APMO data: Fix script bug and manually fix data errors

APMO/md/en-apmo1990_sol.md

@@ -56,7 +56,9 @@ $$
 (which by (1) is equivalent to $A D^{2} / 3=C D^{2}$, i.e. $A D=C D \cdot \sqrt{3}$ ).
 Let, without loss of generality, $C D=1$. Then $A$ lies on the circle of radius $\sqrt{3}$ with centre $D$. If $C D$ and $D A$ are perpendicular, the angle $B A C$ is $60^{\circ}$, otherwise it must be less.
 In this case, for each angle $B A C$ there are two solutions, which are congruent.
-${ }^{\top} \mathrm{n}$ the figure as shown below, we first show that it is necessary that $\angle A$ is less than $90^{\circ}$ if the quadrilateral $A E G F$ ; cyclic.
+
+THIRD SOLUTION
+in the figure as shown below, we first show that it is necessary that $\angle A$ is less than $90^{\circ}$ if the quadrilateral $A E G F$ ; cyclic.
 ![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-02.jpg?height=418&width=638&top_left_y=344&top_left_x=630)
 
 Now, since $E F \| B C$, we get

APMO/md/en-apmo1995_sol.md

@@ -1,5 +1,3 @@
-# SOLUTIONS 
-
 Note: On the left side of the page the maximum number of points that may be awarded for every part of the solution is indicated in brackets.
 
 Question 1. Suppose that $\left(a_{1}, a_{2}, \ldots, a_{1995}\right)$ is a solution of the given system of incqualities. Then

APMO/md/en-apmo2020_sol.md

@@ -1,6 +1,6 @@
 # APMO 2020 Solution 
 
-1. Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
+Problem 1. Let $\Gamma$ be the circumcircle of $\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.
 Solution 1 From the conditions, we have
 ![](https://cdn.mathpix.com/cropped/2024_11_22_207f52578b01f443eb43g-1.jpg?height=960&width=743&top_left_y=772&top_left_x=710)
 
@@ -23,7 +23,8 @@ Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\ang
 
 Let $P$ be the intersection of $B F$ and $D E$. From $\angle A F P=\angle A C B=\angle A E P$, the points $A, P, F, E$ are concyclic. In addition, from $\angle E P A=\angle E F A=\angle D B A$, points $A, B, D, P$ are concyclic.
 By considering the radical centre of $(B D F E),(A P F E)$ and $(B D P A)$, we find that the lines $B D, A P, E F$ are concurrent at $C$. The result follows.
-2. Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
+
+Problem 2. Show that $r=2$ is the largest real number $r$ which satisfies the following condition:
 
 If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
 
@@ -87,7 +88,8 @@ $$
 
 hence $a_{n} \leq a_{n+2} \leq a_{n+1}$.
 Now let $m$ be a positive integer such that $a_{m+2}>a_{m}$. By the observations above, we must have $a_{m}<a_{m+2} \leq a_{m+1}$. Thus the assertion ( $\dagger$ ) is true for $k=1$. Assume that the assertion holds for some positive integer $k$. Using observation (a), we get $a_{m+2 k+1}=a_{m+2 k-1}=a_{m+1}$. Thus $a_{m+2 k} \leq a_{m+2 k+1}$, and then using observation (b), we get $a_{m+2 k+2} \leq a_{m+2 k+1}=a_{m+1}$, which proves the assertion ( $\dagger$ ).
-3. Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways.
+
+Problem 3. Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways.
 
 ## Solution:
 
@@ -123,7 +125,7 @@ A_{t}(x)=A_{t-1}(x)+x^{s_{t}} u(x)+k\left(x^{s_{t}+m+1}+\cdots x^{2 s_{t}-1}\rig
 $$
 
 Recall that the coefficent of $x^{s_{t+1}}$ in $A_{t}(x)$ is $k-1$. But if $s_{t}+m+1<s_{t+1}<s_{2 t}$, then the coefficient of $x^{s_{t+1}}$ in $A_{t}(x)$ is at least $k$, which is a contradiction. Therefore $s_{t+1} \geq 2 s_{t}$.
-4. Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
+Problem 4. Let $\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:
 For any infinite sequence $a_{1}, a_{2}, \ldots$ of integers in which each integer in $\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{i}+a_{i+1}+\cdots+a_{j}=P(k)$.
 
 ## Solution:
@@ -147,7 +149,8 @@ $$
 which is given by $a_{3 n+1}=2 n, a_{3 n+2}=2 n+1, a_{3 n+3}=-(n+1)$ for all $n \geq 0$. Notice that for any $i<j$ we have $a_{i}+\cdots+a_{j} \geq 0$. Then for the sequence $\left\{b_{n}\right\}$ defined by $b_{n}=a_{n}+c$, clearly $b_{i}+\cdots+b_{j} \geq\left(a_{i}+\cdots+a_{j}\right)+2 c>c$ which is out side the range of $P(x)$.
 
 Now if $P$ is bounded from below, there exist a positive integer $c$ such that $P(x)>-c$. In this case, take $b_{n}$ to be $b_{n}=-a_{n}-c$. Then for all $i<j$ we have $b_{i}+\cdots+b_{j} \leq-\left(a_{1}+\cdots a_{n}\right)-2 c<-c$ which is again out side the range of $P(x)$.
-5. Let $n \geq 3$ be a fixed integer. The number 1 is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A move consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers 1 and $a+b$, then adding one stone to the first bucket and $\operatorname{gcd}(a, b)$ stones to the second bucket. After some finite number of moves, there are $s$ stones in the first bucket and $t$ stones in the second bucket, where $s$ and $t$ are positive integers. Find all possible values of the ratio $\frac{t}{s}$.
+
+Problem 5. Let $n \geq 3$ be a fixed integer. The number 1 is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A move consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers 1 and $a+b$, then adding one stone to the first bucket and $\operatorname{gcd}(a, b)$ stones to the second bucket. After some finite number of moves, there are $s$ stones in the first bucket and $t$ stones in the second bucket, where $s$ and $t$ are positive integers. Find all possible values of the ratio $\frac{t}{s}$.
 
