add segmented Balkan_MO 2008-2024 problem and solution

Balkan_MO/download_script/download.py

@@ -69,7 +69,7 @@ def main():
         "en-2012-BMO-problems": "https://artofproblemsolving.com/downloads/printable_post_collections/4084.pdf",
         "en-2022-BMO": "https://cdn.b3web.xyz/web/cms/optimizedBMO_2022_Problems_Solutions.pdf1652174033.pdf",
         "en-2023-BMO": "https://drive.google.com/uc?id=1lpcx_M-JLVLFMtZjFRngsAACtuSMPyfI&export=download",
-        "en-2024-BMO": "https://bmo2024.org/wp-content/uploads/2024/05/BOM_english.pdf"
+        "en-2024-BMO": "https://bmo2024.org/wp-content/uploads/2024/05/BOM_englishSolutions.pdf"
     }
 
     for title, url in tqdm(urls.items(), desc="Downloading "):

Balkan_MO/md/{en-2008-BMO.md → en-2008-BMO-type1.md}

@@ -35,7 +35,7 @@ $$
 
 Therefore the triangles XFD, OEF are similar and we get:
 $\angle O F X=\angle O F C+\angle L F X=\angle F O E+\angle F X D=\angle X F D+\angle F X D=90^{\circ}$, so $\mathrm{O} F \perp F X$.
-![](https://cdn.mathpix.com/cropped/2024_12_10_5cbb10a7bf0c4c933f1bg-2.jpg?height=892&width=995&top_left_y=356&top_left_x=536)
+![](https://cdn.mathpix.com/cropped/2024_12_12_cb71d810a84e2c573281g-2.jpg?height=892&width=995&top_left_y=356&top_left_x=536)
 
 ## Problem 2
 

Balkan_MO/md/en-2012-BMO-problems.md

@@ -1,18 +0,0 @@
-## Balkan MO 2012
-
-1 \quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ with the line $A C$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$.
-Prove that the circumcircles of triangles $B F E$ and $C F D$ are tangent at $F$.
-
-2 Prove that
-
-$$
-\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
-$$
-
-for all positive real numbers $x, y$ and $z$.
-
-3 Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
-
-4 \quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
-(i) $f(n!)=f(n)$ ! for every positive integer $n$,
-(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.

Balkan_MO/md/{en-2012-BMO-solutions.md → en-2012-BMO-type3.md}

@@ -1,4 +1,7 @@
-## Problem 1.
+## Balkan MO 2012
+
+1. \quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ with the line $A C$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$.
+Prove that the circumcircles of triangles $B F E$ and $C F D$ are tangent at $F$.
 
 Solution. Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=1006&width=675&top_left_y=531&top_left_x=696)
@@ -6,7 +9,13 @@ Solution. Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diamet
 As $\angle C D F=\angle G D K=\angle G A C=\angle G F C, F G$ is tangent to the circumcircle of triangle $C F D$ at $F$. As $\angle F B E=\angle F B G=\angle F A G=\angle G F K=\angle G F E, F G$ is also tangent to the circumcircle of $B F E$ at $F$. Hence the circumcircles of the triangles $C F D$ and $B F E$ are tangent at $F$.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=412&width=417&top_left_y=2350&top_left_x=1462)
 
-## Problem 2.
+2. Prove that
+
+$$
+\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
+$$
+
+for all positive real numbers $x, y$ and $z$.
 
 Solution 1. We will obtain the inequality by adding the inequalities
 
@@ -69,7 +78,8 @@ $$
 to become the Euler inequality $R \geq 2 r$.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-2.jpg?height=418&width=429&top_left_y=2352&top_left_x=1456)
 
-## Problem 3.
+
+3. Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
 
 Solution 1. Let $\alpha=3 / 2$ so $1+\alpha>\alpha^{2}$.
 Given $y$, we construct $Y$ algorithmically. Let $Y=\varnothing$ and of course $S_{\varnothing}=0$. For $i=0$ to $m$, perform the following operation:
@@ -101,7 +111,9 @@ $$
 Therefore $0<\left(x-a^{m+1}\right) \leq 1+a+a^{2}+\cdots+a^{m}$. Again by the induction hypothesis, there exists a subset $X$ of $Q_{m}$ satisfying $0 \leq\left(x-a^{m+1}\right)-S_{X}<1$. Hence $0 \leq x-S_{X^{\prime}}<1$ where $X^{\prime}=X \cup\left\{a^{m+1}\right\} \subset Q_{m+1}$.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-3.jpg?height=418&width=423&top_left_y=2347&top_left_x=1459)
 
-## Problem 4.
+4. \quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
+(i) $f(n!)=f(n)$ ! for every positive integer $n$,
+(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.
 
 Solution 1. There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.
 
@@ -117,4 +129,3 @@ If $f$ is not constant, then there exist positive integers $m, n$ with $\{f(n),
 
 Therefore the functions satisfying the conditions are $f \equiv 1, f \equiv 2, f=\mathrm{id}_{\mathbf{z}^{+}}$.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-4.jpg?height=418&width=427&top_left_y=2347&top_left_x=1454)
-

Balkan_MO/md/{en-2017-BMO.md → en-2017-BMO-type3.md}

@@ -1,35 +1,10 @@
-## The problems
-
 1. Find all ordered pairs $(x, y)$ of positive integers such that:
 
 $$
 x^{3}+y^{3}=x^{2}+42 x y+y^{2} \text {. }
 $$
 
-2. Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ intersects $t_{B}$ at $E$. The circumcircle of triangle $B D C$ meets the side $A C$ at $T$ where $T$ lies between $A$ and $C$. The circumcircle of triangle $B E C$ meets the line $A B$ at $S$ where $B$ lies between $A$ and $S$.
-Prove that the lines $S T, B C$ and $A L$ are concurrent.
-3. Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
-
-$$
-n+f(m) \text { divides } f(n)+n f(m)
-$$
-
-for all $m, n \in \mathbb{N}$.
-4. There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
-(a) Pass one candy to the student on their left or the student on their right.
-(b) Divide all their candies into two, possibly empty, sets and pass one set to the student on their left and the other to the student on their right.
-
-At each step the students perform their chosen operations simultaneously.
-An arrangement of candies is legal if it can be obtained in a finite number of steps.
-Find the number of legal arrangements.
-(Two arrangements are different if there is a student who has different numbers of candies in each one.)
-
-## Solutions
-
-The outline solutions below are not full formal write ups, I have omitted some (hopefully) easy details, and also tried to indicated how the different solutions might have been arrived at naturally.
-
-## Problem 1
-
+Solution.
 Possible initial thoughts about this equation might include:
 (i) I can factorise the sum of cubes on the left.
 (ii) How can I use the 42?
@@ -52,8 +27,10 @@ This leaves a mere 484 cases to check! The modulo 7 magic is not really enough t
 
 Other possible approaches, such as substituting $u=x+y, v=x-y$, seem to lie somewhere between the two described above in terms of the amount of fortitude needed to pull them off.
 
-## Problem 2
+2. Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ intersects $t_{B}$ at $E$. The circumcircle of triangle $B D C$ meets the side $A C$ at $T$ where $T$ lies between $A$ and $C$. The circumcircle of triangle $B E C$ meets the line $A B$ at $S$ where $B$ lies between $A$ and $S$.
+Prove that the lines $S T, B C$ and $A L$ are concurrent.
 
+Solution.
 How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.
 
 We might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also divides $B C$ in this ratio.
@@ -78,7 +55,13 @@ Clearly knowing the standard symmedian configuration and corresponding ratio is
 
 Finally it is worth noting that, to the right sort of mind, the problem screams out for areal coordinates. These turn out to kill it fairly easily, not least because all three circles pass through at least two vertices of $\triangle A B C$.
 
-## Problem 3
+3. Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
+
+$$
+n+f(m) \text { divides } f(n)+n f(m)
+$$
+
+Solution.
 
 The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key idea:
 'Eliminate one of the variables from the right hand side.'
@@ -125,7 +108,17 @@ Similarly, setting $(n, m)=(4,3)$ in $(\star)$ gives $4+9 \mid f(4)-16$ while $n
 
 Now we are ready to try induction. Assume $f(n-1)=(n-1)^{2}$ and use $(\star)$ and $(\dagger)$ to obtain $n+(n-1)^{2} \mid f(n)-n^{2}$ and $n+f(n) \mid n+n^{2}$. The latter implies $f(n) \leq n^{2}$ so the former becomes $n^{2}-n+1 \mid n^{2}-f(n)$. If $f(n) \neq n^{2}$ then $n^{2}-f(n)=1 \times\left(n^{2}-n+1\right)$ since any other multiple would be too large. However, putting $f(n)=n-1$ into $n+f(n) \mid n+n^{2}$ implies $2 n-1 \mid n(1+n)$. This is a contradiction since $2 n-1$ is coprime to $n$ and clearly cannot divide $1+n$.
 
-## Problem 4
+for all $m, n \in \mathbb{N}$.
+4. There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
+(a) Pass one candy to the student on their left or the student on their right.
+(b) Divide all their candies into two, possibly empty, sets and pass one set to the student on their left and the other to the student on their right.
+
+At each step the students perform their chosen operations simultaneously.
+An arrangement of candies is legal if it can be obtained in a finite number of steps.
+Find the number of legal arrangements.
+(Two arrangements are different if there is a student who has different numbers of candies in each one.)
+
+Solution.
 
 One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on the next step, then no caramels move. (This is rather too little movement.) Let us see what a small change to this sequence can accomplish. We choose a student with at least one caramel. At the first step, she passes one caramel to the right and any others she has to the left. Every one else passes everything left. At the next step everybody passes everything right. The effect of this is that exactly one caramel has moved exactly two places to the right. Similarly, there is a double step which moves exactly one caramel to places to the left.
 
@@ -163,4 +156,3 @@ If $n$ is even it is possible to start from any balanced position and reversibly
 
 [^0]
 [^0]:    ${ }^{2}$ The word 'candy' was a little too grating for my delicate British ears. I am grateful to the Italians for suggesting the more elegant alternative.
-

Balkan_MO/md/{en-2018-BMO.md → en-2018-BMO-type3.md}

@@ -1,28 +1,6 @@
-# 35 ${ }^{\text {th }}$ BALKAN MATHEMATICAL OLYMPIAD <br> Belgrade, Serbia (May 9, 2018) 
-
-## Problem 1.
-
-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 the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diameter of the circle $k$.
-
-## Problem 2.
-
-Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
-
-## Problem 3.
-
-Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
-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.
-
-Problem 4.
-Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
-
-Time allowed: 4 hours and 30 minutes.
-Each problem is worth 10 points.
-
-## S O L UTIONS
-
-## Problem 1.
+1. 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 the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diameter of the circle $k$.
 
+Solution.
 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
 
 $$
@@ -34,8 +12,9 @@ Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B
 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 $\varangle A C B=90^{\circ}$.
 ![](https://cdn.mathpix.com/cropped/2024_12_10_a48035b438f3db5360fbg-2.jpg?height=827&width=798&top_left_y=1117&top_left_x=635)
 
-## Problem 2.
+2. Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
 
+Solution 1.
 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
 
@@ -67,7 +46,7 @@ $$
 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.
 
-## Solution 2.
+Solution 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
@@ -84,7 +63,10 @@ $$
 
 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 $a=b=1$.
 
-## Problem 3.
+3. Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
+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.
+
+Solution.
 
 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 $k$.
@@ -99,7 +81,12 @@ Therefore, if the starting position is $k$-happy, after $k$ moves they will get
 
 Therefore a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.
 
-## Problem 4.
+4. Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
+
+Time allowed: 4 hours and 30 minutes.
+Each problem is worth 10 points.
+
+Solution.
 
 Answer: $(p, q)=(3,3)$.
 For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
@@ -120,4 +107,3 @@ $$
 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.
-

Balkan_MO/md/{en-2020-BMO.md → en-2020-BMO-type1.md}

@@ -187,37 +187,3 @@ $$
 $$
 
 so $2 e_{2}(n)-1 \leqslant e_{1}(n)$. Combine this with the inequality in the previous paragraph to write $4 e_{2}(n)-2 \leqslant 2 e_{1}(n) \leqslant 3 e_{2}(n)+7$ and infer that $e_{2}(n) \leqslant 9$. Consequently, $2 e_{1}(n) \leqslant 3 e_{2}(n)+7 \leqslant 34$, showing that $e=17$ is suitable for (1) to hold. This establishes (1) and completes the solution.
-
-Problem 1. Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the line $A C$.
-
-Show that, as $X, Y$ vary on $\omega$, the line $X Y$ passes through a fixed point.
-
-Problem 2. Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
-
-$$
-f(x+f(x)+f(y))=2 f(x)+y
-$$
-
-holds for all $x, y \in(0,+\infty)$.
-
-Problem 3. Let $a, b$ and $c$ be positive integers satisfying the equation
-
-$$
-(a, b)+[a, b]=2021^{c} .
-$$
-
-If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
-Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$, and $[a, b]$ denotes the least common multiple of $a$ and $b$.
-
-Problem 4. Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
-(a) He clears every piece of rubbish from a single pile.
-(b) He clears one piece of rubbish from each pile.
-
-However, every evening, a demon sneaks into the warehouse and performs exactly one of the following moves:
-(a) He adds one piece of rubbish to each non-empty pile.
-(b) He creates a new pile with one piece of rubbish.
-
-What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
-
-Time: 4 hours and 30 minutes
-

