Substituting into the original equation we see that $c^{2}=1$, so $f(x)=x-1$ or $f(x)=-x-1$ for $x \in \mathbb{Z}$.
In the first case, let $x=1$. Then $f(y)=1+f(f(y))$, which implies $f(x)=x-1$ for all $x$ as $f$ is surjective. In the second case, set $x=-1$, so $f(-y)=-1+f(f(y))$. However from above we have $f(a+1)+f(-a-1)=2$, so $f(f(y))-1=f(-y)=-f(y)-2$, and we have $f(x)=-x-1$ by surjectivity.
|
proof
|
Yes
|
Incomplete
|
proof
|
Number Theory
|
Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a positive integer $M$ such that $a_{n+1}=a_{n}$ for all $n \geq M$.
|
Firstly, considering $(x, y)=(1,1)$ we get $f(0)=-1$.
Then, setting $y=1$, we see that $-x=f(f(1)-f(x))$, so $f$ must be surjective.
Now let $(x, y)=(a, 0)$ and $(0, a)$ to get
$$
-1=a+f(-1-f(a)) \quad \text { and } \quad-1=-a+f(f(a)+1)
$$
Since $f$ is surjective, for any real $z$ we may write $z=f(a)+1$ and then adding these two results gives $f(z)+f(-z)=-2$.
Letting $(x, y)=(a, 1)$ and $(1, a)$ we get
$$
-a=f(f(1)-f(a)) \quad \text { and } \quad f(a)=a f(1)+1+f(f(a)-f(1)) .
$$
Adding these, and using the previous result with $z=f(a)-f(1)$ gives
$$
f(a)=a f(1)+a-1 .
$$
So $f(x)=k x-1$ for all $x$, for some fixed $k$. Substituting back into the original equation we see that 1 and -1 are the only possibilities for $k$ and that both of these values do give a function that works.
Alternative solution. We prove that $f(x)=x-1$ and $f(x)=-x-1$ are the only solutions. Let $x=y=1$; this gives $f(1)=f(1)+1+f(0)$, so $f(0)=-1$. Then let $(x, y)=(0, a+1),(-a-1,0)$, and $(-a, 1)$ to give the three equalities
$$
\begin{aligned}
f(0)=(a+1) f(0)+f(f(a+1)-f(0)) & \Rightarrow a=f(f(a+1)+1) \\
f(0)=-a-1+f(f(0)-f(-a-1)) & \Rightarrow a=f(-f(-a-1)-1) \\
f(-a)=f(-a)-a+f(f(1)-f(a)) & \Rightarrow a=f(f(1)-f(-a)) .
\end{aligned}
$$
The last of these three implies $f$ is bijective, hence we have
$$
f(a+1)+1=-f(-a-1)-1=f(1)-f(-a)
$$
From the second of these equalities we can deduce the recurrence relation $f(x)=$ $f(x-1)+f(1)+1$, so if $c=f(1)+1$, we have $f(x)=c x-1$ for all $x \in \mathbb{Z}$. Substituting into the original equation we see that $c^{2}=1$, so $f(x)=x-1$ or $f(x)=-x-1$ for $x \in \mathbb{Z}$.
In the first case, let $x=1$. Then $f(y)=1+f(f(y))$, which implies $f(x)=x-1$ for all $x$ as $f$ is surjective. In the second case, set $x=-1$, so $f(-y)=-1+f(f(y))$. However from above we have $f(a+1)+f(-a-1)=2$, so $f(f(y))-1=f(-y)=-f(y)-2$, and we have $f(x)=-x-1$ by surjectivity.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nA1b. ${ }^{4}$",
"solution_match": "\nSolution."
}
|
53e6e720-4e7b-5d9d-b994-6b4bcb391e16
| 605,821
|
Let $a, b, c$ be real numbers such that $0 \leq a \leq b \leq c$. Prove that if
$$
a+b+c=a b+b c+c a>0,
$$
then $\sqrt{b c}(a+1) \geq 2$. When does the equality hold?
|
Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The constraint gives us
$$
a=\frac{b+c-b c}{b+c-1}=1-\frac{b c-1}{b+c-1} \geq 1-\frac{b c-1}{2 \sqrt{b c}-1}=\frac{\sqrt{b c}(2-\sqrt{b c})}{2 \sqrt{b c}-1}
$$
For $\sqrt{b c}=2$ condition $a \geq 0$ gives $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=0$ and $b=c=2$.
For $\sqrt{b c}<2$, taking into account the estimation for $a$, we get
$$
a \sqrt{b c} \geq \frac{b c(2-\sqrt{b c})}{2 \sqrt{b c}-1}=\frac{b c}{2 \sqrt{b c}-1}(2-\sqrt{b c}) .
$$
Since $\frac{b c}{2 \sqrt{b c}-1} \geq 1$, with equality for $b c=1$, we get $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=b=c=1$.
For $\sqrt{b c}>2$ we have $\sqrt{b c}(a+1)>2(a+1) \geq 2$.
The proof is complete.
The equality holds iff $a=b=c=1$ or $a=0$ and $b=c=2$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be real numbers such that $0 \leq a \leq b \leq c$. Prove that if
$$
a+b+c=a b+b c+c a>0,
$$
then $\sqrt{b c}(a+1) \geq 2$. When does the equality hold?
|
Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The constraint gives us
$$
a=\frac{b+c-b c}{b+c-1}=1-\frac{b c-1}{b+c-1} \geq 1-\frac{b c-1}{2 \sqrt{b c}-1}=\frac{\sqrt{b c}(2-\sqrt{b c})}{2 \sqrt{b c}-1}
$$
For $\sqrt{b c}=2$ condition $a \geq 0$ gives $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=0$ and $b=c=2$.
For $\sqrt{b c}<2$, taking into account the estimation for $a$, we get
$$
a \sqrt{b c} \geq \frac{b c(2-\sqrt{b c})}{2 \sqrt{b c}-1}=\frac{b c}{2 \sqrt{b c}-1}(2-\sqrt{b c}) .
$$
Since $\frac{b c}{2 \sqrt{b c}-1} \geq 1$, with equality for $b c=1$, we get $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=b=c=1$.
For $\sqrt{b c}>2$ we have $\sqrt{b c}(a+1)>2(a+1) \geq 2$.
The proof is complete.
The equality holds iff $a=b=c=1$ or $a=0$ and $b=c=2$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nA3.",
"solution_match": "\nSolution."
}
|
2ae7d15e-fbec-5b34-a032-5af90d4dc817
| 605,843
|
Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1} \leq\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1} .
$$
When does the equality hold?
|
We will use the following
Lemma. If $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ are positive real numbers then
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\ldots=\frac{a_{n}}{b_{n}}$.
Proof. Set $x_{j}=\frac{1}{a_{j}}$ and $y_{j}=\frac{1}{b_{j}}$ for each $j=1,2, \ldots, n$. Then we have to prove that
$$
\frac{1}{\sum_{j=1}^{n} x_{j}}+\frac{1}{\sum_{j=1}^{n} y_{j}} \leq \frac{1}{\sum_{j=1}^{n} \frac{x_{j} y_{j}}{x_{j}+y_{j}}} \quad \text { or } \quad \sum_{j=1}^{n} \frac{x_{j} y_{j}}{x_{j}+y_{j}} \leq \frac{\left(\sum_{j=1}^{n} x_{j}\right)\left(\sum_{j=1}^{n} y_{j}\right)}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
Subtract $\sum_{j=1}^{n} x_{j}$, and we have to prove that
$$
\sum_{j=1}^{n}\left(x_{j}-\frac{x_{j} y_{j}}{x_{j}+y_{j}}\right) \geq \sum_{j=1}^{n} x_{j}-\frac{\left(\sum_{j=1}^{n} x_{j}\right)\left(\sum_{j=1}^{n} y_{j}\right)}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
or
$$
\sum_{j=1}^{n}\left(\frac{x_{j}^{2}}{x_{j}+y_{j}}\right) \geq \frac{\left(\sum_{j=1}^{n} x_{j}\right)^{2}}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
The last one is a consequence of Cauchy-Schwarz inequality and thus the lemma is proved.
We will now prove that repeating the lemma we will get the desired inequality. For example, if $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}, c_{1}, c_{2}, \ldots, c_{n}$ are positive reals then by repeating lemma two times we get
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{c_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{c_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{\left(a_{j}+b_{j}\right)+c_{j}}}=\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}+c_{j}}}
$$
Using similar reasoning we can prove by induction that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1}=\sum_{i=1}^{m} \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{i j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{\sum_{i=1}^{m} a_{i j}}}=\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1}
$$
which is the desired result.
The equality holds iff
$$
\frac{a_{i 1}}{a_{11}}=\frac{a_{i 2}}{a_{12}}=\ldots=\frac{a_{i n}}{a_{1 n}}
$$
for all $i=1,2, \ldots, m$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1} \leq\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1} .
$$
When does the equality hold?
|
We will use the following
Lemma. If $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ are positive real numbers then
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\ldots=\frac{a_{n}}{b_{n}}$.
Proof. Set $x_{j}=\frac{1}{a_{j}}$ and $y_{j}=\frac{1}{b_{j}}$ for each $j=1,2, \ldots, n$. Then we have to prove that
$$
\frac{1}{\sum_{j=1}^{n} x_{j}}+\frac{1}{\sum_{j=1}^{n} y_{j}} \leq \frac{1}{\sum_{j=1}^{n} \frac{x_{j} y_{j}}{x_{j}+y_{j}}} \quad \text { or } \quad \sum_{j=1}^{n} \frac{x_{j} y_{j}}{x_{j}+y_{j}} \leq \frac{\left(\sum_{j=1}^{n} x_{j}\right)\left(\sum_{j=1}^{n} y_{j}\right)}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
Subtract $\sum_{j=1}^{n} x_{j}$, and we have to prove that
$$
\sum_{j=1}^{n}\left(x_{j}-\frac{x_{j} y_{j}}{x_{j}+y_{j}}\right) \geq \sum_{j=1}^{n} x_{j}-\frac{\left(\sum_{j=1}^{n} x_{j}\right)\left(\sum_{j=1}^{n} y_{j}\right)}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
or
$$
\sum_{j=1}^{n}\left(\frac{x_{j}^{2}}{x_{j}+y_{j}}\right) \geq \frac{\left(\sum_{j=1}^{n} x_{j}\right)^{2}}{\sum_{j=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}}
$$
The last one is a consequence of Cauchy-Schwarz inequality and thus the lemma is proved.
We will now prove that repeating the lemma we will get the desired inequality. For example, if $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}, c_{1}, c_{2}, \ldots, c_{n}$ are positive reals then by repeating lemma two times we get
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{c_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{c_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{\left(a_{j}+b_{j}\right)+c_{j}}}=\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}+c_{j}}}
$$
Using similar reasoning we can prove by induction that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1}=\sum_{i=1}^{m} \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{i j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{\sum_{i=1}^{m} a_{i j}}}=\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1}
$$
which is the desired result.
The equality holds iff
$$
\frac{a_{i 1}}{a_{11}}=\frac{a_{i 2}}{a_{12}}=\ldots=\frac{a_{i n}}{a_{1 n}}
$$
for all $i=1,2, \ldots, m$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nA4.",
"solution_match": "\nSolution."
}
|
e5eb2fdf-21b3-5fb2-ab76-e6bf0373514f
| 605,854
|
Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 .
$$
|
The inequality is equivalent with
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 \Leftrightarrow\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq(a+1)(b+1)(c+1) .
$$
Thus:
$$
\begin{gathered}
\prod_{c y c}\left(a^{3}+1\right)-\prod_{c y c}(a+1)=\sum_{c y c} a^{3}+\sum_{c y c}(a b)^{3}+(a b c)^{3}-\sum_{c y c} a-\sum_{c y c} a b-a b c= \\
\sum_{c y c}\left(a^{3}+a\right)+\sum_{c y c}\left(a^{3} b^{3}+a b\right)+\left[(a b c)^{3}+1+1\right]-2 \sum_{c y c} a-2 \sum_{c y c} a b-a b c-2^{A M \geq G M} \geq^{\sum_{c y c} a^{2} b^{2}=3} \\
2 \sum_{c y c} a^{2}+2 \sum_{c y c} a^{2} b^{2}+2 a b c-2 \sum_{c y c} a-2 \sum_{c y c} a b-2^{c y c}= \\
\sum_{c y c}\left(a^{2}-2 a+1\right)+\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right)= \\
\sum_{c y c}(a-1)^{2}+\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right) \geq\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right) .
\end{gathered}
$$
We will show that $\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b \geq 0 \quad$ (1) for every $a, b, c \geq 0$.
Firstly, let us observe that
$$
(1+2 a b c)(a+b+c)=(1+a b c+a b c)(a+b+c) \geq 9 \sqrt[3]{a^{2} b^{2} c^{2} a b c}=9 a b c
$$
implying
$$
1+2 a b c \geq \frac{9 a b c}{a+b+c}
$$
Then, using Schur's Inequality, (i.e. $\sum_{c y c} a(a-b)(a-c) \geq 0$, for any $a, b, c \geq 0$ ) we obtain that
$$
\sum_{c y c} a^{2} \geq 2 \sum_{c y c} a b-\frac{9 a b c}{a+b+c}
$$
Returning to (1), we get:
$$
\begin{gathered}
\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b \geq\left(2 \sum_{c y c} a b-\frac{9 a b c}{a+b+c}\right)+2 a b c+1-2 \sum_{c y c} a b= \\
(1+2 a b c)-\frac{9 a b c}{a+b+c} \geq 0
\end{gathered}
$$
which gives us $\prod_{\text {cyc }}\left(a^{3}+1\right)-\prod_{c y c}(a+1) \geq 0$ and, respectively, $\prod_{\text {cyc }}\left(a^{2}-a+1\right) \geq 1$.
## GEOMETRY
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 .
$$
|
The inequality is equivalent with
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 \Leftrightarrow\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq(a+1)(b+1)(c+1) .
$$
Thus:
$$
\begin{gathered}
\prod_{c y c}\left(a^{3}+1\right)-\prod_{c y c}(a+1)=\sum_{c y c} a^{3}+\sum_{c y c}(a b)^{3}+(a b c)^{3}-\sum_{c y c} a-\sum_{c y c} a b-a b c= \\
\sum_{c y c}\left(a^{3}+a\right)+\sum_{c y c}\left(a^{3} b^{3}+a b\right)+\left[(a b c)^{3}+1+1\right]-2 \sum_{c y c} a-2 \sum_{c y c} a b-a b c-2^{A M \geq G M} \geq^{\sum_{c y c} a^{2} b^{2}=3} \\
2 \sum_{c y c} a^{2}+2 \sum_{c y c} a^{2} b^{2}+2 a b c-2 \sum_{c y c} a-2 \sum_{c y c} a b-2^{c y c}= \\
\sum_{c y c}\left(a^{2}-2 a+1\right)+\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right)= \\
\sum_{c y c}(a-1)^{2}+\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right) \geq\left(\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b\right) .
\end{gathered}
$$
We will show that $\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b \geq 0 \quad$ (1) for every $a, b, c \geq 0$.
Firstly, let us observe that
$$
(1+2 a b c)(a+b+c)=(1+a b c+a b c)(a+b+c) \geq 9 \sqrt[3]{a^{2} b^{2} c^{2} a b c}=9 a b c
$$
implying
$$
1+2 a b c \geq \frac{9 a b c}{a+b+c}
$$
Then, using Schur's Inequality, (i.e. $\sum_{c y c} a(a-b)(a-c) \geq 0$, for any $a, b, c \geq 0$ ) we obtain that
$$
\sum_{c y c} a^{2} \geq 2 \sum_{c y c} a b-\frac{9 a b c}{a+b+c}
$$
Returning to (1), we get:
$$
\begin{gathered}
\sum_{c y c} a^{2}+2 a b c+1-2 \sum_{c y c} a b \geq\left(2 \sum_{c y c} a b-\frac{9 a b c}{a+b+c}\right)+2 a b c+1-2 \sum_{c y c} a b= \\
(1+2 a b c)-\frac{9 a b c}{a+b+c} \geq 0
\end{gathered}
$$
which gives us $\prod_{\text {cyc }}\left(a^{3}+1\right)-\prod_{c y c}(a+1) \geq 0$ and, respectively, $\prod_{\text {cyc }}\left(a^{2}-a+1\right) \geq 1$.
## GEOMETRY
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nA5.",
"solution_match": "\nSolution."
}
|
7ea726a6-17db-541e-9734-5667d9bb1015
| 605,868
|
Let $A B C D$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $C M$ and $B D$, and let $S$ be the intersection of $M O$ and $A E$. Show that $S O$ is the angle bisector of $\angle E S B$.
|
We have
$$
\left\{\begin{array}{l}
D C \equiv D A \\
\angle E D C \equiv \angle E D A \quad \Rightarrow \triangle D E C \equiv \triangle D E A \Rightarrow \angle D A E \equiv \angle D C E(*) . \\
D E \equiv D E
\end{array}\right.
$$
Let $C M \cap A D=\{P\}$, then follows $\triangle C D P \equiv \triangle B A P$ and $\angle P C D \equiv \angle P B A(* *)$.
Figure 1: G1
From (*) and (**) follows $\angle D C P \equiv \angle D A E \equiv \angle P B A$.
Now, let $S^{\prime}=A E \cap P B$.
In the triangle $S^{\prime} A B$ we have
$$
m\left(\angle S^{\prime} A B\right)+m\left(\angle S^{\prime} B A\right)=m\left(\angle S^{\prime} A B\right)+m\left(\angle P A S^{\prime}\right)=m(\angle P A B)=90^{\circ},
$$
so $m\left(\angle B S^{\prime} A\right)=90^{\circ}$.
We show that $A E, B P$ and $M O$ are concurrent.
In the triangle $\triangle E M B$ we apply the Ceva theorem, so
$$
\frac{E P}{P M} \cdot \frac{M A}{A B} \cdot \frac{B O}{O E}=1 \Leftrightarrow \frac{E P}{P M}=\frac{O E}{B O}
$$
is true because $P O$ is a midsegment in the triangle $D A B(P O \| A B)$.
According to the Thales theorem in the triangle $E M B, \frac{E P}{P M}=\frac{E O}{O B}$ and $A E, B P$, $M O$ are concurrent in $S^{\prime}$, which is in fact $S$.
Let $P B \cap C A=\{N\}$. Because ESNO has $m(\angle E O N)+m(\angle E S N)=180^{\circ}$, it follows $E S N O$ cyclic and $m(\angle E S O)=m(\angle E N O)=m(\angle D A O)=45^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $C M$ and $B D$, and let $S$ be the intersection of $M O$ and $A E$. Show that $S O$ is the angle bisector of $\angle E S B$.
|
We have
$$
\left\{\begin{array}{l}
D C \equiv D A \\
\angle E D C \equiv \angle E D A \quad \Rightarrow \triangle D E C \equiv \triangle D E A \Rightarrow \angle D A E \equiv \angle D C E(*) . \\
D E \equiv D E
\end{array}\right.
$$
Let $C M \cap A D=\{P\}$, then follows $\triangle C D P \equiv \triangle B A P$ and $\angle P C D \equiv \angle P B A(* *)$.
Figure 1: G1
From (*) and (**) follows $\angle D C P \equiv \angle D A E \equiv \angle P B A$.
Now, let $S^{\prime}=A E \cap P B$.
In the triangle $S^{\prime} A B$ we have
$$
m\left(\angle S^{\prime} A B\right)+m\left(\angle S^{\prime} B A\right)=m\left(\angle S^{\prime} A B\right)+m\left(\angle P A S^{\prime}\right)=m(\angle P A B)=90^{\circ},
$$
so $m\left(\angle B S^{\prime} A\right)=90^{\circ}$.
We show that $A E, B P$ and $M O$ are concurrent.
In the triangle $\triangle E M B$ we apply the Ceva theorem, so
$$
\frac{E P}{P M} \cdot \frac{M A}{A B} \cdot \frac{B O}{O E}=1 \Leftrightarrow \frac{E P}{P M}=\frac{O E}{B O}
$$
is true because $P O$ is a midsegment in the triangle $D A B(P O \| A B)$.
According to the Thales theorem in the triangle $E M B, \frac{E P}{P M}=\frac{E O}{O B}$ and $A E, B P$, $M O$ are concurrent in $S^{\prime}$, which is in fact $S$.
Let $P B \cap C A=\{N\}$. Because ESNO has $m(\angle E O N)+m(\angle E S N)=180^{\circ}$, it follows $E S N O$ cyclic and $m(\angle E S O)=m(\angle E N O)=m(\angle D A O)=45^{\circ}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG1.",
"solution_match": "\nSolution."
}
|
a6d394a0-ba81-5e3a-8718-66c53e035a8c
| 605,876
|
Let be a triangle $\triangle A B C$ with $m(\angle A B C)=75^{\circ}$ and $m(\angle A C B)=45^{\circ}$. The angle bisector of $\angle C A B$ intersects $C B$ at the point $D$. We consider the point $E \in(A B)$, such that $D E=D C$. Let $P$ be the intersection of the lines $A D$ and $C E$. Prove that $P$ is the midpoint of the segment $A D$.
|
Let $P^{\prime}$ be the midpoint of the segment $A D$. We will prove that $P^{\prime}=P$. Let $F \in A C$ such that $D F \perp A C$. The triangle $C D F$ is isosceles with $F D=F C$ and the triangle $D P^{\prime} F$ is equilateral as $m(\angle A D F)=60^{\circ}$. Thus, the triangle $F C P^{\prime}$ is isosceles $\left(F P^{\prime}=F C\right)$ and $m\left(\angle F C P^{\prime}\right)=m\left(\angle F P^{\prime} C\right)=15^{\circ}$.
Figure 2: G2
We prove now that $m(\angle F C E)=15^{\circ}$.
Let $M$ be the point on $[A B$ such that the triangle $A C M$ is equilateral. As $\triangle A D C \equiv$ $\triangle A D M(S A S) \Rightarrow D C=D M(=D E)$ and $m(\angle A M D)=m(\angle A C D)=45^{\circ}$. It follows that the triangle $\triangle D M E$ is isosceles with $m(\angle D M E)=m(\angle D E M)=45^{\circ}$. In the triangle $\triangle B D E$ we have $m(\angle B D E)=60^{\circ}$ and thus $m(\angle C D E)=120^{\circ}$.As the triangle $D C E$ is isoscel with $m(\angle D C E)=m(\angle D E C)=30^{\circ}$. Finaly $m(\angle A C E)=m(\angle A C B)-$ $m(\angle B C E)=45^{\circ}-30^{\circ}=15^{\circ}$.
Thus $m\left(\angle F C P^{\prime}\right)=15^{\circ}=m(\angle F C E)$, and therefore $P^{\prime} \in C E$ and $P^{\prime}=P$, which means that $P$ is the midpoint of the segment $A D$.
Alternative solution: In the way as above we prove that $m(\angle B C E)=15^{\circ}$.
So the quadrilateral $A C D E$ is inscribed in a circle. Now, applying the sine rules to $\triangle D P E$ and $\triangle A P E$ we get
$$
\begin{gathered}
\frac{D P}{\sin 30^{\circ}}=\frac{P E}{\sin 15^{\circ}}, \quad \frac{A P}{\sin 105^{\circ}}=\frac{P E}{\sin 30^{\circ}} \Rightarrow \frac{D P}{\sin 30^{\circ}} \cdot \frac{\sin 105^{\circ}}{A P}=\frac{P E}{\sin 15^{\circ}} \cdot \frac{\sin 30^{\circ}}{P E}, \\
\frac{D P}{A P}=\frac{1}{\sin 30^{\circ}} 105^{\circ} \cdot \sin 15^{\circ} \\
=\frac{1}{4 \cdot \sin 105^{\circ} \cdot \sin 15^{\circ}}=\frac{1}{2 \cdot\left(\cos 90^{\circ}-\cos 120^{\circ}\right)}=\frac{1}{2 \cdot \frac{1}{2}}=1 .
\end{gathered}
$$
Thus, $Q P=A P$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let be a triangle $\triangle A B C$ with $m(\angle A B C)=75^{\circ}$ and $m(\angle A C B)=45^{\circ}$. The angle bisector of $\angle C A B$ intersects $C B$ at the point $D$. We consider the point $E \in(A B)$, such that $D E=D C$. Let $P$ be the intersection of the lines $A D$ and $C E$. Prove that $P$ is the midpoint of the segment $A D$.
|
Let $P^{\prime}$ be the midpoint of the segment $A D$. We will prove that $P^{\prime}=P$. Let $F \in A C$ such that $D F \perp A C$. The triangle $C D F$ is isosceles with $F D=F C$ and the triangle $D P^{\prime} F$ is equilateral as $m(\angle A D F)=60^{\circ}$. Thus, the triangle $F C P^{\prime}$ is isosceles $\left(F P^{\prime}=F C\right)$ and $m\left(\angle F C P^{\prime}\right)=m\left(\angle F P^{\prime} C\right)=15^{\circ}$.
Figure 2: G2
We prove now that $m(\angle F C E)=15^{\circ}$.
Let $M$ be the point on $[A B$ such that the triangle $A C M$ is equilateral. As $\triangle A D C \equiv$ $\triangle A D M(S A S) \Rightarrow D C=D M(=D E)$ and $m(\angle A M D)=m(\angle A C D)=45^{\circ}$. It follows that the triangle $\triangle D M E$ is isosceles with $m(\angle D M E)=m(\angle D E M)=45^{\circ}$. In the triangle $\triangle B D E$ we have $m(\angle B D E)=60^{\circ}$ and thus $m(\angle C D E)=120^{\circ}$.As the triangle $D C E$ is isoscel with $m(\angle D C E)=m(\angle D E C)=30^{\circ}$. Finaly $m(\angle A C E)=m(\angle A C B)-$ $m(\angle B C E)=45^{\circ}-30^{\circ}=15^{\circ}$.