 ## Solution:
 

APMO/md/en-apmo2021_sol.md

@@ -1,10 +1,11 @@
 # APMO 2021 Solution 
 
-1. Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
+Problem 1. Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\lfloor x\rfloor$.
 Note: $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$
 Solution Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\lfloor x\rfloor$ with $\lfloor x\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \leq x<k+1$, we get $k^{2} \leq x^{2}=r k<$ $k^{2}+2 k+1 \leq k^{2}+3 k$, hence $k \leq r<k+3$, or $r-3<k \leq r$. There are at most three positive integers in the interval $(r-3, r]$. Thus there are at most three possible values for $k$. Consequently, there are at most three positive solutions to the given equation.
 Now suppose that $k$ is a positive integer in the interval $[r-2, r]$. There are at least two such positive integer. Observe that $k \leq \sqrt{r k} \leq \sqrt{(k+2) k}<k+1$ and so $r k=r\lfloor\sqrt{r k}\rfloor$. We conclude that the equation $x^{2}=r\lfloor x\rfloor$ has at least two positive solutions, namely $x=\sqrt{r k}$ with $k \in[r-2, r]$.
-2. For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
+
+Problem 2. For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.
 Determine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \leq 2021$.
 Solution There are two possible families of solutions:
 
@@ -31,7 +32,8 @@ Solution 2 Suppose $c \geq 2$. Let $n$ be a positive integer such that $n>2 c M,
 
 If $d \leq-2023$, then there are at least 2022 pairs $a<b$ such that $P(a)=P(b)$, namely $(a, b)=$ $(1,-2 d-1),(2,-2 d-2), \ldots,(-d-1,-d+1)$. This implies that $d \geq-2022$.
 Finally, we verify that $P(x)=x+d$ satisfies the condition for any $d \geq-2022$. Fix a positive integer $n$. Note that $\| P(b)|-|P(a)||<n$ for all positive integers $a<b \leq n$, so the only pairs $a, b$ for which $|P(b)|-|P(a)|$ could be divisible by $n$ are those for which $|P(a)|=|P(b)|$. When $d \geq-2022$, there are indeed at most 2021 such pairs.
-3. Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
+
+Problem 3. Let $A B C D$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.
 
 ## Solution 1
 
@@ -114,7 +116,8 @@ $$
 
 Denote by $X$ and $Y$ the second intersections of lines $B M$ and $C M$ with the circumcircle of $B C L$, respectively. Since $\angle M B C=\angle M C B$, we have $B C \| X Y$. It suffices to show that $B N \| X L$ and $C N \| Y L$. Indeed, from this it follows that $\triangle B C N \sim \triangle X Y L$, and therefore a homothety with center $M$ that maps $B$ to $X$ and $C$ to $Y$ also maps $N$ to $L$, implying that $N$ lies on the line $L M$.
 By symmetry, it suffices to show that $C N \| Y L$, which is equivalent to showing that $\angle B C N=\angle X Y L$. But we have $\angle B C N=\angle M B L=\angle X B L=\angle X Y L$, completing the proof.
-4. Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:
+
+Problem 4. Given a $32 \times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:
 (a) No good subset consists of 888 cells.
 (b) There exists a good subset consisting of at least 666 cells.
 
@@ -133,7 +136,8 @@ For a tile $T$, let $|T|$ be the number of pieces of cheese in it. By straightfo
 ## Another proof of (a).
 
 Let $X_{N}$ be the largest possible density of cheese-cells in a good subset on an $N \times N$ table. We will show that $X_{N} \leq 4 / 5+o(1)$. Specifically, this gives $X_{32} \leq 817 / 1024$. We look at the (discrete analogue) of the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a different empty square, hence $X_{N} \leq 4 / 5$.
-5. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(a)-b)+b f(2 a)$ is a perfect square for all integers $a$ and $b$.
+
+Problem 5. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(a)-b)+b f(2 a)$ is a perfect square for all integers $a$ and $b$.
 
 ## Solution 1.
 

APMO/md/en-apmo2022_sol.md

@@ -1,8 +1,6 @@
 # APMO 2022 Solution 
 
-1. Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.
-
-## Solution
+Problem 1. Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.
 
 ## Solution 1.1
 
@@ -60,10 +58,9 @@ a<2\left(a^{\frac{3}{2}}\right)^{\frac{1}{3}}=2 \sqrt{a} \Longrightarrow a<4 .
 $$
 
 Extracting $a=2,3$ by hand yields no additional solution.
-2. Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
-![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)
 
-## Solution
+Problem 2. Let $A B C$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\triangle A C D$ and the circumcircle of $\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.
+![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)
 
 ## Solution 2.1
 
@@ -107,7 +104,8 @@ $$
 
 and the other solution $D=(-d, 0)$.
 From this we obtain the equation of the line $E F$ which is $a x+(d-2 c) y+a c=0$. It passes through $P(-c, 0)$ which is independent of $d$.
-3. Find all positive integers $k<202$ for which there exists a positive integer $n$ such that
+
+Problem 3. Find all positive integers $k<202$ for which there exists a positive integer $n$ such that
 
 $$
 \left\{\frac{n}{202}\right\}+\left\{\frac{2 n}{202}\right\}+\cdots+\left\{\frac{k n}{202}\right\}=\frac{k}{2}
@@ -140,7 +138,8 @@ Since $n$ is not divisible by 101 , which is prime, it follows that $101 \mid k(
 - If $k=101$, we may choose $n=51$. To see this, note that the first four remainders are $51,102,153$, 2 , which have an average of 77 . The next four remainders $(53,104,155,4)$ are shifted upwards from the first four remainders by 2 each, and so on, until the 25 th set of the remainders ( 99 , $150,201,50)$ which have an average of 125 . Hence, the first 100 remainders have an average of $\frac{77+125}{2}=101$. The 101th remainder is also 101 , meaning that the average of all 101 remainders is 101 .
 
 In conclusion, all values $k \in\{1,100,101,201\}$ satisfy the initial condition.
-4. Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the box containing marble $i+1$. Cathy wins if at any point there is a box containing only marble $n$.
+
+Problem 4. Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the box containing marble $i+1$. Cathy wins if at any point there is a box containing only marble $n$.
 Determine all pairs of integers $(n, k)$ such that Cathy can win this game.
 
 ## Solution
@@ -155,7 +154,8 @@ a victory is possible if $n=2^{k-1}$ or smaller.
 We now prove by induction that Cathy loses if $n=2^{k-1}+1$. The base case of $n=2$ and $k=1$ is trivial. Assume a victory is impossible for $m$ boxes and $2^{m-1}+1$ marbles. For the sake of contradiction, suppose that victory is possible for $m+1$ boxes and $2^{m}+1$ marbles. In a winning sequence of moves, consider the last time a marble $2^{m-1}+1$ leaves the starting box, call this move $X$. After $X$, there cannot be a time when marbles $1, \ldots, 2^{m-1}+1$ are all in the same box. Otherwise, by reversing these moves after $X$ and deleting marbles greater than $2^{m-1}+1$, it gives us a winning sequence of moves for $2^{m-1}+1$ marbles and $m$ boxes (as the original starting box is not used here), contradicting the inductive hypothesis. Hence starting from $X$, marbles 1 will never be in the same box as any marbles greater than or equal to $2^{m-1}+1$.
 