Balkan_MO/md/en-2024-BMO-type1.md

@@ -0,0 +1,297 @@
+![](https://cdn.mathpix.com/cropped/2024_12_12_c0c23d4a5db8995f86cag-1.jpg?height=343&width=329&top_left_y=284&top_left_x=280)
+
+## Problems with Solutions
+
+Language: English
+Monday, April 29, 2024
+
+Problem 1. Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle E D G$ and $\triangle F D H$ are tangent to each other.
+
+Solution 1. Let $X$ and $Y$ lie on the tangent to the circumcircle of $\triangle E D G$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $A C>A B$ ensures $E$ lies on extension of $C A$ beyond $A$, and $F$ lies on extension of $A B$ beyond $B$. The condition on $\ell$ means the points lie in the orders $E, A, G, C$ and $A, B, H, F$.
+![](https://cdn.mathpix.com/cropped/2024_12_12_c0c23d4a5db8995f86cag-1.jpg?height=695&width=1188&top_left_y=1273&top_left_x=471)
+
+Using the alternate segment theorem, the condition that $\odot E D G$ and $\odot F D H$ are tangent at $D$ can be rewritten as
+
+$$
+\Varangle H F D=\Varangle Y D H .
+$$
+
+But using the same theorem, we get $\Varangle Y D H=\Varangle X D G=\Varangle D E G$. So we can remove $G, H$ from the figure, and it is sufficient to prove that $\Varangle D E A=\Varangle D F B$.
+
+The reflection property means that $A D$ and $B D$ are external angle bisectors in $\triangle E A B$ and hence $D$
+is the $E$-excentre of this triangle. Thus $D E$ (internally) bisects $\Varangle B E A$, giving
+
+$$
+\Varangle D E A=\Varangle D E B .
+$$
+
+Now observe that the pairs of lines $(B E, C E)$ and $(B F, C F)$ are reflections in $B C$ thus $E, F$ are reflections in $B C$. Also $D$ is its own reflection in $B C$. Hence $\Varangle D E B=\Varangle D F B$ and so
+
+$$
+\Varangle D E A=\Varangle D E B=\Varangle D F B,
+$$
+
+as required.
+
+Problem 2. Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
+(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
+(ii) all the positive $b_{i}$ are distinct,
+(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
+
+Solution 1. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$ and $a_{t}$ be the largest number not belonging to $A n s$. Clearly the set $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ has cardinality $t$; let its members be $c_{1}>c_{2}>\cdots>c_{t}$. Define $b_{j}:=c_{j}-a_{j}$ for $1 \leq j \leq t$ or zero otherwise. Since $\left\{c_{j}\right\}$ is decreasing and $\left\{a_{j}\right\}$ is increasing, all $b_{j}$ are distinct and clearly $b_{1}<n$. After we add $b_{j}$ to $a_{j}$ we get a permutation of Ans as desired.
+
+Solution 2. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$.
+We proceed with the following reduction. Let $\delta$ be the smallest $b$ we used before (in the beginning it is $n$ ). While $a_{1} \notin A n s$ we map $a_{1}$ to the largest element $q$ of $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ and put $\delta_{\text {new }}:=b_{1}:=q-a_{1}$. Now we rearrange the sequence of $a$-s. We do not touch $\operatorname{Ans} \cap\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ so every $b$ is defined at most once (in the end undefined $b$-s become zeros). Also $b<\delta$ and $\delta$ decreases at each step, because $q$ decreases and $a_{1}$ grows, and hence all nonzero $b$-s are distinct.
+
+Problem 3. Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
+
+Solution 1. Obviously we have $a>b$. Let $a=b q+r$, where $0 \leq r<b$. Then
+
+$$
+3^{a} \equiv 3^{b q+r} \equiv(-2)^{q} \cdot 3^{r} \equiv-2 \quad\left(\bmod 3^{b}+2\right)
+$$
+
+So $3^{b}+2$ divides $A=(-2)^{q} .3^{r}+2$ and it follows that
+
+$$
+\left|(-2)^{q} \cdot 3^{r}+2\right| \geq 3^{b}+2 \text { or }(-2)^{q} \cdot 3^{r}+2=0
+$$
+
+We make case distinction:
+
+1. $(-2)^{q} .3^{r}+2=0$. Then $q=1$ and $r=0$ or $a=b$, a contradiction.
+2. $q$ is even. Then
+
+$$
+A=2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) \cdot k
+$$
+
+Consider both sides of the last equation modulo $3^{r}$. Since $b>r$ :
+
+$$
+2 \equiv 2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) k \equiv 2 k \quad\left(\bmod 3^{r}\right)
+$$
+
+so it follows that $3^{r} \mid k-1$. If $k=1$ then $2^{q} .3^{r}=3^{b}$, a contradiction. So $k \geq 3^{r}+1$, and therefore:
+
+$$
+A=2^{q} .3^{r}+2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}+1\right)>3^{b} .3^{r}+2
+$$
+
+It follows that
+
+$$
+2^{q} .3^{r}>3^{b} .3^{r} \text {, i.e. } 2^{q}>3^{b} \text {, which implies } 3^{b^{2}}<2^{b q}<3^{b q} \leq 3^{b q+r}=3^{a} \text {. }
+$$
+
+Consequently $a>b^{2}$.
+3. If $q$ is odd. Then
+
+$$
+2^{q} \cdot 3^{r}-2=\left(3^{b}+2\right) k
+$$
+
+Considering both sides of the last equation modulo $3^{r}$, and since $b>r$, we get: $k+1$ is divisible by $3^{r}$ and therefore $k \geq 3^{r}-1$. Thus $r>0$ because $k>0$, and:
+
+$$
+\begin{array}{r}
+2^{q} .3^{r}-2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}-1\right), \text { and therefore } \\
+2^{q} .3^{r}>\left(3^{b}+2\right)\left(3^{r}-1\right)>3^{b}\left(3^{r}-1\right)>3^{b} \frac{3^{r}}{2}, \text { which shows } \\
+2^{q+1}>3^{b} .
+\end{array}
+$$
+
+But for $q>1$ we have $2^{q+1}<3^{q}$, which combined with the above inequality, implies that $3^{b^{2}}<2^{(q+1) b}<3^{q b} \leq 3^{a}$, q.e.d. Finally, If $q=1$ then $2^{q} .3^{r}-2=\left(3^{b}+2\right) k$ and consequently $2.3^{r}-2 \geq 3^{b}+2 \geq 3^{r+1}+2>2.3^{r}-2$, a contradiction.
+
+Solution 2. $D=a-b$, and we shall show $D>b^{2}-b$. We have $3^{b}+2 \mid 3^{a}+2$, so $3^{b}+2 \mid 3^{D}-1$. Let $D=b q+r$ where $r<b$. First suppose that $r \neq 0$. We have
+
+$$
+1 \equiv 3^{D} \equiv 3^{b q+r} \equiv(-2)^{q+1} 3^{r-b} \quad\left(\bmod 3^{b}+2\right) \Longrightarrow 3^{b-r} \equiv(-2)^{q+1} \quad\left(\bmod 3^{b}+2\right)
+$$
+
+Therefore
+
+$$
+3^{b}+2 \leq\left|(-2)^{q+1}-3^{b-r}\right| \leq 2^{q+1}+3^{b-r} \leq 2^{q+1}+3^{b-1}
+$$
+
+Hence
+
+$$
+2 \times 3^{b-1}+2 \leq 2^{q+1} \Longrightarrow 3^{b-1}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
+$$
+
+Which yields $D=b q+r>b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$ as desired. Now for the case $r=0,(-2)^{q} \equiv 1$ $\left(\bmod 3^{b}+2\right)$ and so
+
+$$
+3^{b}+2 \leq\left|(-2)^{q}-1\right| \leq 2^{q}+1 \Longrightarrow 3^{b-1}<3^{b}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
+$$
+
+and analogous to the previous case, $D=b q+r=b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$.
+
+Problem 4. Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
+
+$$
+f(f(x)+P(y))=f(x-y)+2 y
+$$
+
+for all real numbers $x>y>0$.
+
+Solution 1. Assume that $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial $P$ with non-negative coefficients and $P(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $Q(x, y)$ for the relation:
+
+$$
+f(f(x)+P(y))=f(x-y)+2 y
+$$
+
+1. Step 1. $f(x) \geq x$. Assume that this is not true. Since $P(0)=0$ and $P$ is with non-negative coefficients, $P(x)+x$ is surjective on positive reals. If $f(x)<x$ for some positive real $x$, then setting $y$ such that $y+P(y)=x-f(x)$ (where obviously $y<x$ ), we shall get $f(x)+P(y)=x-y$ and by $Q(x, y), f(f(x)+P(y))=f(x-y)+2 y$, we get $2 y=0$, a contradiction.
+2. Step 2. $P(x)=c x$ for some non-negative real $c$. We will show $\operatorname{deg} P \leq 1$ and together with $P(0)=0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $P(x) \geq 2 x$ for all $x \geq l$. By Step 1 we get
+
+$$
+\forall x>y \geq l: f(x-y)+2 y=f(f(x)+P(y)) \geq f(x)+P(y) \geq f(x)+2 y
+$$
+
+and therefore $f(x-y) \geq f(x)$. We get $f(y) \geq f(2 y) \geq \cdots \geq f(n y) \geq n y$ for all positive integers $n$, which is a contradiction.
+3. Step 3. If $c \neq 0$, then $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$ for $z>1$. Indeed by $Q(f(x+z)+c z, c)$, we get
+
+$$
+f\left(f(f(x+z)+c z)+c^{2}\right)=f(f(x+z)+c z-c)+2 c=f(x+1)+2(z-1)+2 c
+$$
+
+On the other hand by $Q(x+z, z)$, we have:
+
+$$
+f(x)+2 z+c^{2}=f(f(x+z)+P(z))+c^{2}=f(f(x+z)+c z)+c^{2} .
+$$
+
+Substituting in the LHS of $Q(f(x+z)+c z, c)$, we get $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$.
+4. Step 4. There is $x_{0}$, such that $f(x)$ is linear on $\left(x_{0}, \infty\right)$. If $c \neq 0$, then by Step 3 , fixing $x=1$, we get $f\left(f(1)+2 z+c^{2}\right)=f(2)+2(z-1)+2 c$ which implies that $f$ is linear for $z>f(1)+2+c^{2}$. As for the case $c=0$, consider $y, z \in(0, \infty)$. Pick $x>\max (y, z)$, then by $Q(x, x-y)$ and $Q(x, x-z)$ we get:
+
+$$
+f(y)+2(x-y)=f(f(x))=f(z)+2(x-z)
+$$
+
+which proves that $f(y)-2 y=f(z)-2 z$ and there fore $f$ is linear on $(0, \infty)$.
+5. Step 5. $P(y)=y$ and $f(x)=x$ on $\left(x_{0}, \infty\right)$. By Step 4 , let $f(x)=a x+b$ on $\left(x_{0}, \infty\right)$. Since $f$ takes only positive values, $a \geq 0$. If $a=0$, then by $Q(x+y, y)$ for $y>x_{0}$ we get:
+
+$$
+2 y+f(x)=f(f(x+y)+P(y))=f(b+c y)
+$$
+
+Since the LHS is not constant, we conclude $c \neq 0$, but then for $y>x_{0} / c$, we get that the RHS equals $b$ which is a contradiction.
+
+Hence $a>0$. Now for $x>x_{0}$ and $x>\left(x_{0}-b\right) / a$ large enough by $P(x+y, y)$ we get:
+$a x+b+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(a x+a y+b+c y)=a(a x+a y+b+c y)+b$.
+Comparing the coefficients before $x$, we see $a^{2}=a$ and since $a \neq 0, a=1$. Now $2 b=b$ and thus $b=0$. Finally, equalising the coefficients before $y$, we conclude $2=1+c$ and therefore $c=1$.
+
+Now we know that $f(x)=x$ on $\left(x_{0}, \infty\right)$ and $P(y)=y$. Let $y>x_{0}$. Then by $Q(x+y, x)$ we conclude:
+
+$$
+f(x)+2 y=f(f(x+y)+P(y))=f(x+y+y)=x+2 y .
+$$
+
+Therefore $f(x)=x$ for every $x$. Conversely, it is straightforward that $f(x)=x$ and $P(y)=y$ do indeed satisfy the conditions of the problem.
+
+Solution 2. Assume that the function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial with non-negative coefficients $P(y)=y P_{1}(y)$ satisfy the given equation. Fix $x=x_{0}>0$ and note that:
+
+$$
+f\left(f\left(x_{0}+y\right)+P(y)\right)=f\left(x_{0}+y-y\right)+2 y=f\left(x_{0}\right)+2 y
+$$
+
+Assume that $g=0$. Then $f(f(x+y))=f(x)+2 y$ for $x, y>0$. Let $x>0$ and $z>0$. Pick $y>0$. Then:
+
+$$
+2 y+f(x+z)=f(f(x+y+z))=f(f(x+z+y))=f(x)+2(z+y) .
+$$
+
+Therefore $f(x+z)=f(x)+2 z$ for any $x>0$ and $z>0$. Setting $c=f(1)$, we see that $f(z+1)=c+2 z$ for all positive $z$. Therefore if $x, y>1$ we have that $f(x+y)=c+2(x+y-1)>1$. This shows that:
+
+$$
+f(f(x+y))=c+2(f(x+y)-1)=3 c+4(x+y)-4
+$$
+
+On the other hand $f(x)+2 y=c+2 x+2 y$. Therefore the equality $f(f(x+y))=f(x)+2 y$ is not universally satisfied.
+
+From now on, we assume that $g \neq 0$. Therefore $P$ is strictly increasing with $P(0)=0, \lim _{y \rightarrow \infty} P(y)=$ $\infty$, i.e. $g$ is bijective on $[0, \infty)$ and $P(0)=0$.
+
+Let $x>0, y>0$ and set $u=f(x+y), v=P(y)$. From above, we have $u>0$ and $v>0$. Therefore:
+
+$$
+f(f(u+v)+P(v))=f(u)+2 v=f(f(x+y))+2 P(y)
+$$
+
+On the other hand $f(u+v)=f(f(x+y)+P(y))=f(x)+2 y$. Therefore we obtain that:
+
+$$
+f(f(x)+2 y+P(P(y)))=f(f(x+y))+2 P(y)
+$$
+
+Since $g$ is bijective from $(0, \infty)$ to $(0, \infty)$ for any $z>0$ there is $t$ such that $P(t)=z$. Applying this observation to $z=P(P(y))+2 y$ and setting $x^{\prime}=x+t$, we obtain that:
+$f(f(x+t+y))+2 P(y)=f\left(f\left(x^{\prime}+y\right)\right)+2 P(y)=f\left(f\left(x^{\prime}\right)+P(P(y))+2 y\right)=f(f(x+t)+P(t))=f(x)+2 t$.
+
+Thus if we denote $h(y)=P(P(y))+2 y$, then $t=P^{(-1)}(h(y))$ and the above equality can be rewritten as:
+$f\left(f\left(x+P^{(-1)}(h(y))+y\right)\right)=f(x)+2 P^{(-1)}(h(y))-2 P(y)=f(x)+2 P^{(-1)}(h(y))+2 y-2 y-2 P(y)$.
+
+Let $s(y)=P^{(-1)}(h(y))+y$ and note that since $h$ is continuous and monotone increasing, $g$ is continuous and monotone increasing, then so are $P^{(-1)}$ and consequently $P^{(-1)} \circ h$ and $s$. It is also clear, that $\lim _{y \rightarrow 0} s(y)=0$ and $\lim _{y \rightarrow \infty} s(y)=\infty$. Therefore $s$ is continuously bijective from $[0, \infty)$ to $[0, \infty)$ with $s(0)=0$.
+
+Thus we have:
+
+$$
+f(f(x+s(y)))=f(x)+2 s(y)-2 y-2 P(y)
+$$
+
+and using that $s$ is invertible, we obtain:
+
+$$
+f(f(x+y))=f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right)
+$$
+
+Now fix $x_{0}$, then for any $x>x_{0}$ and any $y>0$ we have:
+
+$$
+\begin{aligned}
+f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right) & =f(f(x+y))=f\left(f\left(x_{0}+x+y-x_{0}\right)\right) \\
+& =f\left(x_{0}\right)+2\left(x+y-x_{0}\right)-2 s^{(-1)}\left(x+y-x_{0}\right)-2 P\left(s^{(-1)}\left(x+y-x_{0}\right)\right) .
+\end{aligned}
+$$
+
+Setting $y=x_{0}$, we get:
+
+$$
+f(x)+2 x_{0}-2 s^{(-1)}\left(x_{0}\right)-2 P\left(s^{(-1)}\left(x_{0}\right)\right)=f\left(x_{0}\right)+2 x-2 s^{(-1)}(x)-2 P\left(s^{(-1)}(x)\right) .
+$$
+
+Since this equality is valid for any $x>x_{0}$ we actually have that:
+
+$$
+f(x)-2 x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)=c \text { for some fixed constant } c \in \mathbb{R} \text { and all } x \in \mathbb{R}^{+} .
+$$
+
+Let $\phi(x)=-x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)$. Then $f(x)=x-\phi(x)+c$ and since $\phi$ is a sum of continuous functions that are continuous at 0 . Therefore $f$ is continuous and can be extended to a continuous function on $[0, \infty)$. Back in the original equation we fix $x>0$ and let $y$ tend to 0 . Using the continuity of $f$ and $g$ on $[0, \infty)$ and $P(0)=0$ we obtain:
+
+$$
+f(f(x))=\lim _{y \rightarrow 0+} f(f(x)+P(y))=\lim _{y \rightarrow 0+}(f(x-y)+P(y))=f(x)+P(0)=f(x) .
+$$
+
+Finally, fixing $x=1$ and varying $y>0$, we obtain:
+
+$$
+f(f(1+y)+P(y))=f(1)+2 y
+$$
+
+It follows that $f$ takes every value on $(f(1), \infty)$. Therefore for any $y \in(f(1), \infty)$ there is $z$ such that $f(z)=y$. Using that $f(f(z))=f(z)$ we conclude that $f(y)=y$ for all $y \in(f(1), \infty)$.
+
+Now fix $x$ and take $y>f(1)$. Hence
+
+$$
+f(x)+2 y=f(f(x+y)+P(y))=f(x+y+P(y))=x+y+P(y)
+$$
+
+We conclude $f(x)-x=P(y)-y$ for every $x$ an $y>f(1)$. In particular $f\left(x_{1}\right)-x_{1}=f\left(x_{2}\right)-x_{2}$ for all $x_{1}, x_{2} \in(0, \infty)$ and since $f(x)=x$ for $x \in(f(1), \infty)$, we get $f(x)=x$ on $(0, \infty)$.
+Finally, $x+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(x+y)+P(y)=x+y+P(y)$, which shows that $P(y)=y$ for every $y \in(0, \infty)$.
+
+It is also straightforward to check that $f(x)=x$ and $P(y)=y$ satisfy the equality:
+
+$$
+f(f(x+y)+P(y))=f(x+2 y)=x+2 y=f(x)+2 y
+$$
+

Balkan_MO/md/en-2024-BMO.md

@@ -1,30 +0,0 @@
-![](https://cdn.mathpix.com/cropped/2024_12_10_8049efac594e6bfa5b3ag-1.jpg?height=340&width=337&top_left_y=285&top_left_x=276)
-
-41st Balkan Mathematical Olympiad
-27th April - 2nd May 2024
-Varna, Bulgaria
-
-## Language: English
-
-Monday, April 29, 2024
-
-Problem 1. Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle E D G$ and $\triangle F D H$ are tangent to each other.
-
-Problem 2. Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
-(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
-(ii) all the positive $b_{i}$ are distinct,
-(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
-
-Problem 3. Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
-
-Problem 4. Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
-
-$$
-f(f(x)+P(y))=f(x-y)+2 y
-$$
-
-for all real numbers $x>y>0$.
-
-Time is 4 hours and 30 minutes
-Each problem is worth 10 points
-

Balkan_MO/raw/{en-2020-BMO.pdf → en-2020-BMO-type1.pdf}

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+size 565393

Balkan_MO/raw/{en-2024-BMO.pdf → en-2024-BMO-type1.pdf}

@@ -1,3 +1,3 @@
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+size 320507

Balkan_MO/segment_script/segment_type1 .py

@@ -0,0 +1,121 @@
+import re
+import json
+
+from tqdm import tqdm
+from loguru import logger
+
+from pathlib import Path
+from typing import Tuple, List
+
+
+problem_tag = 'Problem'
+solution_tag = 'Solution'
+
+
+def analyze(text: str) -> Tuple[List, int]:
+    """
+    Analyze the text and return the tags and problem number.
+
+    Args:
+        text (str): The markdown text to analyze.
+
+    Returns:
+        Tuple[List, int]: A tuple containing the tags and problem number.
+    """
+    problem_pattern = re.compile(r'(?:\n|# |## 2020 BMO\, )Problem\s+(\d+)(.)?', re.IGNORECASE)
+    solution_pattern = re.compile(r'(?:\n|# )Solution(?:\s+(\d+)|\.|\n|\:)', re.IGNORECASE)
+
+    tags = []
+    tags.extend([(x, problem_tag) for x in problem_pattern.finditer(text)])
+    problem_num = len(tags)
+
+    tags.extend([(x, solution_tag) for x in solution_pattern.finditer(text)])
+    tags.sort(key=lambda x: x[0].start())
+    return tags, problem_num
+
+
+def segment(text: str, tags):
+    starts = []
+    ends = []
+
+    for i in range(len(tags)):
+        starts.append(tags[i][0].end())
+        if i + 1 < len(tags):
+            ends.append(tags[i + 1][0].start())
+        else:
+            ends.append(len(text))
+
+    return [text[start:end].strip().strip('#').strip() for start, end in zip(starts, ends)]
+
+
+def join(tags, segments):
+    problem, solution = '', ''
+    problem_label, problem_match, solution_match = '', '', ''
+    pairs = []
+
+    for tag, segment in zip(tags, segments):
+        if tag[1] == problem_tag:
+            problem = segment
+            problem_match = tag[0].group(0)
+            problem_label = tag[0].group(1)
+        else:
+            solution = segment
+            solution_match = tag[0].group(0)
+            pairs.append((problem, solution, problem_label, problem_match, solution_match))
+
+    return pairs
+
+
+def write_pairs(output_file: Path, pairs):
+    year = re.search(r'(\d{4})', output_file.stem).group(1)
+
+    output_jsonl_text = ""
+    for problem, solution, problem_label, problem_match, solution_match in pairs:
+        output_jsonl_text += json.dumps(
+            {
+                'year': year,
+                'problem_label': problem_label,
+                'tier': 1,
+                'problem': problem,
+                'solution': solution,
+                'problem_match': problem_match,
+                'solution_match': solution_match
+            },
+            ensure_ascii=False
+        ) + '\n'
+
+    output_file.write_text(output_jsonl_text, encoding="utf-8")
+
+
+def main():
+    compet_base_path = Path(__file__).resolve().parent.parent
+    compet_md_path = compet_base_path / "md"
+    seg_output_path = compet_base_path / "segmented"
+
+    total_problem_count = 0
+    total_solution_count = 0
+
+    for apmo_md in tqdm(list(compet_md_path.glob('**/*type1.md')), desc='Segmenting'):
+        output_file = seg_output_path / apmo_md.relative_to(compet_md_path).with_suffix('.jsonl')
+        output_file.parent.mkdir(parents=True, exist_ok=True)
+
+        text = '\n' + apmo_md.read_text(encoding="utf-8")
+
+        tags, problem_num = analyze(text)
+
+        segments = segment(text, tags)
+        pairs = join(tags, segments)     
+        if pairs and problem_num > 0:
+            write_pairs(output_file, pairs)
+
+            total_problem_count += problem_num
+            total_solution_count += len(pairs)
+        else:
+            logger.warning(f"No problem found in {apmo_md}")
+        
+    logger.info(f"Total problem count: {total_problem_count}")
+    logger.info(f"Total solution count: {total_solution_count}")
+
+
+if __name__ == '__main__':
+    main()