Thus $m\left(\angle F C P^{\prime}\right)=15^{\circ}=m(\angle F C E)$, and therefore $P^{\prime} \in C E$ and $P^{\prime}=P$, which means that $P$ is the midpoint of the segment $A D$.
Alternative solution: In the way as above we prove that $m(\angle B C E)=15^{\circ}$.
So the quadrilateral $A C D E$ is inscribed in a circle. Now, applying the sine rules to $\triangle D P E$ and $\triangle A P E$ we get
$$
\begin{gathered}
\frac{D P}{\sin 30^{\circ}}=\frac{P E}{\sin 15^{\circ}}, \quad \frac{A P}{\sin 105^{\circ}}=\frac{P E}{\sin 30^{\circ}} \Rightarrow \frac{D P}{\sin 30^{\circ}} \cdot \frac{\sin 105^{\circ}}{A P}=\frac{P E}{\sin 15^{\circ}} \cdot \frac{\sin 30^{\circ}}{P E}, \\
\frac{D P}{A P}=\frac{1}{\sin 30^{\circ}} 105^{\circ} \cdot \sin 15^{\circ} \\
=\frac{1}{4 \cdot \sin 105^{\circ} \cdot \sin 15^{\circ}}=\frac{1}{2 \cdot\left(\cos 90^{\circ}-\cos 120^{\circ}\right)}=\frac{1}{2 \cdot \frac{1}{2}}=1 .
\end{gathered}
$$
Thus, $Q P=A P$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG2.",
"solution_match": "\nSolution."
}
|
6d618284-fecc-50cd-b819-0863b5f55635
| 605,887
|
Let $A B C$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $B C$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $F G \perp A O, \angle B F D=\angle D G C$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes with respect to the line $A D$. Show that the tangents to $\omega$ at $F$ and $G$ meet on the circumcircle of $A B C$.
|
Consider any two points $F, G$ on $\omega$ such that $\angle B F D=\angle D G C$. Exploiting the isosceles triangles $\triangle A F G, \triangle A F D$, and $\triangle A D G$, we deduce (using directed angles throughout):
$$
\begin{gathered}
\angle D B F-\angle G C D=180^{\circ}-\angle B F D-\angle B D F-\left(180^{\circ}-\angle D G C-\angle C D G\right) \stackrel{(*)}{=} \\
\angle C D G-\angle F D B=\frac{1}{2} \cdot(\angle D A G-\angle D A F)=\frac{1}{2} \cdot\left[\left(180^{\circ}-2 \cdot \angle A D G\right)-\left(180^{\circ}-2 \cdot \angle A D F\right)\right]= \\
\angle A D F-\angle G D A=\angle D F A-\angle A G D=\angle D F G-\angle F G D \stackrel{(*)}{=} \angle B F G-\angle F G C,
\end{gathered}
$$
where we use $\angle B F D=\angle D G C$ at $\left(^{*}\right)$. Thus $B F G C$ is cyclic.
Figure 3: G3
Now, if in addition $F G \perp A O$, then since $A$ is the centre of $\omega$, in fact $A O$ is the perpendicular bisector of $F G$. But by definition, since $A B C$ is scalene, $A O$ meets the perpendicular bisector of $B C$ at $O$. Hence $O$ is the centre of $B F G C$, and thus in fact $B F A G C$ is cyclic. But then the lines perpendicular to $A F$ at $F$, and $A G$ at $G$ (the tangents to $\omega$ ) must intersect at $E$, the point antipodal to $A$ on $\odot B F A G C$.
Alternative solution: Let the circumcircle of $A B C$ be $\Gamma$. From the conditions, $G$ is the reflection of $F$ in the line $A O$. Let $B^{\prime}, D^{\prime}$ be the reflections of $B, D$ across this same line $A O$. Clearly $D^{\prime}$ also lies on $\omega$ and $B^{\prime}$ lies on $\Gamma$.
Then, using directed angles, $\angle C G D=\angle D F B=\angle B^{\prime} G D^{\prime}$ so
$$
\angle B^{\prime} G C=\angle B^{\prime} G D^{\prime}-\angle C G D^{\prime}=\angle C G D-\angle C G D^{\prime}=\angle D^{\prime} G D=\frac{1}{2} \angle D^{\prime} A D=\angle O A D .
$$
Then, exploiting the isogonality property that $\angle D A B=\angle C A O$, we have $\angle O A D=\angle C A B-2 \angle D A B=\angle A B C-\angle B C A=\angle A B C-\angle B^{\prime} B A=\angle B^{\prime} B C$.
So $G$ lies on $\Gamma$, and by the reflection property so does $F$.
But then, as in the previous solution, the tangents at $F$ and $G$ to $\omega$ must intersect at $E$, the point antipodal to $A$ on $\Gamma$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $B C$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $F G \perp A O, \angle B F D=\angle D G C$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes with respect to the line $A D$. Show that the tangents to $\omega$ at $F$ and $G$ meet on the circumcircle of $A B C$.
|
Consider any two points $F, G$ on $\omega$ such that $\angle B F D=\angle D G C$. Exploiting the isosceles triangles $\triangle A F G, \triangle A F D$, and $\triangle A D G$, we deduce (using directed angles throughout):
$$
\begin{gathered}
\angle D B F-\angle G C D=180^{\circ}-\angle B F D-\angle B D F-\left(180^{\circ}-\angle D G C-\angle C D G\right) \stackrel{(*)}{=} \\
\angle C D G-\angle F D B=\frac{1}{2} \cdot(\angle D A G-\angle D A F)=\frac{1}{2} \cdot\left[\left(180^{\circ}-2 \cdot \angle A D G\right)-\left(180^{\circ}-2 \cdot \angle A D F\right)\right]= \\
\angle A D F-\angle G D A=\angle D F A-\angle A G D=\angle D F G-\angle F G D \stackrel{(*)}{=} \angle B F G-\angle F G C,
\end{gathered}
$$
where we use $\angle B F D=\angle D G C$ at $\left(^{*}\right)$. Thus $B F G C$ is cyclic.
Figure 3: G3
Now, if in addition $F G \perp A O$, then since $A$ is the centre of $\omega$, in fact $A O$ is the perpendicular bisector of $F G$. But by definition, since $A B C$ is scalene, $A O$ meets the perpendicular bisector of $B C$ at $O$. Hence $O$ is the centre of $B F G C$, and thus in fact $B F A G C$ is cyclic. But then the lines perpendicular to $A F$ at $F$, and $A G$ at $G$ (the tangents to $\omega$ ) must intersect at $E$, the point antipodal to $A$ on $\odot B F A G C$.
Alternative solution: Let the circumcircle of $A B C$ be $\Gamma$. From the conditions, $G$ is the reflection of $F$ in the line $A O$. Let $B^{\prime}, D^{\prime}$ be the reflections of $B, D$ across this same line $A O$. Clearly $D^{\prime}$ also lies on $\omega$ and $B^{\prime}$ lies on $\Gamma$.
Then, using directed angles, $\angle C G D=\angle D F B=\angle B^{\prime} G D^{\prime}$ so
$$
\angle B^{\prime} G C=\angle B^{\prime} G D^{\prime}-\angle C G D^{\prime}=\angle C G D-\angle C G D^{\prime}=\angle D^{\prime} G D=\frac{1}{2} \angle D^{\prime} A D=\angle O A D .
$$
Then, exploiting the isogonality property that $\angle D A B=\angle C A O$, we have $\angle O A D=\angle C A B-2 \angle D A B=\angle A B C-\angle B C A=\angle A B C-\angle B^{\prime} B A=\angle B^{\prime} B C$.
So $G$ lies on $\Gamma$, and by the reflection property so does $F$.
But then, as in the previous solution, the tangents at $F$ and $G$ to $\omega$ must intersect at $E$, the point antipodal to $A$ on $\Gamma$.
|
{
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"problem_match": "\nG3.",
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|
68066e38-8d2b-5f29-8406-e91fa64e18f0
| 605,899
|
Given an acute triangle $A B C$, let $M$ be the midpoint of $B C$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $H M$, and let $X, Y$ be distinct points on $\Gamma$ such that $A X, A Y$ are tangent to $\Gamma$. Prove that $B X Y C$ is cyclic.
|
Let $D$ be the foot of the altitude from $A$ to $B C$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle A B C$. Since $\angle H D M=90^{\circ}$, note that rays $H D$ and $H M$ meet the circumcircle at points which are reflections in $O M$. Then, since $\angle B A D=\angle O A C$, we recover the well-known fact that ray $H M$ meets the circumcircle at $A^{\prime}$, the point antipodal to $A$. Therefore, the ray $M H$ meets the circumcircle at a point $T$ such that $\angle M T A=90^{\circ}$. Note that $T, D$ lie on the circle with diameter $A M$.
Figure 4: G4
Now, study $K$, the centre of $\Gamma$. Clearly $A X K Y$ is cyclic, with diameter $A K$, so $T$ also lies on this circle. We can now apply the radical axis theorem to the three circles $\odot A T X K Y, \odot A T D M, \odot H X D M Y$ to deduce that $A T, X Y, D M$ concur at a point, $Z$.
Then, by power of a point in $\odot A T X Y$, we have $Z X \cdot Z Y=Z T \cdot Z A$; but also by power of a point in the circumcircle, we have $Z A \cdot Z T=Z B \cdot Z C$. Therefore
$$
Z X \cdot Z Y=Z B \cdot Z C
$$
and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given an acute triangle $A B C$, let $M$ be the midpoint of $B C$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $H M$, and let $X, Y$ be distinct points on $\Gamma$ such that $A X, A Y$ are tangent to $\Gamma$. Prove that $B X Y C$ is cyclic.
|
Let $D$ be the foot of the altitude from $A$ to $B C$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle A B C$. Since $\angle H D M=90^{\circ}$, note that rays $H D$ and $H M$ meet the circumcircle at points which are reflections in $O M$. Then, since $\angle B A D=\angle O A C$, we recover the well-known fact that ray $H M$ meets the circumcircle at $A^{\prime}$, the point antipodal to $A$. Therefore, the ray $M H$ meets the circumcircle at a point $T$ such that $\angle M T A=90^{\circ}$. Note that $T, D$ lie on the circle with diameter $A M$.
Figure 4: G4
Now, study $K$, the centre of $\Gamma$. Clearly $A X K Y$ is cyclic, with diameter $A K$, so $T$ also lies on this circle. We can now apply the radical axis theorem to the three circles $\odot A T X K Y, \odot A T D M, \odot H X D M Y$ to deduce that $A T, X Y, D M$ concur at a point, $Z$.
Then, by power of a point in $\odot A T X Y$, we have $Z X \cdot Z Y=Z T \cdot Z A$; but also by power of a point in the circumcircle, we have $Z A \cdot Z T=Z B \cdot Z C$. Therefore
$$
Z X \cdot Z Y=Z B \cdot Z C
$$
and the result follows.
|
{
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"problem_match": "\nG4.",
"solution_match": "\nSolution."
}
|
62b785c3-5b66-50d2-a3bf-92eb76756538
| 605,907
|
Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K \in(P Q)$. The line $P D$ intersects the circumcircle of triangle $B K Q$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $B Q$. Prove that $T B=T Q$.
|
As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C D=\angle Q C D=\angle Q B A$ (the latter is due to $Q B C D$ being cyclic). We deduce that $P X \| Q B$ and that $P$ is the midpoint of $A Q$, so $A P=P Q$.
Figure 5: G5
Now let $T_{1}$ be the midpoint of the arc $B Q$, not containing $K$, from the circumcircle of $\triangle B K Q$, then $T_{1} B=T_{1} Q$. Due to $\angle D P O=90^{\circ}$, it suffices to show that $\angle O P T_{1}=90^{\circ}$ - indeed, $T \equiv T_{1}$ and $T B=T Q$ would follow.
Denote by $Y$ the midpoint of $B Q$. Then $\angle O X B=\angle T_{1} Y B=90^{\circ}$. The quadrilateral $Q K B T_{1}$ is inscribed in a circle, hence $\angle B T_{1} Q=180-\angle B K Q=\angle A K B$. Then $\angle X B O=$ $\frac{1}{2} \angle A K B=\frac{1}{2} \angle B T_{1} Q=\angle B T_{1} Y$ and thus $\triangle O X B \sim \triangle B Y T_{1}$. The quadrilaterals $P X B Y$
and $A X Y P$ are paralellograms, since $X Y$ and $P Y$ are middle lines of the triangle $A Q B$. Consequently,
$$
\frac{O X}{X P}=\frac{O X}{B Y}=\frac{X B}{T_{1} Y}=\frac{P Y}{T_{1} Y}
$$
which along with $\angle P X B=\angle P Y B$ and $\angle O X B=\angle T_{1} Y B$ gives $\angle O X P=\angle P Y T_{1}$ and $\triangle O X P \sim \triangle P Y T_{1}$. Thus $\angle X P O=\angle Y T_{1} P$ and $\angle P O X=\angle T_{1} P Y$.
In conclusion,
$$
\angle O P T_{1}=\angle X P Y+\angle X P O+\angle Y P T_{1}=\angle P X A+\angle X P O+\angle X O P=90^{\circ}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K \in(P Q)$. The line $P D$ intersects the circumcircle of triangle $B K Q$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $B Q$. Prove that $T B=T Q$.
|
As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C D=\angle Q C D=\angle Q B A$ (the latter is due to $Q B C D$ being cyclic). We deduce that $P X \| Q B$ and that $P$ is the midpoint of $A Q$, so $A P=P Q$.
Figure 5: G5
Now let $T_{1}$ be the midpoint of the arc $B Q$, not containing $K$, from the circumcircle of $\triangle B K Q$, then $T_{1} B=T_{1} Q$. Due to $\angle D P O=90^{\circ}$, it suffices to show that $\angle O P T_{1}=90^{\circ}$ - indeed, $T \equiv T_{1}$ and $T B=T Q$ would follow.
Denote by $Y$ the midpoint of $B Q$. Then $\angle O X B=\angle T_{1} Y B=90^{\circ}$. The quadrilateral $Q K B T_{1}$ is inscribed in a circle, hence $\angle B T_{1} Q=180-\angle B K Q=\angle A K B$. Then $\angle X B O=$ $\frac{1}{2} \angle A K B=\frac{1}{2} \angle B T_{1} Q=\angle B T_{1} Y$ and thus $\triangle O X B \sim \triangle B Y T_{1}$. The quadrilaterals $P X B Y$
and $A X Y P$ are paralellograms, since $X Y$ and $P Y$ are middle lines of the triangle $A Q B$. Consequently,
$$
\frac{O X}{X P}=\frac{O X}{B Y}=\frac{X B}{T_{1} Y}=\frac{P Y}{T_{1} Y}
$$
which along with $\angle P X B=\angle P Y B$ and $\angle O X B=\angle T_{1} Y B$ gives $\angle O X P=\angle P Y T_{1}$ and $\triangle O X P \sim \triangle P Y T_{1}$. Thus $\angle X P O=\angle Y T_{1} P$ and $\angle P O X=\angle T_{1} P Y$.
In conclusion,
$$
\angle O P T_{1}=\angle X P Y+\angle X P O+\angle Y P T_{1}=\angle P X A+\angle X P O+\angle X O P=90^{\circ}
$$
|
{
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"problem_match": "\nG5.",
"solution_match": "\nSolution."
}
|
1ec7dc96-a6ee-5c58-a087-06ff0d570c7a
| 605,918
|
Let $A B C$ be an acute triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$.
|
Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi,
$$
so, due to the 90-degree angles formed, we have $\widehat{K S L}=\widehat{K T L}$. Thus, KLST is cyclic.
Figure 6: G6
Consider $M$ to be the midpoint of $B C$ and $K^{\prime}$ to be the symmetric point of $K$ with respect to $M$. Then, $B K C K^{\prime}$ is a parallelogram, and so $B K \| C K^{\prime}$. But $B K \| C T$, because they are both perpendicular to $A X$. So, $K^{\prime}$ lies on $C T$ and, as $\widehat{K T K^{\prime}}=90$ and $M$ is the midpoint of $K K^{\prime}, M K=M T$. In a similar way, we have that $M S=M L$. Thus, the center of $(K L S T)$ is $M$.
Consider $D$ to be the foot of altitude from $A$ to $B C$. Then, $D$ belongs in both $(A B K S)$ and ( $A C L T)$. So,
$$
\widehat{A D T}+\widehat{A C T}=180^{\circ}=\widehat{A B S}+\widehat{A D S}=\widehat{A D T}+90^{\circ}-\alpha=\widehat{A D S}+90^{\circ}-\alpha,
$$
and $A D$ is the bisector of $\widehat{S D T}$.
Because $D M$ is perpendicular to $A D, D M$ is the external bisector of this angle, and, as $M S=M T$, it follows that $D M S T$ is cyclic. In a similar way, we have that $D M L K$ is also cyclic.
So, we have that $S T, K L$ and $D M$ are the radical axes of these three circles, $(K L S T)$, $(D M S T),(D M K L)$. These lines are, therefore, concurrent, and we have proved the desired result.
Alternative solution. We continue after proving that $M$ is the center of $(K L S T)$. If $D$ is the foot of perpendicular from $A$ to $B C$, then $A S D K B$ is cyclic, as well as $A T D L C$. The radical axes of those two circles and ( $K L S T)$ are concurrent, thus $K S$ and $L T$ intersect on point $Q \in A D$. So, if $P$ is the intersection point of $K L$ and $T S$, due to Brokard's theorem, $A Q$ is perpendicular to $M P$. This is, of course, equivalent to proving that $P$ belongs on $B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$.
|
Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi,
$$
so, due to the 90-degree angles formed, we have $\widehat{K S L}=\widehat{K T L}$. Thus, KLST is cyclic.
Figure 6: G6
Consider $M$ to be the midpoint of $B C$ and $K^{\prime}$ to be the symmetric point of $K$ with respect to $M$. Then, $B K C K^{\prime}$ is a parallelogram, and so $B K \| C K^{\prime}$. But $B K \| C T$, because they are both perpendicular to $A X$. So, $K^{\prime}$ lies on $C T$ and, as $\widehat{K T K^{\prime}}=90$ and $M$ is the midpoint of $K K^{\prime}, M K=M T$. In a similar way, we have that $M S=M L$. Thus, the center of $(K L S T)$ is $M$.
Consider $D$ to be the foot of altitude from $A$ to $B C$. Then, $D$ belongs in both $(A B K S)$ and ( $A C L T)$. So,
$$
\widehat{A D T}+\widehat{A C T}=180^{\circ}=\widehat{A B S}+\widehat{A D S}=\widehat{A D T}+90^{\circ}-\alpha=\widehat{A D S}+90^{\circ}-\alpha,
$$
and $A D$ is the bisector of $\widehat{S D T}$.
Because $D M$ is perpendicular to $A D, D M$ is the external bisector of this angle, and, as $M S=M T$, it follows that $D M S T$ is cyclic. In a similar way, we have that $D M L K$ is also cyclic.
So, we have that $S T, K L$ and $D M$ are the radical axes of these three circles, $(K L S T)$, $(D M S T),(D M K L)$. These lines are, therefore, concurrent, and we have proved the desired result.
Alternative solution. We continue after proving that $M$ is the center of $(K L S T)$. If $D$ is the foot of perpendicular from $A$ to $B C$, then $A S D K B$ is cyclic, as well as $A T D L C$. The radical axes of those two circles and ( $K L S T)$ are concurrent, thus $K S$ and $L T$ intersect on point $Q \in A D$. So, if $P$ is the intersection point of $K L$ and $T S$, due to Brokard's theorem, $A Q$ is perpendicular to $M P$. This is, of course, equivalent to proving that $P$ belongs on $B C$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG6.",
"solution_match": "\nSolution."
}
|
06444c5c-1fd0-5295-8e41-660d3d201748
| 605,926
|
Let $A D, B E$, and $C F$ denote the altitudes of triangle $\triangle A B C$. Points $E^{\prime}$ and $F^{\prime}$ are the reflections of $E$ and $F$ over $A D$, respectively. The lines $B F^{\prime}$ and $C E^{\prime}$ intersect at $X$, while the lines $B E^{\prime}$ and $C F^{\prime}$ intersect at the point $Y$. Prove that if $H$ is the orthocenter of $\triangle A B C$, then the lines $A X, Y H$, and $B C$ are concurrent.
|
We will prove that the desired point of concurrency is the midpoint of $B C$. Assume that $\triangle A B C$ is acute. Let $(A B C)^{5}$ intersect $(A E F)$ at the point $Y^{\prime}$; we will prove that $Y=Y^{\prime}$.
Figure 7: G7
Using the fact that $H$ is the incenter of $\triangle D E F$ we get that $D, E^{\prime}, F$ and $D, F^{\prime}, E$ are triples of collinear points. Furthermore,
$$
90^{\circ}=\angle{ }^{6} A E H=\angle A F^{\prime} H=\angle A E^{\prime} H=\angle A F H \Rightarrow F^{\prime}, E^{\prime}, H \in\left(A E F Y^{\prime}\right) .
$$
We will now prove that the points $Y^{\prime}, B, D, F^{\prime}$ are concyclic. Indeed,
$$
\angle Y^{\prime} B D=\angle Y^{\prime} B C=\angle Y^{\prime} A C=\angle Y^{\prime} A E=\angle Y^{\prime} F^{\prime} E \Rightarrow\left(Y^{\prime}, B, D, F^{\prime}\right) .
$$
Now, as
$$
\angle F^{\prime} Y^{\prime} B=\angle F^{\prime} D C=\angle E D C=\angle C A B=\angle C Y^{\prime} B,
$$
the points $C, F^{\prime}, Y^{\prime}$ are collinear. Similarly we get that $B, E^{\prime}, Y^{\prime}$ are collinear, which implies
$$
Y^{\prime}=Y=(A B C) \cap(A E F) .
$$
[^4]Since we proved this property using directed angles, we know that it is also true for obtuse triangles.
Notice that the points $A, B, C, H$ form an orthocentric system; in other words $H$ is the orthocenter of $\triangle A B C$ and $A$ is the orthocenter $\triangle H B C$. Furthermore, notice that $F^{\prime}$ is to $\triangle A B C$ as $E^{\prime}$ is to $\triangle H B C$ and that $E^{\prime}$ is to $\triangle A B C$ as $F^{\prime}$ is to $\triangle H B C$. This means that $X$ is to $\triangle H B C$ as $Y$ is to $\triangle A B C$ and, as we know the proven property is also true for obtuse triangles, we get
$$
X=(H B C) \cap(A E F) .
$$
By Reflecting the Orthocenter Lemma we know that in a triangle $A B C$, the reflection of its orthocenter over the midpoint of $B C$ is the antipode of $A$ w.r.t. ( $A B C$ ). Applying this Lemma on the triangles $A B C$ and $H B C$ we get that $Y H$ and $A X$ both go through the midpoint of $B C$, thus finishing the solution.
Remark 1: The crucial part of this solution is defining the points $X, Y$ as intersections of circles. This can also be achieved directly by using similar triangles or by using the Spiral Similarity Lemma on $\triangle H B C, \triangle H F^{\prime} E^{\prime}$ and $\triangle A B C, \triangle A E^{\prime} F^{\prime}$.
Remark 2: We can also invert around $A$ with radius $\sqrt{A H \cdot A D}$ or around $H$ with radius $\sqrt{H A \cdot H D}$ to prove that $X$ or $Y$ invert to the midpoint of $B C$ by using the existence of the nine-point circle.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A D, B E$, and $C F$ denote the altitudes of triangle $\triangle A B C$. Points $E^{\prime}$ and $F^{\prime}$ are the reflections of $E$ and $F$ over $A D$, respectively. The lines $B F^{\prime}$ and $C E^{\prime}$ intersect at $X$, while the lines $B E^{\prime}$ and $C F^{\prime}$ intersect at the point $Y$. Prove that if $H$ is the orthocenter of $\triangle A B C$, then the lines $A X, Y H$, and $B C$ are concurrent.
|
We will prove that the desired point of concurrency is the midpoint of $B C$. Assume that $\triangle A B C$ is acute. Let $(A B C)^{5}$ intersect $(A E F)$ at the point $Y^{\prime}$; we will prove that $Y=Y^{\prime}$.
Figure 7: G7
Using the fact that $H$ is the incenter of $\triangle D E F$ we get that $D, E^{\prime}, F$ and $D, F^{\prime}, E$ are triples of collinear points. Furthermore,
$$
90^{\circ}=\angle{ }^{6} A E H=\angle A F^{\prime} H=\angle A E^{\prime} H=\angle A F H \Rightarrow F^{\prime}, E^{\prime}, H \in\left(A E F Y^{\prime}\right) .
$$
We will now prove that the points $Y^{\prime}, B, D, F^{\prime}$ are concyclic. Indeed,
$$
\angle Y^{\prime} B D=\angle Y^{\prime} B C=\angle Y^{\prime} A C=\angle Y^{\prime} A E=\angle Y^{\prime} F^{\prime} E \Rightarrow\left(Y^{\prime}, B, D, F^{\prime}\right) .
$$
Now, as
$$
\angle F^{\prime} Y^{\prime} B=\angle F^{\prime} D C=\angle E D C=\angle C A B=\angle C Y^{\prime} B,
$$
the points $C, F^{\prime}, Y^{\prime}$ are collinear. Similarly we get that $B, E^{\prime}, Y^{\prime}$ are collinear, which implies
$$
Y^{\prime}=Y=(A B C) \cap(A E F) .
$$
[^4]Since we proved this property using directed angles, we know that it is also true for obtuse triangles.
Notice that the points $A, B, C, H$ form an orthocentric system; in other words $H$ is the orthocenter of $\triangle A B C$ and $A$ is the orthocenter $\triangle H B C$. Furthermore, notice that $F^{\prime}$ is to $\triangle A B C$ as $E^{\prime}$ is to $\triangle H B C$ and that $E^{\prime}$ is to $\triangle A B C$ as $F^{\prime}$ is to $\triangle H B C$. This means that $X$ is to $\triangle H B C$ as $Y$ is to $\triangle A B C$ and, as we know the proven property is also true for obtuse triangles, we get
$$
X=(H B C) \cap(A E F) .
$$
By Reflecting the Orthocenter Lemma we know that in a triangle $A B C$, the reflection of its orthocenter over the midpoint of $B C$ is the antipode of $A$ w.r.t. ( $A B C$ ). Applying this Lemma on the triangles $A B C$ and $H B C$ we get that $Y H$ and $A X$ both go through the midpoint of $B C$, thus finishing the solution.