 Now delete marbles $2, \ldots, 2^{m-1}$ and consider the winning moves starting from $X$. Marble 1 would only move from one empty box to another, while blocking other marbles from entering its box. Thus we effectively have a sequence of moves for $2^{m-1}+1$ marbles, while only able to use $m$ boxes. This again contradicts the inductive hypothesis. Therefore, a victory is not possible if $n=2^{k-1}+1$ or greater.
-5. Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.
+
+Problem 5. Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.
 
 ## Solution
 

APMO/segment_script/segment.py

@@ -23,8 +23,8 @@ def analyze(text: str) -> Tuple[List, int]:
     Returns:
         Tuple[List, int]: A tuple containing the tags and problem number.
     """
-    problem_pattern = re.compile(r'(?:\n|# )Problem\s+(\d+)(.)?')
-    solution_pattern = re.compile(r'(?:\n|# |\()Solution(?:\s+(\d+)|\)|\.|\n|:)')
+    problem_pattern = re.compile(r'(?:\n|# )(?:Problem|Question)\s+(\d+)(.)?', re.IGNORECASE)
+    solution_pattern = re.compile(r'(?:\n|# |\()(?:Solution|FIRST SOLUTION|SECOND SOLUTION|THIRD SOLUTION|Solution and Marking Scheme|First Solution and Marking Scheme|Second Solution and Marking Scheme)(?:\s+(\d*)|\)|\.|\n|:|\s+\(.+\)(:|.))', re.IGNORECASE)
 
     tags = []
     tags.extend([(x, problem_tag) for x in problem_pattern.finditer(text)])
@@ -93,6 +93,9 @@ def main():
     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('**/*.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)
@@ -101,12 +104,18 @@ def main():
 
         tags, problem_num = analyze(text)
 
-        if problem_num > 0:
-            segments = segment(text, tags)
-            pairs = join(tags, segments)        
+        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__':