Balkan_MO/segment_script/segment_type2.py

@@ -0,0 +1,130 @@
+import re
+import json
+
+from tqdm import tqdm
+from loguru import logger
+
+from pathlib import Path
+from typing import Tuple, List
+
+
+problem_tag = 'Problem'
+solution_tag = 'Solution'
+
+
+def analyze(text: str) -> Tuple[List, int]:
+    """
+    Analyze the text and return the tags and problem number.
+
+    Args:
+        text (str): The markdown text to analyze.
+
+    Returns:
+        Tuple[List, int]: A tuple containing the tags and problem number.
+    """
+    separator_pattern = re.compile(r"\n\#*\s*Solutions\n", re.IGNORECASE)
+    separator_match = separator_pattern.search(text)
+
+    section_pattern = re.compile(r"\n(\d+)\.")
+
+    tags = []
+    tags.extend(
+        [
+            (x, solution_tag if x.start() > separator_match.end() else problem_tag)
+            for x in section_pattern.finditer(text)
+        ]
+    )
+
+    problem_num = len([_ for _ in tags if _[-1] == problem_tag])
+    tags.sort(key=lambda x: x[0].start())
+    return tags, problem_num
+
+
+def segment(text: str, tags):
+    starts = []
+    ends = []
+
+    for i in range(len(tags)):
+        starts.append(tags[i][0].end())
+        if i + 1 < len(tags):
+            ends.append(tags[i + 1][0].start())
+        else:
+            ends.append(len(text))
+
+    return [text[start:end].strip().strip('#').strip() for start, end in zip(starts, ends)]
+
+
+def join(tags, segments):
+    pairs = []
+    problems = {}
+    solutions = {}
+
+    for tag, segment in zip(tags, segments):
+        if tag[1] == "Problem":
+            problems[tag[0].group(1)] = (tag, segment)
+        else:
+            solutions[tag[0].group(1)] = (tag, segment)
+
+    for problem_label, (problem_tag, problem) in problems.items():
+        for solution_label, (solution_tag, solution) in solutions.items():
+            if problem_label == solution_label:
+                break
+
+        pairs.append((problem, solution, problem_label, problem_tag[0].group(0), solution_tag[0].group(0)))
+
+    return pairs
+
+
+def write_pairs(output_file: Path, pairs):
+    year = re.search(r'(\d{4})', output_file.stem).group(1)
+
+    output_jsonl_text = ""
+    for problem, solution, problem_label, problem_match, solution_match in pairs:
+        output_jsonl_text += json.dumps(
+            {
+                'year': year,
+                'problem_label': problem_label,
+                'tier': 1,
+                'problem': problem,
+                'solution': solution,
+                'problem_match': problem_match,
+                'solution_match': solution_match
+            },
+            ensure_ascii=False
+        ) + '\n'
+
+    output_file.write_text(output_jsonl_text, encoding="utf-8")
+
+
+def main():
+    compet_base_path = Path(__file__).resolve().parent.parent
+    compet_md_path = compet_base_path / "md"
+    seg_output_path = compet_base_path / "segmented"
+
+    total_problem_count = 0
+    total_solution_count = 0
+
+    for apmo_md in tqdm(list(compet_md_path.glob('**/*type2.md')), desc='Segmenting'):
+        output_file = seg_output_path / apmo_md.relative_to(compet_md_path).with_suffix('.jsonl')
+        output_file.parent.mkdir(parents=True, exist_ok=True)
+
+        text = '\n' + apmo_md.read_text(encoding="utf-8")
+
+        tags, problem_num = analyze(text)
+
+        segments = segment(text, tags)
+        pairs = join(tags, segments)     
+        if pairs and problem_num > 0:
+            write_pairs(output_file, pairs)
+
+            total_problem_count += problem_num
+            total_solution_count += len(pairs)
+        else:
+            logger.warning(f"No problem found in {apmo_md}")
+        
+    logger.info(f"Total problem count: {total_problem_count}")
+    logger.info(f"Total solution count: {total_solution_count}")
+
+
+if __name__ == '__main__':
+    main()

Balkan_MO/segment_script/segment_type3.py

@@ -0,0 +1,121 @@
+import re
+import json
+
+from tqdm import tqdm
+from loguru import logger
+
+from pathlib import Path
+from typing import Tuple, List
+
+
+problem_tag = 'Problem'
+solution_tag = 'Solution'
+
+
+def analyze(text: str) -> Tuple[List, int]:
+    """
+    Analyze the text and return the tags and problem number.
+
+    Args:
+        text (str): The markdown text to analyze.
+
+    Returns:
+        Tuple[List, int]: A tuple containing the tags and problem number.
+    """
+    problem_pattern = re.compile(r'\n(\d+)\.\s', re.IGNORECASE)
+    solution_pattern = re.compile(r'\nSolution\s*\d*\.', re.IGNORECASE)
+
+    tags = []
+    tags.extend([(x, problem_tag) for x in problem_pattern.finditer(text)])
+    problem_num = len(tags)
+
+    tags.extend([(x, solution_tag) for x in solution_pattern.finditer(text)])
+    tags.sort(key=lambda x: x[0].start())
+    return tags, problem_num
+
+
+def segment(text: str, tags):
+    starts = []
+    ends = []
+
+    for i in range(len(tags)):
+        starts.append(tags[i][0].end())
+        if i + 1 < len(tags):
+            ends.append(tags[i + 1][0].start())
+        else:
+            ends.append(len(text))
+
+    return [text[start:end].strip().strip('#').strip() for start, end in zip(starts, ends)]
+
+
+def join(tags, segments):
+    problem, solution = '', ''
+    problem_label, problem_match, solution_match = '', '', ''
+    pairs = []
+
+    for tag, segment in zip(tags, segments):
+        if tag[1] == problem_tag:
+            problem = segment
+            problem_match = tag[0].group(0)
+            problem_label = tag[0].group(1)
+        else:
+            solution = segment
+            solution_match = tag[0].group(0)
+            pairs.append((problem, solution, problem_label, problem_match, solution_match))
+
+    return pairs
+
+
+def write_pairs(output_file: Path, pairs):
+    year = re.search(r'(\d{4})', output_file.stem).group(1)
+
+    output_jsonl_text = ""
+    for problem, solution, problem_label, problem_match, solution_match in pairs:
+        output_jsonl_text += json.dumps(
+            {
+                'year': year,
+                'problem_label': problem_label,
+                'tier': 1,
+                'problem': problem,
+                'solution': solution,
+                'problem_match': problem_match,
+                'solution_match': solution_match
+            },
+            ensure_ascii=False
+        ) + '\n'
+
+    output_file.write_text(output_jsonl_text, encoding="utf-8")
+
+
+def main():
+    compet_base_path = Path(__file__).resolve().parent.parent
+    compet_md_path = compet_base_path / "md"
+    seg_output_path = compet_base_path / "segmented"
+
+    total_problem_count = 0
+    total_solution_count = 0
+
+    for apmo_md in tqdm(list(compet_md_path.glob('**/*type3.md')), desc='Segmenting'):
+        output_file = seg_output_path / apmo_md.relative_to(compet_md_path).with_suffix('.jsonl')
+        output_file.parent.mkdir(parents=True, exist_ok=True)
+
+        text = '\n' + apmo_md.read_text(encoding="utf-8")
+
+        tags, problem_num = analyze(text)
+
+        segments = segment(text, tags)
+        pairs = join(tags, segments)     
+        if pairs and problem_num > 0:
+            write_pairs(output_file, pairs)
+
+            total_problem_count += problem_num
+            total_solution_count += len(pairs)
+        else:
+            logger.warning(f"No problem found in {apmo_md}")
+        
+    logger.info(f"Total problem count: {total_problem_count}")
+    logger.info(f"Total solution count: {total_solution_count}")
+
+
+if __name__ == '__main__':
+    main()

Balkan_MO/segmented/en-2008-BMO-type1.jsonl

@@ -0,0 +1,4 @@
+{"year": "2008", "problem_label": "1", "tier": 1, "problem": "An acute-angled scalene triangle $A B C$ is given, with $A C>B C$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$ ) on the line $A B$ such that $A F=P F$, and $M$ be the midpoint of $A C$. We denote the intersection of $P H$ and $B C$ by $X$, the intersection of $O M$ and $F X$ by $Y$, and the intersection of $O F$ and $A C$ by $Z$. Prove that the points $F, M, Y$ and $Z$ are concyclic.", "solution": "It is enough to show that $\\mathrm{O} F \\perp F X$.\nLet $\\mathrm{OE} \\perp \\mathrm{AB}$, then it is trivial that :\n\n$$\nC \\mathrm{H}=2 \\mathrm{OE} .\n$$\n\nSince from the hypothesis we have $\\mathrm{P} F=\\mathrm{A} F$ then we take $\\mathrm{PB}=\\mathrm{P} F-\\mathrm{B} F$ or\n\n$$\n\\mathrm{PB}=\\mathrm{A} F-\\mathrm{B} F\n$$\n\nAlso, $\\angle X P B=\\angle H A P$ and $\\angle H A P=\\angle H C X$ since AFGC in inscribable (where G is the foot of the altidude from A),\nso $\\angle X P B=\\angle H C X$ and since $\\angle B X P=\\angle H X C$, the triangles XHC and XBP are similar.\nIf XL and XD are respectively the heights of the triangles XHC and XBP we have:\n\n$$\n\\frac{X D}{X L}=\\frac{P B}{C H},\n$$\n\nand from (1) and (2) we get:\n\n$$\n\\frac{X D}{X L}=\\frac{A F-B F}{2 O E}=\\frac{F E}{O E} \\Rightarrow \\frac{X D}{F D}=\\frac{F E}{O E}\n$$\n\nTherefore the triangles XFD, OEF are similar and we get:\n$\\angle O F X=\\angle O F C+\\angle L F X=\\angle F O E+\\angle F X D=\\angle X F D+\\angle F X D=90^{\\circ}$, so $\\mathrm{O} F \\perp F X$.\n![](https://cdn.mathpix.com/cropped/2024_12_12_cb71d810a84e2c573281g-2.jpg?height=892&width=995&top_left_y=356&top_left_x=536)", "problem_match": "# Problem 1", "solution_match": "# Solution:"}
+{"year": "2008", "problem_label": "2", "tier": 1, "problem": "Does there exist a sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ of positive real numbers satisfying both of the following conditions:\n(i) $\\sum_{i=1}^{n} a_{i} \\leq n^{2}$, for every positive integer $n$;\n(ii) $\\sum_{i=1}^{n} \\frac{1}{a_{i}} \\leq 2008$, for every positive integer $n$ ?", "solution": "The answer is no.\nIt is enough to show that\nif $\\sum_{i=1}^{n} a_{i} \\leq n^{2}$ for any $n$, then $\\sum_{i=2}^{2^{n}} \\frac{1}{a_{i}}>\\frac{n}{4}$. (or any other precise estimate)\nFor this, we use that $\\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \\sum_{i=2^{k}+1}^{2^{k+1}} \\frac{1}{a_{i}} \\geq 2^{2 k}$ for any $k \\geq 0$ by the arithmetic-harmonic mean inequality.\nSince $\\sum_{i=2^{k}+1}^{2^{k+1}} a_{i}<\\sum_{i=1}^{2^{k+1}} a_{i} \\leq 2^{2 k+2}$, it follows that $\\sum_{i=2^{k}+1}^{2^{k+1}} \\frac{1}{a_{i}}>\\frac{1}{4}$ and hence $\\sum_{i=2}^{2^{n}} \\frac{1}{a^{i}}>\\sum_{k=0}^{n-1} \\sum_{i=2^{k}+1}^{2^{k+1}} \\frac{1}{a_{i}}>\\frac{n}{4}$. (it can be stated in words)\n\n## Remark: no points for using some inequality, that doesn't lead to solution", "problem_match": "# Problem 2", "solution_match": "# Solution."}
+{"year": "2008", "problem_label": "3", "tier": 1, "problem": "Let $n$ be a positive integer. The rectangle $A B C D$ with side lengths $A B=90 n+1$ and $B C=90 n+5$ is partitioned into unit squares with sides parallel to the sides of $A B C D$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points from $S$ is divisible by 4.", "solution": "Denote $90 n+1=m$. We investigate the number of the lines modulo 4 consecutively reducing different types of lines.\nThe vertical and horizontal lines are\n$(m+5)+(m+1)=2(m+3)$ which is divisible to 4.\nMoreover, every line which makes an acute angle to the axe $O x$ (i.e. that line has a positive angular coefficient) corresponds to unique line with an obtuse angle (consider the symmetry with respect to the line through the midpoints of $A B$ and $C D$ ). Therefore it is enough to prove that the lines with acute angles are an even number.\nEvery line which does not pass through the center $O$ of the rectangle corresponds to another line with the same angular coefficent(consider the symmetry with respect to $O$ ). Therefore it is enough to consider the lines through $O$.\nEvery line through $O$ has an angular coefficient $\\frac{p}{q}$, where $(p, q)=1, p$ and $q$ are odd positive integers. (To see this, consider the two nearest, from the two sides, to $O$ points of the line).\nIf $p \\neq 1, q \\neq 1, \\quad p \\leq m$ and $q \\leq m$, the line with angular coefficient $\\frac{p}{q}$, uniquely corresponds to the line with angular coefficient $\\frac{q}{p}$. It remains to prove that the number of the remaining lines is even.\n\nThe last number is\n\n$$\n1+\\frac{\\varphi(m+2)}{2}+\\frac{\\varphi(m+4)}{2}-1=\\frac{\\varphi(m+2)+\\varphi(m+4)}{2}\n$$\n\nbecause we have:\n\n1) one line with $p=q=1$;\n2) $\\frac{\\varphi(m+2)}{2}$ lines with angular coefficient $\\frac{p}{m+2}, p \\leq m$ is odd and $(p, m+2)=1$;\n3) $\\frac{\\varphi(m+4)}{2}-1$ lines with angular coefficient $\\frac{p}{m+4}, p \\leq m$ is odd and $(p, m+4)=1$.\nNow the assertion follows from the fact that the number $\\varphi(m+2)+\\varphi(m+4)=\\varphi(90 n+3)+\\varphi(90 n+5)$ is divisible to 4.", "problem_match": "# Problem 3", "solution_match": "# Solution."}
+{"year": "2008", "problem_label": "4", "tier": 1, "problem": "Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \\ldots, a_{n}, \\ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \\geq 1$ and $m \\geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\\text {th }}$ power of some positive integer.", "solution": "First, notice:\n\n$$\na_{n+1}^{2}+c^{3}=\\left(a_{n}^{2}+a_{n}+c^{3}\\right)^{2}+c^{3}=\\left(a_{n}^{2}+c^{3}\\right)\\left(a_{n}^{2}+2 a_{n}+1+c^{3}\\right)\n$$\n\nWe first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.\nWe prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \\geq 1$.\nLet $n=1$ and $p$ be a prime divisor of $4 c^{3}+1$ and $2 a_{1}+1=2 c+1$. Then $p$ divides $2\\left(4 c^{3}+1\\right)=(2 c+1)\\left(4 c^{2}-2 c+1\\right)+1$, hence $p$ divides 1 , a contradiction. Assume now that $\\left(4 c^{3}+1,2 a_{n}+1\\right)=1$ for some $n \\geq 1$ and the prime $p$ divides $4 c^{3}+1$ and $2 a_{n+1}+1$. Then $p$ divides $4 a_{n+1}+2=\\left(2 a_{n}+1\\right)^{2}+4 c^{3}+1$, which gives a contradiction.\n\nAssume that for some $n \\geq 1$ the number\n\n$$\na_{n+1}^{2}+c^{3}=\\left(a_{n}^{2}+a_{n}+c^{3}\\right)^{2}+c^{3}=\\left(a_{n}^{2}+c^{3}\\right)\\left(a_{n}^{2}+2 a_{n}+1+c^{3}\\right)\n$$\n\nis a power. Since $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime, than $a_{n}^{2}+c^{3}$ is a power as well.\nThe same argument can be further applied giving that $a_{1}^{2}+c^{3}=c^{2}+c^{3}=c^{2}(c+1)$ is a power.\nIf $a^{2}(a+1)=t^{m}$ with odd $m \\geq 3$, then $a=t_{1}^{m}$ and $a+1=t_{2}^{m}$, which is impossible. If $a^{2}(a+1)=t^{2 m_{1}}$ with $m_{1} \\geq 2$, then $a=t_{1}^{m_{1}}$ and $a+1=t_{2}^{m_{1}}$, which is impossible.\n\nTherefore $a^{2}(a+1)=t^{2}$ whence we obtain the solutions $a=s^{2}-1, s \\geq 2, s \\in \\mathbb{N}$.", "problem_match": "# Problem 4", "solution_match": "# Solution."}