Remark 1: The crucial part of this solution is defining the points $X, Y$ as intersections of circles. This can also be achieved directly by using similar triangles or by using the Spiral Similarity Lemma on $\triangle H B C, \triangle H F^{\prime} E^{\prime}$ and $\triangle A B C, \triangle A E^{\prime} F^{\prime}$.
Remark 2: We can also invert around $A$ with radius $\sqrt{A H \cdot A D}$ or around $H$ with radius $\sqrt{H A \cdot H D}$ to prove that $X$ or $Y$ invert to the midpoint of $B C$ by using the existence of the nine-point circle.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG7.",
"solution_match": "\nSolution."
}
|
7c062336-7226-594d-baa9-74a600ef2018
| 605,933
|
Given an acute triangle $A B C,(c)$ is circumcircle with center $O$ and $H$ the orthocenter of the triangle $A B C$. The line $A O$ intersects $(c)$ at the point $D$. Let $D_{1}, D_{2}$ and $H_{2}, H_{3}$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $A B, A C$ respectively. Let $\left(c_{1}\right)$ be the circumcircle of the triangle $A D_{1} D_{2}$. Suppose that the line $A H$ intersects again $\left(c_{1}\right)$ at the point $U$, the line $H_{2} H_{3}$ intersects the segment $D_{1} D_{2}$ at the point $K_{1}$ and the line $D H_{3}$ intersects the segment $U D_{2}$ at the point $L_{1}$. Prove that one of the intersection points of the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$ lies on the line $K_{1} L_{1}$.
|
It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
Figure 8: G8
Let $L$ be the second point of intersection of $(c)$ and $\left(c_{1}\right)$. First we will prove that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ pass through the point $L$.
Suppose that the line $A H$ intersects the side $B C$ at the point $Z$. Since $H_{1} D\|B C\| D_{1} D_{2}$ and $B, C$ are the midpoints of the segments $D_{1} D, D_{2} D$ respectively, we get that $Z$ is the midpoint of the segment $H H_{1}$, so the point $H$ lies on $D_{1} D_{2}$. Therefore, $A H \perp D_{1} D_{2}$ and $A U$ is a diameter of $\left(c_{1}\right)$. Thus, $A L \perp U L$ and $A L \perp D L$. We have that the points $U, D, L$ are collinear. (1)
Now, $\angle A L D_{1}=\angle A D_{2} D_{1}, \angle A L H_{2}=\angle A C H_{2}$. Since $A H C D_{2}$ is cyclic we get
$\angle A C H_{2}=\angle A D_{2} D_{1}$. Therefore, $\angle A L H_{2}=\angle A L D_{1}$. So the points $D_{1}, H_{2}, L$ are collinear. (2)
Similarly,
$$
\angle D_{1} L D_{2}=\angle D_{1} A D_{2}=180^{\circ}-2\left(\angle A D_{1} H\right) .
$$
Since $A D_{1} B H$ is cyclic we have $\angle A D_{1} H=\angle A B H=\angle A B H_{3}$. Therefore, we get
$$
\angle D_{1} L D_{2}=180^{\circ}-2\left(\angle A B H_{3}\right)=180^{\circ}-2\left(\angle A D H_{3}\right)=180^{\circ}-\angle H_{2} D H_{3} .
$$
Thus,
$$
\angle D_{1} L D_{2}+\angle H_{2} D H_{3}=180^{\circ} \quad \text { or } \angle D_{1} L D_{2}+\angle H_{3} L H_{2}=180^{\circ} .
$$
So the points $H_{3}, L, D_{2}$ are collinear. (3)
From (1), (2), (3) we have that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ are concurrent at the point $L$.
Also we have
$$
\angle H_{3} D A=\angle D_{2} D A-\angle C D H_{3}=\angle A D_{2} D-\angle C B H_{3}
$$
and because $B \mathrm{HD}_{2} \mathrm{C}$ is a parallelogram, we get $\angle C B H_{3}=\angle H D_{2} C$. So
$$
\angle H_{3} D A=\angle A D_{2} D-\angle H D_{2} C=\angle A D_{2} D_{1}=\angle A D_{1} D_{2}=\angle A U D_{2} .
$$
Therefore, the circumcircle of the triangle $U D L_{1}$ passes through the point $A$. Also, $\angle A D_{1} K_{1}=\angle D_{2} D_{1} A=\angle D_{2} U A$. But $A U L_{1} D$ is cyclic and we have $\angle D_{2} U A=\angle H_{3} D A=$ $\angle H_{3} B A=\angle H_{3} H_{2} A$. Therefore, $\angle A D_{1} K_{1}=\angle H_{3} H_{2} A$. Thus, the circumcircle of the triangle $D_{1} K_{1} H_{2}$ passes through the point $A$.
Because the points $H_{3}, L, D_{2}$ are collinear by the Desargues theorem, the lines $U D_{1}$, $L_{1} K_{1}, D H_{2}$ are concurrent, let say in the point $M$.
From the similarity of the triangles $U D L_{1}$ and $D_{1} K_{1} H_{2}$ we conclude that $M$ is the center of unique spiral similarity and because the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$ intersect at the point $A$, then the second point of intersection is $M$. Therefore, $M$ lies on the line $K_{1} L_{1}$.
Comment. We can prove the last part in a different way.
Let $M$ be the point of intersection of the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$. Now, we have
$$
\angle K_{1} M A=\angle H_{3} H_{2} A=\angle H_{3} B A=\angle A D H_{3}=\angle L_{1} U A=\angle L_{1} M A .
$$
Therefore, the points $L_{1}, K_{1}, M$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given an acute triangle $A B C,(c)$ is circumcircle with center $O$ and $H$ the orthocenter of the triangle $A B C$. The line $A O$ intersects $(c)$ at the point $D$. Let $D_{1}, D_{2}$ and $H_{2}, H_{3}$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $A B, A C$ respectively. Let $\left(c_{1}\right)$ be the circumcircle of the triangle $A D_{1} D_{2}$. Suppose that the line $A H$ intersects again $\left(c_{1}\right)$ at the point $U$, the line $H_{2} H_{3}$ intersects the segment $D_{1} D_{2}$ at the point $K_{1}$ and the line $D H_{3}$ intersects the segment $U D_{2}$ at the point $L_{1}$. Prove that one of the intersection points of the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$ lies on the line $K_{1} L_{1}$.
|
It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
Figure 8: G8
Let $L$ be the second point of intersection of $(c)$ and $\left(c_{1}\right)$. First we will prove that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ pass through the point $L$.
Suppose that the line $A H$ intersects the side $B C$ at the point $Z$. Since $H_{1} D\|B C\| D_{1} D_{2}$ and $B, C$ are the midpoints of the segments $D_{1} D, D_{2} D$ respectively, we get that $Z$ is the midpoint of the segment $H H_{1}$, so the point $H$ lies on $D_{1} D_{2}$. Therefore, $A H \perp D_{1} D_{2}$ and $A U$ is a diameter of $\left(c_{1}\right)$. Thus, $A L \perp U L$ and $A L \perp D L$. We have that the points $U, D, L$ are collinear. (1)
Now, $\angle A L D_{1}=\angle A D_{2} D_{1}, \angle A L H_{2}=\angle A C H_{2}$. Since $A H C D_{2}$ is cyclic we get
$\angle A C H_{2}=\angle A D_{2} D_{1}$. Therefore, $\angle A L H_{2}=\angle A L D_{1}$. So the points $D_{1}, H_{2}, L$ are collinear. (2)
Similarly,
$$
\angle D_{1} L D_{2}=\angle D_{1} A D_{2}=180^{\circ}-2\left(\angle A D_{1} H\right) .
$$
Since $A D_{1} B H$ is cyclic we have $\angle A D_{1} H=\angle A B H=\angle A B H_{3}$. Therefore, we get
$$
\angle D_{1} L D_{2}=180^{\circ}-2\left(\angle A B H_{3}\right)=180^{\circ}-2\left(\angle A D H_{3}\right)=180^{\circ}-\angle H_{2} D H_{3} .
$$
Thus,
$$
\angle D_{1} L D_{2}+\angle H_{2} D H_{3}=180^{\circ} \quad \text { or } \angle D_{1} L D_{2}+\angle H_{3} L H_{2}=180^{\circ} .
$$
So the points $H_{3}, L, D_{2}$ are collinear. (3)
From (1), (2), (3) we have that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ are concurrent at the point $L$.
Also we have
$$
\angle H_{3} D A=\angle D_{2} D A-\angle C D H_{3}=\angle A D_{2} D-\angle C B H_{3}
$$
and because $B \mathrm{HD}_{2} \mathrm{C}$ is a parallelogram, we get $\angle C B H_{3}=\angle H D_{2} C$. So
$$
\angle H_{3} D A=\angle A D_{2} D-\angle H D_{2} C=\angle A D_{2} D_{1}=\angle A D_{1} D_{2}=\angle A U D_{2} .
$$
Therefore, the circumcircle of the triangle $U D L_{1}$ passes through the point $A$. Also, $\angle A D_{1} K_{1}=\angle D_{2} D_{1} A=\angle D_{2} U A$. But $A U L_{1} D$ is cyclic and we have $\angle D_{2} U A=\angle H_{3} D A=$ $\angle H_{3} B A=\angle H_{3} H_{2} A$. Therefore, $\angle A D_{1} K_{1}=\angle H_{3} H_{2} A$. Thus, the circumcircle of the triangle $D_{1} K_{1} H_{2}$ passes through the point $A$.
Because the points $H_{3}, L, D_{2}$ are collinear by the Desargues theorem, the lines $U D_{1}$, $L_{1} K_{1}, D H_{2}$ are concurrent, let say in the point $M$.
From the similarity of the triangles $U D L_{1}$ and $D_{1} K_{1} H_{2}$ we conclude that $M$ is the center of unique spiral similarity and because the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$ intersect at the point $A$, then the second point of intersection is $M$. Therefore, $M$ lies on the line $K_{1} L_{1}$.
Comment. We can prove the last part in a different way.
Let $M$ be the point of intersection of the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$. Now, we have
$$
\angle K_{1} M A=\angle H_{3} H_{2} A=\angle H_{3} B A=\angle A D H_{3}=\angle L_{1} U A=\angle L_{1} M A .
$$
Therefore, the points $L_{1}, K_{1}, M$ are collinear.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG8.",
"solution_match": "\nSolution."
}
|
63a076a4-353e-52de-8346-addf28d94bc9
| 605,942
|
Given semicircle (c) with diameter $A B$ and center $O$. On the (c) we take point $C$ such that the tangent at the $C$ intersects the line $A B$ at the point $E$. The perpendicular line from $C$ to $A B$ intersects the diameter $A B$ at the point $D$. On the (c) we get the points $H, Z$ such that $C D=C H=C Z$. The line $H Z$ intersects the lines $C O, C D, A B$ at the points $S, I, K$ respectively and the parallel line from $I$ to the line $A B$ intersects the lines $C O, C K$ at the points $L, M$ respectively. We consider the circumcircle $(k)$ of the triangle $L M D$, which intersects again the lines $A B, C K$ at the points $P, U$ respectively. Let $\left(e_{1}\right)$, $\left(e_{2}\right),\left(e_{3}\right)$ be the tangents of the $(k)$ at the points $L, M, P$ respectively and $R=\left(e_{1}\right) \cap\left(e_{2}\right)$, $X=\left(e_{2}\right) \cap\left(e_{3}\right), T=\left(e_{1}\right) \cap\left(e_{3}\right)$. Prove that if $Q$ is the center of $(k)$, the lines $R D, T U$, $X S$ pass through the same point, which lies in the line $I Q$.
|
Since $C H=C Z$ we have $O C \perp H Z$. So from the cyclic quadrilateral $S O D I$ we get
$$
C S \cdot C O=C I \cdot C D .
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H C$ at the point $H$. Let $J$ be the intersection point of lines $(v)$ and $C O$. Then $C J$ is diameter of the circle $(O, O A)$ and
$$
C J=2 C O
$$
From the right triangle $J H C$ we have
$$
H C^{2}=C S \cdot C J
$$
Therefore, from (1), (2) and (3) we get
$$
C S \cdot \frac{1}{2} C J=C I \cdot C D \quad \text { or } \quad H C^{2}=2 C I \cdot C D \text {. }
$$
However $H C=C D$ and thus $C D=2 C I$. Thus, $I$ is the midpoint of the segment $C D$. Nevertheless, $L M \| O K$, so the points $L, M$ are the midpoints of the sides $C O$ and $C K$ respectively. Therefore, the circumcircle $(k)$ of the triangle $L M D$ is the Euler circle of the $C O K$ and thus it passes through the point $S$.
We have $Q S=Q U$ and from the right triangles $O S K, O U K$ we get $P S=P U=\frac{O K}{2}$.
Therefore, the points $P, Q$ are located on the perpendicular bisector of the segment $S U$. Now, we conclude that $S U \| T X$, because $Q P \perp\left(e_{3}\right)$. Similarly, we prove that $D U \| R T$ and $S D \| R X$.
Since the triangles $S U D$ and $X T R$ are homothetic we get that the lines $R D, T U, X S$ are concurrent at the center $\mathcal{M}$ of homothety.
The points $I$ and $Q$ atre the incenters of homothetic triangles $S U D$ and $X T R$, respectively. Thus, the line $I Q$ passes through the point $\mathcal{M}$.
## NUMBER THEORY
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Given semicircle (c) with diameter $A B$ and center $O$. On the (c) we take point $C$ such that the tangent at the $C$ intersects the line $A B$ at the point $E$. The perpendicular line from $C$ to $A B$ intersects the diameter $A B$ at the point $D$. On the (c) we get the points $H, Z$ such that $C D=C H=C Z$. The line $H Z$ intersects the lines $C O, C D, A B$ at the points $S, I, K$ respectively and the parallel line from $I$ to the line $A B$ intersects the lines $C O, C K$ at the points $L, M$ respectively. We consider the circumcircle $(k)$ of the triangle $L M D$, which intersects again the lines $A B, C K$ at the points $P, U$ respectively. Let $\left(e_{1}\right)$, $\left(e_{2}\right),\left(e_{3}\right)$ be the tangents of the $(k)$ at the points $L, M, P$ respectively and $R=\left(e_{1}\right) \cap\left(e_{2}\right)$, $X=\left(e_{2}\right) \cap\left(e_{3}\right), T=\left(e_{1}\right) \cap\left(e_{3}\right)$. Prove that if $Q$ is the center of $(k)$, the lines $R D, T U$, $X S$ pass through the same point, which lies in the line $I Q$.
|
Since $C H=C Z$ we have $O C \perp H Z$. So from the cyclic quadrilateral $S O D I$ we get
$$
C S \cdot C O=C I \cdot C D .
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H C$ at the point $H$. Let $J$ be the intersection point of lines $(v)$ and $C O$. Then $C J$ is diameter of the circle $(O, O A)$ and
$$
C J=2 C O
$$
From the right triangle $J H C$ we have
$$
H C^{2}=C S \cdot C J
$$
Therefore, from (1), (2) and (3) we get
$$
C S \cdot \frac{1}{2} C J=C I \cdot C D \quad \text { or } \quad H C^{2}=2 C I \cdot C D \text {. }
$$
However $H C=C D$ and thus $C D=2 C I$. Thus, $I$ is the midpoint of the segment $C D$. Nevertheless, $L M \| O K$, so the points $L, M$ are the midpoints of the sides $C O$ and $C K$ respectively. Therefore, the circumcircle $(k)$ of the triangle $L M D$ is the Euler circle of the $C O K$ and thus it passes through the point $S$.
We have $Q S=Q U$ and from the right triangles $O S K, O U K$ we get $P S=P U=\frac{O K}{2}$.
Therefore, the points $P, Q$ are located on the perpendicular bisector of the segment $S U$. Now, we conclude that $S U \| T X$, because $Q P \perp\left(e_{3}\right)$. Similarly, we prove that $D U \| R T$ and $S D \| R X$.
Since the triangles $S U D$ and $X T R$ are homothetic we get that the lines $R D, T U, X S$ are concurrent at the center $\mathcal{M}$ of homothety.
The points $I$ and $Q$ atre the incenters of homothetic triangles $S U D$ and $X T R$, respectively. Thus, the line $I Q$ passes through the point $\mathcal{M}$.
## NUMBER THEORY
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nG9.",
"solution_match": "\nSolution."
}
|
cdb7cad7-0964-526d-b9e5-6a7590356336
| 605,951
|
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
|
Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2} .
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $f(p)=2$ for any odd prime number $p$.
Taking any two different odd prime numbers $p, q$ we have
$$
2^{2}+q^{p}=2^{2}+p^{q} \Rightarrow p^{q}=q^{p} \Rightarrow p=q,
$$
contradiction. Hence, $f(2)=2$.
So for any odd prime number $p$ we have
$$
f(p)^{2}+2^{p}=2^{f(p)}+p^{2} .
$$
Copy this relation as
$$
2^{p}-p^{2}=2^{f(p)}-f(p)^{2}
$$
Let $T$ be the set of all positive integers greater than 2 , i.e. $T=\{3,4,5, \ldots\}$. The function $g: T \rightarrow \mathbb{Z}, g(n)=2^{n}-n^{2}$, is strictly increasing, i.e.
$$
g(n+1)-g(n)=2^{n}-2 n-1>0
$$
for all $n \in T$. We show this by induction. Indeed, for $n=3$ it is true, $2^{3}-2 \cdot 3-1>0$. Assume that $2^{k}-2 k-1>0$. It follows that for $n=k+1$ we have
$$
2^{k+1}-2(k+1)-1=\left(2^{k}-2 k-1\right)+\left(2^{k}-2\right)>0
$$
for any $k \geq 3$. Therefore, (2) is true for all $n \in T$.
As consequence, (1) holds if and only if $f(p)=p$ for all odd prime numbers $p$, as well as for $p=2$.
Therefore, the only function that satisfies the given relation is $f(p)=p$, for all $p \in \mathbb{P}$.
|
f(p)=p
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
|
Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2} .
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $f(p)=2$ for any odd prime number $p$.
Taking any two different odd prime numbers $p, q$ we have
$$
2^{2}+q^{p}=2^{2}+p^{q} \Rightarrow p^{q}=q^{p} \Rightarrow p=q,
$$
contradiction. Hence, $f(2)=2$.
So for any odd prime number $p$ we have
$$
f(p)^{2}+2^{p}=2^{f(p)}+p^{2} .
$$
Copy this relation as
$$
2^{p}-p^{2}=2^{f(p)}-f(p)^{2}
$$
Let $T$ be the set of all positive integers greater than 2 , i.e. $T=\{3,4,5, \ldots\}$. The function $g: T \rightarrow \mathbb{Z}, g(n)=2^{n}-n^{2}$, is strictly increasing, i.e.
$$
g(n+1)-g(n)=2^{n}-2 n-1>0
$$
for all $n \in T$. We show this by induction. Indeed, for $n=3$ it is true, $2^{3}-2 \cdot 3-1>0$. Assume that $2^{k}-2 k-1>0$. It follows that for $n=k+1$ we have
$$
2^{k+1}-2(k+1)-1=\left(2^{k}-2 k-1\right)+\left(2^{k}-2\right)>0
$$
for any $k \geq 3$. Therefore, (2) is true for all $n \in T$.
As consequence, (1) holds if and only if $f(p)=p$ for all odd prime numbers $p$, as well as for $p=2$.
Therefore, the only function that satisfies the given relation is $f(p)=p$, for all $p \in \mathbb{P}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nTN1.",
"solution_match": "\nSolution."
}
|
01a57e24-70d9-5918-8e35-f22a36f64ad3
| 605,960
|
Let $S \subset\{1, \ldots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than 1 . Let $T \subset\{1, \ldots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1+\left[\frac{n}{d}\right]$. Prove that the greatest common divisor of the elements in $T$ is 1 .
|
Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t / s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1+\left[\frac{n}{d}\right]$, we infer that we can find at least $1+\left[\frac{n}{d}\right]$ elements in $T$. All of them will be divisible by $t$, and the largest of them, which we shall denote by $M$, will be at least $t \cdot\left(1+\left[\frac{n}{d}\right]\right)$. On the other hand, $t \geq d$, hence
$$
M \geq t \cdot\left(1+\left[\frac{n}{d}\right]\right) \geq d \cdot\left(1+\left[\frac{n}{d}\right]\right)>d \cdot \frac{n}{d}=n
$$
Therefore, $M>n$, which contradicts the fact that $M \in\{1, \ldots, n\}$.
In conclusion, $t=1$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $S \subset\{1, \ldots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than 1 . Let $T \subset\{1, \ldots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1+\left[\frac{n}{d}\right]$. Prove that the greatest common divisor of the elements in $T$ is 1 .
|
Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t / s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1+\left[\frac{n}{d}\right]$, we infer that we can find at least $1+\left[\frac{n}{d}\right]$ elements in $T$. All of them will be divisible by $t$, and the largest of them, which we shall denote by $M$, will be at least $t \cdot\left(1+\left[\frac{n}{d}\right]\right)$. On the other hand, $t \geq d$, hence
$$
M \geq t \cdot\left(1+\left[\frac{n}{d}\right]\right) \geq d \cdot\left(1+\left[\frac{n}{d}\right]\right)>d \cdot \frac{n}{d}=n
$$
Therefore, $M>n$, which contradicts the fact that $M \in\{1, \ldots, n\}$.
In conclusion, $t=1$, as desired.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nTN2.",
"solution_match": "\nSolution."
}
|
53b6cdf9-32ee-5ff8-a1aa-b829b396fb2e
| 605,967
|
Let $n(n \geq 1)$ be a positive integer and $U=\{1, \ldots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d(d \neq 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the greatest common divisor of all elements of $T$ is equal to 1 .
|
We will show that $k_{\min }=1+\left[\frac{n}{d}\right]$ (here [.] denotes the integer part).
Obviously, the number of elements of $S$ is not grater than $\left[\frac{n}{d}\right]$, i.e. $|S| \leq\left[\frac{n}{d}\right]$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to 1 , then $|T| \geq|S|+1$.
1) Assume that $|S|<\left[\frac{n}{d}\right]$. Let $T$ be the subset of $U$, consisting of all multiples of $d$ in $U$. Thus, $|T|=\left[\frac{n}{d}\right]$ and $S \subset T$. Therefore, the greatest common divisor of all elements of $T$ is $d>1$. Thus, $k \geq 1+\left[\frac{n}{d}\right]$.
2) Assume $|S|=\left[\frac{n}{d}\right]$. Let $T$ be any subset of $U$ with $S \subset T, S \neq T$. Therefore, $|T| \geq 1+\left[\frac{n}{d}\right]$. Let $q$ be the greatest common divisor of all elements of $T$. Assume that $q>1$. Therefore, $q$ is a common divisor of all elements of $S$ as well. Hence, $q \geq d$. It follows that $|T| \leq\left[\frac{n}{q}\right] \leq\left[\frac{n}{d}\right]$, contradiction. Hence, $q=1$.
Therefore, the minimal possible value of $k$ is $1+\left[\frac{n}{d}\right]$.
[^5]
## COMBINATORICS
|
1+\left[\frac{n}{d}\right]
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $n(n \geq 1)$ be a positive integer and $U=\{1, \ldots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d(d \neq 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the greatest common divisor of all elements of $T$ is equal to 1 .
|
We will show that $k_{\min }=1+\left[\frac{n}{d}\right]$ (here [.] denotes the integer part).
Obviously, the number of elements of $S$ is not grater than $\left[\frac{n}{d}\right]$, i.e. $|S| \leq\left[\frac{n}{d}\right]$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to 1 , then $|T| \geq|S|+1$.
1) Assume that $|S|<\left[\frac{n}{d}\right]$. Let $T$ be the subset of $U$, consisting of all multiples of $d$ in $U$. Thus, $|T|=\left[\frac{n}{d}\right]$ and $S \subset T$. Therefore, the greatest common divisor of all elements of $T$ is $d>1$. Thus, $k \geq 1+\left[\frac{n}{d}\right]$.
2) Assume $|S|=\left[\frac{n}{d}\right]$. Let $T$ be any subset of $U$ with $S \subset T, S \neq T$. Therefore, $|T| \geq 1+\left[\frac{n}{d}\right]$. Let $q$ be the greatest common divisor of all elements of $T$. Assume that $q>1$. Therefore, $q$ is a common divisor of all elements of $S$ as well. Hence, $q \geq d$. It follows that $|T| \leq\left[\frac{n}{q}\right] \leq\left[\frac{n}{d}\right]$, contradiction. Hence, $q=1$.
Therefore, the minimal possible value of $k$ is $1+\left[\frac{n}{d}\right]$.
[^5]
## COMBINATORICS
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nTN2b. ${ }^{7}$",
"solution_match": "\nSolution."
}
|
9c634123-ce97-511d-833f-16a7312272c2
| 605,977
|
100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ steps.
|
With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1, \sqrt{2,1}, 2
$$
The brackets indicate how to arrive at a suitable final ordering with one step. Obviously one step is necessary in the second and third cases.
Upper bound: First we show $C(N) \leq N-1$, by induction. The base case $N=2$ has already been seen. Now suppose the claim is true for $N-1$, and consider an initial arrangement of $N$ couples. Suppose the types of the left-most couples in line are $a$ and $b$. If $a \neq b$, then in the first step, swap the $b$ in place two with the other person with type $a$. If $a=b$, skip this. In both cases, we now have $N-1$ couples distributed among the final $2 N-2$ places, and we know that $N-2$ steps suffices to order them appropriately, by induction. So $N-1$ steps suffices for $N$ couples.