APMO/segmented/en-apmo1989_sol.jsonl

@@ -1,6 +1,6 @@
 {"year": "1989", "problem_label": "1", "tier": 1, "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive real numbers, and let\n\n$$\nS=x_{1}+x_{2}+\\cdots+x_{n}\n$$\n\nProve that\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq 1+S+\\frac{S^{2}}{2!}+\\frac{S^{3}}{3!}+\\cdots+\\frac{S^{n}}{n!}\n$$", "solution": "Let $\\sigma_{k}$ be the $k$ th symmetric polynomial, namely\n\n$$\n\\sigma_{k}=\\sum_{\\substack{|S|=k \\\\ S \\subseteq\\{1,2, \\ldots, n\\}}} \\prod_{i \\in S} x_{i},\n$$\n\nand more explicitly\n\n$$\n\\sigma_{1}=S, \\quad \\sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\\cdots+x_{n-1} x_{n}, \\quad \\text { and so on. }\n$$\n\nThen\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right)=1+\\sigma_{1}+\\sigma_{2}+\\cdots+\\sigma_{n}\n$$\n\nThe expansion of\n\n$$\nS^{k}=\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)^{k}=\\underbrace{\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right) \\cdots\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)}_{k \\text { times }}\n$$\n\nhas at least $k$ ! occurrences of $\\prod_{i \\in S} x_{i}$ for each subset $S$ with $k$ indices from $\\{1,2, \\ldots, n\\}$. In fact, if $\\pi$ is a permutation of $S$, we can choose each $x_{\\pi(i)}$ from the $i$ th factor of $\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)^{k}$. Then each term appears at least $k$ ! times, and\n\n$$\nS^{k} \\geq k!\\sigma_{k} \\Longleftrightarrow \\sigma_{k} \\leq \\frac{S^{k}}{k!}\n$$\n\nSumming the obtained inequalities for $k=1,2, \\ldots, n$ yields the result.", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}
 {"year": "1989", "problem_label": "1", "tier": 1, "problem": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be positive real numbers, and let\n\n$$\nS=x_{1}+x_{2}+\\cdots+x_{n}\n$$\n\nProve that\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq 1+S+\\frac{S^{2}}{2!}+\\frac{S^{3}}{3!}+\\cdots+\\frac{S^{n}}{n!}\n$$", "solution": "By AM-GM,\n\n$$\n\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right) \\leq\\left(\\frac{\\left(1+x_{1}\\right)+\\left(1+x_{2}\\right)+\\cdots+\\left(1+x_{n}\\right)}{n}\\right)^{n}=\\left(1+\\frac{S}{n}\\right)^{n}\n$$\n\nBy the binomial theorem,\n\n$$\n\\left(1+\\frac{S}{n}\\right)^{n}=\\sum_{k=0}^{n}\\binom{n}{k}\\left(\\frac{S}{n}\\right)^{k}=\\sum_{k=0}^{n} \\frac{1}{k!} \\frac{n(n-1) \\ldots(n-k+1)}{n^{k}} S^{k} \\leq \\sum_{k=0}^{n} \\frac{S^{k}}{k!}\n$$\n\nand the result follows.\nComment: Maclaurin's inequality states that\n\n$$\n\\frac{\\sigma_{1}}{n} \\geq \\sqrt{\\frac{\\sigma_{2}}{\\binom{n}{2}}} \\geq \\cdots \\geq \\sqrt[k]{\\frac{\\sigma_{k}}{\\binom{n}{k}}} \\geq \\cdots \\geq \\sqrt[n]{\\frac{\\sigma_{n}}{\\binom{n}{n}}}\n$$\n\nThen $\\sigma_{k} \\leq\\binom{ n}{k} \\frac{S^{k}}{n^{k}}=\\frac{1}{k!} \\frac{n(n-1) \\ldots(n-k+1)}{n^{k}} S^{k} \\leq \\frac{S^{k}}{k!}$.", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}
-{"year": "1989", "problem_label": "2", "tier": 1, "problem": "Prove that the equation\n\n$$\n6\\left(6 a^{2}+3 b^{2}+c^{2}\\right)=5 n^{2}\n$$\n\nhas no solutions in integers except $a=b=c=n=0$.", "solution": "We can suppose without loss of generality that $a, b, c, n \\geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to\n\n$$\n6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}\n$$\n\nThe number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to\n\n$$\n2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}\n$$\n\nNow look at the equation modulo 8:\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 2\\left(n_{0}^{2}-a^{2}\\right) \\quad(\\bmod 8)\n$$\n\nIntegers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\\left(n_{0}-a\\right)\\left(n_{0}+a\\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\\left(n_{0}^{2}-a^{2}\\right)$ is a multiple of 8 , and\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 0 \\quad(\\bmod 8)\n$$\n\nIf $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \\equiv 4(\\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find\n\n$$\na^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}\n$$\n\nLook at the last equation modulo 8:\n\n$$\na^{2}+3 n_{0}^{2} \\equiv 2\\left(c_{1}^{2}-b_{0}^{2}\\right) \\quad(\\bmod 8)\n$$\n\nA similar argument shows that $a$ and $n_{0}$ are both even.\nWe have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find\n\n$$\n6\\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\\right)=5(n / 2)^{2}\n$$\n\nand we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
+{"year": "1989", "problem_label": "2", "tier": 1, "problem": "Prove that the equation\n\n$$\n6\\left(6 a^{2}+3 b^{2}+c^{2}\\right)=5 n^{2}\n$$\n\nhas no solutions in integers except $a=b=c=n=0$.", "solution": "We can suppose without loss of generality that $a, b, c, n \\geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to\n\n$$\n6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}\n$$\n\nThe number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to\n\n$$\n2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}\n$$\n\nNow look at the equation modulo 8:\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 2\\left(n_{0}^{2}-a^{2}\\right) \\quad(\\bmod 8)\n$$\n\nIntegers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\\left(n_{0}-a\\right)\\left(n_{0}+a\\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\\left(n_{0}^{2}-a^{2}\\right)$ is a multiple of 8 , and\n\n$$\nb^{2}+3 c_{0}^{2} \\equiv 0 \\quad(\\bmod 8)\n$$\n\nIf $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \\equiv 4(\\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find\n\n$$\na^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}\n$$\n\nLook at the last equation modulo 8:\n\n$$\na^{2}+3 n_{0}^{2} \\equiv 2\\left(c_{1}^{2}-b_{0}^{2}\\right) \\quad(\\bmod 8)\n$$\n\nA similar argument shows that $a$ and $n_{0}$ are both even.\nWe have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find\n\n$$\n6\\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\\right)=5(n / 2)^{2}\n$$\n\nand we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