Balkan_MO/segmented/en-2009-BMO-type2.jsonl

@@ -0,0 +1,4 @@
+{"year": "2009", "problem_label": "1", "tier": 1, "problem": "We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \\equiv(-1)^{x}-1(\\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\\left(3^{t}-z\\right)\\left(3^{t}+z\\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonnegative integer $k$. Since $5^{k}+5^{y-k}=2 \\cdot 3^{t}$ is not divisible by 5 , it follows that $k=0$ and\n\n$$\n2 \\cdot 3^{t}=5^{y}+1\n$$\n\nSuppose that $t \\geq 2$. Then $5^{y}+1$ is divisible by 9 , which is only possible if $y \\equiv 3$ $(\\bmod 6)$. However, in this case $5^{y}+1 \\equiv 5^{3}+1 \\equiv 0(\\bmod 7)$, so $5^{y}+1$ is also divisible by 7 , which is impossible.\nTherefore we must have $t \\leq 1$, which yields a (unique) solution $(x, y, z)=(2,1,2)$.", "solution": "We start by observing that $f$ is injective. From the known identity\n\n$$\n\\left(a^{2}+2 b^{2}\\right)\\left(c^{2}+2 d^{2}\\right)=(a c \\pm 2 b d)^{2}+2(a d \\mp b c)^{2}\n$$\n\nwe obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \\geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to\n\n$$\ng(n)=A(-1)^{n}+B+C n+D n^{2} .\n$$\n\nSubstituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\\dagger)$ gives us\n\n$$\n\\begin{aligned}\nL=9 n^{4} & =A(-1)^{3\\left(A(-1)^{n}+B+C n+D n^{2}\\right)}+B+\\underbrace{3 C\\left[A(-1)^{n}+B+C n+D n^{2}\\right]} \\\\\n& +9 D\\left[A(-1)^{n}+B+C n+D n^{2}\\right]^{2}=R .\n\\end{aligned}\n$$\n\nSince $0=\\lim _{n \\rightarrow \\infty} \\frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \\neq 0$ ); similarly, $0=\\lim _{n \\rightarrow \\infty} \\frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$\nrespectively we obtain $0=\\lim _{k \\rightarrow \\infty} \\frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\\lim _{k \\rightarrow \\infty} \\frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.\nFinally, $g(n)=D n^{2}, D^{3}=1$ and $g: \\mathbb{N} \\rightarrow \\mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.\n\nRemark. Using limits can be avoided. Since the rigth-hand side in ( $\\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.", "problem_match": "\n1.", "solution_match": "\n4."}
+{"year": "2009", "problem_label": "2", "tier": 1, "problem": "In a triangle $A B C$, points $M$ and $N$ on the sides $A B$ and $A C$ respectively are such that $M N \\| B C$. Let $B N$ and $C M$ intersect at point $P$. The circumcircles of triangles $B M P$ and $C N P$ intersect at two distinct points $P$ and $Q$. Prove that $\\angle B A Q=\\angle C A P$.\n(Moldova)", "solution": "Since the quadrilaterals $B M P Q$ and $C N P Q$ are cyclic, we have $\\angle B Q N=\\angle B Q P+$ $\\angle P Q N=\\angle A M C+\\angle M C A=180^{\\circ}-$ $\\angle C A B$, so $A B Q N$ is cyclic as well. Hence $\\frac{\\sin \\angle B A Q}{\\sin \\angle N A Q}=\\frac{B Q}{N Q}$. Moreover, triangles $M B Q$ and $C N Q$ are similar, so\n\n$$\n\\frac{\\sin \\angle B A Q}{\\sin \\angle C A Q}=\\frac{B Q}{N Q}=\\frac{B M}{C N}=\\frac{A B}{A C}\n$$\n\nOn the other hand, if $A P$ meets $B C$ at $A_{1}$, then by the Cheva theorem $\\frac{B A_{1}}{A_{1} C}=$\n![](https://cdn.mathpix.com/cropped/2024_12_10_458e228b03e5ba87f05cg-2.jpg?height=464&width=475&top_left_y=1118&top_left_x=1096)\n$\\frac{B M}{M A} \\cdot \\frac{A N}{N C}=1$, so $A_{1}$ is the midpoint of $B C$ and\n\n$$\n\\frac{\\sin \\angle C A P}{\\sin \\angle B A P}=\\frac{A B}{A C} \\cdot \\frac{A C \\cdot A A_{1} \\sin \\angle C A P}{A B \\cdot A A_{1} \\sin \\angle B A P}=\\frac{A B}{A C} \\cdot \\frac{S_{\\triangle C A A_{1}}}{S_{\\triangle B A A_{1}}}=\\frac{A B}{A C}\n$$\n\nTherefore, if we denote $\\angle C A P=\\varphi, \\angle B A Q=\\psi$ and $\\angle B A C=\\alpha$, we have $\\frac{\\sin \\psi}{\\sin (\\alpha-\\psi)}=\\frac{\\sin \\varphi}{\\sin (\\alpha-\\varphi)}$, which is equivalent to $\\sin \\psi \\sin (\\alpha-\\varphi)=\\sin \\varphi \\sin (\\alpha-\\psi)$. The addition formulas reduce the last equality to $0=\\sin \\alpha(\\sin \\varphi \\cos \\psi-\\sin \\psi \\cos \\varphi)=$ $\\sin \\alpha \\sin (\\varphi-\\psi)$, from which we conclude that $\\psi=\\varphi$, as desired.", "problem_match": "\n2.", "solution_match": "\n2."}
+{"year": "2009", "problem_label": "3", "tier": 1, "problem": "A $9 \\times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \\ldots, C_{96}$ so that the following two conditions are fulfilled:\n$1^{\\circ}$ All segments $C_{1} C_{2}, C_{2} C_{3}, \\ldots C_{95} C_{96}, C_{96} C_{1}$ have the length $\\sqrt{13}$;\n$2^{\\circ}$ The poligonal line $C_{1} C_{2} \\ldots C_{96} C_{1}$ is centrally symmetric?\n(Bulgaria)", "solution": "Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\\sqrt{13}$, passing through each red point exactly once. Note that points $(i, j)$ and $(k, l)$ are adjacent in the path if and only if $\\{|i-k|,|j-l|\\}=\\{2,3\\}$.\n\nThe center of symmetry must be at point $C\\left(5 \\frac{1}{2}, 5\\right)$. Consider the points $A(2,2)$, $B(11,8)$. These two points are symmetric with respect to $C$ and divide the path into two parts $\\gamma_{1}$ and $\\gamma_{2}$. Note that, if the rectangular board is colored alternately white and black (like a chessboard), $A$ and $B$ are of different colors, and each segment connects two squares of different colors. It follows that each of $\\gamma_{1}, \\gamma_{2}$ consists of an odd number of segments. Thus these two parts are of different lengths and cannot be symmetric to each other. Therefore each\n![](https://cdn.mathpix.com/cropped/2024_12_10_458e228b03e5ba87f05cg-3.jpg?height=443&width=589&top_left_y=429&top_left_x=1256)\nof $\\gamma_{1}, \\gamma_{2}$ is centrally symmetric itself.\nBeing of an odd length, each of the parts $\\gamma_{1}, \\gamma_{2}$ must contain a segment which is centrally symmetric with respect to $C$. There are only two such segments one connecting $(5,4)$ and $(8,6)$, and one connecting $(5,6)$ and $(8,4)$, so these two segments must be parts of our path. Moreover, point $(2,2)$ is connected with only two points, namely $(4,5)$ and $(5,4)$, so these three points are directly connected. Analogous conclusions can be made about points $(2,8),(11,2)$ and $(11,8)$, so the closed path $(5,4)-(2,2)-(4,5)-(2,8)-(5,6)-(8,4)-(11,2)-(9,5)-(11,8)-$ $(8,6)-(5,4)$ is entirely contained in our path, which is clearly a contradiction.", "problem_match": "\n3.", "solution_match": "\n3."}
+{"year": "2009", "problem_label": "4", "tier": 1, "problem": "Determine all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfying\n\n$$\nf\\left(f(m)^{2}+2 f(n)^{2}\\right)=m^{2}+2 n^{2} \\quad \\text { for all } m, n \\in \\mathbb{N} . \\quad \\text { (Bulgaria) }\n$$\n\nTime allowed: 270 minutes.\nEach problem is worth 10 points.\n\n## SOLUTIONS", "solution": "We start by observing that $f$ is injective. From the known identity\n\n$$\n\\left(a^{2}+2 b^{2}\\right)\\left(c^{2}+2 d^{2}\\right)=(a c \\pm 2 b d)^{2}+2(a d \\mp b c)^{2}\n$$\n\nwe obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \\geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to\n\n$$\ng(n)=A(-1)^{n}+B+C n+D n^{2} .\n$$\n\nSubstituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\\dagger)$ gives us\n\n$$\n\\begin{aligned}\nL=9 n^{4} & =A(-1)^{3\\left(A(-1)^{n}+B+C n+D n^{2}\\right)}+B+\\underbrace{3 C\\left[A(-1)^{n}+B+C n+D n^{2}\\right]} \\\\\n& +9 D\\left[A(-1)^{n}+B+C n+D n^{2}\\right]^{2}=R .\n\\end{aligned}\n$$\n\nSince $0=\\lim _{n \\rightarrow \\infty} \\frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \\neq 0$ ); similarly, $0=\\lim _{n \\rightarrow \\infty} \\frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$\nrespectively we obtain $0=\\lim _{k \\rightarrow \\infty} \\frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\\lim _{k \\rightarrow \\infty} \\frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.\nFinally, $g(n)=D n^{2}, D^{3}=1$ and $g: \\mathbb{N} \\rightarrow \\mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.\n\nRemark. Using limits can be avoided. Since the rigth-hand side in ( $\\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.", "problem_match": "\n4.", "solution_match": "\n4."}

Balkan_MO/segmented/en-2010-BMO-type2.jsonl

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+{"year": "2010", "problem_label": "1", "tier": 1, "problem": "The left-hand side is equal to\n\n$$\n\\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}\n$$\n\nso it is enough to show that $a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \\geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b$. The AM-GM inequality gives us $a^{3} b^{3}+a^{3} b^{3}+a^{3} c^{3} \\geq 3 \\sqrt[3]{a^{3} b^{3} \\cdot a^{3} b^{3} \\cdot a^{3} c^{3}}=3 a^{3} b^{2} c$; summing this inequality and its cyclic analogs yields the desired inequality. Equality holds if and only if $a=b=c$.", "solution": "There are $n+1-\\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\\frac{1}{2} n(n+1-\\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \\in \\mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\\varphi(k p))=(k+1)(k p+p+$ $1-\\varphi(k p+p)$ ), so\n\n$$\nk p+1-\\varphi(k p)=(k+1) x \\quad \\text { and } \\quad k p+p+1-\\varphi(k p+p)=k x\n$$\n\nfor some $x \\in \\mathbb{N}, x<p$. Subtraction gives us $x=\\varphi(k p+p)-\\varphi(k p)-p$. Since $\\varphi(k p)$ and $\\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\\varphi(n)$ ), we obtain $x \\equiv-1(\\bmod p-1)$.\nIf $p=2$ then $x=1$ and $\\varphi(2 k+2)=k+3$, which is impossible because $\\varphi(2 k+2) \\leq$ $k+1$. If $p=3$ then $x=1$ and $\\varphi(3 k+3)=2 k+4$, again impossible because $\\varphi(3 k+3) \\leq 2 k+2$. Therefore $p \\geq 5$, so $x \\equiv-1(\\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to\n\n$$\n\\varphi(k p)=2 k+3-p \\quad \\text { and } \\quad \\varphi(k p+p)=2 k+1+p\n$$\n\nIf $k$ is divisible by $p$, then $\\varphi(k p)$ is also divisible by $p$, so $p \\mid 2 k+3$ and hence $p \\mid 3$, a contradiction. Similarly, $p \\nmid k+1$. It follows that $\\varphi(k p)=(p-1) \\varphi(k)$ and $\\varphi(k p+p)=(p-1) \\varphi(k+1)$ which together with (1) yields\n\n$$\n\\varphi(k)=\\frac{2 k+2}{p-1}-1 \\quad \\text { and } \\quad \\varphi(k+1)=\\frac{2 k+2}{p-1}+1\n$$\n\nFrom here we see that $\\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \\in \\mathbb{N}$, or $t \\in\\{1,2,4\\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \\varphi\\left(q^{i}\\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\\frac{2}{3}\\left(2 q^{i}+2\\right)$. The other three cases are similarly shown to be impossible.", "problem_match": "\n1.", "solution_match": "\n4."}
+{"year": "2010", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be an acute-angled triangle with orthocenter $H$ and let $M$ be the midpoint of $A C$. The foot of the altitude from $C$ is $C_{1}$. Point $H_{1}$ is symmetric to $H$ in $A B$. The projections of $C_{1}$ on lines $A H_{1}, A C$ and $B C$ are $P, Q$ and $R$ respectively. If $M_{1}$ is the circumcenter of triangle $P Q R$, prove that the point symmetric to $M$ with respect to $M_{1}$ lies on line $B H_{1}$.\n(Serbia)", "solution": "We shall use the following simple statement.\n\nLemma. Let $A_{1} A_{2} A_{3} A_{4}$ be a convex cyclic quadrilateral whose diagonals are orthogonal and meet at $X$. If $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$ and $X_{i}$ the projection of $X$ on this side $\\left(A_{5}=A_{1}\\right)$, then the eight points $B_{i}, X_{i}$ $(i=1,2,3,4)$ lie on a circle.\n\nProof. Quadrilateral $B_{1} B_{2} B_{3} B_{4}$ is a rectangle because $B_{1} B_{2}\\left\\|B_{3} B_{4}\\right\\| A_{1} A_{3}$ and $B_{2} B_{3}\\left\\|B_{4} B_{1}\\right\\| A_{2} A_{4}$. Denote by $k$ the circumcircle of $B_{1} B_{2} B_{3} B_{4}$. Since $\\angle B_{3} X A_{3}=\\angle A_{4} A_{3} A_{1}=\\angle A_{4} A_{2} A_{1}=\\angle A_{1} X X_{1}$, points $B_{3}, X, X_{1}$ are collinear, so $X_{1}$ lies on the circle with diameter $B_{1} B_{3}$, i.e. $k$. Similarly, $X_{2}, X_{3}, X_{4}$ lie on $k$.\n\nIt is known that $H_{1}$ lies on the circumcircle of $A B C$. By the lemma, points $P, Q, R$ all lie on the circle with diameter $M N$, where $N$ is the midpoint of $B H_{1}$. Therefore $N$ is symmetric to $M$ with respect to $M_{1}$ and lies on $B H_{1}$ as desired.\n![](https://cdn.mathpix.com/cropped/2024_12_10_d6c03e4497ab5252661ag-2.jpg?height=494&width=543&top_left_y=1412&top_left_x=1062)", "problem_match": "\n2.", "solution_match": "\n2."}
+{"year": "2010", "problem_label": "3", "tier": 1, "problem": "We define a $w$-strip as the set of all points in the plane that are between or on two parallel lines on a mutual distance $w$. Let $S$ be a set of $n$ points in the plane such that any three points from $S$ can be covered by a 1 -strip. Show that the entire set $S$ can be covered by a 2 -strip.\n(Romania)", "solution": "Of all triangles with the vertices in $S$, consider one with a maximum area, say $\\triangle A B C$. Let $A^{\\prime}, B^{\\prime}, C^{\\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$, respectively. We claim that all points from $S$ must lie inside or on the boundary of $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$. Indeed, if $X \\in S$ is outside $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$, we can assume without loss of generality that $X$ and $B C$ are on different sides of $B^{\\prime} C^{\\prime}$, and then $\\triangle B C X$ has an area greater than $\\triangle A B C$, a contradiction.\nThe triangle $A B C$ can be covered by a 1-strip, so the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, being similar to $A B C$ with ratio 2 , can be covered by a 2 -strip, also covering all of $S$.", "problem_match": "\n3.", "solution_match": "\n3."}
+{"year": "2010", "problem_label": "4", "tier": 1, "problem": "For every integer $n \\geq 2$, denote by $f(n)$ the sum of positive integers not exceeding $n$ that are not coprime to $n$. Prove that $f(n+p) \\neq f(n)$ for any such $n$ and any prime number $p$.\n(Turkey)\n\nTime allowed: 270 minutes.\nEach problem is worth 10 points.\n\n## SOLUTIONS", "solution": "There are $n+1-\\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\\frac{1}{2} n(n+1-\\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \\in \\mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\\varphi(k p))=(k+1)(k p+p+$ $1-\\varphi(k p+p)$ ), so\n\n$$\nk p+1-\\varphi(k p)=(k+1) x \\quad \\text { and } \\quad k p+p+1-\\varphi(k p+p)=k x\n$$\n\nfor some $x \\in \\mathbb{N}, x<p$. Subtraction gives us $x=\\varphi(k p+p)-\\varphi(k p)-p$. Since $\\varphi(k p)$ and $\\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\\varphi(n)$ ), we obtain $x \\equiv-1(\\bmod p-1)$.\nIf $p=2$ then $x=1$ and $\\varphi(2 k+2)=k+3$, which is impossible because $\\varphi(2 k+2) \\leq$ $k+1$. If $p=3$ then $x=1$ and $\\varphi(3 k+3)=2 k+4$, again impossible because $\\varphi(3 k+3) \\leq 2 k+2$. Therefore $p \\geq 5$, so $x \\equiv-1(\\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to\n\n$$\n\\varphi(k p)=2 k+3-p \\quad \\text { and } \\quad \\varphi(k p+p)=2 k+1+p\n$$\n\nIf $k$ is divisible by $p$, then $\\varphi(k p)$ is also divisible by $p$, so $p \\mid 2 k+3$ and hence $p \\mid 3$, a contradiction. Similarly, $p \\nmid k+1$. It follows that $\\varphi(k p)=(p-1) \\varphi(k)$ and $\\varphi(k p+p)=(p-1) \\varphi(k+1)$ which together with (1) yields\n\n$$\n\\varphi(k)=\\frac{2 k+2}{p-1}-1 \\quad \\text { and } \\quad \\varphi(k+1)=\\frac{2 k+2}{p-1}+1\n$$\n\nFrom here we see that $\\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \\in \\mathbb{N}$, or $t \\in\\{1,2,4\\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \\varphi\\left(q^{i}\\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\\frac{2}{3}\\left(2 q^{i}+2\\right)$. The other three cases are similarly shown to be impossible.", "problem_match": "\n4.", "solution_match": "\n4."}