Lower bound: We need to exhibit an example of an initial order for which $N-1$ steps are necessary. Consider
$$
\mathcal{A}_{N}:=1,2,2,3,3, \ldots, N-1, N-1, N, N, 1
$$
Proceed by induction, with the base case $N=2$ trivial. Suppose there is a sequence of at most $N-2$ steps which works. In any suitable final arrangement, a given type must be in positions (odd, even), whereas they start in positions (even, odd). So each type must be involved in at least one step. However, each step involves at most two types, so by the pigeonhole principle, at least four types are involved in at most one step. Pick one such type $a \neq 1$. The one step involving $a$ must be one of
$$
\ldots, ?, a, a, ?, \ldots .
$$
Neither of these steps affects the relative order of the $2 N-2$ other people. So by ignoring this step involving the $a$, we have a sequence of at most $N-3$ steps acting on the other $2 N-2$ people which appropriately sorts them. By induction, this is a contradiction.
Alternative lower bound I: Consider the graph with vertices given by pairs of positions $\{(1,2),(3,4), \ldots,(2 N-1,2 N)\}$. We add an edge between pairs of (different)
vertices if we ever swap two people in places corresponding to those vertices. In particular, at the end, the two people with type $k$ end up in places corresponding to a single vertex.
Suppose we start from the ordering (1) and have some number of steps leading to an ordering where everyone is next to their partner. Then, in the induced graph, there is a path between the vertices corresponding to the places $(2 k-3,2 k-2)$ and $(2 k-1,2 k)$ for each $2 \leq k \leq N$, and also between $(1,2)$ and $(2 N-1,2 N)$. In other words, the graph is connected, and so must have at least $N-1$ edges.
Alternative lower bound II: Consider a bipartite multigraph with vertex classes $\left(v_{1}, \ldots, v_{n}\right)$ and $\left(w_{1}, \ldots, w_{n}\right)$. Connect $v_{i}$ to $w_{j}$ if a person of type $j$ is in positions ( $2 i-1,2 i$ ) (if both positions are taken by the type $j$ couple, then add two edges).
Each step in the dance consists of replacing edges $\left.E=\left\{v_{a} \leftrightarrow w_{c}, v_{b} \leftrightarrow w_{d}\right)\right\}$ with $E^{\prime}=\left\{v_{a} \leftrightarrow w_{d}, v_{b} \leftrightarrow w_{c}\right\}$. However, both before and after the step, the number of components in the graph which include $\left\{v_{a}, v_{b}, w_{c}, w_{d}\right\}$ is either one or two. The structure of other components which do not include these vertices is unaffected by the move.
Therefore, the number of connected components increases by at most 1 in each step.
Starting from configuration (1), the graph initially consists of a single (cyclic) component, so one requires at least $n-1$ steps to get to the final configuration for which there are $n$ connected components.
|
N-1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ steps.
|
With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1, \sqrt{2,1}, 2
$$
The brackets indicate how to arrive at a suitable final ordering with one step. Obviously one step is necessary in the second and third cases.
Upper bound: First we show $C(N) \leq N-1$, by induction. The base case $N=2$ has already been seen. Now suppose the claim is true for $N-1$, and consider an initial arrangement of $N$ couples. Suppose the types of the left-most couples in line are $a$ and $b$. If $a \neq b$, then in the first step, swap the $b$ in place two with the other person with type $a$. If $a=b$, skip this. In both cases, we now have $N-1$ couples distributed among the final $2 N-2$ places, and we know that $N-2$ steps suffices to order them appropriately, by induction. So $N-1$ steps suffices for $N$ couples.
Lower bound: We need to exhibit an example of an initial order for which $N-1$ steps are necessary. Consider
$$
\mathcal{A}_{N}:=1,2,2,3,3, \ldots, N-1, N-1, N, N, 1
$$
Proceed by induction, with the base case $N=2$ trivial. Suppose there is a sequence of at most $N-2$ steps which works. In any suitable final arrangement, a given type must be in positions (odd, even), whereas they start in positions (even, odd). So each type must be involved in at least one step. However, each step involves at most two types, so by the pigeonhole principle, at least four types are involved in at most one step. Pick one such type $a \neq 1$. The one step involving $a$ must be one of
$$
\ldots, ?, a, a, ?, \ldots .
$$
Neither of these steps affects the relative order of the $2 N-2$ other people. So by ignoring this step involving the $a$, we have a sequence of at most $N-3$ steps acting on the other $2 N-2$ people which appropriately sorts them. By induction, this is a contradiction.
Alternative lower bound I: Consider the graph with vertices given by pairs of positions $\{(1,2),(3,4), \ldots,(2 N-1,2 N)\}$. We add an edge between pairs of (different)
vertices if we ever swap two people in places corresponding to those vertices. In particular, at the end, the two people with type $k$ end up in places corresponding to a single vertex.
Suppose we start from the ordering (1) and have some number of steps leading to an ordering where everyone is next to their partner. Then, in the induced graph, there is a path between the vertices corresponding to the places $(2 k-3,2 k-2)$ and $(2 k-1,2 k)$ for each $2 \leq k \leq N$, and also between $(1,2)$ and $(2 N-1,2 N)$. In other words, the graph is connected, and so must have at least $N-1$ edges.
Alternative lower bound II: Consider a bipartite multigraph with vertex classes $\left(v_{1}, \ldots, v_{n}\right)$ and $\left(w_{1}, \ldots, w_{n}\right)$. Connect $v_{i}$ to $w_{j}$ if a person of type $j$ is in positions ( $2 i-1,2 i$ ) (if both positions are taken by the type $j$ couple, then add two edges).
Each step in the dance consists of replacing edges $\left.E=\left\{v_{a} \leftrightarrow w_{c}, v_{b} \leftrightarrow w_{d}\right)\right\}$ with $E^{\prime}=\left\{v_{a} \leftrightarrow w_{d}, v_{b} \leftrightarrow w_{c}\right\}$. However, both before and after the step, the number of components in the graph which include $\left\{v_{a}, v_{b}, w_{c}, w_{d}\right\}$ is either one or two. The structure of other components which do not include these vertices is unaffected by the move.
Therefore, the number of connected components increases by at most 1 in each step.
Starting from configuration (1), the graph initially consists of a single (cyclic) component, so one requires at least $n-1$ steps to get to the final configuration for which there are $n$ connected components.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nC1.",
"solution_match": "\nSolution."
}
|
379abfeb-7d99-5ab7-af81-c8663566089b
| 605,985
|
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$.
|
We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.
Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there are some 4 squares (namely all the intersections of an even column with an even row) that we don't take in such pairs.
It follows that the maximal total sum over all $3^{2}$ choices of an odd column with an odd row is
$$
5 \times(17+18+\cdots+25)+3 \times(5+6+\cdots+16)=1323
$$
So, by an averaging argument, there exists a pair of an odd column with an odd row with sum at most $\frac{1323}{9}=147$.
Then all the other squares of the array will have sum at least
$$
(1+2+\cdots+25)-147=178
$$
But for these squares there is a tiling with $2 \times 2$ arrays, which are 4 in total. So there is an $2 \times 2$ array, whose numbers have a sum at least $\frac{178}{4}>44$. So, there is a $2 \times 2$ array whose numbers have a sum at least 45 . This argument gives that
$$
k_{\max } \geq 45 .
$$
We are going now to give an example of an array, in which 45 is the best possible. We fill the rows of the array as follows:
| 25 | 5 | 24 | 6 | 23 |
| :---: | :---: | :---: | :---: | :---: |
| 11 | 4 | 12 | 3 | 13 |
| 22 | 7 | 21 | 8 | 20 |
| 14 | 2 | 15 | 1 | 16 |
| 19 | 9 | 18 | 10 | 17 |
We are going now to even rows:
In the above array, every $2 \times 2$ subarray has a sum, which is less or equal to 45 . This gives that
$$
k_{\max } \leq 45 .
$$
A combination of (1) and (2) gives that $k_{\max }=45$.
[^6]C3. Anna and Bob play a game on the set of all points of the form $(m, n)$ where $m, n$ are integers with $|m|,|n| \leqslant 2019$. Let us call the lines $x= \pm 2019$ and $y= \pm 2019$ the boundary lines of the game. The points of these lines are called the boundary points. The neighbours of point $(m, n)$ are the points $(m+1, n),(m-1, n),(m, n+1),(m, n-1)$.
Anna starts with a token at the origin ( 0,0 ). With Bob playing first, they alternately perform the following steps: At his turn, Bob deletes two points on each boundary line. On her turn Anna makes a sequences of three moves of the token, where a move of the token consists of picking up the token from its current position and placing it in one of its neighbours.
To win the game Anna must place her token on a boundary point before it is deleted by Bob. Does Anna have a winning strategy?
[Note: At every turn except perhaps her last, Anna must make exactly three moves.]
|
45
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$.
|
We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.
Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there are some 4 squares (namely all the intersections of an even column with an even row) that we don't take in such pairs.
It follows that the maximal total sum over all $3^{2}$ choices of an odd column with an odd row is
$$
5 \times(17+18+\cdots+25)+3 \times(5+6+\cdots+16)=1323
$$
So, by an averaging argument, there exists a pair of an odd column with an odd row with sum at most $\frac{1323}{9}=147$.
Then all the other squares of the array will have sum at least
$$
(1+2+\cdots+25)-147=178
$$
But for these squares there is a tiling with $2 \times 2$ arrays, which are 4 in total. So there is an $2 \times 2$ array, whose numbers have a sum at least $\frac{178}{4}>44$. So, there is a $2 \times 2$ array whose numbers have a sum at least 45 . This argument gives that
$$
k_{\max } \geq 45 .
$$
We are going now to give an example of an array, in which 45 is the best possible. We fill the rows of the array as follows:
| 25 | 5 | 24 | 6 | 23 |
| :---: | :---: | :---: | :---: | :---: |
| 11 | 4 | 12 | 3 | 13 |
| 22 | 7 | 21 | 8 | 20 |
| 14 | 2 | 15 | 1 | 16 |
| 19 | 9 | 18 | 10 | 17 |
We are going now to even rows:
In the above array, every $2 \times 2$ subarray has a sum, which is less or equal to 45 . This gives that
$$
k_{\max } \leq 45 .
$$
A combination of (1) and (2) gives that $k_{\max }=45$.
[^6]C3. Anna and Bob play a game on the set of all points of the form $(m, n)$ where $m, n$ are integers with $|m|,|n| \leqslant 2019$. Let us call the lines $x= \pm 2019$ and $y= \pm 2019$ the boundary lines of the game. The points of these lines are called the boundary points. The neighbours of point $(m, n)$ are the points $(m+1, n),(m-1, n),(m, n+1),(m, n-1)$.
Anna starts with a token at the origin ( 0,0 ). With Bob playing first, they alternately perform the following steps: At his turn, Bob deletes two points on each boundary line. On her turn Anna makes a sequences of three moves of the token, where a move of the token consists of picking up the token from its current position and placing it in one of its neighbours.
To win the game Anna must place her token on a boundary point before it is deleted by Bob. Does Anna have a winning strategy?
[Note: At every turn except perhaps her last, Anna must make exactly three moves.]
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nC2b. ${ }^{8}$",
"solution_match": "\nSolution."
}
|
9dd429f2-67e5-51d1-b358-a81a695571c6
| 605,995
|
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$.
|
Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her step, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly 1 . In that step, Bob deletes the next available point to the left and the next available point to the right.
Bob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x>0$ and the exact opposite sequence of moves in the horizontal direction reaching $(x, y)$ then Bob deletes at least as many points to the left of $(0,2019)$ in the first sequence than points to the right of $(0,2019)$ in the second sequence.
So we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k>0$.
Define $\Delta=3 m-(2 x+y)$ where $m$ is the total number of points deleted by Bob to the right of $(0,2019)$, and $(x, y)$ is the position of Anna's token.
For each sequence of steps performed first by Anna and then by Bob, $\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in $3 m$ and $2 x+y$. We have excluded the cases where $2 x+y<0$.
| Step | $(0,3)$ | $(1,2)$ | $(-1,2)$ | $(2,1)$ | $(0,1)$ | $(3,0)$ | $(1,0)$ | $(2,-1)$ | $(1,-2)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m$ | 1 | 2 | 0 (or 1$)$ | 2 | 1 | 2 | 2 | 2 | 2 |
| $3 m$ | 3 | 6 | 0 (or 3$)$ | 6 | 3 | 6 | 6 | 6 | 6 |
| $2 x+y$ | 3 | 4 | 0 | 5 | 1 | 6 | 2 | 3 | 0 |
The table also shows that if in this sequence of steps Anna changes $y$ by +1 or -2 then $\Delta$ is increased by 1 . Also, if Anna changes $y$ by +2 or -1 then the first time this happens $\Delta$ is increased by 2 . (This also holds if her move is $(0,-1)$ or $(-2,-1)$ which are not shown in the table.)
Since Anna wins by placing her token at $(k, 2019)$ we must have $m \leqslant k-1$ and $k \leqslant 2018$. So at that exact moment we have:
$$
\Delta=3 m-(2 k+2019)=k-2022 \leqslant-4 .
$$
So in her last turn she must have decreased $\Delta$ by at least 4 . So her last step must have been $(1,2)$ or $(2,1)$ which give a decrease of 4 and 5 respectively. (It could not be $(3,0)$ because then she must have already won. Also she could not have done just one or two moves in her last turn since this is not enough for the required decrease in $\Delta$.)
If her last step was $(1,2)$ then just before doing it we had $y=2017$ and $\Delta=0$. This means that in one of her steps the total change in $y$ was not $0 \bmod 3$. However in that case we have seen that $\Delta>0$, a contradiction.
If her last step was $(2,1)$ then just before doing it we had $y=2018$ and $\Delta=0$ or $\Delta=1$. So she must have made at least two steps with the change of $y$ being +1 or -2 or at least one step with the change of $y$ being +2 or -1 . In both cases, consulting the table, we get an increase of at least 2 in $\Delta$, a contradiction.
Note 1: If Anna is allowed to make at most three moves at each step, then she actually has a winning strategy.
Note 2: If 2019 is replaced by $N>1$ then Bob has a winning strategy if and only if $3 \mid N$.
|
not found
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$.
|
Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her step, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly 1 . In that step, Bob deletes the next available point to the left and the next available point to the right.
Bob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x>0$ and the exact opposite sequence of moves in the horizontal direction reaching $(x, y)$ then Bob deletes at least as many points to the left of $(0,2019)$ in the first sequence than points to the right of $(0,2019)$ in the second sequence.
So we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k>0$.
Define $\Delta=3 m-(2 x+y)$ where $m$ is the total number of points deleted by Bob to the right of $(0,2019)$, and $(x, y)$ is the position of Anna's token.
For each sequence of steps performed first by Anna and then by Bob, $\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in $3 m$ and $2 x+y$. We have excluded the cases where $2 x+y<0$.
| Step | $(0,3)$ | $(1,2)$ | $(-1,2)$ | $(2,1)$ | $(0,1)$ | $(3,0)$ | $(1,0)$ | $(2,-1)$ | $(1,-2)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m$ | 1 | 2 | 0 (or 1$)$ | 2 | 1 | 2 | 2 | 2 | 2 |
| $3 m$ | 3 | 6 | 0 (or 3$)$ | 6 | 3 | 6 | 6 | 6 | 6 |
| $2 x+y$ | 3 | 4 | 0 | 5 | 1 | 6 | 2 | 3 | 0 |
The table also shows that if in this sequence of steps Anna changes $y$ by +1 or -2 then $\Delta$ is increased by 1 . Also, if Anna changes $y$ by +2 or -1 then the first time this happens $\Delta$ is increased by 2 . (This also holds if her move is $(0,-1)$ or $(-2,-1)$ which are not shown in the table.)
Since Anna wins by placing her token at $(k, 2019)$ we must have $m \leqslant k-1$ and $k \leqslant 2018$. So at that exact moment we have:
$$
\Delta=3 m-(2 k+2019)=k-2022 \leqslant-4 .
$$
So in her last turn she must have decreased $\Delta$ by at least 4 . So her last step must have been $(1,2)$ or $(2,1)$ which give a decrease of 4 and 5 respectively. (It could not be $(3,0)$ because then she must have already won. Also she could not have done just one or two moves in her last turn since this is not enough for the required decrease in $\Delta$.)
If her last step was $(1,2)$ then just before doing it we had $y=2017$ and $\Delta=0$. This means that in one of her steps the total change in $y$ was not $0 \bmod 3$. However in that case we have seen that $\Delta>0$, a contradiction.
If her last step was $(2,1)$ then just before doing it we had $y=2018$ and $\Delta=0$ or $\Delta=1$. So she must have made at least two steps with the change of $y$ being +1 or -2 or at least one step with the change of $y$ being +2 or -1 . In both cases, consulting the table, we get an increase of at least 2 in $\Delta$, a contradiction.
Note 1: If Anna is allowed to make at most three moves at each step, then she actually has a winning strategy.
Note 2: If 2019 is replaced by $N>1$ then Bob has a winning strategy if and only if $3 \mid N$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nC2b. ${ }^{8}$",
"solution_match": "\nSolution."
}
|
9dd429f2-67e5-51d1-b358-a81a695571c6
| 605,995
|
A town-planner has built an isolated city whose road network consists of 2 N roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his driving test, and is nervous about roundabouts. He starts driving from his house, and always takes the first exit at each roundabout he encounters. It turns out his journey includes every road in the town in both directions before he arrives back at the starting point in the starting direction. For what values of $N$ is this possible?
|
$N$ odd. In fact, the number of trajectories has the same parity as $N$.
The setting is a (multi)graph where every vertex has degree three. Each vertex has an orientation, an ordering of its incident edges. We call Vlad's possible paths trajectories, and a complete trajectory if he traverses every edge in both directions. We may assume the multigraph is connected, as otherwise a complete trajectory is certainly not possible.
N odd (construction): There is an example when $N=1$, as shown in Figure 10.
Figure 10: C4: $\quad N=1$
There are two 3-regular graphs on two vertices, the handcuffs and theta. The handcuffs fail since each self-loop has its own trajectory, but the theta does work for two of the four possible orientations.
We now construct examples for $N \geq 3$ odd by induction. Suppose we have a valid 3 -regular graph on $2(N-2)$ vertices, such that Vlad's trajectory is complete. This has at least two (undirected) edges, so pick two of them, $e$ and $e^{\prime}$. (It does not matter if they share incident vertices.) Split both $e$ and $e^{\prime}$ into three, by adding two new vertices to each, and connect as in Figure 11.
New vertices have degree three; other degrees are unchanged, so the graph is still 3regular. For each edge $e$ and $e^{\prime}$, pick a direction. (Both $u p$ in the figure.) These directed edges are part of the complete trajectory given by the induction hypothesis. Choose the orientations of the new vertices to preserve these two sections of the trajectory. The remaining two directed edges in the original graph will end up as partial trajectories in the new graph (see Figure 11).
However, because all the new partial trajectories start and finish at the same places and in the same directions in the original graph, and no other directed edges are changed, the trajectory remains complete. The result for $N$ odd follows by induction.
$\mathbf{N}$ even: Split each edge $e$ in the graph into two directed edges $\overleftarrow{e}$ and $\vec{e}$. Let $D$ be the set of the $6 N$ directed edges. Let $\alpha$ be the permutation of $D$ which exchanges $\overleftarrow{e}$
Figure 11: C4: Trajectories in the old and new graphs
and $\vec{e}$.
Now, for each roundabout $v$, let $\overleftarrow{e_{1}}, \overleftarrow{e_{2}}, \overleftarrow{e_{3}}$ be the three directed edges into $v$. The roundabout has a cyclic orientation, either $\left(\overleftarrow{e_{1}}, \overleftarrow{e_{2}}, \overleftarrow{e_{3}}\right)$ or $\left(\overleftarrow{e_{1}}, \overleftarrow{e_{3}}, \overleftarrow{e_{2}}\right)$. Let $\theta\left(\overleftarrow{e_{1}}\right)$ describe the directed edge after $\overleftarrow{e_{1}}$ in this orientation. By considering all roundabouts, $\theta$ is also a permutation of $D$.
Note that $\theta\left(\overleftarrow{e_{1}}\right)$ is directed towards $v$, so the directed edge after $\overleftarrow{e_{1}}$ in a trajectory is $\alpha\left(\theta\left(\overleftarrow{e_{1}}\right)\right)$. So Vlad makes a complete trajectory precisely if $\alpha \theta$ is a cyclic permutation of $D$. Note that the cycle type of $\theta$ is $(3,3, \ldots, 3)$, and the cycle type of $\alpha$ is $(2,2, \ldots, 2)$. So $\theta$ is always an even permutation, while $\alpha$ is an even permutation precisely when $N$ is even.
However, a cyclic permutation of $D$ is always odd, since $|D|=6 N$ is even. So there is certainly no complete trajectory when $N$ is even.
Alternative I: We claim that in a graph with $E$ edges, and $V$ vertices, the number of trajectories, $T$, has the same parity as $V+E$. We allow degenerate cases of this statement, for example graphs that are disconnected, or trajectories that consist of only a single vertex, so that the graph that consists of $V$ vertices and no edges has precisely $V$ trajectories, and thus satisfies the given claim. This shows that $N$ cannot be even.
We prove the claim by induction on $E$. Suppose we are given a graph with $E \geq 1$ edges and $T$ trajectories. Then consider any edge $e$, and its two directions $\vec{e}, \overleftarrow{e}$. Let $A$ be the sequence of directed edges starting from the one after $\vec{e}$ in its trajectory, ending at the edge before $\overleftarrow{e}$ or $\vec{e}$, whichever appears first. Similarly define $B$ starting after $\overleftarrow{e}$. $A$ and $B$ are disjoint, and may be empty.
Figure 12: C4: (a) Initial trajectories.
(b) After removing $e$
We consider removing $e$, but otherwise keep the orientations at its incident vertices the same. Then if $\vec{e}, \overleftarrow{e}$ are in different trajectories, these are the concatenations ( $\vec{e}, A)$ and $(\overleftarrow{e}, B)$. After removing $e$, for each direction $\overleftarrow{e}, \vec{e}$, instead of proceeding onto this directed edge, the relevant trajectory moves to the other trajectory. In other words, the resulting trajectory is the concatenation $(A, B)$. So $T$ decreases by one.
Similarly, if both directions of $e$ are part of the same trajectory, this is the concatenation $(\vec{e}, A, \overleftarrow{e}, B)$. Then when we remove $e$, this splits into the two trajectories $(A)$ and $(B)$, by an essentially identical argument. So $T$ increases by one. Thus in both cases, removing one edge changes the parity of $T$, and so the claim follows by induction on $E$.
In the original setting we have $V=2 N, E=3 N$, so $T$ must have the same parity as $5 N$. Thus $T=1$ is impossible when $N$ is even.
Alternative II: An alternative is to induct on $N$, using the following stronger claim.
Claim: You can't have exactly one trajectory for $N$ even; nor exactly two trajectories for a connected graph with $N$ odd.
Proof of claim: We have to check that the claim is true for $N=1,2$. Checking $N=2$ requires a couple of case. Alternatively, one can argue that a single cyclic edge with no vertices (!) counts as the case $N=0$. Now use strong induction by contradiction. If $N$ is even, but has exactly one trajectory, then there are no self-loops, so pick any edge $e$, connecting vertices $v \neq w$. Remove $e$, then remove $v$, and connect $v$ 's other two incident edges (which are distinct from each other and $e$ ) to form a single edge. Do the same for $w$.
Figure 13: C4: Trajectories in the old and new graphs
The effect on the trajectories is shown in Figure 13. Note that the new graph is still 3-regular. We then argue as in Proof I that this operation splits the trajectory into two. So if the new graph is connected, this contradicts the hypothesis for $N-1$. Alternately, the new graph might consist of two components. Since it is 3 -regular, each component has an even number of vertices. The total number of vertices is $2(N-1)$, which is 2 modulo 4 , and so one of the components has a number of vertices which is a multiple of four, and a complete trajectory of this component, which also contradicts the induction hypothesis.
Now suppose $N$ is odd, but the original oriented graph has exactly two trajectories. If there is a self-loop at some vertex $v$, then one of the trajectories involves only this self-loop. So remove this vertex, and consider the other vertex $w$ connected to $v$. Remove $w$ and join up its other two incident edges. The resulting graph corresponds to $N$ even, and has a complete trajectory, which is a contradiction.
Otherwise, there are no self-loops, but the graph is connected hence there must be one edge $e$ connecting vertices $v \neq w$ which has one trajectory in one direction, and the other trajectory in the other direction. Collapse this edge as in Figure 13, and again by the same argument as in Proof I, this merges the two trajectories, giving a complete trajectory for $N$ even, and a contradiction.
[^0]: ${ }^{1}$ Proposed by PSC.
[^1]: ${ }^{2}$ Proposed by PSC.
[^2]: ${ }^{3}$ Proposed by PSC.
[^3]: ${ }^{4}$ Proposed by PSC.
[^4]: ${ }^{5}(X Y Z)$ denotes the circumcircle of $\triangle X Y Z$
${ }^{6} \angle$ denotes a directed angle modulo $\pi$
[^5]: ${ }^{7}$ Proposed by PSC.
[^6]: ${ }^{8}$ Proposed by PSC.
|
N \text{ odd}
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
A town-planner has built an isolated city whose road network consists of 2 N roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his driving test, and is nervous about roundabouts. He starts driving from his house, and always takes the first exit at each roundabout he encounters. It turns out his journey includes every road in the town in both directions before he arrives back at the starting point in the starting direction. For what values of $N$ is this possible?
|
$N$ odd. In fact, the number of trajectories has the same parity as $N$.
The setting is a (multi)graph where every vertex has degree three. Each vertex has an orientation, an ordering of its incident edges. We call Vlad's possible paths trajectories, and a complete trajectory if he traverses every edge in both directions. We may assume the multigraph is connected, as otherwise a complete trajectory is certainly not possible.