 {"year": "1989", "problem_label": "3", "tier": 1, "problem": "Let $A_{1}, A_{2}, A_{3}$ be three points in the plane, and for convenience,let $A_{4}=A_{1}, A_{5}=A_{2}$. For $n=1,2$, and 3 , suppose that $B_{n}$ is the midpoint of $A_{n} A_{n+1}$, and suppose that $C_{n}$ is the midpoint of $A_{n} B_{n}$. Suppose that $A_{n} C_{n+1}$ and $B_{n} A_{n+2}$ meet at $D_{n}$, and that $A_{n} B_{n+1}$ and $C_{n} A_{n+2}$ meet at $E_{n}$. Calculate the ratio of the area of triangle $D_{1} D_{2} D_{3}$ to the area of triangle $E_{1} E_{2} E_{3}$.\nAnswer: $\\frac{25}{49}$.", "solution": "Let $G$ be the centroid of triangle $A B C$, and also the intersection point of $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ 。\n![](https://cdn.mathpix.com/cropped/2024_11_22_831b46bbbb7a3a492868g-3.jpg?height=441&width=449&top_left_y=750&top_left_x=769)\n\nBy Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,\n\n$$\n\\frac{A_{1} B_{1}}{A_{1} A_{2}} \\cdot \\frac{D_{1} A_{3}}{D_{1} B_{1}} \\cdot \\frac{C_{2} A_{2}}{C_{2} A_{3}}=1 \\Longleftrightarrow \\frac{D_{1} A_{3}}{D_{1} B_{1}}=2 \\cdot 3=6 \\Longleftrightarrow \\frac{D_{1} B_{1}}{A_{3} B_{1}}=\\frac{1}{7}\n$$\n\nSince $A_{3} G=\\frac{2}{3} A_{3} B_{1}$, if $A_{3} B_{1}=21 t$ then $G A_{3}=14 t, D_{1} B_{1}=\\frac{21 t}{7}=3 t, A_{3} D_{1}=18 t$, and $G D_{1}=A_{3} D_{1}-A_{3} G=18 t-14 t=4 t$, and\n\n$$\n\\frac{G D_{1}}{G A_{3}}=\\frac{4}{14}=\\frac{2}{7}\n$$\n\nSimilar results hold for the other medians, therefore $D_{1} D_{2} D_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $-\\frac{2}{7}$.\nBy Menelao's theorem on triangle $A_{1} A_{2} B_{2}$ and line $C_{1} E_{1} A_{3}$,\n\n$$\n\\frac{C_{1} A_{1}}{C_{1} A_{2}} \\cdot \\frac{E_{1} B_{2}}{E_{1} A_{1}} \\cdot \\frac{A_{3} A_{2}}{A_{3} B_{2}}=1 \\Longleftrightarrow \\frac{E_{1} B_{2}}{E_{1} A_{1}}=3 \\cdot \\frac{1}{2}=\\frac{3}{2} \\Longleftrightarrow \\frac{A_{1} E_{1}}{A_{1} B_{2}}=\\frac{2}{5}\n$$\n\nIf $A_{1} B_{2}=15 u$, then $A_{1} G=\\frac{2}{3} \\cdot 15 u=10 u$ and $G E_{1}=A_{1} G-A_{1} E_{1}=10 u-\\frac{2}{5} \\cdot 15 u=4 u$, and\n\n$$\n\\frac{G E_{1}}{G A_{1}}=\\frac{4}{10}=\\frac{2}{5}\n$$\n\nSimilar results hold for the other medians, therefore $E_{1} E_{2} E_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $\\frac{2}{5}$.\nThen $D_{1} D_{2} D_{3}$ and $E_{1} E_{2} E_{3}$ are homothetic with center $G$ and ratio $-\\frac{2}{7}: \\frac{2}{5}=-\\frac{5}{7}$, and the ratio of their area is $\\left(\\frac{5}{7}\\right)^{2}=\\frac{25}{49}$.", "problem_match": "# Problem 3", "solution_match": "\nSolution\n"}
-{"year": "1989", "problem_label": "4", "tier": 1, "problem": "Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \\leq a<$ $b \\leq n$. Show that there are at least\n\n$$\n4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n}\n$$\n\ntriples $(a, b, c)$ such that $(a, b),(a, c)$, and $(b, c)$ belong to $S$.", "solution": "Call a triple $(a, b, c)$ good if and only if $(a, b),(a, c)$, and $(b, c)$ all belong to $S$. For $i$ in $\\{1,2, \\ldots, n\\}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $\\left|D_{i}\\right|=d_{i}$ ). Consider a pair $(i, j) \\in S$. Our goal is to estimate the number of integers $k$ such that any permutation of $\\{i, j, k\\}$ is good, that is, $\\left|D_{i} \\cap D_{j}\\right|$. Note that $i \\notin D_{i}$ and $j \\notin D_{j}$, so $i, j \\notin D_{i} \\cap D_{j}$; thus any $k \\in D_{i} \\cap D_{j}$ is different from both $i$ and $j$, and $\\{i, j, k\\}$ has three elements as required. Now, since $D_{i} \\cup D_{j} \\subseteq\\{1,2, \\ldots, n\\}$,\n\n$$\n\\left|D_{i} \\cap D_{j}\\right|=\\left|D_{i}\\right|+\\left|D_{j}\\right|-\\left|D_{i} \\cup D_{j}\\right| \\leq d_{i}+d_{j}-n\n$$\n\nSumming all the results, and having in mind that each good triple is counted three times (one for each two of the three numbers), the number of good triples $T$ is at least\n\n$$\nT \\geq \\frac{1}{3} \\sum_{(i, j) \\in S}\\left(d_{i}+d_{j}-n\\right)\n$$\n\nEach term $d_{i}$ appears each time $i$ is in a pair from $S$, that is, $d_{i}$ times; there are $m$ pairs in $S$, so $n$ is subtracted $m$ times. By the Cauchy-Schwartz inequality\n\n$$\nT \\geq \\frac{1}{3}\\left(\\sum_{i=1}^{n} d_{i}^{2}-m n\\right) \\geq \\frac{1}{3}\\left(\\frac{\\left(\\sum_{i=1}^{n} d_{i}\\right)^{2}}{n}-m n\\right) .\n$$\n\nFinally, the sum $\\sum_{i=1}^{n} d_{i}$ is $2 m$, since $d_{i}$ counts the number of pairs containing $i$, and each pair $(i, j)$ is counted twice: once in $d_{i}$ and once in $d_{j}$. Therefore\n\n$$\nT \\geq \\frac{1}{3}\\left(\\frac{(2 m)^{2}}{n}-m n\\right)=4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n} .\n$$\n\nComment: This is a celebrated graph theory fact named Goodman's bound, after A. M. Goodman's method published in 1959. The generalized version of the problem is still studied to this day.", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