Balkan_MO/segmented/en-2011-BMO-type1.jsonl

@@ -0,0 +1,4 @@
+{"year": "2011", "problem_label": "1", "tier": 1, "problem": "Let $A B C D$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $A B$ and $C D$ are $F$ and $G$ respectively, and $\\ell$ is the line through $G$ parallel to $A B$. The feet of the perpendiculars from $E$ onto the lines $\\ell$ and $C D$ are $H$ and $K$, respectively. Prove that the lines $E F$ and $H K$ are perpendicular.", "solution": "The points $E, K, H, G$ are on the circle of diameter $G E$, so\n\n$$\n\\angle E H K=\\angle E G K\n$$\n\nAlso, from $\\angle D C A=\\angle D B A$ and $\\frac{C E}{C D}=\\frac{B E}{B A}$ it follows\n\n$$\n\\frac{C E}{C G}=\\frac{2 C E}{C D}=\\frac{2 B E}{B A}=\\frac{B E}{B F},\n$$\n\ntherefore $\\triangle C G E \\sim \\triangle B F E$. In particular, $\\angle E G C=\\angle B F E$, so by $(\\dagger)$\n\n$$\n\\angle E H K=\\angle B F E .\n$$\n\nBut $H E \\perp F B$ and so, since $F E$ and $H K$ are obtained by rotations of these lines by the same (directed) angle, $F E \\perp H K$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-1.jpg?height=570&width=609&top_left_y=1317&top_left_x=758)\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-1.jpg?height=293&width=1413&top_left_y=1942&top_left_x=383)", "problem_match": "# PROBLEM 1", "solution_match": "\nSolution."}
+{"year": "2011", "problem_label": "2", "tier": 1, "problem": "Given real numbers $x, y, z$ such that $x+y+z=0$, show that\n\n$$\n\\frac{x(x+2)}{2 x^{2}+1}+\\frac{y(y+2)}{2 y^{2}+1}+\\frac{z(z+2)}{2 z^{2}+1} \\geq 0 .\n$$\n\nWhen does equality hold?", "solution": "The inequality is clear if $x y z=0$, in which case equality holds if and only if $x=y=z=0$.\n\nHenceforth assume $x y z \\neq 0$ and rewrite the inequality as\n\n$$\n\\frac{(2 x+1)^{2}}{2 x^{2}+1}+\\frac{(2 y+1)^{2}}{2 y^{2}+1}+\\frac{(2 z+1)^{2}}{2 z^{2}+1} \\geq 3 .\n$$\n\nNotice that (exactly) one of the products $x y, y z, z x$ is positive, say $y z>0$, to get\n\n$$\n\\begin{array}{rlr}\n\\frac{(2 y+1)^{2}}{2 y^{2}+1}+\\frac{(2 z+1)^{2}}{2 z^{2}+1} & \\geq \\frac{2(y+z+1)^{2}}{y^{2}+z^{2}+1} & \\text { (by Jensen) } \\\\\n& =\\frac{2(x-1)^{2}}{x^{2}-2 y z+1} & (\\text { for } x+y+z=0) \\\\\n& \\geq \\frac{2(x-1)^{2}}{x^{2}+1} . & (\\text { for } y z>0)\n\\end{array}\n$$\n\nHere equality holds if and only if $x=1$ and $y=z=-1 / 2$. Finally, since\n\n$$\n\\frac{(2 x+1)^{2}}{2 x^{2}+1}+\\frac{2(x-1)^{2}}{x^{2}+1}-3=\\frac{2 x^{2}(x-1)^{2}}{\\left(2 x^{2}+1\\right)\\left(x^{2}+1\\right)} \\geq 0, \\quad x \\in \\mathbb{R},\n$$\n\nthe conclusion follows. Clearly, equality holds if and only if $x=1$, so $y=z=-1 / 2$. Therefore, if $x y z \\neq 0$, equality holds if and only if one of the numbers is 1 , and the other two are $-1 / 2$.\n\nMarking Scheme. Proving the inequality and identifying the equality case when one of the variables vanishes\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-2.jpg?height=50&width=68&top_left_y=1845&top_left_x=1721)\n\nApplying Jensen or Cauchy-Schwarz inequality to the fractions involving the pair of numbers of the same sign 3p\nProducing the corresponding lower bound in the third variable ................... 3p\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-2.jpg?height=47&width=1397&top_left_y=2036&top_left_x=391)\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-2.jpg?height=50&width=1399&top_left_y=2081&top_left_x=390)\nRemark. Any partial or equivalent approach should be marked accordingly.", "problem_match": "# PROBLEM 2", "solution_match": "\nSolution."}
+{"year": "2011", "problem_label": "3", "tier": 1, "problem": "Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is good if whenever $x, y \\in T$ and $x<y$, the ratio $y / x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever $x, y \\in T$ and $x<y$, the ratio $y / x$ is not a power of a prime number. We agree that a singleton subset of $S$ is both good and bad. Let $k$ be the largest possible size of a good subset of $S$. Prove that $k$ is also the smallest number of pairwise-disjoint bad subsets whose union is $S$.", "solution": "Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subsets has at least as many members as a maximal good subset.\n\nNotice further that the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime: if $x<y<z$ are elements of a good subset of $S$, then $y=x p^{\\alpha}$ and $z=y q^{\\beta}=x p^{\\alpha} q^{\\beta}$ for some primes $p$ and $q$ and some positive integers $\\alpha$ and $\\beta$, so $p=q$ for $z / x$ to be a power of a prime.\nNext, let $P=\\{2,3,5,7,11, \\cdots\\}$ denote the set of all primes, let\n\n$$\nm=\\max \\left\\{\\exp _{p} x: x \\in S \\text { and } p \\in P\\right\\}\n$$\n\nwhere $\\exp _{p} x$ is the exponent of the prime $p$ in the canonical decomposition of $x$, and notice that a maximal good subset of $S$ must be of the form $\\left\\{a, a p, \\cdots, a p^{m}\\right\\}$ for some prime $p$ and some positive integer $a$ which is not divisible by $p$. Consequently, a maximal good subset of $S$ has $m+1$ elements, so a partition of $S$ into bad subsets has at least $m+1$ members.\n\nFinally, notice by maximality of $m$ that the sets\n\n$$\nS_{k}=\\left\\{x: x \\in S \\text { and } \\sum_{p \\in P} \\exp _{p} x \\equiv k(\\bmod m+1)\\right\\}, \\quad k=0,1, \\cdots, m\n$$\n\nform a partition of $S$ into $m+1$ bad subsets. The conclusion follows.\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-3.jpg?height=47&width=1399&top_left_y=1839&top_left_x=390) Considering the maximal exponent $m$ of a prime and deriving $k=m+1 \\ldots \\ldots \\mathbf{1 p}$ Noticing that the intersection of a bad set and a good set contains at most one element and infering that a partition of $S$ into bad sets has at least $k$ members.....2p\n\nProducing a partition of $S$ into $k$ bad subsets . . . . . . . . . . . . . . . . . . . . . . .\nRemark. Any partial or equivalent approach should be marked accordingly.", "problem_match": "# PROBLEM 3", "solution_match": "\nSolution."}
+{"year": "2011", "problem_label": "4", "tier": 1, "problem": "Let $A B C D E F$ be a convex hexagon of area 1 , whose opposite sides are parallel. The lines $A B, C D$ and $E F$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $B C, D E$ and $F A$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3 / 2$.", "solution": "Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6 . Label the vertices of the hexagon in circular order, $A_{0}, A_{1}, \\cdots, A_{5}$, and let the lines of support of the alternate sides $A_{i} A_{i+1}$ and $A_{i+2} A_{i+3}$ meet at $B_{i}$. To show that the area of at least one of the triangles $B_{0} B_{2} B_{4}, B_{1} B_{3} B_{5}$ is greater than or equal to $3 / 2$, it is sufficient to prove that the total area of the six triangles $A_{i+1} B_{i} A_{i+2}$ is at least 1:\n\n$$\n\\sum_{i=0}^{5} \\operatorname{area} A_{i+1} B_{i} A_{i+2} \\geq 1\n$$\n\nTo begin with, reflect each $B_{i}$ through the midpoint of the segment $A_{i+1} A_{i+2}$ to get the points $B_{i}^{\\prime}$. We shall prove that the six triangles $A_{i+1} B_{i}^{\\prime} A_{i+2}$ cover the hexagon. To this end, reflect $A_{2 i+1}$ through the midpoint of the segment $A_{2 i} A_{2 i+2}$ to get the points $A_{2 i+1}^{\\prime}$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\\prime}, i=0,1,2$, and a (possibly degenerate) triangle, $A_{1}^{\\prime} A_{3}^{\\prime} A_{5}^{\\prime}$. Notice first that each parallelogram $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\\prime}$ is covered by the pair of triangles $\\left(A_{2 i} B_{2 i+5}^{\\prime} A_{2 i+1}, A_{2 i+1} B_{2 i}^{\\prime} A_{2 i+2}\\right)$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A_{1}^{\\prime} A_{3}^{\\prime} A_{5}^{\\prime}$. To this end, it is sufficient to prove that $A_{2 i} B_{2 i+5}^{\\prime} \\geq A_{2 i} A_{2 i+5}^{\\prime}$ and $A_{2 j+2} B_{2 j}^{\\prime} \\geq A_{2 j+2} A_{2 j+3}^{\\prime}$ for some indices $i, j \\in\\{0,1,2\\}$. To establish the first inequality, notice that\n\n$$\n\\begin{gathered}\nA_{2 i} B_{2 i+5}^{\\prime}=A_{2 i+1} B_{2 i+5}, \\quad A_{2 i} A_{2 i+5}^{\\prime}=A_{2 i+4} A_{2 i+5}, \\quad i=0,1,2, \\\\\n\\\\\n\\frac{A_{1} B_{5}}{A_{4} A_{5}}=\\frac{A_{0} B_{5}}{A_{5} B_{3}} \\quad \\text { and } \\quad \\frac{A_{3} B_{1}}{A_{0} A_{1}}=\\frac{A_{2} A_{3}}{A_{0} B_{5}},\n\\end{gathered}\n$$\n\nto get\n\n$$\n\\prod_{i=0}^{2} \\frac{A_{2 i} B_{2 i+5}^{\\prime}}{A_{2 i} A_{2 i+5}^{\\prime}}=1\n$$\n\nSimilarly,\n\n$$\n\\prod_{j=0}^{2} \\frac{A_{2 j+2} B_{2 j}^{\\prime}}{A_{2 j+2} A_{2 j+3}^{\\prime}}=1\n$$\n\nwhence the conclusion.\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-4.jpg?height=362&width=828&top_left_y=1963&top_left_x=646)\n\nMarking Scheme. Stating that the total area of the small triangles $\\geq 1 \\ldots \\ldots . . \\mathbf{1 p}$\n![](https://cdn.mathpix.com/cropped/2024_12_10_17b2c7bd84b20d05d38bg-4.jpg?height=58&width=1399&top_left_y=2405&top_left_x=390)\nDecomposition of the hexagon into three adequate parallelograms and a triangle $\\mathbf{1 p}$\nProving that each pair of triangles adjacent to a parallelogram covers that parallelogram\n\nProving the central triangle also covered 5p\nRemark. Any partial or equivalent approach should be marked accordingly.", "problem_match": "# PROBLEM 4", "solution_match": "\nSolution."}

Balkan_MO/segmented/en-2012-BMO-type3.jsonl

@@ -0,0 +1,7 @@
+{"year": "2012", "problem_label": "1", "tier": 1, "problem": "\\quad$ Let $A, B$ and $C$ be points lying on a circle $\\Gamma$ with centre $O$. Assume that $\\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ with the line $A C$, and let $F$ be the point of intersection of $\\Gamma$ with $l$ that lies between $D$ and $E$.\nProve that the circumcircles of triangles $B F E$ and $C F D$ are tangent at $F$.", "solution": "Let $\\ell \\cap A O=\\{K\\}$ and $G$ be the other end point of the diameter of $\\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \\perp A D$ and $G, E, B$ are collinear.\n![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=1006&width=675&top_left_y=531&top_left_x=696)\n\nAs $\\angle C D F=\\angle G D K=\\angle G A C=\\angle G F C, F G$ is tangent to the circumcircle of triangle $C F D$ at $F$. As $\\angle F B E=\\angle F B G=\\angle F A G=\\angle G F K=\\angle G F E, F G$ is also tangent to the circumcircle of $B F E$ at $F$. Hence the circumcircles of the triangles $C F D$ and $B F E$ are tangent at $F$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-1.jpg?height=412&width=417&top_left_y=2350&top_left_x=1462)", "problem_match": "\n1. ", "solution_match": "\nSolution."}
+{"year": "2012", "problem_label": "2", "tier": 1, "problem": "Prove that\n\n$$\n\\sum_{c y c}(x+y) \\sqrt{(z+x)(z+y)} \\geq 4(x y+y z+z x),\n$$\n\nfor all positive real numbers $x, y$ and $z$.", "solution": "We will obtain the inequality by adding the inequalities\n\n$$\n(x+y) \\sqrt{(z+x)(z+y)} \\geq 2 x y+y z+z x\n$$\n\nfor cyclic permutation of $x, y, z$.\nSquaring both sides of this inequality we obtain\n\n$$\n(x+y)^{2}(z+x)(z+y) \\geq 4 x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+4 x y^{2} z+4 x^{2} y z+2 x y z^{2}\n$$\n\nwhich is equivalent to\n\n$$\nx^{3} y+x y^{3}+z\\left(x^{3}+y^{3}\\right) \\geq 2 x^{2} y^{2}+x y z(x+y)\n$$\n\nwhich can be rearranged to\n\n$$\n(x y+y z+z x)(x-y)^{2} \\geq 0\n$$\n\nwhich is clearly true.", "problem_match": "\n2. ", "solution_match": "\nSolution 1."}
+{"year": "2012", "problem_label": "2", "tier": 1, "problem": "Prove that\n\n$$\n\\sum_{c y c}(x+y) \\sqrt{(z+x)(z+y)} \\geq 4(x y+y z+z x),\n$$\n\nfor all positive real numbers $x, y$ and $z$.", "solution": "For positive real numbers $x, y, z$ there exists a triangle with the side lengths $\\sqrt{x+y}, \\sqrt{y+z}, \\sqrt{z+x}$ and the area $K=\\sqrt{x y+y z+z x} / 2$.\n\nThe existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have\n\n$$\nK=\\frac{1}{2} \\sqrt{x+y} \\sqrt{z+x} \\sin \\alpha\n$$\n\nwhere $\\alpha$ is the angle between the sides of length $\\sqrt{x+y}$ and $\\sqrt{z+x}$. On the other hand, from the law of cosines we have\n\n$$\n\\cos \\alpha=\\frac{x+y+z+x-y-z}{2 \\sqrt{(x+y)(z+x)}}=\\frac{x}{\\sqrt{(x+y)(z+x)}}\n$$\n\nand\n\n$$\n\\sin \\alpha=\\sqrt{1-\\cos ^{2} \\alpha}=\\frac{\\sqrt{x y+y z+z x}}{\\sqrt{(x+y)(z+x)}}\n$$\n\nNow the inequality is equivalent to\n\n$$\n\\sqrt{x+y} \\sqrt{y+z} \\sqrt{z+x} \\sum_{c y c} \\sqrt{x+y} \\geq 16 K^{2}\n$$\n\nThis can be rewritten as\n\n$$\n\\frac{\\sqrt{x+y} \\sqrt{y+z} \\sqrt{z+x}}{4 K} \\geq 2 \\frac{K}{\\sum_{c y c} \\sqrt{x+y} / 2}\n$$\n\nto become the Euler inequality $R \\geq 2 r$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-2.jpg?height=418&width=429&top_left_y=2352&top_left_x=1456)", "problem_match": "\n2. ", "solution_match": "\nSolution 2."}
+{"year": "2012", "problem_label": "3", "tier": 1, "problem": "Let $n$ be a positive integer. Let $P_{n}=\\left\\{2^{n}, 2^{n-1} \\cdot 3,2^{n-2} \\cdot 3^{2}, \\ldots, 3^{n}\\right\\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\\emptyset}=0$ where $\\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \\leq y \\leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \\leq y-S_{Y}<2^{n}$", "solution": "Let $\\alpha=3 / 2$ so $1+\\alpha>\\alpha^{2}$.\nGiven $y$, we construct $Y$ algorithmically. Let $Y=\\varnothing$ and of course $S_{\\varnothing}=0$. For $i=0$ to $m$, perform the following operation:\n\n$$\n\\text { If } S_{Y}+2^{i} 3^{m-i} \\leq y \\text {, then replace } Y \\text { by } Y \\cup\\left\\{2^{i} 3^{m-i}\\right\\}\n$$\n\nWhen this process is finished, we have a subset $Y$ of $P_{m}$ such that $S_{Y} \\leq y$.\nNotice that the elements of $P_{m}$ are in ascending order of size as given, and may alternatively be described as $2^{m}, 2^{m} \\alpha, 2^{m} \\alpha^{2}, \\ldots, 2^{m} \\alpha^{m}$. If any member of this list is not in $Y$, then no two consecutive members of the list to the left of the omitted member can both be in $Y$. This is because $1+\\alpha>\\alpha^{2}$, and the greedy nature of the process used to construct $Y$.\n\nTherefore either $Y=P_{m}$, in which case $y=3^{m+1}-2^{m+1}$ and all is well, or at least one of the two leftmost elements of the list is omitted from $Y$.\n\nIf $2^{m}$ is not omitted from $Y$, then the algorithmic process ensures that $\\left(S_{Y}-2^{m}\\right)+2^{m-1} 3>y$, and so $y-S_{Y}<2^{m}$. On the other hand, if $2^{m}$ is omitted from $Y$, then $y-S_{Y}<2^{m}$ ).", "problem_match": "\n3. ", "solution_match": "\nSolution 1."}
+{"year": "2012", "problem_label": "3", "tier": 1, "problem": "Let $n$ be a positive integer. Let $P_{n}=\\left\\{2^{n}, 2^{n-1} \\cdot 3,2^{n-2} \\cdot 3^{2}, \\ldots, 3^{n}\\right\\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\\emptyset}=0$ where $\\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \\leq y \\leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \\leq y-S_{Y}<2^{n}$", "solution": "Note that $3^{m+1}-2^{m+1}=(3-2)\\left(3^{m}+3^{m-1} \\cdot 2+\\cdots+3 \\cdot 2^{m-1}+2^{m}\\right)=S_{P_{m}}$. Dividing every element of $P_{m}$ by $2^{m}$ gives us the following equivalent problem:\n\nLet $m$ be a positive integer, $a=3 / 2$, and $Q_{m}=\\left\\{1, a, a^{2}, \\ldots, a^{m}\\right\\}$. Show that for any real number $x$ satisfying $0 \\leq x \\leq 1+a+a^{2}+\\cdots+a^{m}$, there exists a subset $X$ of $Q_{m}$ such that $0 \\leq x-S_{X}<1$.\n\nWe will prove this problem by induction on $m$. When $m=1, S_{\\varnothing}=0, S_{\\{1\\}}=1, S_{\\{a\\}}=3 / 2$, $S_{\\{1, a\\}}=5 / 2$. Since the difference between any two consecutive of them is at most 1, the claim is true.\n\nSuppose that the statement is true for positive integer $m$. Let $x$ be a real number with $0 \\leq$ $x \\leq 1+a+a^{2}+\\cdots+a^{m+1}$. If $0 \\leq x \\leq 1+a+a^{2}+\\cdots+a^{m}$, then by the induction hypothesis there exists a subset $X$ of $Q_{m} \\subset Q_{m+1}$ such that $0 \\leq x-S_{X}<1$.\nIf $\\frac{a^{m+1}-1}{a-1}=1+a+a^{2}+\\cdots+a^{m}<x$, then $x>a^{m+1}$ as\n\n$$\n\\frac{a^{m+1}-1}{a-1}=2\\left(a^{m+1}-1\\right)=a^{m+1}+\\left(a^{m+1}-2\\right) \\geq a^{m+1}+a^{2}-2=a^{m+1}+\\frac{1}{4} .\n$$\n\nTherefore $0<\\left(x-a^{m+1}\\right) \\leq 1+a+a^{2}+\\cdots+a^{m}$. Again by the induction hypothesis, there exists a subset $X$ of $Q_{m}$ satisfying $0 \\leq\\left(x-a^{m+1}\\right)-S_{X}<1$. Hence $0 \\leq x-S_{X^{\\prime}}<1$ where $X^{\\prime}=X \\cup\\left\\{a^{m+1}\\right\\} \\subset Q_{m+1}$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-3.jpg?height=418&width=423&top_left_y=2347&top_left_x=1459)", "problem_match": "\n3. ", "solution_match": "\nSolution 2."}
+{"year": "2012", "problem_label": "4", "tier": 1, "problem": "\\quad$ Let $\\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \\mathbb{Z}^{+} \\rightarrow \\mathbb{Z}^{+}$such that the following conditions both hold:\n(i) $f(n!)=f(n)$ ! for every positive integer $n$,\n(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.", "solution": "There are three such functions: the constant functions 1, 2 and the identity function $\\mathrm{id}_{\\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.\n\nConsider such a function $f$ and suppose that it has a fixed point $a \\geq 3$, that is $f(a)=a$. Then $a!,(a!)!, \\cdots$ are all fixed points of $f$, hence the function $f$ has a strictly increasing sequence $a_{1}<a_{2}<\\cdots<a_{k}<\\cdots$ of fixed points. For a positive integer $n$, $a_{k}-n$ divides $a_{k}-f(n)=$ $f\\left(a_{k}\\right)-f(n)$ for every $k \\in \\mathbf{Z}^{+}$. Also $a_{k}-n$ divides $a_{k}-n$, so it divides $a_{k}-f(n)-\\left(a_{k}-n\\right)=$ $n-f(n)$. This is possible only if $f(n)=n$, hence in this case we get $f=\\mathrm{id}_{\\mathbf{Z}^{+}}$.\n\nNow suppose that $f$ has no fixed points greater than 2 . Let $p \\geq 5$ be a prime and notice that by Wilson's Theorem we have $(p-2)!\\equiv 1(\\bmod p)$. Therefore $p$ divides $(p-2)!-1$. But $(p-2)!-1$ divides $f((p-2)!)-f(1)$, hence $p$ divides $f((p-2)!)-f(1)=(f(p-2))!-f(1)$. Clearly we have $f(1)=1$ or $f(1)=2$. As $p \\geq 5$, the fact that $p$ divides $(f(p-2))!-f(1)$ implies that $f(p-2)<p$. It is easy to check, again by Wilson's Theorem, that $p$ does not divide $(p-1)!-1$ and $(p-1)!-2$, hence we deduce that $f(p-2) \\leq p-2$. On the other hand, $p-3=(p-2)-1$ divides $f(p-2)-f(1) \\leq(p-2)-1$. Thus either $f(p-2)=f(1)$ or $f(p-2)=p-2$. As $p-2 \\geq 3$, the last case is excluded, since the function $f$ has no fixed points greater than 2 . It follows $f(p-2)=f(1)$ and this property holds for all primes $p \\geq 5$. Taking $n$ any positive integer, we deduce that $p-2-n$ divides $f(p-2)-f(n)=f(1)-f(n)$ for all primes $p \\geq 5$. Thus $f(n)=f(1)$, hence $f$ is the constant function 1 or 2 .", "problem_match": "\n4. ", "solution_match": "\nSolution 1."}
+{"year": "2012", "problem_label": "4", "tier": 1, "problem": "\\quad$ Let $\\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \\mathbb{Z}^{+} \\rightarrow \\mathbb{Z}^{+}$such that the following conditions both hold:\n(i) $f(n!)=f(n)$ ! for every positive integer $n$,\n(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.", "solution": "Note first that if $f\\left(n_{0}\\right)=n_{0}$, then $m-n_{0} \\mid f(m)-m$ for all $m \\in \\mathbf{Z}^{+}$. If $f\\left(n_{0}\\right)=n_{0}$ for infinitely many $n_{0} \\in \\mathbf{Z}^{+}$, then $f(m)-m$ has infinitely many divisors, hence $f(m)=m$ for all $m \\in \\mathbf{Z}^{+}$. On the other hand, if $f\\left(n_{0}\\right)=n_{0}$ for some $n_{0} \\geq 3$, then $f$ fixes each term of the sequence $\\left(n_{k}\\right)_{k=0}^{\\infty}$, which is recursively defined by $n_{k}=n_{k-1}!$. Hence if $f(3)=3$, then $f(n)=n$ for all $n \\in \\mathbf{Z}^{+}$.\n\nWe may assume that $f(3) \\neq 3$. Since $f(1)=f(1)!$, and $f(2)=f(2)!, f(1), f(2) \\in\\{1,2\\}$. We have $4=3!-2 \\mid f(3)!-f(2)$. This together with $f(3) \\neq 3$ implies that $f(3) \\in\\{1,2\\}$. Let $n>3$, then $n!-3 \\mid f(n)!-f(3)$ and $3 \\nmid f(n)!$, i.e. $f(n)!\\in\\{1,2\\}$. Hence we conclude that $f(n) \\in\\{1,2\\}$ for all $n \\in \\mathbf{Z}^{+}$.\n\nIf $f$ is not constant, then there exist positive integers $m, n$ with $\\{f(n), f(m)\\}=\\{1,2\\}$. Let $k=2+\\max \\{m, n\\}$. If $f(k) \\neq f(m)$, then $k-m \\mid f(k)-f(m)$. This is a contradiction as $|f(k)-f(m)|=1$ and $k-m \\geq 2$.\n\nTherefore the functions satisfying the conditions are $f \\equiv 1, f \\equiv 2, f=\\mathrm{id}_{\\mathbf{z}^{+}}$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_ead8d0bf173f1f5b4aafg-4.jpg?height=418&width=427&top_left_y=2347&top_left_x=1454)", "problem_match": "\n4. ", "solution_match": "\nSolution 2."}