N odd (construction): There is an example when $N=1$, as shown in Figure 10.
Figure 10: C4: $\quad N=1$
There are two 3-regular graphs on two vertices, the handcuffs and theta. The handcuffs fail since each self-loop has its own trajectory, but the theta does work for two of the four possible orientations.
We now construct examples for $N \geq 3$ odd by induction. Suppose we have a valid 3 -regular graph on $2(N-2)$ vertices, such that Vlad's trajectory is complete. This has at least two (undirected) edges, so pick two of them, $e$ and $e^{\prime}$. (It does not matter if they share incident vertices.) Split both $e$ and $e^{\prime}$ into three, by adding two new vertices to each, and connect as in Figure 11.
New vertices have degree three; other degrees are unchanged, so the graph is still 3regular. For each edge $e$ and $e^{\prime}$, pick a direction. (Both $u p$ in the figure.) These directed edges are part of the complete trajectory given by the induction hypothesis. Choose the orientations of the new vertices to preserve these two sections of the trajectory. The remaining two directed edges in the original graph will end up as partial trajectories in the new graph (see Figure 11).
However, because all the new partial trajectories start and finish at the same places and in the same directions in the original graph, and no other directed edges are changed, the trajectory remains complete. The result for $N$ odd follows by induction.
$\mathbf{N}$ even: Split each edge $e$ in the graph into two directed edges $\overleftarrow{e}$ and $\vec{e}$. Let $D$ be the set of the $6 N$ directed edges. Let $\alpha$ be the permutation of $D$ which exchanges $\overleftarrow{e}$
Figure 11: C4: Trajectories in the old and new graphs
and $\vec{e}$.
Now, for each roundabout $v$, let $\overleftarrow{e_{1}}, \overleftarrow{e_{2}}, \overleftarrow{e_{3}}$ be the three directed edges into $v$. The roundabout has a cyclic orientation, either $\left(\overleftarrow{e_{1}}, \overleftarrow{e_{2}}, \overleftarrow{e_{3}}\right)$ or $\left(\overleftarrow{e_{1}}, \overleftarrow{e_{3}}, \overleftarrow{e_{2}}\right)$. Let $\theta\left(\overleftarrow{e_{1}}\right)$ describe the directed edge after $\overleftarrow{e_{1}}$ in this orientation. By considering all roundabouts, $\theta$ is also a permutation of $D$.
Note that $\theta\left(\overleftarrow{e_{1}}\right)$ is directed towards $v$, so the directed edge after $\overleftarrow{e_{1}}$ in a trajectory is $\alpha\left(\theta\left(\overleftarrow{e_{1}}\right)\right)$. So Vlad makes a complete trajectory precisely if $\alpha \theta$ is a cyclic permutation of $D$. Note that the cycle type of $\theta$ is $(3,3, \ldots, 3)$, and the cycle type of $\alpha$ is $(2,2, \ldots, 2)$. So $\theta$ is always an even permutation, while $\alpha$ is an even permutation precisely when $N$ is even.
However, a cyclic permutation of $D$ is always odd, since $|D|=6 N$ is even. So there is certainly no complete trajectory when $N$ is even.
Alternative I: We claim that in a graph with $E$ edges, and $V$ vertices, the number of trajectories, $T$, has the same parity as $V+E$. We allow degenerate cases of this statement, for example graphs that are disconnected, or trajectories that consist of only a single vertex, so that the graph that consists of $V$ vertices and no edges has precisely $V$ trajectories, and thus satisfies the given claim. This shows that $N$ cannot be even.
We prove the claim by induction on $E$. Suppose we are given a graph with $E \geq 1$ edges and $T$ trajectories. Then consider any edge $e$, and its two directions $\vec{e}, \overleftarrow{e}$. Let $A$ be the sequence of directed edges starting from the one after $\vec{e}$ in its trajectory, ending at the edge before $\overleftarrow{e}$ or $\vec{e}$, whichever appears first. Similarly define $B$ starting after $\overleftarrow{e}$. $A$ and $B$ are disjoint, and may be empty.
Figure 12: C4: (a) Initial trajectories.
(b) After removing $e$
We consider removing $e$, but otherwise keep the orientations at its incident vertices the same. Then if $\vec{e}, \overleftarrow{e}$ are in different trajectories, these are the concatenations ( $\vec{e}, A)$ and $(\overleftarrow{e}, B)$. After removing $e$, for each direction $\overleftarrow{e}, \vec{e}$, instead of proceeding onto this directed edge, the relevant trajectory moves to the other trajectory. In other words, the resulting trajectory is the concatenation $(A, B)$. So $T$ decreases by one.
Similarly, if both directions of $e$ are part of the same trajectory, this is the concatenation $(\vec{e}, A, \overleftarrow{e}, B)$. Then when we remove $e$, this splits into the two trajectories $(A)$ and $(B)$, by an essentially identical argument. So $T$ increases by one. Thus in both cases, removing one edge changes the parity of $T$, and so the claim follows by induction on $E$.
In the original setting we have $V=2 N, E=3 N$, so $T$ must have the same parity as $5 N$. Thus $T=1$ is impossible when $N$ is even.
Alternative II: An alternative is to induct on $N$, using the following stronger claim.
Claim: You can't have exactly one trajectory for $N$ even; nor exactly two trajectories for a connected graph with $N$ odd.
Proof of claim: We have to check that the claim is true for $N=1,2$. Checking $N=2$ requires a couple of case. Alternatively, one can argue that a single cyclic edge with no vertices (!) counts as the case $N=0$. Now use strong induction by contradiction. If $N$ is even, but has exactly one trajectory, then there are no self-loops, so pick any edge $e$, connecting vertices $v \neq w$. Remove $e$, then remove $v$, and connect $v$ 's other two incident edges (which are distinct from each other and $e$ ) to form a single edge. Do the same for $w$.
Figure 13: C4: Trajectories in the old and new graphs
The effect on the trajectories is shown in Figure 13. Note that the new graph is still 3-regular. We then argue as in Proof I that this operation splits the trajectory into two. So if the new graph is connected, this contradicts the hypothesis for $N-1$. Alternately, the new graph might consist of two components. Since it is 3 -regular, each component has an even number of vertices. The total number of vertices is $2(N-1)$, which is 2 modulo 4 , and so one of the components has a number of vertices which is a multiple of four, and a complete trajectory of this component, which also contradicts the induction hypothesis.
Now suppose $N$ is odd, but the original oriented graph has exactly two trajectories. If there is a self-loop at some vertex $v$, then one of the trajectories involves only this self-loop. So remove this vertex, and consider the other vertex $w$ connected to $v$. Remove $w$ and join up its other two incident edges. The resulting graph corresponds to $N$ even, and has a complete trajectory, which is a contradiction.
Otherwise, there are no self-loops, but the graph is connected hence there must be one edge $e$ connecting vertices $v \neq w$ which has one trajectory in one direction, and the other trajectory in the other direction. Collapse this edge as in Figure 13, and again by the same argument as in Proof I, this merges the two trajectories, giving a complete trajectory for $N$ even, and a contradiction.
[^0]: ${ }^{1}$ Proposed by PSC.
[^1]: ${ }^{2}$ Proposed by PSC.
[^2]: ${ }^{3}$ Proposed by PSC.
[^3]: ${ }^{4}$ Proposed by PSC.
[^4]: ${ }^{5}(X Y Z)$ denotes the circumcircle of $\triangle X Y Z$
${ }^{6} \angle$ denotes a directed angle modulo $\pi$
[^5]: ${ }^{7}$ Proposed by PSC.
[^6]: ${ }^{8}$ Proposed by PSC.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2019_bmo_shortlist.jsonl",
"problem_match": "\nC4.",
"solution_match": "\nSolution."
}
|
f5d82b7b-42d3-5737-a5ca-e8f813c62999
| 606,009
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y^{2}\right)=g(x y)
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
Given any $u \geqslant 2$, take $a, b \in \mathbb{R}^{+}$such that $a+b=u$ and $a b=1$. This is possible as the equation $x^{2}-u x+1$ for $u \geqslant 2$ has two positive real solutions. (Discriminant is $u^{2}-4 \geqslant 0$, sum and product of solutions are positive.) Now taking $x=\sqrt{a}, y=\sqrt{b}$ we get $f(u)=g(1)$. Now given any $t \in \mathbb{R}^{+}$, taking $x=t / 2, y=2$ we have
$$
g(t)=f\left(\frac{t^{2}}{4}+4\right)=g(1)
$$
as $\frac{t^{2}}{4}+4 \geqslant 2$. So $g$ is constant. But since any real number can be written as a sum of two squares, then $f$ is constant as well. So there is a $c \in \mathbb{R}$ such that $f(x)=c$ and $g(x)=c$ for every $x \in \mathbb{R}^{+}$. Obviosuly any such pair of functions satisfies the equation.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y^{2}\right)=g(x y)
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
Given any $u \geqslant 2$, take $a, b \in \mathbb{R}^{+}$such that $a+b=u$ and $a b=1$. This is possible as the equation $x^{2}-u x+1$ for $u \geqslant 2$ has two positive real solutions. (Discriminant is $u^{2}-4 \geqslant 0$, sum and product of solutions are positive.) Now taking $x=\sqrt{a}, y=\sqrt{b}$ we get $f(u)=g(1)$. Now given any $t \in \mathbb{R}^{+}$, taking $x=t / 2, y=2$ we have
$$
g(t)=f\left(\frac{t^{2}}{4}+4\right)=g(1)
$$
as $\frac{t^{2}}{4}+4 \geqslant 2$. So $g$ is constant. But since any real number can be written as a sum of two squares, then $f$ is constant as well. So there is a $c \in \mathbb{R}$ such that $f(x)=c$ and $g(x)=c$ for every $x \in \mathbb{R}^{+}$. Obviosuly any such pair of functions satisfies the equation.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA1.",
"solution_match": "\nSolution."
}
|
ecb3b536-a365-5123-8258-2101b6ee4dea
| 606,020
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y\right) \geqslant\left(\frac{1}{x}+1\right) f(y)
$$
holds for all $x \in \mathbb{R} \backslash\{0\}$ and all $y \in \mathbb{R}$.
## Proposed by Uzbekistan
|
We will show that $f(x)=0$ for all $x \in \mathbb{R}$ which obviously satisfies the equation.
For $x=-1$ and $y=t+1$ we get $f(t) \geqslant 0$ for every $t \in \mathbb{R}$.
For $x=\frac{1}{n}$, we get that
$$
f\left(y+\frac{1}{n^{2}}\right) \geqslant(n+1) f(y) .
$$
Therefore
$$
f\left(y+\frac{2}{n^{2}}\right) \geqslant(n+1) f\left(y+\frac{1}{n^{2}}\right) \geqslant(n+1)^{2} f(y)
$$
and inductively we have
$$
f\left(y+\frac{k}{n^{2}}\right) \geqslant(n+1)^{k} f(y)
$$
This holds for each $k, n \in \mathbb{N}$ and each $y \in \mathbb{R}$. In particular, for $k=n^{2}$ we get
$$
f(y+1) \geqslant(n+1)^{n^{2}} f(y) .
$$
Now if $f(y)>0$, then letting $n$ tend to infinity we obtain a contradiction. (E.g. taking $n>$ $f(y+1) / f(y)$ we get $f(y+1) \geqslant(n+1)^{n^{2}} f(y) \geqslant(n+1) f(y)>f(y+1)$, a contradiction.)
So $f(x)=0$ for every $x \in \mathbb{R}$.
|
f(x)=0
|
Yes
|
Yes
|
proof
|
Algebra
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y\right) \geqslant\left(\frac{1}{x}+1\right) f(y)
$$
holds for all $x \in \mathbb{R} \backslash\{0\}$ and all $y \in \mathbb{R}$.
## Proposed by Uzbekistan
|
We will show that $f(x)=0$ for all $x \in \mathbb{R}$ which obviously satisfies the equation.
For $x=-1$ and $y=t+1$ we get $f(t) \geqslant 0$ for every $t \in \mathbb{R}$.
For $x=\frac{1}{n}$, we get that
$$
f\left(y+\frac{1}{n^{2}}\right) \geqslant(n+1) f(y) .
$$
Therefore
$$
f\left(y+\frac{2}{n^{2}}\right) \geqslant(n+1) f\left(y+\frac{1}{n^{2}}\right) \geqslant(n+1)^{2} f(y)
$$
and inductively we have
$$
f\left(y+\frac{k}{n^{2}}\right) \geqslant(n+1)^{k} f(y)
$$
This holds for each $k, n \in \mathbb{N}$ and each $y \in \mathbb{R}$. In particular, for $k=n^{2}$ we get
$$
f(y+1) \geqslant(n+1)^{n^{2}} f(y) .
$$
Now if $f(y)>0$, then letting $n$ tend to infinity we obtain a contradiction. (E.g. taking $n>$ $f(y+1) / f(y)$ we get $f(y+1) \geqslant(n+1)^{n^{2}} f(y) \geqslant(n+1) f(y)>f(y+1)$, a contradiction.)
So $f(x)=0$ for every $x \in \mathbb{R}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA2.",
"solution_match": "\nSolution."
}
|
1f98e4ec-f061-54a9-ba38-831c7f2c80d8
| 606,027
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
1. We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a))=f(1+f(1)+f(b))=2 f(1)+b
$$
and therefore $a=b$.
Let $A=\left\{x \in \mathbb{R}^{+}: f(x)=x\right\}$. It is enough to show that $A=\mathbb{R}^{+}$.
$P(x, x)$ shows that $x+2 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Now $P(x, x+2 f(x))$ gives that
$$
f(2 x+3 f(x))=x+4 f(x)
$$
for every $x \in \mathbb{R}^{+}$. Therefore $P(x, 2 x+3 f(x))$ gives that $2 x+5 f(x) \in A$ for every $x \in \mathbb{R}^{+}$.
Suppose $x, y \in \mathbb{R}^{+}$such that $x, 2 x+y \in A$. Then $P(x, y)$ gives that
$$
f(2 x+f(y))=f(x+f(x)+f(y))=2 f(x)+y=2 x+y=f(2 x+y)
$$
and by the injectivity of $f$ we have that $2 x+f(y)=2 x+y$. We conlude that $y \in A$ as well.
Now since $x+2 f(x) \in A$ and $2 x+5 f(x)=2(x+2 f(x))+f(x) \in A$ we deduce that $f(x) \in A$ for every $x \in \mathbb{R}^{+}$. I.e. $f(f(x))=f(x)$ for every $x \in \mathbb{R}^{+}$.
By injectivity of $f$ we now conclude that $f(x)=x$ for every $x \in \mathbb{R}^{+}$.
|
f(x)=x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
1. We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a))=f(1+f(1)+f(b))=2 f(1)+b
$$
and therefore $a=b$.
Let $A=\left\{x \in \mathbb{R}^{+}: f(x)=x\right\}$. It is enough to show that $A=\mathbb{R}^{+}$.
$P(x, x)$ shows that $x+2 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Now $P(x, x+2 f(x))$ gives that
$$
f(2 x+3 f(x))=x+4 f(x)
$$
for every $x \in \mathbb{R}^{+}$. Therefore $P(x, 2 x+3 f(x))$ gives that $2 x+5 f(x) \in A$ for every $x \in \mathbb{R}^{+}$.
Suppose $x, y \in \mathbb{R}^{+}$such that $x, 2 x+y \in A$. Then $P(x, y)$ gives that
$$
f(2 x+f(y))=f(x+f(x)+f(y))=2 f(x)+y=2 x+y=f(2 x+y)
$$
and by the injectivity of $f$ we have that $2 x+f(y)=2 x+y$. We conlude that $y \in A$ as well.
Now since $x+2 f(x) \in A$ and $2 x+5 f(x)=2(x+2 f(x))+f(x) \in A$ we deduce that $f(x) \in A$ for every $x \in \mathbb{R}^{+}$. I.e. $f(f(x))=f(x)$ for every $x \in \mathbb{R}^{+}$.
By injectivity of $f$ we now conclude that $f(x)=x$ for every $x \in \mathbb{R}^{+}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA3.",
"solution_match": "\nSolution"
}
|
e06b7421-5221-5931-a916-96cee40bfad4
| 606,037
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
2. As in Solution $1, f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=$ $a+f(a)-b-f(b)$. It is enough to show that $c=d$. By interchanging the roles of $a$ and $b$ in necessary, we may assume that $d \geqslant 0$.
From $P(a, y)$ and $P(b, y)$, after subtraction, we get
$$
f(a+f(a)+f(y))-f(b+f(b)+f(y))=2 f(a)-2 f(b)=c .
$$
so for any $t>m$ (picking $y$ such that $f(y)=t$ in (1)) we get
$$
f(a+f(a)+t)-f(b+f(b)+t)=2 f(a)-2 f(b)=c .
$$
Now for any $z>m+b+f(b)$, taking $t=z-b-f(b)$ in (2) we get
$$
f(z+d)-f(z)=c
$$
Now for any $x>m+b+f(b)$ from (3) we get that
$$
2 f(x+d)+y=2 f(x)+y+2 c .
$$
Also, for any $x$ large enough, $(x>\max \{m+b+f(b), m+b+f(b)+c-d\}$ will do), by repeated application of (3), we have
$$
\begin{aligned}
f(x+d+f(x+d)+f(y)) & =f(x+f(x+d)+y)+c \\
& =f(x+f(x)+y+c)+c \\
& =f(x+f(x)+y+c-d)+2 c .
\end{aligned}
$$
(In the first equality we applied (3) with $z=x+f(x+d)+y>x>m+b+f(b)$, in the second with $z=x>m+b+f(b)$ and in the third with $z=x+f(x)+y-c+d>x+c-d>m+b+f(b)$.
In particular, now $P(x+d, y)$ implies that
$$
f(x+f(x)+y+c-d)=2 f(x)+y=f(x+f(x)+y)
$$
for every large enough $x$. By injectivity of $f$ we deduce that $x+f(x)+y+c-d=x+f(x)+y$ and therefore $c=d$ as required.
It now follows that $f(x)=x+k$ for every $x \in \mathbb{R}^{+}$and some fixed constant $k$. Substituting in the initial equation we get $k=0$.
|
f(x) = x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by Greece
|
2. As in Solution $1, f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=$ $a+f(a)-b-f(b)$. It is enough to show that $c=d$. By interchanging the roles of $a$ and $b$ in necessary, we may assume that $d \geqslant 0$.
From $P(a, y)$ and $P(b, y)$, after subtraction, we get
$$
f(a+f(a)+f(y))-f(b+f(b)+f(y))=2 f(a)-2 f(b)=c .
$$
so for any $t>m$ (picking $y$ such that $f(y)=t$ in (1)) we get
$$
f(a+f(a)+t)-f(b+f(b)+t)=2 f(a)-2 f(b)=c .
$$
Now for any $z>m+b+f(b)$, taking $t=z-b-f(b)$ in (2) we get
$$
f(z+d)-f(z)=c
$$
Now for any $x>m+b+f(b)$ from (3) we get that
$$
2 f(x+d)+y=2 f(x)+y+2 c .
$$
Also, for any $x$ large enough, $(x>\max \{m+b+f(b), m+b+f(b)+c-d\}$ will do), by repeated application of (3), we have
$$
\begin{aligned}
f(x+d+f(x+d)+f(y)) & =f(x+f(x+d)+y)+c \\
& =f(x+f(x)+y+c)+c \\
& =f(x+f(x)+y+c-d)+2 c .
\end{aligned}
$$
(In the first equality we applied (3) with $z=x+f(x+d)+y>x>m+b+f(b)$, in the second with $z=x>m+b+f(b)$ and in the third with $z=x+f(x)+y-c+d>x+c-d>m+b+f(b)$.
In particular, now $P(x+d, y)$ implies that
$$
f(x+f(x)+y+c-d)=2 f(x)+y=f(x+f(x)+y)
$$
for every large enough $x$. By injectivity of $f$ we deduce that $x+f(x)+y+c-d=x+f(x)+y$ and therefore $c=d$ as required.
It now follows that $f(x)=x+k$ for every $x \in \mathbb{R}^{+}$and some fixed constant $k$. Substituting in the initial equation we get $k=0$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA3.",
"solution_match": "\nSolution"
}
|
e06b7421-5221-5931-a916-96cee40bfad4
| 606,037
|
Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight possibilities in total. Is it always possible that Vlad can choose his jumps to return to his initial location $(0,0)$ infinitely many times when
(a) $f, g$ are polynomials with integer coefficients?
(b) $f, g$ are any pair of functions from the positive integers to the integers?
## Proposed by United Kingdom
|
1.
(a) Yes it is always possible. The key idea is the following: Let $b(n)$ be the number of 1 's in the binary expansion of $n=0,1,2, \ldots$.
Lemma: Given a polynomial $f$ with integer coefficients and degree at most $d$, then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=f(n)-f(n+1)-f(n+2)+\cdots \pm f\left(n+\left(2^{d+1}-1\right)\right)=0
$$
Proof of Lemma: The result is clear for $d=0$. For $d \geqslant 1$, we have
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=\sum_{k=0}^{2^{d}-1}(-1)^{b(k)}\left[f(n+k)-f\left(n+k+2^{d}\right)\right] .
$$
So set $\tilde{f}(n)=f(n)-f\left(n+2^{d}\right)$, which is a polynomial of degree at most $d-1$. Then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=\sum_{k=0}^{2^{d}-1} \tilde{f}(n+k)=0
$$
by induction, completing the proof of the lemma.
In particular, if we take
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=\left((-1)^{b(n)} f(n),(-1)^{b(n)} g(n)\right),
$$
then $\mathbf{x}_{D}=\mathbf{0}$ whenever $D$ is a multiple of $2^{1+\max (\operatorname{deg}(f), \operatorname{deg}(g))}$.
(b) No, it is not always possible. Let $g$ be any suitable function. Then, we construct $f$ inductively. There are at most $8^{n-1}$ possibilities for $\mathbf{x}_{n-1}$, so choose $f(n)$ to be greater than the magnitude of all of them. Consequently $\mathbf{x}_{n}$ cannot be $\mathbf{0}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight possibilities in total. Is it always possible that Vlad can choose his jumps to return to his initial location $(0,0)$ infinitely many times when
(a) $f, g$ are polynomials with integer coefficients?
(b) $f, g$ are any pair of functions from the positive integers to the integers?
## Proposed by United Kingdom
|
1.
(a) Yes it is always possible. The key idea is the following: Let $b(n)$ be the number of 1 's in the binary expansion of $n=0,1,2, \ldots$.
Lemma: Given a polynomial $f$ with integer coefficients and degree at most $d$, then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=f(n)-f(n+1)-f(n+2)+\cdots \pm f\left(n+\left(2^{d+1}-1\right)\right)=0
$$
Proof of Lemma: The result is clear for $d=0$. For $d \geqslant 1$, we have
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=\sum_{k=0}^{2^{d}-1}(-1)^{b(k)}\left[f(n+k)-f\left(n+k+2^{d}\right)\right] .
$$
So set $\tilde{f}(n)=f(n)-f\left(n+2^{d}\right)$, which is a polynomial of degree at most $d-1$. Then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=\sum_{k=0}^{2^{d}-1} \tilde{f}(n+k)=0
$$
by induction, completing the proof of the lemma.
In particular, if we take
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=\left((-1)^{b(n)} f(n),(-1)^{b(n)} g(n)\right),
$$
then $\mathbf{x}_{D}=\mathbf{0}$ whenever $D$ is a multiple of $2^{1+\max (\operatorname{deg}(f), \operatorname{deg}(g))}$.
(b) No, it is not always possible. Let $g$ be any suitable function. Then, we construct $f$ inductively. There are at most $8^{n-1}$ possibilities for $\mathbf{x}_{n-1}$, so choose $f(n)$ to be greater than the magnitude of all of them. Consequently $\mathbf{x}_{n}$ cannot be $\mathbf{0}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA4.",
"solution_match": "\n## Solution"
}
|
153677a8-db7b-513b-8cdb-7edfc6d686a1
| 606,050
|
Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight possibilities in total. Is it always possible that Vlad can choose his jumps to return to his initial location $(0,0)$ infinitely many times when
(a) $f, g$ are polynomials with integer coefficients?
(b) $f, g$ are any pair of functions from the positive integers to the integers?
## Proposed by United Kingdom
|
2.
(a) Given a polynomial $f$ of degree at most $d$ and integers $n, r$, we claim that
$$
\sum_{k=0}^{2^{d+1}-1} \varepsilon_{k} f\left(2^{d} n+r+k\right)=0
$$
for some choice of $\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{2^{d+1}-1} \in\{-1,1\}$. (Which are allowed to depend on $d$ and f.)
We proceed by induction on $d$, the case $d=0$ being immediate. For the inductive step we define the polynomial $g(n)=f(2 n+r+1)-f(2 n+r)$ which is a polynomial of degree at most $d-1$. Then
$$
\sum_{k=0}^{2^{d}-1} \varepsilon_{k} g\left(2^{d-1} n+k\right)=0
$$
for some choice of the $\varepsilon_{k}$ 's giving
$$
\sum_{k=0}^{2^{d+1}-1} \varepsilon_{k}^{\prime} f\left(2^{d} n+r+k\right)=0
$$
where $\varepsilon_{2 k}^{\prime}=-\varepsilon_{k}$ and $\varepsilon_{2 k+1}^{\prime}=\varepsilon_{k}$. This completes the proof of the claim.
Now the proof can be completed as in Solution 1.
(b) Apart from magnitude arguments, one could also use modulo arguments. For example, taking $f(0), g(0)$ to be odd and $f(n), g(n)$ to be even for every $n \geqslant 1$ works.
## Comments.
(1) We propose to omit part (b) as it is easy and furthermore it suggests that the answer to (a) is most likely affirmative.
(2) Giving a precise self-contained characterisation of $b(n)$ in Solution 1 is not necessary for the lemma. It could instead be phrased as:
There exists a sequence $\beta(k) \in\{-1,+1\}^{\mathbb{N}}$ such that $\sum \beta(k) f(n+k)=0$.