-{"year": "1989", "problem_label": "5", "tier": 1, "problem": "Determine all functions $f$ from the reals to the reals for which\n(1) $f(x)$ is strictly increasing,\n(2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.\n(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)\n\nAnswer: $f(x)=x+c, c \\in \\mathbb{R}$ constant.", "solution": "Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\\underbrace{f(f(\\ldots f}_{n \\text { times }}(x)))$.\nPlug $x \\rightarrow f_{n+1}(x)$ in (2): since $g\\left(f_{n+1}(x)\\right)=g\\left(f\\left(f_{n}(x)\\right)\\right)=f_{n}(x)$,\n\n$$\nf_{n+2}(x)+f_{n}(x)=2 f_{n+1}(x)\n$$\n\nthat is,\n\n$$\nf_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_{n}(x)\n$$\n\nTherefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find\n\n$$\nf_{n}(x)-x=n(f(x)-x) .\n$$\n\nSince $g$ has the same properties as $f$,\n\n$$\ng_{n}(x)-x=n(g(x)-x)=-n(f(x)-x) .\n$$\n\nFinally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \\Longrightarrow f(g(x))>$ $f(g(y)) \\Longrightarrow x>y$. An induction proves that $f_{n}$ and $g_{n}$ are also increasing functions.\nLet $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,\n\n$$\nx+n(f(x)-x)>y+n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x\n$$\n\nand\n\n$$\nx-n(f(x)-x)>y-n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y\n$$\n\nSumming it up,\n\n$$\n|n[(f(x)-x)-(f(y)-y)]|<x-y \\quad \\text { for all } n \\in \\mathbb{Z}_{>0}\n$$\n\nSuppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,\n\n$$\n|n(a-b)|<x-y\n$$\n\nwhich is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \\in \\mathbb{R}$, that is, $f(x)=x+c$.\nIt is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "1989", "problem_label": "4", "tier": 1, "problem": "Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \\leq a<$ $b \\leq n$. Show that there are at least\n\n$$\n4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n}\n$$\n\ntriples $(a, b, c)$ such that $(a, b),(a, c)$, and $(b, c)$ belong to $S$.", "solution": "Call a triple $(a, b, c)$ good if and only if $(a, b),(a, c)$, and $(b, c)$ all belong to $S$. For $i$ in $\\{1,2, \\ldots, n\\}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $\\left|D_{i}\\right|=d_{i}$ ). Consider a pair $(i, j) \\in S$. Our goal is to estimate the number of integers $k$ such that any permutation of $\\{i, j, k\\}$ is good, that is, $\\left|D_{i} \\cap D_{j}\\right|$. Note that $i \\notin D_{i}$ and $j \\notin D_{j}$, so $i, j \\notin D_{i} \\cap D_{j}$; thus any $k \\in D_{i} \\cap D_{j}$ is different from both $i$ and $j$, and $\\{i, j, k\\}$ has three elements as required. Now, since $D_{i} \\cup D_{j} \\subseteq\\{1,2, \\ldots, n\\}$,\n\n$$\n\\left|D_{i} \\cap D_{j}\\right|=\\left|D_{i}\\right|+\\left|D_{j}\\right|-\\left|D_{i} \\cup D_{j}\\right| \\leq d_{i}+d_{j}-n\n$$\n\nSumming all the results, and having in mind that each good triple is counted three times (one for each two of the three numbers), the number of good triples $T$ is at least\n\n$$\nT \\geq \\frac{1}{3} \\sum_{(i, j) \\in S}\\left(d_{i}+d_{j}-n\\right)\n$$\n\nEach term $d_{i}$ appears each time $i$ is in a pair from $S$, that is, $d_{i}$ times; there are $m$ pairs in $S$, so $n$ is subtracted $m$ times. By the Cauchy-Schwartz inequality\n\n$$\nT \\geq \\frac{1}{3}\\left(\\sum_{i=1}^{n} d_{i}^{2}-m n\\right) \\geq \\frac{1}{3}\\left(\\frac{\\left(\\sum_{i=1}^{n} d_{i}\\right)^{2}}{n}-m n\\right) .\n$$\n\nFinally, the sum $\\sum_{i=1}^{n} d_{i}$ is $2 m$, since $d_{i}$ counts the number of pairs containing $i$, and each pair $(i, j)$ is counted twice: once in $d_{i}$ and once in $d_{j}$. Therefore\n\n$$\nT \\geq \\frac{1}{3}\\left(\\frac{(2 m)^{2}}{n}-m n\\right)=4 m \\frac{\\left(m-\\frac{n^{2}}{4}\\right)}{3 n} .\n$$\n\nComment: This is a celebrated graph theory fact named Goodman's bound, after A. M. Goodman's method published in 1959. The generalized version of the problem is still studied to this day.", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
+{"year": "1989", "problem_label": "5", "tier": 1, "problem": "Determine all functions $f$ from the reals to the reals for which\n(1) $f(x)$ is strictly increasing,\n(2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.\n(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)\n\nAnswer: $f(x)=x+c, c \\in \\mathbb{R}$ constant.", "solution": "Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\\underbrace{f(f(\\ldots f}_{n \\text { times }}(x)))$.\nPlug $x \\rightarrow f_{n+1}(x)$ in (2): since $g\\left(f_{n+1}(x)\\right)=g\\left(f\\left(f_{n}(x)\\right)\\right)=f_{n}(x)$,\n\n$$\nf_{n+2}(x)+f_{n}(x)=2 f_{n+1}(x)\n$$\n\nthat is,\n\n$$\nf_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_{n}(x)\n$$\n\nTherefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find\n\n$$\nf_{n}(x)-x=n(f(x)-x) .\n$$\n\nSince $g$ has the same properties as $f$,\n\n$$\ng_{n}(x)-x=n(g(x)-x)=-n(f(x)-x) .\n$$\n\nFinally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \\Longrightarrow f(g(x))>$ $f(g(y)) \\Longrightarrow x>y$. An induction proves that $f_{n}$ and $g_{n}$ are also increasing functions.\nLet $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,\n\n$$\nx+n(f(x)-x)>y+n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x\n$$\n\nand\n\n$$\nx-n(f(x)-x)>y-n(f(y)-y) \\Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y\n$$\n\nSumming it up,\n\n$$\n|n[(f(x)-x)-(f(y)-y)]|<x-y \\quad \\text { for all } n \\in \\mathbb{Z}_{>0}\n$$\n\nSuppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,\n\n$$\n|n(a-b)|<x-y\n$$\n\nwhich is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \\in \\mathbb{R}$, that is, $f(x)=x+c$.\nIt is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo1990_sol.jsonl