Balkan_MO/segmented/en-2013-BMO-type2.jsonl

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+{"year": "2013", "problem_label": "1", "tier": 1, "problem": "We denote the angles of the triangle by $\\alpha, \\beta$ and $\\gamma$ as usual. Since $\\angle K M P=90^{\\circ}-\\frac{\\beta}{2}$, it suffices to prove that $\\angle K L P=90^{\\circ}+\\frac{\\beta}{2}$, which is equivalent to $\\angle K L C=\\frac{\\beta}{2}$.\nLet $I$ be the incenter of triangle $A B C$ and let $D$ be the tangency point of the incircle with $A B$. Since $C K \\| I B$ and $C L \\| I A$, it holds that $\\angle K C L=\\angle A I B$. Moreover, from $C N=A D=\\frac{b+c-a}{2}$ and $\\angle K C N=\\frac{\\beta}{2}$ we obtain $C K=C N \\cos \\frac{\\beta}{2}=$ $A D \\cos \\frac{\\beta}{2}=A I \\cos \\frac{\\alpha}{2} \\cos \\frac{\\beta}{2}$ and analogously $C L=B I \\cos \\frac{\\alpha}{2} \\cos \\frac{\\beta}{2}$, which imply $\\frac{C K}{C L}=\\frac{A I}{B I}$. Hence the triangles $K C L$ and $A I B$ are similar, and thus $\\angle K L C=$ $\\angle A B I=\\frac{\\beta}{2}$.", "solution": "Consider the graph $\\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.\n\nLemma. There is a vertex in graph $\\mathcal{G}$ with degree at most 2 .\nProof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \\ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \\geq j$. Then $u_{0}, u_{1}, \\ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.\n\nWe now prove the problem statement by induction on the number $n$ of vertices in $\\mathcal{G}$. For $n \\leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\\mathcal{G}$ of degree at most two. Graph $\\mathcal{G}^{\\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.", "problem_match": "\n1.", "solution_match": "\n4."}
+{"year": "2013", "problem_label": "2", "tier": 1, "problem": "Determine all positive integers $x, y$ and $z$ such that\n\n$$\nx^{5}+4^{y}=2013^{z}\n$$", "solution": "Reducing modulo 11 yields $x^{5}+4^{y} \\equiv 0(\\bmod 11)$, where $x^{5} \\equiv \\pm 1(\\bmod 11)$, so we also have $4^{y} \\equiv \\pm 1(\\bmod 11)$. Congruence $4^{y} \\equiv-1(\\bmod 11)$ does not hold for any $y$, whereas $4^{y} \\equiv 1(\\bmod 11)$ holds if and only if $5 \\mid y$.\nSetting $t=4^{y / 5}$, the equation becomes $x^{5}+t^{5}=A \\cdot B=2013^{z}$, where $(x, t)=1$ and $A=x+t, B=x^{4}-x^{3} t+x^{2} t^{2}-x t^{3}+t^{4}$. Furthermore, from $B=A\\left(x^{3}-\\right.$ $\\left.2 x^{2} t+3 x t^{2}-4 t^{3}\\right)+5 t^{4}$ we deduce $(A, B)=\\left(A, 5 t^{4}\\right) \\mid 5$, but $5 \\nmid 2013^{z}$, so we must have $(A, B)=1$. Therefore $A=a^{z}$ and $B=b^{z}$ for some positive integers $a$ and $b$ with $a \\cdot b=2013$.\nOn the other hand, from $\\frac{1}{16} A^{4} \\leq B \\leq A^{4}$ (which is a simple consequence of the mean inequality) we obtain $\\frac{1}{16} a^{4} \\leq b \\leq a^{4}$, i.e. $\\frac{1}{16} a^{5} \\leq a b=2013 \\leq a^{5}$. Therefore $5 \\leq a \\leq 8$, which is impossible because 2013 has no divisors in the interval [5,8].", "problem_match": "\n2.", "solution_match": "\n2."}
+{"year": "2013", "problem_label": "3", "tier": 1, "problem": "Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \\rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:\n(i) $x f(x, y, z)=z f(z, y, x)$;\n(ii) $f\\left(x, k y, k^{2} z\\right)=k f(x, y, z)$;\n(iii) $f(1, k, k+1)=k+1$.\n(United Kingdom)", "solution": "It follows from the properties of function $f$ that, for all $x, y, z, a, b>0$, $f\\left(a^{2} x, a b y, b^{2} z\\right)=b f\\left(a^{2} x, a y, z\\right)=b \\cdot \\frac{z}{a^{2} x} f\\left(z, a y, a^{2} x\\right)=\\frac{b z}{a x} f(z, y, x)=\\frac{b}{a} f(x, y, z)$.\n\nWe shall choose $a$ and $b$ in such a way that the triple $\\left(a^{2} x, a b y, b^{2} z\\right)$ is of the form $(1, k, k+1)$ for some $k$ : namely, we take $a=\\frac{1}{\\sqrt{x}}$ and $b$ satisfying $b^{2} z-a b y=1$, which upon solving the quadratic equation yields $b=\\frac{y+\\sqrt{y^{2}+4 x z}}{2 z \\sqrt{x}}$ and $k=\\frac{y\\left(y+\\sqrt{y^{2}+4 x z}\\right)}{2 x z}$. Now we easily obtain\n\n$$\nf(x, y, z)=\\frac{a}{b} f\\left(a^{2} x, a b y, b^{2} z\\right)=\\frac{a}{b} f(1, k, k+1)=\\frac{a}{b}(k+1)=\\frac{y+\\sqrt{y^{2}+4 x z}}{2 x} .\n$$\n\nIt is directly verified that $f$ satisfies the problem conditions.", "problem_match": "\n3.", "solution_match": "\n3."}
+{"year": "2013", "problem_label": "4", "tier": 1, "problem": "In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \\geq 3$ different competitors $A_{1}, A_{2}, \\ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$ for $1 \\leq i \\leq n\\left(A_{n+1}=A_{1}\\right)$, and there are no other pairs of non-friends among the components of this cycle.\nThe following property is satisfied:\nfor every competitor $C$, and every weakly-friendly cycle $\\mathscr{S}$ of competitors not including $C$, the set of competitors $D$ in $\\mathscr{S}$ which are not friends of $C$ has at most one element.\n\nProve that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors that are in the same room are friends.\n(Serbia)\nTime allowed: 270 minutes.\nEach problem is worth 10 points.\n\n## SOLUTIONS", "solution": "Consider the graph $\\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.\n\nLemma. There is a vertex in graph $\\mathcal{G}$ with degree at most 2 .\nProof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \\ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \\geq j$. Then $u_{0}, u_{1}, \\ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.\n\nWe now prove the problem statement by induction on the number $n$ of vertices in $\\mathcal{G}$. For $n \\leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\\mathcal{G}$ of degree at most two. Graph $\\mathcal{G}^{\\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.", "problem_match": "\n4.", "solution_match": "\n4."}

Balkan_MO/segmented/en-2014-BMO-type1.jsonl

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+{"year": "2014", "problem_label": "1", "tier": 1, "problem": "Let $x, y$ and $z$ be positive real numbers such that $x y+y z+z x=3 x y z$. Prove that\n\n$$\nx^{2} y+y^{2} z+z^{2} x \\geq 2(x+y+z)-3\n$$\n\nand determine when equality holds.", "solution": "The given condition can be rearranged to $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=3$. Using this, we obtain:\n\n$$\n\\begin{aligned}\nx^{2} y+y^{2} z+z^{2} x-2(x+y+z)+3 & =x^{2} y-2 x+\\frac{1}{y}+y^{2} z-2 y+\\frac{1}{z}+z^{2} x-2 x+\\frac{1}{x}= \\\\\n& =y\\left(x-\\frac{1}{y}\\right)^{2}+z\\left(y-\\frac{1}{z}\\right)^{2}+x\\left(z-\\frac{1}{z}\\right)^{2} \\geq 0\n\\end{aligned}\n$$\n\nEquality holds if and only if we have $x y=y z=z x=1$, or, in other words, $x=y=z=1$.\nAlternative solution. It follows from $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=3$ and Cauchy-Schwarz inequality that\n\n$$\n\\begin{aligned}\n3\\left(x^{2} y+y^{2} z+z^{2} x\\right) & =\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right)\\left(x^{2} y+y^{2} z+z^{2} x\\right) \\\\\n& \\left.=\\left(\\left(\\frac{1}{\\sqrt{y}}\\right)^{2}+\\left(\\frac{1}{\\sqrt{z}}\\right)^{2}+\\left(\\frac{1}{\\sqrt{x}}\\right)^{2}\\right)\\left((x \\sqrt{y})^{2}\\right)+(y \\sqrt{z})^{2}+(z \\sqrt{x})^{2}\\right) \\\\\n& \\geq(x+y+z)^{2}\n\\end{aligned}\n$$\n\nTherefore, $x^{2} y+y^{2} z+z^{2} x \\geq \\frac{(x+y+z)^{2}}{3}$ and if $x+y+z=t$ it suffices to show that $\\frac{t^{2}}{3} \\geq 2 t-3$. The latter is equivalent to $(t-3)^{2} \\geq 0$. Equality holds when\n\n$$\nx \\sqrt{y} \\sqrt{y}=y \\sqrt{z} \\sqrt{z}=z \\sqrt{x} \\sqrt{x},\n$$\n\ni.e. $x y=y z=z x$ and $t=x+y+z=3$. Hence, $x=y=z=1$.\n\nComment. The inequality is true with the condition $x y+y z+z x \\leq 3 x y z$.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}
+{"year": "2014", "problem_label": "2", "tier": 1, "problem": "A special number is a positive integer $n$ for which there exist positive integers $a, b, c$ and $d$ with\n\n$$\nn=\\frac{a^{3}+2 b^{3}}{c^{3}+2 d^{3}}\n$$\n\nProve that:\n(a) there are infinitely many special numbers;\n(b) 2014 is not a special number.", "solution": "(a) Every perfect cube $k^{3}$ of a positive integer is special because we can write\n\n$$\nk^{3}=k^{3} \\frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}\n$$\n\nfor some positive integers $a, b$.\n(b) Observe that $2014=2.19 .53$. If 2014 is special, then we have,\n\n$$\nx^{3}+2 y^{3}=2014\\left(u^{3}+2 v^{3}\\right)\n$$\n\nfor some positive integers $x, y, u, v$. We may assume that $x^{3}+2 y^{3}$ is minimal with this property. Now, we will use the fact that if 19 divides $x^{3}+2 y^{3}$, then it divides both $x$ and $y$. Indeed, if 19 does not divide $x$, then it does not divide $y$ too. The relation $x^{3} \\equiv-2 y^{3}(\\bmod 19)$ implies $\\left(x^{3}\\right)^{6} \\equiv\\left(-2 y^{3}\\right)^{6}(\\bmod 19)$. The latter congruence is equivalent to $x^{18} \\equiv 2^{6} y^{18}(\\bmod 19)$. Now, according to the Fermat's Little Theorem, we obtain $1 \\equiv 2^{6}(\\bmod 19)$, that is 19 divides 63 , not possible.\nIt follows $x=19 x_{1}, y=19 y_{1}$, for some positive integers $x_{1}$ and $y_{1}$. Replacing in (1) we get\n\n$$\n19^{2}\\left(x_{1}^{3}+2 y_{1}^{3}\\right)=2.53\\left(u^{3}+2 v^{3}\\right)\n$$\n\ni.e. $19 \\mid u^{3}+2 v^{3}$. It follows $u=19 u_{1}$ and $v=19 v_{1}$, and replacing in (2) we get\n\n$$\nx_{1}^{3}+2 y_{1}^{3}=2014\\left(u_{1}^{3}+2 v_{1}^{3}\\right) .\n$$\n\nClearly, $x_{1}^{3}+2 y_{1}^{3}<x^{3}+2 y^{3}$, contradicting the minimality of $x^{3}+2 y^{3}$.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}
+{"year": "2014", "problem_label": "3", "tier": 1, "problem": "Let $A B C D$ be a trapezium inscribed in a circle $\\Gamma$ with diameter $A B$. Let $E$ be the intersection point of the diagonals $A C$ and $B D$. The circle with center $B$ and radius $B E$ meets $\\Gamma$ at the points $K$ and $L$, where $K$ is on the same side of $A B$ as $C$. The line perpendicular to $B D$ at $E$ intersects $C D$ at $M$.\n\nProve that $K M$ is perpendicular to $D L$.", "solution": "Since $A B \\| C D$, we have that $A B C D$ is isosceles trapezium. Let $O$ be the center of $k$ and $E M$ meets $A B$ at point $Q$. Then, from the right angled triangle $B E Q$, we have $B E^{2}=B O \\cdot B Q$. Since $B E=B K$, we get $B K^{2}=B O \\cdot B Q$ (1). Suppose that $K L$ meets $A B$ at $P$. Then, from the right angled triangle $B A K$, we have $B K^{2}=B P . B A$ (2)\n![](https://cdn.mathpix.com/cropped/2024_12_10_86b1467c9ecfa953dbf8g-4.jpg?height=580&width=595&top_left_y=1446&top_left_x=762)\n\nFrom (1) and (2) we get $\\frac{B P}{B Q}=\\frac{B O}{B A}=\\frac{1}{2}$, and therefore $P$ is the midpoint of $B Q$ (3). However, $D M \\| A Q$ and $M Q \\| A D$ (both are perpendicular to $D B$ ). Hence, $A Q M D$ is parallelogram and thus $M Q=A D=B C$. We conclude that $Q B C M$ is isosceles trapezium. It follows from (3) that $K L$ is the perpendicular bisector of $B Q$ and $C M$, that is, $M$ is symmetric to $C$ with respect to $K L$. Finally, we get that $M$ is the orthocenter\nof the triangle $D L K$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vise versa.", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}
+{"year": "2014", "problem_label": "4", "tier": 1, "problem": "Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length 1 by lines parallel to its sides.\nFind the number of regular hexagons all of whose vertices are among the vertices of the equilateral triangles.", "solution": "By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.\n\nYet also there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centres lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons equals\n![](https://cdn.mathpix.com/cropped/2024_12_10_86b1467c9ecfa953dbf8g-5.jpg?height=461&width=513&top_left_y=1437&top_left_x=1314)\n$N=\\sum_{m=1}^{n}(3(n-m)(n-m+1)+1) m=\\left(3 n^{2}+3 n\\right) \\sum_{m=1}^{n} m-3(2 m+1) \\sum_{m=1}^{n} m^{2}+3 \\sum_{m=1}^{n} m^{3}$.\nSince $\\sum_{m=1}^{n} m=\\frac{n(n+1)}{2}, \\sum_{m=1}^{n} m^{2}=\\frac{n(n+1)(2 n+1)}{6}$ and $\\sum_{m=1}^{n} m^{3}=\\left(\\frac{n(n+1)}{2}\\right)^{2}$ it is easily checked that $N=\\left(\\frac{n(n+1)}{2}\\right)^{2}$.", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}