Then, one constructs $\beta(\cdot)$ inductively as part of the proof via $\beta\left(k+2^{d}\right)=-\beta(k)$ for $k<2^{d}$, which coincides with the original definition, ie $\beta(\cdot)=(-1)^{b(\cdot)}$.
(3) The sequence of signs in both solutions are essentially the same. (Either all signs exactly the same or all signs different.)
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight possibilities in total. Is it always possible that Vlad can choose his jumps to return to his initial location $(0,0)$ infinitely many times when
(a) $f, g$ are polynomials with integer coefficients?
(b) $f, g$ are any pair of functions from the positive integers to the integers?
## Proposed by United Kingdom
|
2.
(a) Given a polynomial $f$ of degree at most $d$ and integers $n, r$, we claim that
$$
\sum_{k=0}^{2^{d+1}-1} \varepsilon_{k} f\left(2^{d} n+r+k\right)=0
$$
for some choice of $\varepsilon_{0}, \varepsilon_{1}, \ldots, \varepsilon_{2^{d+1}-1} \in\{-1,1\}$. (Which are allowed to depend on $d$ and f.)
We proceed by induction on $d$, the case $d=0$ being immediate. For the inductive step we define the polynomial $g(n)=f(2 n+r+1)-f(2 n+r)$ which is a polynomial of degree at most $d-1$. Then
$$
\sum_{k=0}^{2^{d}-1} \varepsilon_{k} g\left(2^{d-1} n+k\right)=0
$$
for some choice of the $\varepsilon_{k}$ 's giving
$$
\sum_{k=0}^{2^{d+1}-1} \varepsilon_{k}^{\prime} f\left(2^{d} n+r+k\right)=0
$$
where $\varepsilon_{2 k}^{\prime}=-\varepsilon_{k}$ and $\varepsilon_{2 k+1}^{\prime}=\varepsilon_{k}$. This completes the proof of the claim.
Now the proof can be completed as in Solution 1.
(b) Apart from magnitude arguments, one could also use modulo arguments. For example, taking $f(0), g(0)$ to be odd and $f(n), g(n)$ to be even for every $n \geqslant 1$ works.
## Comments.
(1) We propose to omit part (b) as it is easy and furthermore it suggests that the answer to (a) is most likely affirmative.
(2) Giving a precise self-contained characterisation of $b(n)$ in Solution 1 is not necessary for the lemma. It could instead be phrased as:
There exists a sequence $\beta(k) \in\{-1,+1\}^{\mathbb{N}}$ such that $\sum \beta(k) f(n+k)=0$.
Then, one constructs $\beta(\cdot)$ inductively as part of the proof via $\beta\left(k+2^{d}\right)=-\beta(k)$ for $k<2^{d}$, which coincides with the original definition, ie $\beta(\cdot)=(-1)^{b(\cdot)}$.
(3) The sequence of signs in both solutions are essentially the same. (Either all signs exactly the same or all signs different.)
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA4.",
"solution_match": "\n## Solution"
}
|
153677a8-db7b-513b-8cdb-7edfc6d686a1
| 606,050
|
Find all functions $f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+}$such that
$$
f(x f(x+y))=y f(x)+1
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by North Macedonia
|
1. We will show that that $f(x)=\frac{1}{x}$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x f(x+y))=y f(x)+1$.
We first show that $f$ is injective. So assume $f\left(x_{1}\right)=f\left(x_{2}\right)$ and take any $x<x_{1}, x_{2}$. Then $P\left(x, x_{1}-x\right)$ and $P\left(x, x_{2}-x\right)$ give
$$
\left(x_{1}-x\right) f(x)+1=f\left(x f\left(x_{1}\right)\right)=f\left(x f\left(x_{2}\right)\right)=\left(x_{2}-x\right) f(x)+1
$$
giving $x_{1}=x_{2}$.
It is also immediate that for every $z>1$ there is an $x$ such that $f(x)=z$. Indeed $P\left(x, \frac{z-1}{f(x)}\right)$ gives that
$$
f\left(x f\left(x+\frac{z-1}{f(x)}\right)\right)=z .
$$
Now given $z>1$, take $x$ such that $f(x)=z$. Then $P\left(x, \frac{z-1}{z}\right)$ gives
$$
f\left(x f\left(x+\frac{z-1}{z}\right)\right)=\frac{z-1}{z} f(x)+1=z=f(x) .
$$
Since $f$ is injective, we deduce that $f\left(x+\frac{z-1}{z}\right)=1$.
So there is a $k \in \mathbb{R}^{+}$such that $f(k)=1$. Since $f$ is injective this $k$ is unique. Therefore $x=k+\frac{1}{z}-1$. I.e. for every $z>1$ we have
$$
f\left(k+\frac{1}{z}-1\right)=z
$$
We must have $k+\frac{1}{z}-1 \in \mathbb{R}^{+}$for each $z>1$ and taking the limit as $z$ tends to infinity we deduce that $k \geqslant 1$. (Without mentioning limits, assuming for contradiction that $k<1$, taking $z=\frac{2}{1-k}$ leads to a contradiction.) Set $r=k-1$.
Now $P\left(r+\frac{1}{6}, \frac{1}{3}\right)$ gives
$$
f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{6}+\frac{1}{3}\right)\right)=\frac{1}{3} f\left(r+\frac{1}{6}\right)+1=\frac{6}{3}+1=3=f\left(r+\frac{1}{3}\right) .
$$
But
$$
f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{6}+\frac{1}{3}\right)\right)=f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{2}\right)\right)=f\left(2 r+\frac{1}{3}\right) .
$$
The injectivity of $f$ now shows that $r=0$, i.e. that $f(1)=k=1$.
This shows that $f\left(\frac{1}{z}\right)=z$ for every $z>1$, i.e. $f(x)=\frac{1}{x}$ for every $x<1$. Now for $x>1$ consider $P(1, x-1)$ to get $f(f(x))=(x-1) f(1)+1=x=f\left(\frac{1}{x}\right)$. Injectivity of $f$ shows that $f(x)=\frac{1}{x}$.
So for all possible values of $x$ we have shown that $f(x)=\frac{1}{x}$.
|
f(x)=\frac{1}{x}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+}$such that
$$
f(x f(x+y))=y f(x)+1
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by North Macedonia
|
1. We will show that that $f(x)=\frac{1}{x}$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x f(x+y))=y f(x)+1$.
We first show that $f$ is injective. So assume $f\left(x_{1}\right)=f\left(x_{2}\right)$ and take any $x<x_{1}, x_{2}$. Then $P\left(x, x_{1}-x\right)$ and $P\left(x, x_{2}-x\right)$ give
$$
\left(x_{1}-x\right) f(x)+1=f\left(x f\left(x_{1}\right)\right)=f\left(x f\left(x_{2}\right)\right)=\left(x_{2}-x\right) f(x)+1
$$
giving $x_{1}=x_{2}$.
It is also immediate that for every $z>1$ there is an $x$ such that $f(x)=z$. Indeed $P\left(x, \frac{z-1}{f(x)}\right)$ gives that
$$
f\left(x f\left(x+\frac{z-1}{f(x)}\right)\right)=z .
$$
Now given $z>1$, take $x$ such that $f(x)=z$. Then $P\left(x, \frac{z-1}{z}\right)$ gives
$$
f\left(x f\left(x+\frac{z-1}{z}\right)\right)=\frac{z-1}{z} f(x)+1=z=f(x) .
$$
Since $f$ is injective, we deduce that $f\left(x+\frac{z-1}{z}\right)=1$.
So there is a $k \in \mathbb{R}^{+}$such that $f(k)=1$. Since $f$ is injective this $k$ is unique. Therefore $x=k+\frac{1}{z}-1$. I.e. for every $z>1$ we have
$$
f\left(k+\frac{1}{z}-1\right)=z
$$
We must have $k+\frac{1}{z}-1 \in \mathbb{R}^{+}$for each $z>1$ and taking the limit as $z$ tends to infinity we deduce that $k \geqslant 1$. (Without mentioning limits, assuming for contradiction that $k<1$, taking $z=\frac{2}{1-k}$ leads to a contradiction.) Set $r=k-1$.
Now $P\left(r+\frac{1}{6}, \frac{1}{3}\right)$ gives
$$
f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{6}+\frac{1}{3}\right)\right)=\frac{1}{3} f\left(r+\frac{1}{6}\right)+1=\frac{6}{3}+1=3=f\left(r+\frac{1}{3}\right) .
$$
But
$$
f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{6}+\frac{1}{3}\right)\right)=f\left(\left(r+\frac{1}{6}\right) f\left(r+\frac{1}{2}\right)\right)=f\left(2 r+\frac{1}{3}\right) .
$$
The injectivity of $f$ now shows that $r=0$, i.e. that $f(1)=k=1$.
This shows that $f\left(\frac{1}{z}\right)=z$ for every $z>1$, i.e. $f(x)=\frac{1}{x}$ for every $x<1$. Now for $x>1$ consider $P(1, x-1)$ to get $f(f(x))=(x-1) f(1)+1=x=f\left(\frac{1}{x}\right)$. Injectivity of $f$ shows that $f(x)=\frac{1}{x}$.
So for all possible values of $x$ we have shown that $f(x)=\frac{1}{x}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA5.",
"solution_match": "\nSolution"
}
|
c5a549e7-3457-50e4-b1ca-1c5723d72067
| 606,068
|
Find all functions $f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+}$such that
$$
f(x f(x+y))=y f(x)+1
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by North Macedonia
|
2. $P(1, y)$ shows that $f(f(y+1))=y f(1)+1$. Now $P\left(f(y+1), \frac{y f(1)}{y f(1)+1}\right)$ shows that
$$
f\left(f(y+1) f\left(f(y+1)+\frac{y f(1)}{y f(1)+1}\right)\right)=\frac{y f(1)}{y f(1)+1} f(f(y+1))+1=y f(1)+1
$$
Since $f$ is injective (as in Solution 1) we get that
$$
f(y+1) f\left(f(y+1)+\frac{y f(1)}{y f(1)+1}\right)=f(y+1)
$$
and therefore there is a unique $k$ such that $f(k)=1$. Furtermore, for every $y>0$ we have
$$
f(y+1)=k-\frac{y f(1)}{y f(1)+1}
$$
The right hand side of (1) is always positive. But letting $y$ tend to infinity, the right hand side tends to $k-1$ so we must have $k \geqslant 1$.
If $k>1$, then $P(k-1,1)$ gives
$$
f(k-1)=f((k-1) f(k))=f(k-1)+1
$$
a contradiction. So $f(1)=k=1$.
For $x<1, P(x, 1-x)$ gives
$$
f(x)=f(x f(x+(1-x)))=(1-x) f(x)+1
$$
from which we deduce that $f(x)=\frac{1}{x}$. To show that $f(x)=\frac{1}{x}$ for $x>1$ we can either work as in Solution 1 or take $y=x-1$ in (1) to get that
$$
f(x)=1-\frac{x-1}{(x-1)+1}=\frac{1}{x}
$$
|
f(x)=\frac{1}{x}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+}$such that
$$
f(x f(x+y))=y f(x)+1
$$
holds for all $x, y \in \mathbb{R}^{+}$.
## Proposed by North Macedonia
|
2. $P(1, y)$ shows that $f(f(y+1))=y f(1)+1$. Now $P\left(f(y+1), \frac{y f(1)}{y f(1)+1}\right)$ shows that
$$
f\left(f(y+1) f\left(f(y+1)+\frac{y f(1)}{y f(1)+1}\right)\right)=\frac{y f(1)}{y f(1)+1} f(f(y+1))+1=y f(1)+1
$$
Since $f$ is injective (as in Solution 1) we get that
$$
f(y+1) f\left(f(y+1)+\frac{y f(1)}{y f(1)+1}\right)=f(y+1)
$$
and therefore there is a unique $k$ such that $f(k)=1$. Furtermore, for every $y>0$ we have
$$
f(y+1)=k-\frac{y f(1)}{y f(1)+1}
$$
The right hand side of (1) is always positive. But letting $y$ tend to infinity, the right hand side tends to $k-1$ so we must have $k \geqslant 1$.
If $k>1$, then $P(k-1,1)$ gives
$$
f(k-1)=f((k-1) f(k))=f(k-1)+1
$$
a contradiction. So $f(1)=k=1$.
For $x<1, P(x, 1-x)$ gives
$$
f(x)=f(x f(x+(1-x)))=(1-x) f(x)+1
$$
from which we deduce that $f(x)=\frac{1}{x}$. To show that $f(x)=\frac{1}{x}$ for $x>1$ we can either work as in Solution 1 or take $y=x-1$ in (1) to get that
$$
f(x)=1-\frac{x-1}{(x-1)+1}=\frac{1}{x}
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA5.",
"solution_match": "\nSolution"
}
|
c5a549e7-3457-50e4-b1ca-1c5723d72067
| 606,068
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania
|
1. We will show that $f(x)=0$ for every $x \in \mathbb{R}$ or $f(x)=x-1$ for every $x \in \mathbb{R}$. It is easy to check that both of these functions work.
We write $P(x, y)$ for the assertion that $f(x y)=f(x) f(y)+f(f(x+y))$. For later use we write $Q(x, y)$ for the assertion that $f(x y)=f(x) f(y)$ and $R(x, y)$ for the assertion that $f(x y)=$ $f(x) f(y)+f(x+y-1)$.
Assume first that $f(0)=0$.
For each $t \in \mathbb{R}, P(0, t)$ gives $f(f(t))=0$. Therefore we get that $Q(x, y)$ holds for each $x, y \in \mathbb{R}$. Now $Q(x, 1)$ gives $f(x)=f(x) f(1)$ for each $x \in \mathbb{R}$. But $f(1) \neq 1$ as otherwise we would have $f(f(1))=f(1)=1 \neq 0$, a contradiction. Since $f(1) \neq 1$, then $f(x)=f(x) f(1)$ gives $f(x)=0$. This holds for each $x \in \mathbb{R}$ and gives our first solution.
From now on we assume that $f(0)=a \neq 0$. If $f(1)=1$, then for $t \in \mathbb{R}, P(t-1,1)$ gives $f(f(t))=0$ so we get that $Q(x, y)$ holds for each $x, y \in \mathbb{R}$. Now $Q(x, 0)$ gives $f(0)=f(x) f(0)$ for each $x \in \mathbb{R}$. Since $f(0) \neq 0$, then $f(x)=1$ for each $x \in \mathbb{R}$. This however contradicts the fact that $f(f(t))=0$ for each $t \in \mathbb{R}$.
So from now on we can further assume that $f(1)=b \neq 1$.
Now $P(x, 0)$ gives
$$
f(f(x))=a-a f(x)
$$
and $P(x-1,1)$ gives
$$
f(f(x))=f(x-1)-b f(x-1) .
$$
Therefore, letting $c=\frac{b-1}{a}$, we get
$$
f(x)=c f(x-1)+1
$$
for every $x \in \mathbb{R}$.
Claim 1. There is an integer $n$ such that $n^{2} \geqslant 4 f(n)$.
Proof. If $c=1$, then inductively from (1) we get that $f(n)=f(0)+n=a+n$ for each $n \in \mathbb{N}$. So for $n$ large enough we have $n^{2} \geqslant 4 f(n)$.
If $c \neq 1$, then inductively from (1) we get that
$$
f(n)=\left(a-\frac{1}{1-c}\right) c^{n}+\frac{1}{1-c}
$$
for every $n \in \mathbb{Z}$. (We apply induction once to prove the result for every $n \geqslant 0$ and once to prove the result for every $n<0$.)
For $|c|<1$ we have $\lim _{n \rightarrow \infty} f(n)=\frac{1}{1-c}$ so we can find $n$ large enough such that $4 f(n) \leqslant n^{2}$.
For $|c|>1$ we have $\lim _{n \rightarrow-\infty} f(n)=\frac{1}{1-c}$ so we can find a negative integer $n$ with $|n|$ large enough such that $4 f(n) \leqslant n^{2}$.
For $|c|=1$, we must have $c=-1$, so $f(n)= \pm\left(a-\frac{1}{2}\right)+\frac{1}{2}$ and again for $n$ large enough we have $4 f(n) \leqslant n^{2}$.
Claim 2. $f(1)=0$.
Proof. Let $n$ be as given by Claim 1 and pick $x^{\prime}, y^{\prime} \in \mathbb{R}$ such that $x^{\prime}+y^{\prime}=n$ and $x^{\prime} y^{\prime}=f(n)$. This is possible since $n^{2} \geqslant 4 f(n)$. Now $P\left(x^{\prime}, y^{\prime}\right)$ gives $f\left(x^{\prime}\right) f\left(y^{\prime}\right)=0$.
So there is a $d \in \mathbb{R}$ such that $f(d)=0$.
Putting $x=d+1$ in (1) we get $f(d+1)=1$. Now $P(d, 1)$ gives $f(f(d+1))=0$ and therefore $b=f(1)=0$.
Claim 3. $c \neq-1$.
Proof. If $c=-1$, then $f(x)+f(x-1)=1$ for every $x \in \mathbb{R}$. In particular, for every $x \in \mathbb{R}$, we have
$$
f(x)+f(x+1)=1=f(x+1)+f(x+2)
$$
giving $f(x)=f(x+2)$. So $P\left(\frac{1}{2}, \frac{1}{2}\right)$ and $P\left(\frac{1}{2}, \frac{5}{2}\right)$ give
$$
f\left(\frac{5}{4}\right)=f\left(\frac{1}{2}\right) f\left(\frac{5}{2}\right)+f(f(3))=f\left(\frac{1}{2}\right) f\left(\frac{1}{2}\right)+f(f(1))=f\left(\frac{1}{4}\right) .
$$
But $f\left(\frac{1}{4}\right)+f\left(\frac{5}{4}\right)=1$, therefore $f\left(\frac{1}{4}\right)=f\left(\frac{5}{4}\right)=\frac{1}{2}$. Since $f(1)=0$, then $f(0)=1$ and so
$$
\frac{1}{2}=f\left(\frac{1}{4}\right)=f\left(\frac{1}{2}\right)^{2}+f(f(1)) \geqslant f(f(1))=f(0)=1
$$
a contradiction.
Claim 4. $c=1$.
Proof. From (1) we get that $f(2)=1, f(3)=c+1$ and $f(4)=c^{2}+c+1$. Now $P(3,1)$ and $P(2,2)$ give that
$$
f(f(4))=f(3)-f(3) f(1)=c+1 \quad \text { and } \quad f(f(4))=f(4)-f(2)^{2}=c^{2}+c=c(c+1) .
$$
Since by Claim $3 c \neq-1$, then we must have $c=1$.
Since $f(1)=0$, then $P(x+y-1,1)$ gives $f(x+y-1)=f(f(x+y))$. Thus we have that $R(x, y)$ holds for every $x, y \in \mathbb{R}$.
Now $R(x, y+1)$ gives
$$
f(x y+y)=f(x) f(y+1)+f(x+y)
$$
and from (1) and the fact that $c=1$ we deduce that
$$
\begin{aligned}
f(x y+x) & =f(x) f(y)+f(x)+f(x+y) \\
& =f(x) f(y)+f(x)+f(x+y-1)+1 \\
& =f(x y)+f(x)+1
\end{aligned}
$$
This holds for every $x, y \in \mathbb{R}$. In particular, taking $x \neq 0$ and $y=t / x$, we have
$$
f(t+x)=f(t)+f(x)+1
$$
for every $t \in \mathbb{R}, x \in \mathbb{R} \backslash\{0\}$. Note that (2) holds for $x=0$ as well, since $c=1$ implies that $f(0)=-1$.
Defining $g(x)=f(x)+1$ for each $x \in \mathbb{R}$ then (2) gives that
$$
g(t+x)=g(t)+g(x)
$$
for every $t, x \in \mathbb{R}$. I.e. $g$ is additive. Furthermore $R(x, y)$ implies that
$$
\begin{aligned}
g(x y)-1 & =(g(x)-1)(g(y)-1)+g(x+y-1)-1 \\
& =g(x) g(y)-g(x)-g(y)+g(x+y-1) \\
& =g(x) g(y)-1 .
\end{aligned}
$$
This implies that $g$ is multiplicative.
We know that an additive and multiplicative function is either identically zero or the identity function. [Since $g$ is multiplicative, $g\left(x^{2}\right)=g(x)^{2} \geqslant 0$ giving that $g$ takes non-negative values at non-negative arguments. Since also $g$ is additive we get that $g$ is monotone increasing. Since also $g$ is additive it is know that $g(x)=C x$ for every $x \in \mathbb{R}$ for some contant $C$. The multiplicativity of $g$ now gives that $C=0$ or $C=1$.]
Since $g$ is not identically 0 we get that $g(x)=x$ for every $x \in \mathbb{R}$ giving that $f(x)=x-1$ for every $x \in \mathbb{R}$.
|
proof
|
Yes
|
Incomplete
|
proof
|
Algebra
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania
|
1. We will show that $f(x)=0$ for every $x \in \mathbb{R}$ or $f(x)=x-1$ for every $x \in \mathbb{R}$. It is easy to check that both of these functions work.
We write $P(x, y)$ for the assertion that $f(x y)=f(x) f(y)+f(f(x+y))$. For later use we write $Q(x, y)$ for the assertion that $f(x y)=f(x) f(y)$ and $R(x, y)$ for the assertion that $f(x y)=$ $f(x) f(y)+f(x+y-1)$.
Assume first that $f(0)=0$.
For each $t \in \mathbb{R}, P(0, t)$ gives $f(f(t))=0$. Therefore we get that $Q(x, y)$ holds for each $x, y \in \mathbb{R}$. Now $Q(x, 1)$ gives $f(x)=f(x) f(1)$ for each $x \in \mathbb{R}$. But $f(1) \neq 1$ as otherwise we would have $f(f(1))=f(1)=1 \neq 0$, a contradiction. Since $f(1) \neq 1$, then $f(x)=f(x) f(1)$ gives $f(x)=0$. This holds for each $x \in \mathbb{R}$ and gives our first solution.
From now on we assume that $f(0)=a \neq 0$. If $f(1)=1$, then for $t \in \mathbb{R}, P(t-1,1)$ gives $f(f(t))=0$ so we get that $Q(x, y)$ holds for each $x, y \in \mathbb{R}$. Now $Q(x, 0)$ gives $f(0)=f(x) f(0)$ for each $x \in \mathbb{R}$. Since $f(0) \neq 0$, then $f(x)=1$ for each $x \in \mathbb{R}$. This however contradicts the fact that $f(f(t))=0$ for each $t \in \mathbb{R}$.
So from now on we can further assume that $f(1)=b \neq 1$.
Now $P(x, 0)$ gives
$$
f(f(x))=a-a f(x)
$$
and $P(x-1,1)$ gives
$$
f(f(x))=f(x-1)-b f(x-1) .
$$
Therefore, letting $c=\frac{b-1}{a}$, we get
$$
f(x)=c f(x-1)+1
$$
for every $x \in \mathbb{R}$.
Claim 1. There is an integer $n$ such that $n^{2} \geqslant 4 f(n)$.
Proof. If $c=1$, then inductively from (1) we get that $f(n)=f(0)+n=a+n$ for each $n \in \mathbb{N}$. So for $n$ large enough we have $n^{2} \geqslant 4 f(n)$.
If $c \neq 1$, then inductively from (1) we get that
$$
f(n)=\left(a-\frac{1}{1-c}\right) c^{n}+\frac{1}{1-c}
$$
for every $n \in \mathbb{Z}$. (We apply induction once to prove the result for every $n \geqslant 0$ and once to prove the result for every $n<0$.)
For $|c|<1$ we have $\lim _{n \rightarrow \infty} f(n)=\frac{1}{1-c}$ so we can find $n$ large enough such that $4 f(n) \leqslant n^{2}$.
For $|c|>1$ we have $\lim _{n \rightarrow-\infty} f(n)=\frac{1}{1-c}$ so we can find a negative integer $n$ with $|n|$ large enough such that $4 f(n) \leqslant n^{2}$.
For $|c|=1$, we must have $c=-1$, so $f(n)= \pm\left(a-\frac{1}{2}\right)+\frac{1}{2}$ and again for $n$ large enough we have $4 f(n) \leqslant n^{2}$.
Claim 2. $f(1)=0$.
Proof. Let $n$ be as given by Claim 1 and pick $x^{\prime}, y^{\prime} \in \mathbb{R}$ such that $x^{\prime}+y^{\prime}=n$ and $x^{\prime} y^{\prime}=f(n)$. This is possible since $n^{2} \geqslant 4 f(n)$. Now $P\left(x^{\prime}, y^{\prime}\right)$ gives $f\left(x^{\prime}\right) f\left(y^{\prime}\right)=0$.
So there is a $d \in \mathbb{R}$ such that $f(d)=0$.
Putting $x=d+1$ in (1) we get $f(d+1)=1$. Now $P(d, 1)$ gives $f(f(d+1))=0$ and therefore $b=f(1)=0$.
Claim 3. $c \neq-1$.
Proof. If $c=-1$, then $f(x)+f(x-1)=1$ for every $x \in \mathbb{R}$. In particular, for every $x \in \mathbb{R}$, we have
$$
f(x)+f(x+1)=1=f(x+1)+f(x+2)
$$
giving $f(x)=f(x+2)$. So $P\left(\frac{1}{2}, \frac{1}{2}\right)$ and $P\left(\frac{1}{2}, \frac{5}{2}\right)$ give
$$
f\left(\frac{5}{4}\right)=f\left(\frac{1}{2}\right) f\left(\frac{5}{2}\right)+f(f(3))=f\left(\frac{1}{2}\right) f\left(\frac{1}{2}\right)+f(f(1))=f\left(\frac{1}{4}\right) .
$$
But $f\left(\frac{1}{4}\right)+f\left(\frac{5}{4}\right)=1$, therefore $f\left(\frac{1}{4}\right)=f\left(\frac{5}{4}\right)=\frac{1}{2}$. Since $f(1)=0$, then $f(0)=1$ and so
$$
\frac{1}{2}=f\left(\frac{1}{4}\right)=f\left(\frac{1}{2}\right)^{2}+f(f(1)) \geqslant f(f(1))=f(0)=1
$$
a contradiction.