@@ -0,0 +1,9 @@
+{"year": "1990", "problem_label": "1", "tier": 1, "problem": "In $\\triangle A B C$, let $D, E, F$ be the midpoints of $B C, A C, A B$ respectively and let $G$ be the centroid of the triangle.\nFor each value of $\\angle B A C$, how many non-similar triangles are there in which $A E G F$ is a cyclic quadrilateral?", "solution": "Let $I$ be the intersection of $A G$ and $E F$.\nLet $\\delta=A I . I G-F I$ IE. Then\n\n$$\nA I=A D / 2, \\quad I G=A D / 6, \\quad F I=B C / 4=I E\n$$\n\nFurther, applying the cosine rule to triangles $A B D, A C D$ we get\n\n$$\n\\begin{aligned}\nA B^{2} & =B C^{2} / 4+A D^{2}-A D \\cdot B C \\cdot \\cos \\angle B D A, \\\\\nA C^{2} & =B C^{2} / 4+A D^{2}+A D \\cdot B C \\cdot \\cos \\angle B D A, \\\\\n\\text { so } \\quad A D^{2} & =\\left(A B^{2}+A C^{2}-B C^{2} / 2\\right) / 2\n\\end{aligned}\n$$\n\nHence\n\n$$\n\\begin{aligned}\n\\delta & =\\left(A B^{2}+A C^{2}-2 B C^{2}\\right) / 24 \\\\\n& =\\left(4 A B \\cdot A C \\cdot \\cos \\angle B A C-A B^{2}-A C^{2}\\right)\n\\end{aligned}\n$$\n\nNow $A E F G$ is a cyclic quadrilateral if and only if $\\delta=0$, i.e. if and only if\n\n$$\n\\begin{aligned}\n\\cos \\angle B A C & =\\left(A B^{2}+A B^{2}\\right) /(4 \\cdot A B \\cdot A C) \\\\\n& =(A B / A C+A C / A B) / 4\n\\end{aligned}\n$$\n\n5\nNow $A B / A C+A C / A B \\geq 2$. Hence $\\cos \\angle B A C \\geq 1 / 2$ and so $\\angle B A C \\leq 60^{\\circ}$.\nFor $\\angle B A C>60^{\\circ}$ there is no triangle with the required property.\nFor $\\angle B A C=60^{\\circ}$ there exists, within similarity, precisely one triangle (which is equilateral) having the required property.\nFor $\\angle B A C<60^{\\circ}$ there exists, within similarity, again precisely one triangle having the required property (even though for fixed median $A E$ there are two, but one arises from the other by interchanging point $B$ with point $C$, thus proving them to be similar).", "problem_match": "# Question 1 ", "solution_match": "# FIRST SOLUTION\n\n"}
+{"year": "1990", "problem_label": "1", "tier": 1, "problem": "In $\\triangle A B C$, let $D, E, F$ be the midpoints of $B C, A C, A B$ respectively and let $G$ be the centroid of the triangle.\nFor each value of $\\angle B A C$, how many non-similar triangles are there in which $A E G F$ is a cyclic quadrilateral?", "solution": "(Mr Marcus Brazil, La Trobe University, Bundoora, Melbourne, Australia):\nWe require, as above,\n\n$$\nA I \\cdot I G=E I \\cdot I F\n$$\n\n(which by (1) is equivalent to $A D^{2} / 3=C D^{2}$, i.e. $A D=C D \\cdot \\sqrt{3}$ ).\nLet, without loss of generality, $C D=1$. Then $A$ lies on the circle of radius $\\sqrt{3}$ with centre $D$. If $C D$ and $D A$ are perpendicular, the angle $B A C$ is $60^{\\circ}$, otherwise it must be less.\nIn this case, for each angle $B A C$ there are two solutions, which are congruent.", "problem_match": "# Question 1 ", "solution_match": "\nSECOND SOLUTION "}
+{"year": "1990", "problem_label": "1", "tier": 1, "problem": "In $\\triangle A B C$, let $D, E, F$ be the midpoints of $B C, A C, A B$ respectively and let $G$ be the centroid of the triangle.\nFor each value of $\\angle B A C$, how many non-similar triangles are there in which $A E G F$ is a cyclic quadrilateral?", "solution": "in the figure as shown below, we first show that it is necessary that $\\angle A$ is less than $90^{\\circ}$ if the quadrilateral $A E G F$ ; cyclic.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-02.jpg?height=418&width=638&top_left_y=344&top_left_x=630)\n\nNow, since $E F \\| B C$, we get\n\n$$\n\\begin{aligned}\n\\angle E G F & =180^{\\circ}-\\left(B_{1}+C_{1}\\right) \\\\\n& \\geq 180^{\\circ}-(B+C) \\\\\n& =A .\n\\end{aligned}\n$$\n\n(1)\n\nThus, if $A E G F$ is cyclic, we would have $\\angle E G F+\\angle A=180^{\\circ}$. Therefore it is necessary that $0<\\angle A \\leq 90^{\\circ}$.\n\n## Continuation \"A\"\n\nLet $O$ be the circumcentre of $\\triangle A F E$. Without loss of generality, let the radius of this circle be 1.\nWe then let $A=1, F=z=e^{i \\theta}$ and $E=z e^{2 i \\alpha}=e^{i(\\theta+2 \\alpha)}$.\nThen $\\angle A=\\alpha, 0<\\alpha \\leq 90^{\\circ}$, and $0<\\theta<360^{\\circ}-2 \\alpha$.\nThus,\n\n$$\nB=2 z-1\n$$\n\nand\n\n$$\n\\begin{aligned}\nG & =\\frac{1}{3}(2 z-1)+\\frac{2}{3}\\left(z e^{2 i \\alpha}\\right) \\\\\n& =\\frac{1}{3}\\left(2 e^{i \\theta}+2 e^{i(\\theta+2 \\alpha)}-1\\right)\n\\end{aligned}\n$$\n\nFor quadrilateral $A F G E$ to be cyclic, it is now necessary that\n\n$$\n|G|=1 .\n$$\n\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-03.jpg?height=1123&width=1148&top_left_y=162&top_left_x=456)\n\nFor $|G|=1$, we must have\n\n$$\n\\begin{aligned}\n9= & (2 \\cos (\\theta)+2 \\cos (\\theta+2 \\alpha)-1)^{2}+(2 \\sin (\\theta)+2 \\sin (\\theta+2 \\alpha))^{2} \\\\\n= & 4\\left(\\cos ^{2}(\\theta)+\\sin ^{2}(\\theta)\\right)+4\\left(\\cos ^{2}(\\theta+2 \\alpha)+\\sin ^{2}(\\theta+2 \\alpha)\\right)+1 \\\\\n& +8(\\cos (\\theta) \\cos (\\theta+2 \\alpha)+\\sin (\\theta) \\sin (\\theta+2 \\alpha))-4 \\cos (\\theta)-4 \\cos (\\theta+2 \\alpha) \\\\\n= & 9+8 \\cos (2 \\alpha)-8 \\cos (\\alpha) \\cos (\\theta+\\alpha)\n\\end{aligned}\n$$\n\nso that\n\n$$\n\\cos (\\theta+\\alpha)=\\frac{\\cos (2 \\alpha)}{\\cos (\\alpha)}\n$$\n\n] Now, $\\left|\\frac{\\cos (2 \\alpha)}{\\cos (\\alpha)}\\right| \\leq 1$ if and only if $\\alpha \\in\\left(0,60^{\\circ}\\right]$ in the range of $\\alpha$ under consideration, that is $\\alpha \\in\\left(0,00^{\\circ}\\right]$. There is equality if and only if $\\alpha=60^{\\circ}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-03.jpg?height=350&width=641&top_left_y=1950&top_left_x=661)\n$\\square$ Note there is only one solution. The apparent other solution is the mirror image of the first. We are solving for $\\alpha+\\theta$. The other solution is $360^{\\circ}-\\alpha-\\theta$.\n\n## Continuation \"B\"\n\nLet $O$ be the circumcentre of triangle $A E F$. Let $A P$ be a diameter of this circle. Construct the circle with centre $P$ and radius $A P$. Then $B$ and $C$ lie on this circle.\n\nIt is clear that the problem is solved if we allow the angle $\\angle B A C=\\alpha$ to vary and restrict $B$ and $C$ to the constructed circle.\n\nLet $\\theta$ be the angle from the drawn axis. Then $\\theta$ lies in the range $\\left(0,180^{\\circ}-\\alpha\\right)$. We must not forget the necessary restriction of $\\alpha$, that is $\\alpha \\in\\left(0,90^{\\circ}\\right.$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-04.jpg?height=1091&width=1484&top_left_y=669&top_left_x=478)\n\nNow, $D$ lies on an arc of a circle, centre $P$, radius $P D$ exterior to the circle, centre $O$, radius $A O$.\nBy similarity, $G$, lies on an arc of a circle, centre $Q$, radius $Q G$ where $A Q=\\frac{2}{3} A P$ and $Q G=\\frac{2}{3} P D$.\nFor the quadrilateral $A F G E$ to be cyclic, we must have that the radius $Q G$ is greater than or equal to $Q P$.