Balkan_MO/segmented/en-2015-BMO-type1.jsonl

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+{"year": "2015", "problem_label": "1", "tier": 1, "problem": "Let $a, b$ and $c$ be positive real numbers. Prove that\n\n$$\na^{3} b^{6}+b^{3} c^{6}+c^{3} a^{6}+3 a^{3} b^{3} c^{3} \\geq a b c\\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\\right)+a^{2} b^{2} c^{2}\\left(a^{3}+b^{3}+c^{3}\\right)\n$$", "solution": "After dividing both sides of the given inequality by $a^{3} b^{3} c^{3}$ it becomes\n\n$$\n\\left(\\frac{b}{c}\\right)^{3}+\\left(\\frac{c}{a}\\right)^{3}+\\left(\\frac{a}{b}\\right)^{3}+3 \\geq\\left(\\frac{a}{c} \\cdot \\frac{b}{c}+\\frac{b}{a} \\cdot \\frac{c}{a}+\\frac{c}{b} \\cdot \\frac{a}{b}\\right)+\\left(\\frac{a}{b} \\cdot \\frac{a}{c}+\\frac{b}{a} \\cdot \\frac{b}{c}+\\frac{c}{a} \\cdot \\frac{c}{b}\\right) .\n$$\n\nSet\n\n$$\n\\frac{b}{a}=\\frac{1}{x}, \\quad \\frac{c}{b}=\\frac{1}{y}, \\quad \\frac{a}{c}=\\frac{1}{z} .\n$$\n\nThen we have that $x y z=1$ and by substituting (2) into (1), we find that\n\n$$\nx^{3}+y^{3}+z^{3}+3 \\geq\\left(\\frac{y}{z}+\\frac{z}{x}+\\frac{x}{y}\\right)+\\left(\\frac{x}{z}+\\frac{y}{x}+\\frac{z}{y}\\right) .\n$$\n\nMultiplying the inequality (3) by $x y z$, and using the fact that $x y z=1$, the inequality is equivalent to\n\n$$\nx^{3}+y^{3}+z^{3}+3 x y z-x y^{2}-y z^{2}-z x^{2}-y x^{2}-z y^{2}-x z^{2} \\geq 0 .\n$$\n\nFinally, notice that by the special case of Schur's inequality\n\n$$\nx^{r}(x-y)(x-z)+y^{r}(y-x)(y-z)+z^{r}(z-y)(z-x) \\geq 0, \\quad x, y, z \\geq 0, r>0,\n$$\n\nwith $r=1$ there holds\n\n$$\nx(x-y)(x-z)+y(y-x)(y-z)+z(z-y)(z-x) \\geq 0\n$$\n\nwhich after expansion actually coincides with the congruence (4).\nRemark 1. The inequality (5) immediately follows by supposing (without loss of generality) that $x \\geq y \\geq z$, and then writing the left hand side of the inequality (5) in the form\n\n$$\n(x-y)(x(x-z)-y(y-z))+z(y-z)(x-z)\n$$\n\nwhich is obviously $\\geq 0$.\nRemark 2. One can obtain the relation (4) using also the substitution $x=a b^{2}, y=b c^{2}$ and $z=c a^{2}$.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}
+{"year": "2015", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be a scalene triangle with incentre $I$ and circumcircle ( $\\omega$ ). The lines $A I, B I, C I$ intersect $(\\omega)$ for the second time at the points $D, E, F$, respectively. The lines through $I$ parallel to the sides $B C, A C, A B$ intersect the lines $E F, D F, D E$ at the points $K, L, M$, respectively. Prove that the points $K, L, M$ are collinear.", "solution": "First we will prove that $K A$ is tangent to $(\\omega)$.\nIndeed, it is a well-known fact that $F A=F B=F I$ and $E A=E C=E I$, so $F E$ is the perpendicular bisector of $A I$. It follows that $K A=K I$ and\n\n$$\n\\angle K A F=\\angle K I F=\\angle F C B=\\angle F E B=\\angle F E A,\n$$\n\nso $K A$ is tangent to $(\\omega)$. Similarly we can prove that $L B, M C$ are tangent to $(\\omega)$ as well.\n![](https://cdn.mathpix.com/cropped/2024_12_10_66c2034b30e72958b298g-2.jpg?height=1096&width=1039&top_left_y=837&top_left_x=505)\n\nLet $A^{\\prime}, B^{\\prime}, C^{\\prime}$ the intersections of $A I, B I, C I$ with $B C, C A, A B$ respectively. From Pascal's Theorem on the cyclic hexagon $A A C D E B$ we get $K, C^{\\prime}, B^{\\prime}$ collinear. Similarly $L, C^{\\prime}, A^{\\prime}$ collinear and $M, B^{\\prime}, A^{\\prime}$ collinear.\n\nThen from Desargues' Theorem for $\\triangle D E F, \\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ which are perspective from the point $I$, we get that points $K, L, M$ of the intersection of their corresponding sides are collinear as wanted.\nRemark (P.S.C.). After proving that $K A, L B, M C$ are tangent to ( $\\omega$ ), we can argue as follows:\nIt readily follows that $\\triangle K A F \\sim \\triangle K A E$ and so $\\frac{K A}{K E}=\\frac{K F}{K A}=\\frac{A F}{A E}$, thus $\\frac{K F}{K E}=\\left(\\frac{A F}{A E}\\right)^{2}$. In a similar way we can find that $\\frac{M E}{M D}=\\left(\\frac{C E}{C D}\\right)^{2}$ and $\\frac{L D}{L F}=\\left(\\frac{B D}{B F}\\right)^{2}$. Multiplying we obtain $\\frac{K F}{K E} \\cdot \\frac{M E}{M D} \\cdot \\frac{L D}{L F}=1$, so by the converse of Menelaus theorem applied in the triangle $D E F$ we get that the points $K, L, M$ are collinear.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}
+{"year": "2015", "problem_label": "3", "tier": 1, "problem": "A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for his favourite actor, and a single vote for his favourite actress. It turns out that for every integer $n \\in\\{1,2, \\ldots, 100\\}$ there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics who voted for the same actor and for the same actress.", "solution": "Let us assume that every critic votes for a different pair of actor and actress. We'll arrive at a contradiction proving the required result. Indeed:\n\nCall the vote of each critic, i.e his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i.e. the one for an actor and the other one for an actress. In this terminology, a double-vote corresponds to two single-votes.\n\nFor each $n=34,35, \\ldots, 100$ let us pick out one actor or one actress who has been voted by exactly $n$ critics (i.e. appears in exactly $n$ single-votes) and call $S$ the set of these movie stars. Calling $a, b$ the number of men and women in $S$, we have $a+b=67$.\n\nNow let $S_{1}$ be the set of double-votes, each having exactly one of its two corresponding singlevotes in $S$, and let $S_{2}$ be the set of double-votes with both its single-votes in $S$. If $s_{1}, s_{2}$ are the number of elements in $S_{1}, S_{2}$ respectively, we have that the number of all double-votes with at least one single-vote in $S$ is $s_{1}+s_{2}$, whereas the number of all double-votes with both single-votes in $S$ is $s_{2} \\leq a b$.\n\nSince all double-votes are distinct, there must exist at least $s_{1}+s_{2}$ critics. But the number of all single-votes in $S$ is $s_{1}+2 s_{2}=34+35+\\cdots+100=4489$, and moreover $s \\leq a b$. So there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \\geq 4489-a b$ critics.\n\nNow notice that as $a+b=67$, the maximum value of $a b$ with $a, b$ integers is obtained for $\\{a, b\\}=$ $\\{33,34\\}$, so $a b \\leq 33 \\cdot 34=1122$. A quick proof of this is the following: $a b=\\frac{(a+b)^{2}-(a-b)^{2}}{4}=$ $\\frac{67^{2}-(a-b)^{2}}{4}$ which is maximized (for not equal integers $a, b$ as $a+b=67$ ) whenever $|a-b|=1$, thus for $\\{a, b\\}=\\{33,34\\}$.\n\nThus there exist at least $4489-1122=3367$ critics which is a contradiction and we are done.\nRemark. We are going here to give some motivation about the choice of number 34, used in the above solution.\nLet us assume that every critic votes for a different pair of actor and actress. One can again start by picking out one actor or one actress who has been voted by exactly $n$ critics for $n=k, k+1, \\ldots, 100$. Then $a+b=100-k+1=101-k$ and the number of all single-votes is $s_{1}+2 s_{2}=k+k+1+\\cdots+100=$ $5050-\\frac{k(k-1)}{2}$, so there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \\geq 5050-\\frac{k(k-1)}{2}-a b$ and\n\n$$\na b=\\frac{(a+b)^{2}-(a-b)^{2}}{4}=\\frac{(101-k)^{2}-(a-b)^{2}}{4} \\leq \\frac{(101-k)^{2}-1}{4} .\n$$\n\nAfter all, the number of critics is at least\n\n$$\n5050-\\frac{k(k-1)}{2}-\\frac{(101-k)^{2}-1}{4}\n$$\n\nIn order to arrive at a contradiction we have to choose $k$ such that\n\n$$\n5050-\\frac{k(k-1)}{2}-\\frac{(101-k)^{2}-1}{4} \\geq 3367\n$$\n\nand solving the inequality with respect to $k$, the only value that makes the last one true is $k=34$.", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}
+{"year": "2015", "problem_label": "4", "tier": 1, "problem": "Prove that among any 20 consecutive positive integers there exists an integer $d$ such that for each positive integer $n$ we have the inequality\n\n$$\nn \\sqrt{d}\\{n \\sqrt{d}\\}>\\frac{5}{2}\n$$\n\nwhere $\\{x\\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greatest integer less than or equal to $x$.", "solution": "Among the given numbers there is a number of the form $20 k+15=5(4 k+3)$. We shall prove that $d=5(4 k+3)$ satisfies the statement's condition. Since $d \\equiv-1(\\bmod 4)$, it follows that $d$ is not a perfect square, and thus for any $n \\in \\mathbb{N}$ there exists $a \\in \\mathbb{N}$ such that $a+1>n \\sqrt{d}>a$, that is, $(a+1)^{2}>n^{2} d>a^{2}$. Actually, we are going to prove that $n^{2} d \\geq a^{2}+5$. Indeed:\n\nIt is known that each positive integer of the form $4 s+3$ has a prime divisor of the same form. Let $p \\mid 4 k+3$ and $p \\equiv-1(\\bmod 4)$. Because of the form of $p$, the numbers $a^{2}+1^{2}$ and $a^{2}+2^{2}$ are not divisible by $p$, and since $p \\mid n^{2} d$, it follows that $n^{2} d \\neq a^{2}+1, a^{2}+4$. On the other hand, $5 \\mid n^{2} d$, and since $5 \\nmid a^{2}+2, a^{2}+3$, we conclude $n^{2} d \\neq a^{2}+2, a^{2}+3$. Since $n^{2} d>a^{2}$ we must have $n^{2} d \\geq a^{2}+5$ as claimed. Therefore,\n\n$$\nn \\sqrt{d}\\{n \\sqrt{d}\\}=n \\sqrt{d}(n \\sqrt{d}-a) \\geq a^{2}+5-a \\sqrt{a^{2}+5}>a^{2}+5-\\frac{a^{2}+\\left(a^{2}+5\\right)}{2}=\\frac{5}{2},\n$$\n\nwhich was to be proved.", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}

Balkan_MO/segmented/en-2016-BMO-type1.jsonl

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+{"year": "2016", "problem_label": "1", "tier": 1, "problem": "Find all injective functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ such that for every real number $x$ and every positive integer $n$,\n\n$$\n\\left|\\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\\right|<2016\n$$", "solution": "From the condition of the problem we get\n\n$$\n\\left|\\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\\right|<2016\n$$\n\nThen\n\n$$\n\\begin{aligned}\n& |n(f(x+n+1)-f(f(x+n)))| \\\\\n= & \\left|\\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\\right| \\\\\n< & 2 \\cdot 2016=4032\n\\end{aligned}\n$$\n\nimplying\n\n$$\n|f(x+n+1)-f(f(x+n))|<\\frac{4032}{n}\n$$\n\nfor every real number $x$ and every positive integer $n$.\nLet $y \\in \\mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain\n\n$$\n|f(y+1)-f(f(y))|<\\frac{4032}{n}\n$$\n\nfor every real number $y$ and every positive integer $n$. The last inequality holds for every positive integer $n$ from where $f(y+1)=f(f(y))$ for every $y \\in \\mathbb{R}$ and since the function $f$ is an injection, then $f(y)=y+1$. The function $f(y)=y+1$ satisfies the required condition.", "problem_match": "# Problem 1.", "solution_match": "\nSolution."}
+{"year": "2016", "problem_label": "2", "tier": 1, "problem": "Let $A B C D$ be a cyclic quadrilateral with $A B<C D$. The diagonals intersect at the point $F$ and lines $A D$ and $B C$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $A D$ and $B C$ respectively, and let $M, S$ and $T$ be the midpoints of $E F, C F$ and $D F$ respectively. Prove that the second intersection point of the circumcircles of triangles $M K T$ and $M L S$ lies on the segment $C D$.", "solution": "Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)\nFirst will prove that the circumcircle of $M L S$ passes through $N$.\nLet $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$, so it passes also through $Q .\\left({ }^{*}\\right)(3)$\n![](https://cdn.mathpix.com/cropped/2024_12_10_12f716d0ab3d0178f6c3g-2.jpg?height=995&width=1003&top_left_y=842&top_left_x=532)\n\nWe will prove that\n\n$$\n\\angle S L Q=\\angle Q N S \\quad \\text { or } \\quad \\angle S L Q+\\angle Q N S=180^{\\circ}\n$$\n\nIndeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and\n\n$$\n\\angle S L C=\\angle S C L=\\angle A C B\n$$\n\nIn addition, since $N$ and $S$ are the midpoints of $D C$ and $F C$ we have that $S N \\| F D$ and similarly, since $Q$ and $N$ are the midpoints of $E C$ and $C D$, so $Q N \\| E D$.\nIt 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\n\n$$\n\\angle E D B=\\angle Q N S \\quad \\text { or } \\quad \\angle E D B+\\angle Q N S=180^{\\circ}\n$$\n\nBut, from the cyclic quadrilateral $A B C D$, we get that\n\n$$\n\\angle E D B=\\angle A C B\n$$\n\nNow, from (2),(3) and (4) we obtain immediately (1), so the quadrilateral $L N S Q$ is cyclic. Since from $\\left(^{*}\\right)$, its circumcircle passes also through $M$, we get that the points $M, L, Q, S, N$ are cocylic and this means that the circumcircle of $M L S$ passes through $N$.\nSimilarly, the circumcircle of $M K T$ passes also through $N$ and we have the desired.", "problem_match": "# Problem 2.", "solution_match": "\nSolution."}
+{"year": "2016", "problem_label": "3", "tier": 1, "problem": "Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)!)+1$ for every prime $p>N$ for which $f(p)$ is a positive integer.\nNote: A monic polynomial has leading coefficient equal to 1.", "solution": "If $f$ is a constant polynomial then it's obvious that the condition cannot hold for\n\n$$\np \\geq 5 \\text { since } f(p)=1\n$$\n\nFrom the divisibility relation $p \\mid 2(f(p))$ ! +1 we conclude that:\n\n$$\nf(p)<p, \\text { for all primes } p>N \\quad(*)\n$$\n\nIn 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.\nNow 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 $\\operatorname{deg} f(x)=1$ and $f(x)=x-a$, for some positive integer $a$. (3)\nThus the given condition becomes:\n\n$$\np \\mid 2(p-a)!+1\n$$\n\nBut we have (using Wilsons theorem)\n\n$$\n\\begin{gathered}\n2(p-3)!\\equiv-(p-3)!(p-2) \\equiv-(p-2)!\\equiv-1(\\bmod p) \\\\\n\\Rightarrow p \\mid 2(p-3)!+1\n\\end{gathered}\n$$\n\nFrom (1) and (2) we get\n\n$$\n\\begin{aligned}\n& (p-3)!\\equiv(p-a)!(\\bmod p) \\\\\n& (p-3)!(-1)^{a}(a-1)!\\equiv(p-a)!(-1)^{a}(a-1)!(\\bmod p) \\\\\n& (p-3)!(-1)^{a}(a-1)!\\equiv 1(\\bmod p)\n\\end{aligned}\n$$\n\nSince $-2(p-3)!\\equiv 1(\\bmod p)$, it follows that\n\n$$\n(-1)^{a}(a-1)!\\equiv-2(\\bmod p)\n$$\n\nTaking $p>(a-1)$ !, we conclude that $a=3$ and so $f(x)=x-3$, for all $x$.\nThe function $f(x)=x-3$ satisfies the required condition.", "problem_match": "# Problem 3.", "solution_match": "\nSolution."}
+{"year": "2016", "problem_label": "4", "tier": 1, "problem": "The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \\times 1201$ or $1201 \\times 1$ contains two squares of the same colour.\nNote: Any rectangle is assumed here to have sides contained in the lines of the grid.", "solution": "Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.\nConsider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \\leq 24$. Any integer translate of $D$ is called a diamond.\nSince any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100 , a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^{2}+25^{2}=1201$ unit squares, a diamond must contain every colour exactly once.\n\nChoose one colour, say, green, and let $a_{1}, a_{2}, \\ldots$ be all green unit squares. Let $P_{i}$ be the diamond of center $a_{i}$. We will show that no unit square is covered by two $P$ 's and that every unit square is covered by some $P_{i}$.\nIndeed, suppose first that $P_{i}$ and $P_{j}$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.\n\nLet, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_{i}$. The diamond $P_{i}$ of center $a_{i}$ will then contain $b$.\n\nTherefore, $P_{1}, P_{2}, \\ldots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)\n![](https://cdn.mathpix.com/cropped/2024_12_10_12f716d0ab3d0178f6c3g-4.jpg?height=689&width=555&top_left_y=1683&top_left_x=753)\n\nFigure 1:\nWithout loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24 x+25 y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \\times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.\n![](https://cdn.mathpix.com/cropped/2024_12_10_12f716d0ab3d0178f6c3g-5.jpg?height=758&width=986&top_left_y=178&top_left_x=541)\n\nFigure 2:", "problem_match": "# Problem 4.", "solution_match": "\nSolution."}