Claim 4. $c=1$.
Proof. From (1) we get that $f(2)=1, f(3)=c+1$ and $f(4)=c^{2}+c+1$. Now $P(3,1)$ and $P(2,2)$ give that
$$
f(f(4))=f(3)-f(3) f(1)=c+1 \quad \text { and } \quad f(f(4))=f(4)-f(2)^{2}=c^{2}+c=c(c+1) .
$$
Since by Claim $3 c \neq-1$, then we must have $c=1$.
Since $f(1)=0$, then $P(x+y-1,1)$ gives $f(x+y-1)=f(f(x+y))$. Thus we have that $R(x, y)$ holds for every $x, y \in \mathbb{R}$.
Now $R(x, y+1)$ gives
$$
f(x y+y)=f(x) f(y+1)+f(x+y)
$$
and from (1) and the fact that $c=1$ we deduce that
$$
\begin{aligned}
f(x y+x) & =f(x) f(y)+f(x)+f(x+y) \\
& =f(x) f(y)+f(x)+f(x+y-1)+1 \\
& =f(x y)+f(x)+1
\end{aligned}
$$
This holds for every $x, y \in \mathbb{R}$. In particular, taking $x \neq 0$ and $y=t / x$, we have
$$
f(t+x)=f(t)+f(x)+1
$$
for every $t \in \mathbb{R}, x \in \mathbb{R} \backslash\{0\}$. Note that (2) holds for $x=0$ as well, since $c=1$ implies that $f(0)=-1$.
Defining $g(x)=f(x)+1$ for each $x \in \mathbb{R}$ then (2) gives that
$$
g(t+x)=g(t)+g(x)
$$
for every $t, x \in \mathbb{R}$. I.e. $g$ is additive. Furthermore $R(x, y)$ implies that
$$
\begin{aligned}
g(x y)-1 & =(g(x)-1)(g(y)-1)+g(x+y-1)-1 \\
& =g(x) g(y)-g(x)-g(y)+g(x+y-1) \\
& =g(x) g(y)-1 .
\end{aligned}
$$
This implies that $g$ is multiplicative.
We know that an additive and multiplicative function is either identically zero or the identity function. [Since $g$ is multiplicative, $g\left(x^{2}\right)=g(x)^{2} \geqslant 0$ giving that $g$ takes non-negative values at non-negative arguments. Since also $g$ is additive we get that $g$ is monotone increasing. Since also $g$ is additive it is know that $g(x)=C x$ for every $x \in \mathbb{R}$ for some contant $C$. The multiplicativity of $g$ now gives that $C=0$ or $C=1$.]
Since $g$ is not identically 0 we get that $g(x)=x$ for every $x \in \mathbb{R}$ giving that $f(x)=x-1$ for every $x \in \mathbb{R}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA6.",
"solution_match": "\nSolution"
}
|
c20ca433-9f8e-5102-8f44-7c0071be6282
| 606,085
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania
|
2 (Sketch). One can prove directly Claims 3 and 4 without the use of Claims 1 and 2. To prove Claim 3 we can make use of $P(x+1, y-1)$ which together with $P(x, y)$ and (1) gives
$$
f(x y+y-x)-c f(x y)=f(y)-c f(x) .
$$
Assuming $c=-1$, then (1) and (3) give that $f(x+2)=f(x)$ for every $x \in \mathbb{R}$. It follows that $f(x+2 n)=f(x)$ for every $x \in \mathbb{R}$ and every $n \in \mathbb{Z}$. Now with similar ideas as in the proof of Claim 1, it can be shown that for every $u, v \in \mathbb{R}$ there is $n \in \mathbb{N}$ large enough such that $u=x y+x-y+2 n$ and $v=x y+y-x$. Then using (3) we can get
$$
f(u)=f(x y+x-y+2 n)=f(x y+x-y)=f(x y+y-x)=f(v) .
$$
So $f$ is constant and it must be identically equal to $1 / 2$ which leads to a contradiction.
Now using (3) with $x=y$ and assuming $c \neq 1$ we get $f\left(x^{2}\right)=f(x)$. So $f$ is even. This eventually leads to $f(n)=1 /(1-c)=a=b$ for every integer $n$. Now $P(0,0)$ gives $a=a^{2}+f(a)$ and $P(a,-a)$ gives $f\left(-a^{2}\right)=f(a) f(-a)+f(a)$. Since $f$ is even we eventually get $f(a)=0$ which gives $a=0$ or $a=1$ both contraidicting the facts that $a \neq 0$ and $b \neq 1$.
So $c=1$ and using (1) and (3) one can eventually get $a=-1$. The solution can then finish in the same way as in Solution 1.
## COMBINATORICS
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania
|
2 (Sketch). One can prove directly Claims 3 and 4 without the use of Claims 1 and 2. To prove Claim 3 we can make use of $P(x+1, y-1)$ which together with $P(x, y)$ and (1) gives
$$
f(x y+y-x)-c f(x y)=f(y)-c f(x) .
$$
Assuming $c=-1$, then (1) and (3) give that $f(x+2)=f(x)$ for every $x \in \mathbb{R}$. It follows that $f(x+2 n)=f(x)$ for every $x \in \mathbb{R}$ and every $n \in \mathbb{Z}$. Now with similar ideas as in the proof of Claim 1, it can be shown that for every $u, v \in \mathbb{R}$ there is $n \in \mathbb{N}$ large enough such that $u=x y+x-y+2 n$ and $v=x y+y-x$. Then using (3) we can get
$$
f(u)=f(x y+x-y+2 n)=f(x y+x-y)=f(x y+y-x)=f(v) .
$$
So $f$ is constant and it must be identically equal to $1 / 2$ which leads to a contradiction.
Now using (3) with $x=y$ and assuming $c \neq 1$ we get $f\left(x^{2}\right)=f(x)$. So $f$ is even. This eventually leads to $f(n)=1 /(1-c)=a=b$ for every integer $n$. Now $P(0,0)$ gives $a=a^{2}+f(a)$ and $P(a,-a)$ gives $f\left(-a^{2}\right)=f(a) f(-a)+f(a)$. Since $f$ is even we eventually get $f(a)=0$ which gives $a=0$ or $a=1$ both contraidicting the facts that $a \neq 0$ and $b \neq 1$.
So $c=1$ and using (1) and (3) one can eventually get $a=-1$. The solution can then finish in the same way as in Solution 1.
## COMBINATORICS
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nA6.",
"solution_match": "\nSolution"
}
|
c20ca433-9f8e-5102-8f44-7c0071be6282
| 606,085
|
Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if every three distinct elements of $A$ form a good triple.
Prove that every good subset of $\mathcal{A}_{n}$ has at most $2\left(\frac{3}{2}\right)^{n}$ elements.
## Proposed by Greece
|
1. We proceed by induction on $n$, the case $n=1$ being trivial. Let
$$
A_{0}=\left\{\left(x_{1}, \ldots, x_{n}\right) \in A: x_{n} \neq 0\right\}
$$
and define $A_{1}$ and $A_{2}$ similarly.
Since $A$ is good and $A_{0}$ is a subset of $A$, then $A_{0}$ is also good. Therefore, any three of its elements have a coordinate that differs. This coordinate cannot be the last one since 0 cannot appear as a last coordinate. This means that the set $A_{0}^{\prime}$ obtained from $A_{0}$ by deleting the last coordinate from each of its elements is a good subset of $\mathcal{A}_{n-1}$.
Moreover, if $\left|A_{0}\right| \geqslant 3$ then $\left|A_{0}^{\prime}\right|=\left|A_{0}\right|$. Indeed, if otherwise, then there is an element $a \in A_{0}^{\prime}$ such that $x, y \in A_{0}$, where $x$ and $y$ are obtained from $a$ by adding to it the digits 1 and 2 respectively as the $n$-th coordinate. But then if $z$ is any other element of $A_{0}$ then $x, y, z$ do not form a good triple, a contradiction. So by the inductive hypothesis
$$
\left|A_{0}\right| \leqslant \max \left\{2,\left|A_{0}^{\prime}\right|\right\} \leqslant 2\left(\frac{3}{2}\right)^{n-1}
$$
Similarly,
$$
\left|A_{2}\right|,\left|A_{3}\right| \leqslant 2\left(\frac{3}{2}\right)^{n-1}
$$
On the other hand, each element of $A$ appears in exactly two of $A_{0}, A_{1}, A_{2}$. As a result,
$$
|A|=\frac{1}{2}\left(\left|A_{0}\right|+\left|A_{1}\right|+\left|A_{2}\right|\right) \leqslant 2\left(\frac{3}{2}\right)^{n}
$$
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if every three distinct elements of $A$ form a good triple.
Prove that every good subset of $\mathcal{A}_{n}$ has at most $2\left(\frac{3}{2}\right)^{n}$ elements.
## Proposed by Greece
|
1. We proceed by induction on $n$, the case $n=1$ being trivial. Let
$$
A_{0}=\left\{\left(x_{1}, \ldots, x_{n}\right) \in A: x_{n} \neq 0\right\}
$$
and define $A_{1}$ and $A_{2}$ similarly.
Since $A$ is good and $A_{0}$ is a subset of $A$, then $A_{0}$ is also good. Therefore, any three of its elements have a coordinate that differs. This coordinate cannot be the last one since 0 cannot appear as a last coordinate. This means that the set $A_{0}^{\prime}$ obtained from $A_{0}$ by deleting the last coordinate from each of its elements is a good subset of $\mathcal{A}_{n-1}$.
Moreover, if $\left|A_{0}\right| \geqslant 3$ then $\left|A_{0}^{\prime}\right|=\left|A_{0}\right|$. Indeed, if otherwise, then there is an element $a \in A_{0}^{\prime}$ such that $x, y \in A_{0}$, where $x$ and $y$ are obtained from $a$ by adding to it the digits 1 and 2 respectively as the $n$-th coordinate. But then if $z$ is any other element of $A_{0}$ then $x, y, z$ do not form a good triple, a contradiction. So by the inductive hypothesis
$$
\left|A_{0}\right| \leqslant \max \left\{2,\left|A_{0}^{\prime}\right|\right\} \leqslant 2\left(\frac{3}{2}\right)^{n-1}
$$
Similarly,
$$
\left|A_{2}\right|,\left|A_{3}\right| \leqslant 2\left(\frac{3}{2}\right)^{n-1}
$$
On the other hand, each element of $A$ appears in exactly two of $A_{0}, A_{1}, A_{2}$. As a result,
$$
|A|=\frac{1}{2}\left(\left|A_{0}\right|+\left|A_{1}\right|+\left|A_{2}\right|\right) \leqslant 2\left(\frac{3}{2}\right)^{n}
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC1.",
"solution_match": "\nSolution"
}
|
6f56bcec-c1bb-56cc-a295-9e98d418f65f
| 606,106
|
Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if every three distinct elements of $A$ form a good triple.
Prove that every good subset of $\mathcal{A}_{n}$ has at most $2\left(\frac{3}{2}\right)^{n}$ elements.
## Proposed by Greece
|
2. Let
$$
B=\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathcal{A}_{n}: x_{i} \in\{0,1\}\right\}
$$
Let $A$ be a good subset of $\mathcal{A}_{n}$ and define $f: A \times B \rightarrow \mathcal{A}_{n}$ by $f(a, b)=a+b=\left(a_{1}+b_{1}, \ldots, a_{n}+b_{n}\right)$ where the addition is done modulo 3 .
We claim that if $(a, b),\left(a^{\prime}, b^{\prime}\right)$ and $\left(a^{\prime \prime}, b^{\prime \prime}\right)$ are distinct, then $f(a, b), f\left(a^{\prime}, b^{\prime}\right)$ and $f\left(a^{\prime \prime}, b^{\prime \prime}\right)$ cannot all be equal. Indeed assume $f(a, b)=f\left(a^{\prime}, b^{\prime}\right)=f\left(a^{\prime \prime}, b^{\prime \prime}\right)=\left(x_{1}, \ldots, x_{n}\right)$. So for each $i$ we have $a_{i}+b_{i}=a_{i}^{\prime}+b_{i}^{\prime}=a_{i}^{\prime \prime}+b_{i}^{\prime \prime}=x_{i}$. But then $a_{i}=x_{i}-b_{i} \in\left\{x_{i}, x_{i}-1\right\}$ and similarly $a_{i}^{\prime}, a_{i}^{\prime \prime} \in\left\{x_{i}, x_{i}-1\right\}$. So $\left\{a_{i}, a_{i}^{\prime}, a_{i}^{\prime \prime}\right\} \neq\{0,1,2\}$. Since this holds for each $i$ then $A$ cannot be a good set, contradiction.
Therefore $|A||B| \leqslant 2\left|\mathcal{A}_{n}\right|$ which gives $|A| \leqslant 2\left(\frac{3}{2}\right)^{n}$ as required.
Remark. Writing $f(n)$ for the maximal possible size of a good set, we proved that $f(n) \leqslant$ $2\left(\frac{3}{2}\right)^{n}$. We do not know the best possible asymptotic for $f(n)$ but we offer a corresponding lower bound which can increase the difficulty of the proposed problem.
We pick each element of $\mathcal{A}_{n}$ independently with probability $p$ to form a set $A$. For each bad triple $x, y, z$ of elements of $A$ we arbitrarily remove one of the elements to end up with a good set $B$. Note that there are at most $21^{n}$ bad triples $(x, y, z)$ since for coordinate $i$, out of the 27 triples of the form $\left(x_{i}, y_{i}, z_{i}\right)$, only 6 of them will make the triple $(x, y, z)$ a good triple. (Actually there are less than $21^{n}$ triples since this counts also triples where two or more of the $n$-tuples are the same.) So we get that
$$
\mathbb{E}|B| \geqslant p \cdot 3^{n}-p^{3} \cdot 21^{n} .
$$
Taking $p=\frac{1}{\sqrt{3 \cdot 7^{n}}}$ we get
$$
\mathbb{E}|B| \geqslant \frac{1}{\sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}-\frac{1}{3 \sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}=\frac{2}{3 \sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}=C \alpha^{n}
$$
where $\alpha=1.13389 \ldots$ and $C=0.3849 \ldots$. It follows that there is a good set of size at least $C \alpha^{n}$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if every three distinct elements of $A$ form a good triple.
Prove that every good subset of $\mathcal{A}_{n}$ has at most $2\left(\frac{3}{2}\right)^{n}$ elements.
## Proposed by Greece
|
2. Let
$$
B=\left\{x=\left(x_{1}, \ldots, x_{n}\right) \in \mathcal{A}_{n}: x_{i} \in\{0,1\}\right\}
$$
Let $A$ be a good subset of $\mathcal{A}_{n}$ and define $f: A \times B \rightarrow \mathcal{A}_{n}$ by $f(a, b)=a+b=\left(a_{1}+b_{1}, \ldots, a_{n}+b_{n}\right)$ where the addition is done modulo 3 .
We claim that if $(a, b),\left(a^{\prime}, b^{\prime}\right)$ and $\left(a^{\prime \prime}, b^{\prime \prime}\right)$ are distinct, then $f(a, b), f\left(a^{\prime}, b^{\prime}\right)$ and $f\left(a^{\prime \prime}, b^{\prime \prime}\right)$ cannot all be equal. Indeed assume $f(a, b)=f\left(a^{\prime}, b^{\prime}\right)=f\left(a^{\prime \prime}, b^{\prime \prime}\right)=\left(x_{1}, \ldots, x_{n}\right)$. So for each $i$ we have $a_{i}+b_{i}=a_{i}^{\prime}+b_{i}^{\prime}=a_{i}^{\prime \prime}+b_{i}^{\prime \prime}=x_{i}$. But then $a_{i}=x_{i}-b_{i} \in\left\{x_{i}, x_{i}-1\right\}$ and similarly $a_{i}^{\prime}, a_{i}^{\prime \prime} \in\left\{x_{i}, x_{i}-1\right\}$. So $\left\{a_{i}, a_{i}^{\prime}, a_{i}^{\prime \prime}\right\} \neq\{0,1,2\}$. Since this holds for each $i$ then $A$ cannot be a good set, contradiction.
Therefore $|A||B| \leqslant 2\left|\mathcal{A}_{n}\right|$ which gives $|A| \leqslant 2\left(\frac{3}{2}\right)^{n}$ as required.
Remark. Writing $f(n)$ for the maximal possible size of a good set, we proved that $f(n) \leqslant$ $2\left(\frac{3}{2}\right)^{n}$. We do not know the best possible asymptotic for $f(n)$ but we offer a corresponding lower bound which can increase the difficulty of the proposed problem.
We pick each element of $\mathcal{A}_{n}$ independently with probability $p$ to form a set $A$. For each bad triple $x, y, z$ of elements of $A$ we arbitrarily remove one of the elements to end up with a good set $B$. Note that there are at most $21^{n}$ bad triples $(x, y, z)$ since for coordinate $i$, out of the 27 triples of the form $\left(x_{i}, y_{i}, z_{i}\right)$, only 6 of them will make the triple $(x, y, z)$ a good triple. (Actually there are less than $21^{n}$ triples since this counts also triples where two or more of the $n$-tuples are the same.) So we get that
$$
\mathbb{E}|B| \geqslant p \cdot 3^{n}-p^{3} \cdot 21^{n} .
$$
Taking $p=\frac{1}{\sqrt{3 \cdot 7^{n}}}$ we get
$$
\mathbb{E}|B| \geqslant \frac{1}{\sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}-\frac{1}{3 \sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}=\frac{2}{3 \sqrt{3}}\left(\frac{9}{7}\right)^{n / 2}=C \alpha^{n}
$$
where $\alpha=1.13389 \ldots$ and $C=0.3849 \ldots$. It follows that there is a good set of size at least $C \alpha^{n}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC1.",
"solution_match": "\nSolution"
}
|
6f56bcec-c1bb-56cc-a295-9e98d418f65f
| 606,106
|
Let $K$ and $N>K$ be fixed positive integers. Let $n$ be a positive integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct integers. Suppose that whenever $m_{1}, m_{2}, \ldots, m_{n}$ are integers, not all equal to 0 , such that $\left|m_{i}\right| \leqslant K$ for each $i$, then the sum
$$
\sum_{i=1}^{n} m_{i} a_{i}
$$
is not divisible by $N$. What is the largest possible value of $n$ ?
## Proposed by North Macedonia
|
The answer is $n=\left\lfloor\log _{K+1} N\right\rfloor$.
Note first that for $n \leqslant\left\lfloor\log _{K+1} N\right\rfloor$, taking $a_{i}=(K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_{r} \neq 0$. Then on the one hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \leqslant \sum_{i=1}^{n} K(K+1)^{i-1}=(K+1)^{n}-1<N
$$
On the other hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \geqslant\left|m_{r} a_{r}\right|-\left|\sum_{i=1}^{r-1} m_{i} a_{i}\right| \geqslant(K+1)^{r-1}-\sum_{i=1}^{r-1} K(K+1)^{i-1}=1>0
$$
So the sum is indeed not divisible by $n$.
Assume now that $n \geqslant\left\lfloor\log _{K+1} N\right\rfloor$ and look at all $n$-tuples of the form $\left(t_{1}, \ldots, t_{n}\right)$ where each $t_{i}$ is a non-negative integer with $t_{i} \leqslant K$. There are $(K+1)^{n}>N$ such tuples so there are two of them, say $\left(t_{1}, \ldots, t_{n}\right)$ and $\left(t_{1}^{\prime}, \ldots, t_{n}^{\prime}\right)$ such that
$$
\sum_{i=1}^{n} t_{i} a_{i} \equiv \sum_{i=1}^{n} t_{i}^{\prime} a_{i} \bmod N
$$
Now taking $m_{i}=t_{i}-t_{i}^{\prime}$ for each $i$ satisfies the requirements on the $m_{i}$ 's but $N$ divides the sum
$$
\sum_{i=1}^{n} m_{i} a_{i}
$$
a contradiction.
|
n=\left\lfloor\log _{K+1} N\right\rfloor
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $K$ and $N>K$ be fixed positive integers. Let $n$ be a positive integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct integers. Suppose that whenever $m_{1}, m_{2}, \ldots, m_{n}$ are integers, not all equal to 0 , such that $\left|m_{i}\right| \leqslant K$ for each $i$, then the sum
$$
\sum_{i=1}^{n} m_{i} a_{i}
$$
is not divisible by $N$. What is the largest possible value of $n$ ?
## Proposed by North Macedonia
|
The answer is $n=\left\lfloor\log _{K+1} N\right\rfloor$.
Note first that for $n \leqslant\left\lfloor\log _{K+1} N\right\rfloor$, taking $a_{i}=(K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_{r} \neq 0$. Then on the one hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \leqslant \sum_{i=1}^{n} K(K+1)^{i-1}=(K+1)^{n}-1<N
$$
On the other hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \geqslant\left|m_{r} a_{r}\right|-\left|\sum_{i=1}^{r-1} m_{i} a_{i}\right| \geqslant(K+1)^{r-1}-\sum_{i=1}^{r-1} K(K+1)^{i-1}=1>0
$$
So the sum is indeed not divisible by $n$.
Assume now that $n \geqslant\left\lfloor\log _{K+1} N\right\rfloor$ and look at all $n$-tuples of the form $\left(t_{1}, \ldots, t_{n}\right)$ where each $t_{i}$ is a non-negative integer with $t_{i} \leqslant K$. There are $(K+1)^{n}>N$ such tuples so there are two of them, say $\left(t_{1}, \ldots, t_{n}\right)$ and $\left(t_{1}^{\prime}, \ldots, t_{n}^{\prime}\right)$ such that
$$
\sum_{i=1}^{n} t_{i} a_{i} \equiv \sum_{i=1}^{n} t_{i}^{\prime} a_{i} \bmod N
$$
Now taking $m_{i}=t_{i}-t_{i}^{\prime}$ for each $i$ satisfies the requirements on the $m_{i}$ 's but $N$ divides the sum
$$
\sum_{i=1}^{n} m_{i} a_{i}
$$
a contradiction.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC2.",
"solution_match": "\nSolution."
}
|
ad8af092-1ed0-5fc4-bb1d-cb56c3da5418
| 606,125
|
In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most c.
## Proposed by Romania
|
1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.
We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1000, we deduce that there are at most 1000 such coins. Thus we are able to put (at most) 10 such coins in each of the 100 boxes. Everything so far is all right: $10 \cdot \frac{1000}{1001}<10<c=11-\frac{11}{1001}$.
Next, step by step, we take one of the remaining coins and prove there is a box where it can be added. Suppose that at some point this algorithm fails. It would mean that at a certain point the total sums in the 100 boxes would be $x_{1}, x_{2}, \ldots, x_{100}$ and no matter how we would add the coin $x$, where $x \leqslant \frac{1000}{1001}$, in any of the boxes, that box would be overflowed, i.e., it would have a total sum of more than $11-\frac{11}{1001}$. Therefore,
$$
x_{i}+x>11-\frac{11}{1001}
$$
for all $i=1,2, \ldots, 100$. Then
$$
x_{1}+x_{2}+\cdots+x_{100}+100 x>100 \cdot\left(11-\frac{11}{1001}\right) .
$$
But since $1000 \geqslant x_{1}+x_{2}+\cdots+x_{100}+x$ and $\frac{1000}{1001} \geqslant x$ we obtain the contradiction
$$
1000+99 \cdot \frac{1000}{1001}>100 \cdot\left(11-\frac{11}{1001}\right) \Longleftrightarrow 1000 \cdot \frac{1100}{1001}>100 \cdot 11 \cdot \frac{1000}{1001} .
$$
Thus the algorithm does not fail and since we have finitely many coins, we will eventually reach to a happy end.
Now we show that $c=11-11 \alpha$, with $1>\alpha>\frac{1}{1001}$ does not work.
Take $r \in\left[\frac{1}{1001}, \alpha\right)$ and let $n=\left\lfloor\frac{1000}{1-r}\right\rfloor$. Since $r \geqslant \frac{1}{1001}$, then $\frac{1000}{1-r} \geqslant 1001$, therefore $n \geqslant 1001$.
Now take $n$ coins each of value $1-r$. Their sum is $n(1-r) \leqslant \frac{1000}{1-r} \cdot(1-r)=1000$. Now, no matter how we place them in 100 boxes, as $n \geqslant 1001$, there exist 11 coins in the same box. But $11(1-r)=11-11 r>11-11 \alpha$, so the constant $c=11-11 \alpha$ indeed does not work.
|
11-\frac{11}{1001}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most c.
## Proposed by Romania
|
1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.
We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1000, we deduce that there are at most 1000 such coins. Thus we are able to put (at most) 10 such coins in each of the 100 boxes. Everything so far is all right: $10 \cdot \frac{1000}{1001}<10<c=11-\frac{11}{1001}$.
Next, step by step, we take one of the remaining coins and prove there is a box where it can be added. Suppose that at some point this algorithm fails. It would mean that at a certain point the total sums in the 100 boxes would be $x_{1}, x_{2}, \ldots, x_{100}$ and no matter how we would add the coin $x$, where $x \leqslant \frac{1000}{1001}$, in any of the boxes, that box would be overflowed, i.e., it would have a total sum of more than $11-\frac{11}{1001}$. Therefore,
$$
x_{i}+x>11-\frac{11}{1001}
$$
for all $i=1,2, \ldots, 100$. Then
$$
x_{1}+x_{2}+\cdots+x_{100}+100 x>100 \cdot\left(11-\frac{11}{1001}\right) .
$$
But since $1000 \geqslant x_{1}+x_{2}+\cdots+x_{100}+x$ and $\frac{1000}{1001} \geqslant x$ we obtain the contradiction
$$
1000+99 \cdot \frac{1000}{1001}>100 \cdot\left(11-\frac{11}{1001}\right) \Longleftrightarrow 1000 \cdot \frac{1100}{1001}>100 \cdot 11 \cdot \frac{1000}{1001} .
$$
Thus the algorithm does not fail and since we have finitely many coins, we will eventually reach to a happy end.
Now we show that $c=11-11 \alpha$, with $1>\alpha>\frac{1}{1001}$ does not work.
Take $r \in\left[\frac{1}{1001}, \alpha\right)$ and let $n=\left\lfloor\frac{1000}{1-r}\right\rfloor$. Since $r \geqslant \frac{1}{1001}$, then $\frac{1000}{1-r} \geqslant 1001$, therefore $n \geqslant 1001$.
Now take $n$ coins each of value $1-r$. Their sum is $n(1-r) \leqslant \frac{1000}{1-r} \cdot(1-r)=1000$. Now, no matter how we place them in 100 boxes, as $n \geqslant 1001$, there exist 11 coins in the same box. But $11(1-r)=11-11 r>11-11 \alpha$, so the constant $c=11-11 \alpha$ indeed does not work.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC3.",
"solution_match": "\nSolution"
}
|
e8336fea-3606-55e5-aac0-3111d3c0217c
| 606,138
|
In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most c.
## Proposed by Romania
|
2 (for the upper bound). Amongst all possible arrangements into boxes, pick one where the maximum value inside a box is as small as possible. If there are several arrangements achieving this smallest maximum value, pick one where the number of boxes achieving this value is as small as possible.
Say that the boxes have total values equal to $10+x_{1} \geqslant 10+x_{2} \geqslant \cdots \geqslant 10+x_{100}$. respectively. We must have $x_{1}+\cdots+x_{100} \leqslant 0$. In particular, $0 \geqslant x_{1}+99 x_{100}$.
Assume for contradiction that $x_{1}>\frac{990}{1001}=\frac{90}{91}$. Remove the coin of smallest denomination from the first box and add it into the 100-th box. Since the total value in the first box is greater than 10, the first box has at least 11 coins and therefore it has a coin of value at most $\frac{10+x_{1}}{11}$. The total new value in the last box is at most
$$
10+x_{100}+\frac{10+x_{1}}{11} \leqslant 10-\frac{x_{1}}{99}+\frac{10+x_{1}}{11}=10+x_{1}+\frac{90-91 x_{1}}{99}<10+x_{1}
$$
Remark. If we replace $[0,1]$ with $[0, v]$, the total sum with $s$, and the number of available boxes with $n$, then the answer to the problem is
$$
c=v+\frac{s}{n}-\left(\frac{s}{n}+1\right) \cdot \frac{1}{s+1}=v+\frac{s^{2}-n}{n(s+1)}
$$
|
2
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, can split those coins into 100 boxes, such that the total value inside each box is at most c.
## Proposed by Romania
|
2 (for the upper bound). Amongst all possible arrangements into boxes, pick one where the maximum value inside a box is as small as possible. If there are several arrangements achieving this smallest maximum value, pick one where the number of boxes achieving this value is as small as possible.
Say that the boxes have total values equal to $10+x_{1} \geqslant 10+x_{2} \geqslant \cdots \geqslant 10+x_{100}$. respectively. We must have $x_{1}+\cdots+x_{100} \leqslant 0$. In particular, $0 \geqslant x_{1}+99 x_{100}$.
Assume for contradiction that $x_{1}>\frac{990}{1001}=\frac{90}{91}$. Remove the coin of smallest denomination from the first box and add it into the 100-th box. Since the total value in the first box is greater than 10, the first box has at least 11 coins and therefore it has a coin of value at most $\frac{10+x_{1}}{11}$. The total new value in the last box is at most
$$
10+x_{100}+\frac{10+x_{1}}{11} \leqslant 10-\frac{x_{1}}{99}+\frac{10+x_{1}}{11}=10+x_{1}+\frac{90-91 x_{1}}{99}<10+x_{1}
$$
Remark. If we replace $[0,1]$ with $[0, v]$, the total sum with $s$, and the number of available boxes with $n$, then the answer to the problem is
$$
c=v+\frac{s}{n}-\left(\frac{s}{n}+1\right) \cdot \frac{1}{s+1}=v+\frac{s^{2}-n}{n(s+1)}
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC3.",
"solution_match": "\nSolution"
}
|
e8336fea-3606-55e5-aac0-3111d3c0217c
| 606,138
|
A sequence of $2 n+1$ non-negative integers $a_{1}, a_{2}, \ldots, a_{2 n+1}$ is given. There's also a sequence of $2 n+1$ consecutive cells enumerated from 1 to $2 n+1$ from left to right, such that initially the number $a_{i}$ is written on the $i$-th cell, for $i=1,2, \ldots 2 n+1$. Starting from this initial position, we repeat the following sequence of steps, as long as it's possible:
Step 1: Add up the numbers written on all the cells, denote the sum as $s$.
Step 2: If $s$ is equal to 0 or if it is larger than the current number of cells, the process terminates. Otherwise, remove the $s$-th cell, and shift all cells that are to the right of it one position to the left. Then go to Step 1.
Example: $(1,0,1, \underline{2}, 0) \rightarrow(1, \underline{0}, 1,0) \rightarrow(1, \underline{1}, 0) \rightarrow(\underline{1}, 0) \rightarrow(0)$.
A sequence $a_{1}, a_{2}, \ldots, a_{2 n+1}$ of non-negative integers is called balanced, if at the end of this process there's exactly one cell left, and it's the cell that was initially enumerated by $(n+1)$, i.e. the cell that was initially in the middle.
Find the total number of balanced sequences as a function of $n$.
## Proposed by North Macedonia
|
The answer is: $C_{n} \cdot C_{n}$, where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$-th Catalan number.
We divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells will be called the back cells and the front (leftmost) $n$ cells will be called the front cells. The central, $(n+1)$-st, cell will be called the middle cell.
Claim 1. All the back cells must be removed before any front cell is removed.
Proof. Assume for contradiction that this is not the case. Then there must be a point in time where a front cell is deleted and then immediately after a back cell is deleted. Let us say that the deleted front cell was at position $i$. So all back cells have positions greater or equal to $i+2$. After the cell is deleted all back cells have positions greater or equal to $i+1$. But since we deleted cell $i$, then the total sum is $i$ and this does not increase. So at the next step we delete a cell at position at most $i$, a contradiction.
Claim 2. The middle cell must contain the number 0 , i.e., $a_{n+1}=0$.
Proof. Consider the last step in the process where we have total of 2 cells. One of these is the middle cell, and by Claim 1 the other must be one of the front cells. I.e. we have $\left(x, a_{n+1}\right)$. On the next move, we remove $x$, which means that $x+a_{n+1}=1$. So $a_{n+1}=0$ or $a_{n+1}=1$. But after that we cannot remove $a_{n+1}$, which means that $a_{n+1} \neq 1$. So $a_{n+1}=0$.
Now, let's define a self-destructing sequence to be one with no surviving cells at the end of the process. For example, $(0,1,2)$ is self-destructing because $(0,1,2) \rightarrow(0,1) \rightarrow(1) \rightarrow()$.
Let $\mathcal{S}_{n}$ be the set of self-destructing sequences of length $n$. For example, $\mathcal{S}_{2}=\{(0,1),(1,1)\}$. It is clear that the front cells form a self-destructing sequence, i.e., $\left(a_{1}, a_{2}, \cdots a_{n}\right) \in \mathcal{S}_{n}$. The back cells also have certain self-destructing quality, which is made more precise in Claim 3 below.
Claim 3. Fix the front sequence $\varphi=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$. Let $\mathcal{B}_{\phi}$ be the set of all possible back sequences of length $n$ that can be appended to $\varphi$ (with a 0 between them) to get a balanced sequence. Then there is a bijection $f: \mathcal{S}_{n} \mapsto \mathcal{B}_{\phi}$.
Proof. Let $c=n+1-\sum_{i=1}^{n} a_{i}$ and consider a particular $\sigma=\left(s_{1}, s_{2}, \ldots, s_{n}\right) \in \mathcal{S}_{n}$. Let $\ell$ be the initial index of the last surviving cell in $\sigma$. Then $f(\sigma)=\left(s_{1}, s_{2}, \ldots, s_{\ell}+c, s_{\ell+1}, \ldots, s_{n}\right)$ defines a bijection $\mathcal{S}_{n} \mapsto \mathcal{B}_{\phi}$.
Indeed we claim that the $k$-th deleted cell in $\sigma$ is the $k$-th deleted cell in $\overline{\varphi 0 f(\sigma)}$ for each $k=1, \ldots, n$. Indeed after some deletions let $S$ be the total sum remaining in $\sigma$. Then the total sum remaining in $\overline{\varphi 0 f(\sigma)}$ is $-\sum_{i=1}^{n} a_{i}+0+S+c=S+n+1$. So we delete next the cell in position $S$ in $\sigma$ if and only if we delete the cell in position $S+n+1$ in $\overline{\varphi 0 f(\sigma)}$.
So $\overline{\phi 0 f(\sigma)}$ is clearly a balanced sequence: we first eliminate all cells in the back, then the front. In the same manner it follows that every balanced sequence in of this form.
So far we have shown that the total number of balanced sequences is $\left|\mathcal{S}_{n}\right|^{2}$. It remains to calculate the size $\left|\mathcal{S}_{n}\right|$.
Claim 4. Let $\mathcal{T}_{n}$ be the set of $2 n$-sequences consisting of $n$ zeros and $n$ ones such that in each initial segment the number of 1's does not surpass the number of 0 's. Then $\left|\mathcal{S}_{n}\right|=\left|\mathcal{T}_{n}\right|$.
Proof. Let $[n]=\{1,2, \ldots, n\}$, and let us also consider the set $\mathcal{F}_{n}$ of non-decreasing mappings $f:[n] \rightarrow[n]$ such that $f(i) \leqslant i$ for each $i \in[n]$. The claim will follow once we show that $\left|\mathcal{S}_{n}\right|=\left|\mathcal{F}_{n}\right|$ and that $\left|\mathcal{F}_{n}\right|=\left|\mathcal{T}_{n}\right|$.
In order to demonstrate that $\left|\mathcal{S}_{n}\right|=\left|\mathcal{F}_{n}\right|$, observe that there is an obvious bijective correspondence $a \mapsto f$ between the sets $\mathcal{S}_{n}$ and $\mathcal{F}_{n}$. Indeed, reversing the self-destructing process for an $n$-sequence $a=\left(a_{1}, a_{2}, \ldots, a_{n}\right) \in \mathcal{S}_{n}$, simply define $f(i)$ to be the (partial) sum of the existing terms after the $i$-th backward step.
As for $\left|\mathcal{T}_{n}\right|=\left|\mathcal{F}_{n}\right|$, note the following bijective correspondence $t \mapsto f$ between the sets $\mathcal{T}_{n}$ and $\mathcal{F}_{n}$. Let $f(i)$ equal $1+\#(i)$, where $\#(i)$ is defined to be the total number of $1^{\prime} s$ appearing in $t$ before the $i$-th zero.
Finally, it is a known fact that $\left|\mathcal{B}_{n}\right|$ is the $n$-th Catalan number $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$. (The essential idea of the textbook proof of this fact uses the so-called reflection principle of A. D. André.)
|
C_{n} \cdot C_{n}
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
A sequence of $2 n+1$ non-negative integers $a_{1}, a_{2}, \ldots, a_{2 n+1}$ is given. There's also a sequence of $2 n+1$ consecutive cells enumerated from 1 to $2 n+1$ from left to right, such that initially the number $a_{i}$ is written on the $i$-th cell, for $i=1,2, \ldots 2 n+1$. Starting from this initial position, we repeat the following sequence of steps, as long as it's possible:
Step 1: Add up the numbers written on all the cells, denote the sum as $s$.
Step 2: If $s$ is equal to 0 or if it is larger than the current number of cells, the process terminates. Otherwise, remove the $s$-th cell, and shift all cells that are to the right of it one position to the left. Then go to Step 1.
Example: $(1,0,1, \underline{2}, 0) \rightarrow(1, \underline{0}, 1,0) \rightarrow(1, \underline{1}, 0) \rightarrow(\underline{1}, 0) \rightarrow(0)$.
A sequence $a_{1}, a_{2}, \ldots, a_{2 n+1}$ of non-negative integers is called balanced, if at the end of this process there's exactly one cell left, and it's the cell that was initially enumerated by $(n+1)$, i.e. the cell that was initially in the middle.
Find the total number of balanced sequences as a function of $n$.
## Proposed by North Macedonia
|
The answer is: $C_{n} \cdot C_{n}$, where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$-th Catalan number.
We divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells will be called the back cells and the front (leftmost) $n$ cells will be called the front cells. The central, $(n+1)$-st, cell will be called the middle cell.
Claim 1. All the back cells must be removed before any front cell is removed.
Proof. Assume for contradiction that this is not the case. Then there must be a point in time where a front cell is deleted and then immediately after a back cell is deleted. Let us say that the deleted front cell was at position $i$. So all back cells have positions greater or equal to $i+2$. After the cell is deleted all back cells have positions greater or equal to $i+1$. But since we deleted cell $i$, then the total sum is $i$ and this does not increase. So at the next step we delete a cell at position at most $i$, a contradiction.
Claim 2. The middle cell must contain the number 0 , i.e., $a_{n+1}=0$.
Proof. Consider the last step in the process where we have total of 2 cells. One of these is the middle cell, and by Claim 1 the other must be one of the front cells. I.e. we have $\left(x, a_{n+1}\right)$. On the next move, we remove $x$, which means that $x+a_{n+1}=1$. So $a_{n+1}=0$ or $a_{n+1}=1$. But after that we cannot remove $a_{n+1}$, which means that $a_{n+1} \neq 1$. So $a_{n+1}=0$.
Now, let's define a self-destructing sequence to be one with no surviving cells at the end of the process. For example, $(0,1,2)$ is self-destructing because $(0,1,2) \rightarrow(0,1) \rightarrow(1) \rightarrow()$.
Let $\mathcal{S}_{n}$ be the set of self-destructing sequences of length $n$. For example, $\mathcal{S}_{2}=\{(0,1),(1,1)\}$. It is clear that the front cells form a self-destructing sequence, i.e., $\left(a_{1}, a_{2}, \cdots a_{n}\right) \in \mathcal{S}_{n}$. The back cells also have certain self-destructing quality, which is made more precise in Claim 3 below.
Claim 3. Fix the front sequence $\varphi=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$. Let $\mathcal{B}_{\phi}$ be the set of all possible back sequences of length $n$ that can be appended to $\varphi$ (with a 0 between them) to get a balanced sequence. Then there is a bijection $f: \mathcal{S}_{n} \mapsto \mathcal{B}_{\phi}$.
Proof. Let $c=n+1-\sum_{i=1}^{n} a_{i}$ and consider a particular $\sigma=\left(s_{1}, s_{2}, \ldots, s_{n}\right) \in \mathcal{S}_{n}$. Let $\ell$ be the initial index of the last surviving cell in $\sigma$. Then $f(\sigma)=\left(s_{1}, s_{2}, \ldots, s_{\ell}+c, s_{\ell+1}, \ldots, s_{n}\right)$ defines a bijection $\mathcal{S}_{n} \mapsto \mathcal{B}_{\phi}$.
Indeed we claim that the $k$-th deleted cell in $\sigma$ is the $k$-th deleted cell in $\overline{\varphi 0 f(\sigma)}$ for each $k=1, \ldots, n$. Indeed after some deletions let $S$ be the total sum remaining in $\sigma$. Then the total sum remaining in $\overline{\varphi 0 f(\sigma)}$ is $-\sum_{i=1}^{n} a_{i}+0+S+c=S+n+1$. So we delete next the cell in position $S$ in $\sigma$ if and only if we delete the cell in position $S+n+1$ in $\overline{\varphi 0 f(\sigma)}$.
So $\overline{\phi 0 f(\sigma)}$ is clearly a balanced sequence: we first eliminate all cells in the back, then the front. In the same manner it follows that every balanced sequence in of this form.
So far we have shown that the total number of balanced sequences is $\left|\mathcal{S}_{n}\right|^{2}$. It remains to calculate the size $\left|\mathcal{S}_{n}\right|$.
Claim 4. Let $\mathcal{T}_{n}$ be the set of $2 n$-sequences consisting of $n$ zeros and $n$ ones such that in each initial segment the number of 1's does not surpass the number of 0 's. Then $\left|\mathcal{S}_{n}\right|=\left|\mathcal{T}_{n}\right|$.
Proof. Let $[n]=\{1,2, \ldots, n\}$, and let us also consider the set $\mathcal{F}_{n}$ of non-decreasing mappings $f:[n] \rightarrow[n]$ such that $f(i) \leqslant i$ for each $i \in[n]$. The claim will follow once we show that $\left|\mathcal{S}_{n}\right|=\left|\mathcal{F}_{n}\right|$ and that $\left|\mathcal{F}_{n}\right|=\left|\mathcal{T}_{n}\right|$.
In order to demonstrate that $\left|\mathcal{S}_{n}\right|=\left|\mathcal{F}_{n}\right|$, observe that there is an obvious bijective correspondence $a \mapsto f$ between the sets $\mathcal{S}_{n}$ and $\mathcal{F}_{n}$. Indeed, reversing the self-destructing process for an $n$-sequence $a=\left(a_{1}, a_{2}, \ldots, a_{n}\right) \in \mathcal{S}_{n}$, simply define $f(i)$ to be the (partial) sum of the existing terms after the $i$-th backward step.
As for $\left|\mathcal{T}_{n}\right|=\left|\mathcal{F}_{n}\right|$, note the following bijective correspondence $t \mapsto f$ between the sets $\mathcal{T}_{n}$ and $\mathcal{F}_{n}$. Let $f(i)$ equal $1+\#(i)$, where $\#(i)$ is defined to be the total number of $1^{\prime} s$ appearing in $t$ before the $i$-th zero.
Finally, it is a known fact that $\left|\mathcal{B}_{n}\right|$ is the $n$-th Catalan number $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$. (The essential idea of the textbook proof of this fact uses the so-called reflection principle of A. D. André.)
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC4.",
"solution_match": "\nSolution."
}
|
4a9c85ba-a7cb-505b-86ef-164b80ed4964
| 606,160
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or creates a new pile with one piece. What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
## Proposed by United Kingdom
|
1. We will show that he can do so by the morning of day 199 but not earlier.
If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also denote this position by $(n, m)$. Implicitly we will also write $k$ for the number of piles with exactly two pieces of rubbish.
Angel's strategy is the following:
(i) From position $(0, m)$ remove one piece from each pile to go position $(0,0)$. The game ends.
(ii) From position $(n, 0)$, where $n \geqslant 1$, remove one pile to go to position $(n-1,0)$. Either the game ends, or the demon can move to position $(n-1,0)$ or $(n-1,1)$. In any case $V$ reduces by at least $1 / 2$.
(iii) From position $(n, 1)$, where $n \geqslant 1$, remove one pile with at least two pieces to go to position $(n-1,1)$. The demon can move to position $(n, 0)$ or $(n-1,2)$. In any case $V$ reduces by (at least) $1 / 2$.
(iv) From position $(n, m)$, where $n \geqslant 1$ and $m \geqslant 2$, remove one piece from each pile to go to position $(n-k, k)$. The demon can move to position $(n, 0)$ or $(n-k, k+1)$. In any case $V$ reduces by at least $1 / 2$. (The value of position $(n-k, k+1)$ is $n+\frac{1}{2}$ if $k=0$, and $n-k+1 \leqslant n$ if $k \geqslant 1$.)
So during every day if the game does not end then $V$ is decreased by at least $1 / 2$. So after 198 days if the game did not already end we will have $V \leqslant 1$ and we will be in one of positions $(0, m),(1,0)$. The game can then end on the morning of day 199.
We will now provide a strategy for demon which guarantees that at the end of each day $V$ has decreased by at most $1 / 2$ and furthermore at the end of the day $m \leqslant 1$.
(i) If Angel moves from $(n, 0)$ to $(n-1,0)$ (by removing a pile) then create a new pile with one piece to move to $(n-1,1)$. Then $V$ decreases by $1 / 2$ and and $m=1 \leqslant 1$
(ii) If Angel moves from $(n, 0)$ to $(n-k, k)$ (by removing one piece from each pile) then add one piece back to each pile to move to $(n, 0)$. Then $V$ stays the same and $m=0 \leqslant 1$.
(iii) If Angels moves from $(n, 1)$ to $(n-1,1)$ or $(n, 0)$ (by removing a pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
(iv) If Angel moves from $(n, 1)$ to $(n-k, k)$ (by removing a piece from each pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
Since after every move of demon we have $m \leqslant 1$, in order for Angel to finish the game in the next morning we must have $n=1, m=0$ or $n=0, m=1$ and therefore we must have $V \leqslant 1$.
But now inductively the demon can guarantee that by the end of day $N$, where $N \leqslant 198$ the game has not yet finished and that $V \geqslant 100-N / 2$.
|
199
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or creates a new pile with one piece. What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
## Proposed by United Kingdom
|
1. We will show that he can do so by the morning of day 199 but not earlier.
If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also denote this position by $(n, m)$. Implicitly we will also write $k$ for the number of piles with exactly two pieces of rubbish.
Angel's strategy is the following:
(i) From position $(0, m)$ remove one piece from each pile to go position $(0,0)$. The game ends.
(ii) From position $(n, 0)$, where $n \geqslant 1$, remove one pile to go to position $(n-1,0)$. Either the game ends, or the demon can move to position $(n-1,0)$ or $(n-1,1)$. In any case $V$ reduces by at least $1 / 2$.
(iii) From position $(n, 1)$, where $n \geqslant 1$, remove one pile with at least two pieces to go to position $(n-1,1)$. The demon can move to position $(n, 0)$ or $(n-1,2)$. In any case $V$ reduces by (at least) $1 / 2$.
(iv) From position $(n, m)$, where $n \geqslant 1$ and $m \geqslant 2$, remove one piece from each pile to go to position $(n-k, k)$. The demon can move to position $(n, 0)$ or $(n-k, k+1)$. In any case $V$ reduces by at least $1 / 2$. (The value of position $(n-k, k+1)$ is $n+\frac{1}{2}$ if $k=0$, and $n-k+1 \leqslant n$ if $k \geqslant 1$.)
So during every day if the game does not end then $V$ is decreased by at least $1 / 2$. So after 198 days if the game did not already end we will have $V \leqslant 1$ and we will be in one of positions $(0, m),(1,0)$. The game can then end on the morning of day 199.
We will now provide a strategy for demon which guarantees that at the end of each day $V$ has decreased by at most $1 / 2$ and furthermore at the end of the day $m \leqslant 1$.
(i) If Angel moves from $(n, 0)$ to $(n-1,0)$ (by removing a pile) then create a new pile with one piece to move to $(n-1,1)$. Then $V$ decreases by $1 / 2$ and and $m=1 \leqslant 1$
(ii) If Angel moves from $(n, 0)$ to $(n-k, k)$ (by removing one piece from each pile) then add one piece back to each pile to move to $(n, 0)$. Then $V$ stays the same and $m=0 \leqslant 1$.
(iii) If Angels moves from $(n, 1)$ to $(n-1,1)$ or $(n, 0)$ (by removing a pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
(iv) If Angel moves from $(n, 1)$ to $(n-k, k)$ (by removing a piece from each pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
Since after every move of demon we have $m \leqslant 1$, in order for Angel to finish the game in the next morning we must have $n=1, m=0$ or $n=0, m=1$ and therefore we must have $V \leqslant 1$.
But now inductively the demon can guarantee that by the end of day $N$, where $N \leqslant 198$ the game has not yet finished and that $V \geqslant 100-N / 2$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC5.",
"solution_match": "\nSolution"
}
|
cbcc7643-5a85-5395-8124-721c49e1cc61
| 606,172
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or creates a new pile with one piece. What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
## Proposed by United Kingdom
|
2.
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at least 1 in each day. (Case (i) is trivial as the game ends in one day.)
Now define demon's score $S_{D}$ to be $S_{D}=2 n-1$ if $m=0$ and $S_{D}=2 n$ if $m \geqslant 1$. The claim is the if $(n, m) \neq(0,0)$, then the demon can ensure that Angel requires $S_{D}$ days to clear the rubbish.
Again, demon's strategy is the same as in the Solution by PSC and in each of cases (i)-(iv) one needs to check that $S_{D}$ reduced by at most 1 in each day.
Comment. If we start from position $(n, m)$, then the number $N$ of days required is
$$
N= \begin{cases}2 n-1 & \text { if } m=0 \\ 2 n & \text { if } m=1 \\ 2 n & \text { if } m \geqslant 2, \text { and } k \geqslant 1 \\ 2 n+1 & \text { if } m \geqslant 2, \text { and } k=0\end{cases}
$$
|
200
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or creates a new pile with one piece. What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
## Proposed by United Kingdom
|
2.
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at least 1 in each day. (Case (i) is trivial as the game ends in one day.)
Now define demon's score $S_{D}$ to be $S_{D}=2 n-1$ if $m=0$ and $S_{D}=2 n$ if $m \geqslant 1$. The claim is the if $(n, m) \neq(0,0)$, then the demon can ensure that Angel requires $S_{D}$ days to clear the rubbish.
Again, demon's strategy is the same as in the Solution by PSC and in each of cases (i)-(iv) one needs to check that $S_{D}$ reduced by at most 1 in each day.
Comment. If we start from position $(n, m)$, then the number $N$ of days required is
$$
N= \begin{cases}2 n-1 & \text { if } m=0 \\ 2 n & \text { if } m=1 \\ 2 n & \text { if } m \geqslant 2, \text { and } k \geqslant 1 \\ 2 n+1 & \text { if } m \geqslant 2, \text { and } k=0\end{cases}
$$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2021_bmo_shortlist.jsonl",
"problem_match": "\nC5.",
"solution_match": "\n## Solution"
}
|
cbcc7643-5a85-5395-8124-721c49e1cc61
| 606,172
|
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