Balkan_MO/segmented/en-2017-BMO-type3.jsonl

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+{"year": "2017", "problem_label": "1", "tier": 1, "problem": "Find all ordered pairs $(x, y)$ of positive integers such that:\n\n$$\nx^{3}+y^{3}=x^{2}+42 x y+y^{2} \\text {. }\n$$", "solution": "Possible initial thoughts about this equation might include:\n(i) I can factorise the sum of cubes on the left.\n(ii) How can I use the 42?\n(iii) The left is cubic and the right is quadratic, so if $x$ or $y$ is very large there will be no solutions.\n\nThe first two might lead us to rewrite the equation as $(x+y)\\left(x^{2}-x y+y^{2}\\right)=x^{2}-x y+y^{2}+43 x y$. The number $43=42+1$ is prime which is good news since we have $(x+y-1)\\left(x^{2}-x y+y^{2}\\right)=43 x y$.\n\nNow we can employ some wishful thinking: if $x$ and $y$ happen to be coprime, then $\\left(x^{2}-x y+y^{2}\\right)$ has no factors in common with $x$ or $y$ so it must divide 43 . This feels like a significant step except for the fact that $x$ and $y$ may not be coprime.\n\nThis suggests writing $x=d X$ and $y=d Y$ where $d$ is the highest common factor of $x$ and $y$.\nWe end up with $(d X+d Y-1)\\left(X^{2}-X Y+Y^{2}\\right)=43 X Y$ so $X^{2}-X Y+Y^{2}$ equals 1 or 43.\nThe first of these readily gives $X=Y=1$. A neat way to deal with the second is to assume $Y \\leq X$ so $43=X^{2}-X Y+Y^{2} \\geq Y^{2}$. This gives six cases for $Y$ which can be checked in turn. Alternatively you can solve $X^{2}-X Y+Y^{2}=43$ for $X$ and fuss about the discriminant.\n\nIn the end the only solutions turn out to be $(x, y)=(22,22),(1,7)$ or $(7,1)$.\nAnother reasonable initial plan is to bound $x+y$ (using observation (iii) above) and then use modular arithmetic to eliminate cases until only the correct solutions remain. Working modulo 7 is particularly appealing since $7 \\mid 42$ and the only cubes modulo 7 are 0,1 and -1 .\n\nWe have $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$ and also $x^{2}+42 x y+y^{2}=(x+y)^{2}+40 x y$ so the equation becomes $(x+y)^{3}=(x+y)^{2}+x y(3 x y+40)$. Now using $x y \\leq\\left(\\frac{x+y}{2}\\right)^{2}$ leads to $x+y \\leq 44$.\n\nThis leaves a mere 484 cases to check! The modulo 7 magic is not really enough to cut this down to an attractive number, and although the approach can obviously be made to work, to call it laborious is an understatement.\n\nOther possible approaches, such as substituting $u=x+y, v=x-y$, seem to lie somewhere between the two described above in terms of the amount of fortitude needed to pull them off.", "problem_match": "\n1. ", "solution_match": "\nSolution."}
+{"year": "2017", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be an acute triangle with with $A B<A C$ and let $\\Gamma$ be its circumcircle. Let the tangents to $\\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ intersects $t_{B}$ at $E$. The circumcircle of triangle $B D C$ meets the side $A C$ at $T$ where $T$ lies between $A$ and $C$. The circumcircle of triangle $B E C$ meets the line $A B$ at $S$ where $B$ lies between $A$ and $S$.\nProve that the lines $S T, B C$ and $A L$ are concurrent.", "solution": "How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.\n\nWe might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also divides $B C$ in this ratio.\n\nSince we are working with ratios of distances, Menelaus' theorem may prove useful.\n![](https://cdn.mathpix.com/cropped/2024_12_10_29004945acd3761b458dg-3.jpg?height=835&width=672&top_left_y=539&top_left_x=681)\n\nA key step is to notice (based on a careful diagram) that $A C$ is tangent to circle $C B S$. Once spotted this is easy to prove. The parallels give us $\\angle B=\\angle B C E$, and, since $B E$ is tangent to circle $A B C$, we have $\\angle E B C=\\angle A$ by the alternate segment theorem. Now angles in a triangle give $\\angle C=\\angle C E B$, and we have the converse to the alternate segment theorem. We obtain $A B$ is tangent to circle $C B D$ in the same way.\n\nNow we have some tangencies and want some ratios.\nTangent-secant yields $A T . b=c^{2}$ and $c . A S=b^{2}$ or, equivalently, $b . C T=b^{2}-c^{2}$ and $c . B S=b^{2}-c^{2}$.\nBy Menelaus we know that $S T$ divides $B C$ in the ratio $A T . B S: A S . C T$ and it's all over bar the shouting.\n\nIf we are not lucky enough to have the stuff about the symmedian at our fingertips, we can still get essentially the same solution with a bit more work. We begin with the second half of the proof above, and establish that $S T$ divides $B C$ in the ratio $c^{2}: b^{2}$. Now we need to prove that $A L$ divides $B C$ in the same ratio.\n\nThe next (non-obvious!) step is to draw a parallel to $B C$ through $A$ as shown.\n![](https://cdn.mathpix.com/cropped/2024_12_10_29004945acd3761b458dg-3.jpg?height=372&width=450&top_left_y=2001&top_left_x=803)\n\nNow $\\triangle A B C \\sim \\triangle B^{\\prime} A B \\sim \\triangle C C^{\\prime} A$ and the ratio $B^{\\prime} A: A C^{\\prime}$ which equals $B X: X C$ drops out as $c^{2}: b^{2}$ as required.\n\nClearly knowing the standard symmedian configuration and corresponding ratio is an enormous advantage.\n\nFinally it is worth noting that, to the right sort of mind, the problem screams out for areal coordinates. These turn out to kill it fairly easily, not least because all three circles pass through at least two vertices of $\\triangle A B C$.", "problem_match": "\n2. ", "solution_match": "\nSolution."}
+{"year": "2017", "problem_label": "3", "tier": 1, "problem": "Let $\\mathbb{N}$ be the set of positive integers. Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that:\n\n$$\nn+f(m) \\text { divides } f(n)+n f(m)\n$$", "solution": "The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \\mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key idea:\n'Eliminate one of the variables from the right hand side.'\nClearly $n+f(m) \\mid f(n)+n f(m)-n(n+f(m))$ so for any $n, m$ we have\n\n$$\nn+f(m) \\mid f(n)-n^{2}\n$$\n\nThis feels like a strong condition: if we fix $n$ and let $f(m)$ go to infinity, then $f(n)-n^{2}$ will have arbitrarily large factors, which implies it must be zero.\n\nWe must be careful: this argument is fine, so long as the function $f$ takes arbitrarily large values. (We also need to check that $f(n)=n^{2}$ satisfies the original statement which it does.)\n\nWe are left with the case where $f$ takes only finitely many values.\nIn this case $f$ must take the same value infinitely often, so it is natural to focus on an infinite set $S \\subset \\mathbb{N}$ such that $f(s)=k$ for all $s \\in S$. If $n, m \\in S$ then the original statement gives $n+k \\mid k+n k$ where $k$ is fixed and $n$ can be as large as we like.\n\nNow we recycle our key idea and eliminate $n$ from the right.\n$n+k \\mid k+n k-k(n+k)$ so $n+k \\mid k-k^{2}$ for arbitrarily large $n$. This means that $k-k^{2}=0$ so $k=1$, since it must be positive.\n\nAt this point we suspect that $f(n)=1$ for all $n$ is the only bounded solution, so we pick some $t$ such that $f(t)=L>1$ and try to get a contradiction.\n\nIn the original statement we can set $m=t$ and get $n+L \\mid f(n)+n L$. Eliminating $L$ from the right gives us nothing new, so how can we proceed? Well, we have an infinite set $S$ such that $f$ is constantly 1 on $S$ so we can take $n \\in S$ to obtain $n+L \\mid 1+n L$\n\nUsing our key idea one more time and eliminating $n$ from the right, we get $n+L \\mid 1-L^{2}$ for arbitrarily large $n$ which is impossible if $L>1$.\n\nA rather different solution can be found by playing around with small values of $m$ and $n$.\nAs before it helps to establish $(\\star)$ but now $(n, m)=(1,1)$ gives $1+f(1) \\mid f(1)-1$.\nThe left is bigger than the right, so the right must be zero $-f(1)=1$.\nNow try $(n, m)=(2,1)$ and obtain $2+f(2) \\mid f(2)-4$. Subtracting the left from the right gives $2+f(2) \\mid-6$. Since $f(2) \\in \\mathbb{N}$ the left is a factor of -6 which is bigger than 2 . This gives $f(2)=1$ or $f(2)=4$.\n\nIn the first case we can plug this back into the orginal statement to get $2+f(m) \\mid 1+2 f(m)$. Now taking two copies of the left away from the right we have $2+f(m) \\mid-3$.\n\nThus $2+f(m)$ must a factor of -3 which is bigger than 2 , so $f(m)=1$ for any $m$.\nBefore proceeding with the case $f(2)=4$ we take another look at our strong result $(\\star)$. Setting $n=m$ gives $n+f(n) \\mid f(n)-n^{2}$ so taking $f(n)-n^{2}$ away from $n+f(n)$ shows that\n\n$$\nn+f(n) \\mid n+n^{2}\n$$\n\nLet see if we can use $(\\star)$ and $(\\dagger)$ to pin down the value of $f(3)$, using $f(2)=4$.\nFrom $(\\star)$ we have $3+4 \\mid f(3)-9$ and from $(\\dagger)$ we have $3+f(3) \\mid 12$. The second of these shows $f(3)$ is 1,3 or 9 , but 1 and 3 are too small to work in the first relation.\n\nSimilarly, setting $(n, m)=(4,3)$ in $(\\star)$ gives $4+9 \\mid f(4)-16$ while $n=4$ in $(\\dagger)$ gives $4+f(4) \\mid 20$. The latter shows $f(4) \\leq 16$ so $13 \\mid 16-f(4)$. The only possible multiples of 13 are 0 and 13 , of which only the first one works. Thus $f(4)=16$.\n\nNow we are ready to try induction. Assume $f(n-1)=(n-1)^{2}$ and use $(\\star)$ and $(\\dagger)$ to obtain $n+(n-1)^{2} \\mid f(n)-n^{2}$ and $n+f(n) \\mid n+n^{2}$. The latter implies $f(n) \\leq n^{2}$ so the former becomes $n^{2}-n+1 \\mid n^{2}-f(n)$. If $f(n) \\neq n^{2}$ then $n^{2}-f(n)=1 \\times\\left(n^{2}-n+1\\right)$ since any other multiple would be too large. However, putting $f(n)=n-1$ into $n+f(n) \\mid n+n^{2}$ implies $2 n-1 \\mid n(1+n)$. This is a contradiction since $2 n-1$ is coprime to $n$ and clearly cannot divide $1+n$.\n\nfor all $m, n \\in \\mathbb{N}$.", "problem_match": "\n3. ", "solution_match": "\nSolution."}
+{"year": "2017", "problem_label": "4", "tier": 1, "problem": "There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:\n(a) Pass one candy to the student on their left or the student on their right.\n(b) Divide all their candies into two, possibly empty, sets and pass one set to the student on their left and the other to the student on their right.\n\nAt each step the students perform their chosen operations simultaneously.\nAn arrangement of candies is legal if it can be obtained in a finite number of steps.\nFind the number of legal arrangements.\n(Two arrangements are different if there is a student who has different numbers of candies in each one.)", "solution": "One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on the next step, then no caramels move. (This is rather too little movement.) Let us see what a small change to this sequence can accomplish. We choose a student with at least one caramel. At the first step, she passes one caramel to the right and any others she has to the left. Every one else passes everything left. At the next step everybody passes everything right. The effect of this is that exactly one caramel has moved exactly two places to the right. Similarly, there is a double step which moves exactly one caramel to places to the left.\n\nIf we have not already done so, now is the time to start working through some small values of $n$.\nThe case $n=3$ yields a useful observation. Going two places (let's call this a double jump) to the left on a triangle is the same as going one place to the right. Indeed if $n$ is odd, say $2 k+1$, then $k$ double jumps to the left moves the caramel one place to the right and vice versa.\n\nIt is now clear that, if $n$ is odd, any arrangement of caramels is possible. We simply move them into position one at a time.\n\nIn the case $n=4$ it seems hard to get all the caramels into one place. Indeed, if we limit ourselves to double jumps, then we can only get $(1,1,1,1),(1,2,1,0),(2,2,0,0)$ and rotations of these arrangements. What can we say about these? Well it seems that students one and three always hold two candies between them. Having noticed this, it is not too hard to make a more general observation: if $n$ is even then a double jump cannot change the total number of caramels held by the odd numbered students. However, double jumps are not the only moves available to us. For example, it is possible to go from $(2,2,0,0)$ to $(3,1,0,0)$. A double jump now gives $(2,1,1,0)$ as well.\n\nTo squeeze maximum value out of the $n=4$ case, it is worth looking at the arrangements we have not yet managed to get to. They are (rotations of) $(4,0,0,0),(3,0,1,0)$ and $(2,0,2,0)$. What do these have in common? They are precisely the arrangements where the even numbered students hold all the caramels. Can we prove that these are illegal? Well, what can we say about an arrangement which precedes one of these elusive ones? This question leads to the last big idea in the solution to this problem. If after some step the even numbered students have all the caramels, they cannot have had any at all before the step, else they would have passed at least one caramel to an odd numbered student.\n\nTurning this around gives a crucial lemma for even values of $n$. Let's call a caramel held by an odd numbered student an odd caramel and define even caramels similarly. Let's call an arrangement with at least one odd caramel and at least one even caramel balanced. If the arrangement is balanced before some step, then it will be balanced after the step. The initial position is balanced, so every legal position is balanced.\n\nFinally we are on the home straight. We claim that every balanced position is legal. Using double jumps we can move to $\\left(\\ldots, \\frac{n}{2}, \\frac{n}{2}, 0, \\ldots\\right)$. Now we need to tinker with the numbers of odd and even caramels. There are lots of usable sequences. For example:\n\n$$\n\\begin{gathered}\n(\\ldots, a, b, 0, \\ldots) \\\\\n(\\ldots, a-1,1, b, \\ldots) \\\\\n(\\ldots, a-1, b+1,0, \\ldots)\n\\end{gathered}\n$$\n\ncan be used to change the number of odd caramels provided $a-1 \\geq 1$.\nOnce we have the correct number of odd and even caramels, they can be moved into place using double jumps.\n\nIt remains to observe that there are $\\binom{2 n-1}{n}$ possible arrangements of caramels, and that if $n$ is even, then $2\\left(\\frac{\\frac{3 n}{2}-1}{n}\\right)$ of these are not balanced.\n\nAnother sensible approach is to think about which steps are reversible. It turns out that many are, including all those where the students all use option (b).\n\nIt is possible to argue that if $n$ is odd, then we can start with any position, move to $(\\ldots, n, \\ldots)$ reversibly, then move to the initial position reversibly. Playing the whole tape backwards shows all positions are legal.\n\nIf $n$ is even it is possible to start from any balanced position and reversibly move to $(\\ldots, n-1,1, \\ldots)$ and thence to the initial position so we are done.\n\n[^0]\n[^0]:    ${ }^{2}$ The word 'candy' was a little too grating for my delicate British ears. I am grateful to the Italians for suggesting the more elegant alternative.", "problem_match": "\n4. ", "solution_match": "\nSolution."}

Balkan_MO/segmented/en-2018-BMO-type3.jsonl

@@ -0,0 +1,5 @@
+{"year": "2018", "problem_label": "1", "tier": 1, "problem": "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 the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diameter of the circle $k$.", "solution": "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\n\n$$\n\\frac{M P}{B H}=\\frac{D M}{D B}=\\frac{K M}{A B} \\quad \\text { and } \\quad \\frac{M Q}{A H}=\\frac{C M}{C A}=\\frac{M L}{A B}\n$$\n\nwe obtain $A H / H B=K M / M L$.\nLet 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 $M H$.\nThe 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 $\\varangle A C B=90^{\\circ}$.\n![](https://cdn.mathpix.com/cropped/2024_12_10_a48035b438f3db5360fbg-2.jpg?height=827&width=798&top_left_y=1117&top_left_x=635)", "problem_match": "\n1. ", "solution_match": "\nSolution."}
+{"year": "2018", "problem_label": "2", "tier": 1, "problem": "Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \\ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.", "solution": "Answer: $q=1$.\nLet $x_{A}^{(n)}$ (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\n\n$$\n0=x_{A}^{(n)}-x_{B}^{(n)}=P(q),\n$$\n\nwhere $P$ 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. $q \\in\\left\\{\\frac{1}{2}, 1,2\\right\\}$.\nIt is clearly possible when $q=1$.\n\nWe 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\n\n$$\n\\left|x_{B}^{(k+1)}-x_{A}^{(k+1)}\\right|+\\left|y_{B}^{(k+1)}-y_{A}^{(k+1)}\\right|=2 q^{k} .\n$$\n\nBut 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\n\n$$\n\\left|x_{A}^{(n)}-x_{A}^{(k+1)}\\right|+\\left|y_{A}^{(n)}-y_{A}^{(k+1)}\\right| \\leqslant q^{k+1}+q^{k+2}+\\ldots+q^{n-1}\n$$\n\nand similarly for the second ant. Combining (1) and (2) with the triangle inequality, we obtain for any $n \\geqslant k+1$\n\n$$\n\\left|x_{B}^{(n)}-x_{A}^{(n)}\\right|+\\left|y_{B}^{(n)}-y_{A}^{(n)}\\right| \\geqslant 2 q^{k}-2\\left(q^{k+1}+q^{k+2}+\\ldots+q^{n-1}\\right),\n$$\n\nwhich is strictly positive for $q=\\frac{1}{2}$. So for any $n \\geqslant k+1$, the ants cannot meet after $n$ minutes. Thus $q \\neq \\frac{1}{2}$.\nFinally, 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.", "problem_match": "\n2. ", "solution_match": "\nSolution 1."}
+{"year": "2018", "problem_label": "2", "tier": 1, "problem": "Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \\ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.", "solution": "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 $a_{k}, b_{k} \\in\\{1,-1, i,-i\\}$.\nIf $\\alpha_{n}=\\beta_{n}$ for some $n>0$, then\n\n$$\n\\sum_{k=0}^{n-1}\\left(a_{k}-b_{k}\\right) q^{k}=0, \\quad \\text { where } \\quad a_{k}-b_{k} \\in\\{0, \\pm 1 \\pm i, \\pm 2, \\pm 2 i\\}\n$$\n\nNote that the coefficient $a_{k}-b_{k}$ is always divisible by $1+i$ in Gaussian integers: indeed,\n\n$$\nc_{k}=\\frac{a_{k}-b_{k}}{1+i} \\in\\{0, \\pm 1, \\pm i, \\pm 1 \\pm i\\}\n$$\n\nCanceling $1+i$, 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 $a=b=1$.", "problem_match": "\n2. ", "solution_match": "\nSolution 2."}
+{"year": "2018", "problem_label": "3", "tier": 1, "problem": "Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.\nDetermine 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.", "solution": "By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \\mid n$.\nA 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 $k$.\n\n- Given a 0-happy position, the player in turn is unable to play and loses.\n- 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 ( $\\left.k-1\\right)$-happy because $v_{2}\\left(a+\\frac{1}{2} b\\right)=v_{2}\\left(\\frac{1}{2} b\\right)=v_{2}\\left(b+\\frac{1}{2} a\\right)=v_{2}\\left(\\frac{1}{2} a\\right)=k-1$.\n\nTherefore, 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.\n\n- 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.\n- 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 ( $\\left.k-1\\right)$-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 $v_{2}\\left(a+\\frac{1}{2} b\\right)>v_{2}\\left(\\frac{1}{2} b\\right)=k$.\n\nTherefore a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.", "problem_match": "\n3. ", "solution_match": "\nSolution."}
+{"year": "2018", "problem_label": "4", "tier": 1, "problem": "Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.\n\nTime allowed: 4 hours and 30 minutes.\nEach problem is worth 10 points.", "solution": "Answer: $(p, q)=(3,3)$.\nFor $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.\nObserve 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. $\\operatorname{ord}_{r}(a) \\in\\{2,2 p\\}$.\nNote 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.\n\n$$\n3 p^{q-1}+1=2^{\\alpha} 7^{\\beta} p_{1}^{\\gamma_{1}} \\cdots p_{k}^{\\gamma_{k}}\n$$\n\nwhere $p_{i} \\notin\\{2,7\\}$ are prime divisors with $p_{i} \\equiv 1(\\bmod 2 p)$.\nWe already know that $\\alpha \\leqslant 2$. Also, note that\n\n$$\n\\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\\cdots+17^{p-1} \\equiv p \\cdot 4^{p-1} \\quad(\\bmod 7)\n$$\n\nso $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\\beta \\leqslant 1$.\nIf $q=2$, 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.\nThus $q>2$, 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.", "problem_match": "\n4. ", "solution_match": "\nSolution."}