\nThe easiest way to calulate these radii is to consider the case in which the diameter $A P$ bisects the angle $\\angle B A C$.\nThus we re-draw the diagram as below. Let $A H$ be a diameter of the larger circle.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-05.jpg?height=1126&width=1061&top_left_y=163&top_left_x=472)\n\nThus we have $A H=4$ and by similar triangles,\n\n$$\n\\frac{A D}{A B}=\\frac{A B}{A H}=\\cos \\left(\\frac{\\alpha}{2}\\right)\n$$\n\nso that\n\n$$\n\\begin{aligned}\nA D & =4 \\cos ^{2}\\left(\\frac{\\alpha}{2}\\right) \\\\\n& =2+2 \\cos (\\alpha) .\n\\end{aligned}\n$$\n\nThus $P D=2 \\cos (\\alpha)$ and $Q G=\\frac{2}{3} 2 \\cos (\\alpha)=\\frac{4}{3} \\cos (\\alpha)$.\nThe necessary condition for a cyclic quadrilateral is then\n\n$$\n\\frac{4}{3}(1+\\cos (\\alpha)) \\geq 2\n$$\n\n[5\n\n$$\n\\cos (\\alpha) \\geq \\frac{1}{2}\n$$\n\n：７\nThus it is clear that there is precisely one (up to similarity) solution for $0<\\alpha \\leq 60^{\\circ}$ and no solutions otherwise.", "problem_match": "# Question 1 ", "solution_match": "\nTHIRD SOLUTION\n"}
+{"year": "1990", "problem_label": "2", "tier": 1, "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers, and let $S_{k}$ be the sum of products of $a_{1}, a_{2}, \\ldots, a_{n}$ taken $k$ at a time.\nShow that\n\n$$\nS_{k} S_{n-k} \\geq\\binom{ n}{k}^{2} a_{1} a_{2} \\ldots a_{n}, \\quad \\text { for } \\quad k=1,2, \\ldots, n-1\n$$", "solution": "$$\n\\binom{n}{k} a_{1} a_{2} \\ldots a_{n}\n$$\n\n$2=\\sum_{1 \\leq i_{1}<i_{2}<\\ldots<i_{k} \\leq n} a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}} \\cdot a_{1} a_{2} \\ldots a_{n} / a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}}$\n(and using the Cauchy-Schwarz inequality)\n\n$$\n\\begin{aligned}\n& \\leq\\left(\\sum_{1 \\leq i_{1}<i_{2}<\\ldots<i_{k} \\leq n} a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}}\\right)^{\\frac{1}{2}} \\cdot\\left(\\sum_{1 \\leq i_{1}<i_{2}<\\ldots<i_{k} \\leq n} a_{1} a_{2} \\ldots a_{n} / a_{i_{1}} a_{i_{2}} \\ldots a_{i_{k}}\\right)^{\\frac{1}{2}} \\\\\n& =S_{k}^{\\frac{1}{2}} \\cdot S_{n-k}^{\\frac{1}{2}}\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\binom{n}{k}^{2} a_{1} a_{2} \\ldots a_{n} \\leq S_{k} S_{n-k}\n$$\n\nq.e.d.", "problem_match": "# Question 2", "solution_match": "# FIRST SOLUTION\n\n"}
+{"year": "1990", "problem_label": "2", "tier": 1, "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers, and let $S_{k}$ be the sum of products of $a_{1}, a_{2}, \\ldots, a_{n}$ taken $k$ at a time.\nShow that\n\n$$\nS_{k} S_{n-k} \\geq\\binom{ n}{k}^{2} a_{1} a_{2} \\ldots a_{n}, \\quad \\text { for } \\quad k=1,2, \\ldots, n-1\n$$", "solution": "(provided by the Canadian Problems Committee).\nWrite $S_{k}$ as $\\sum_{i=1}^{\\binom{n}{k}} t_{i}$. Then\n주\n\n$$\nS_{n-k}=\\left(\\prod_{m=1}^{n} a_{m}\\right)\\left(\\sum_{i=1}^{\\binom{n}{k}} \\frac{1}{t_{i}}\\right)\n$$\n\n$$\n\\text { so that } \\left.\\begin{array}{rl}\nS_{k} S_{n-k} & =\\left(\\prod_{m=1}^{n} a_{m}\\right) \\cdot\\left(\\begin{array}{l}\n\\binom{n}{k} \\\\\ni=1\n\\end{array} t_{i}\\right)\\left(\\begin{array}{l}\n\\binom{n}{k} \\\\\n\\sum_{j=1}^{2}\n\\end{array} \\frac{1}{t_{j}}\\right) \\\\\n& =\\left(\\prod_{m=1}^{n} a_{m}\\right)\\left[\\sum_{i=1}^{\\binom{n}{k}} 1+\\sum_{i=1}^{\\binom{n}{k}} \\sum_{j=1}^{n} \\begin{array}{l}\nn \\\\\nk\n\\end{array}\\right) \\\\\n\\frac{t_{i}}{} \\\\\nt_{j}\n\\end{array}\\right] .\n$$\n\nAs there are\n\n$$\n\\frac{\\binom{n}{k}^{2}-\\binom{n}{k}}{2}\n$$\n\nterms in the sum\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-07.jpg?height=98&width=91&top_left_y=355&top_left_x=63)\n\n$$\n\\begin{aligned}\nS_{k} S_{n-k} & \\geq\\left(\\prod_{m=1}^{n} a_{m}\\right)\\left[\\binom{n}{k}+2 \\cdot \\frac{\\binom{n}{k}^{2}-\\binom{n}{k}}{2}\\right] \\\\\n& =\\binom{n}{k}^{2}\\left(\\prod_{m=1}^{n} a_{m}\\right)\n\\end{aligned}\n$$\n\nsince $\\frac{t_{i}}{t_{j}}+\\frac{t_{j}}{t_{i}} \\geq 2$ for $t_{i}, t_{j}>0$.", "problem_match": "# Question 2", "solution_match": "\nSECOND SOLUTION "}
+{"year": "1990", "problem_label": "3", "tier": 1, "problem": "Consider all the triangles $A B C$ which have a fixed base $A B$ and whose altitude from $C$ is a constant $h$. For which of these triangles is the product of its altitudes a maximum?", "solution": "Let $h_{a}$ and $h_{b}$ be the altitudes from $A$ and $B$, respectively. Then\n\n$$\n\\begin{aligned}\nA B \\cdot h \\cdot A C \\cdot h_{b} \\cdot B C \\cdot h_{a} & =8 . \\text { area of } \\triangle A B C)^{3} \\\\\n& =(A B \\cdot h)^{3},\n\\end{aligned}\n$$\n\n园\nwhich is a constant. So the product $h . h_{a} \\cdot h_{b}$ attains its maximum when the product $A C . B C$ attains its minimum.\nSince\n\n$$\n\\begin{aligned}\n(\\sin C) \\cdot A C \\cdot B C & =B C \\cdot h_{a} \\\\\n& =2 \\cdot \\text { area of } \\triangle A B C\n\\end{aligned}\n$$\n\n(3)\nwhich is a constant, $A C \\cdot B C$ attains its minimum when $\\sin C$ reaches its maximum. There are two cases:\n(a) $h \\leq A B / 2$. Then there exists a triangle $A B C$ which has a right angle at $C$, and for precisely such a triangle $\\sin C$ attains its maximum, namely 1 .\n(b) $h>A B / 2$. In this case the angle at $C$ is acute and assumes its maximum when the triangle is isosceles.\n\nNote that a solution using calculus obviously exists.", "problem_match": "# Question 3", "solution_match": "# SOLUTION:"}
+{"year": "1990", "problem_label": "4", "tier": 1, "problem": "A set of 1990 persons is divided into non-intersecting subsets in such a way that\n(a) no one in a subset knows all the others in the subset;\n(b) among any three persons in a subset, there are always at least two who do not know each other; and\n(c) for any two persons in a subset who do not know each other, there is exactly one person in the same subset knowing both of them.\n(i) Prove that within each subset, every person has the same number of acquaintances.\n(ii) Determine the maximum possible number of subsets.\n\nNote: it is understood that if a person $A$ knows person $B$, then person $B$ will know person $A$; an acquaintance is someone who is known. Every person is assumed to know one's self.", "solution": "(i) Let $S$ be a subset of persons satisfying conditions (a), (b) and (c). Let $x \\in S$ be one who knows the maximum number of persons in $S$.\nAssume that $x$ knows $x_{1}, x_{2}, \\ldots, x_{n}$. By (b), $x_{i}$ and $x_{j}$ are strangers if $i \\neq j$. For each $x_{i}$, let $N_{i}$ be the set of persons in $S$ who know $x_{i}$ but not $x$. Note that, for $i \\neq j, N_{i}$ has no person in common with $N_{j}$, otherwise there would be more than one person knowing $x_{i}$ and $x_{j}$, contradicting (c).\nBy (a) we may assume that $N_{1}$ is not empty.) Let $y_{1} \\in N_{1}$. By (c), for each $k>1$, there is exactly one person $y_{k}$ in $N_{k}$ who knows $y_{1}$. This means that $y_{1}$ knows $n$ persons, namely $x_{1}, y_{2}, \\ldots, y_{n}$.\nBecause $n$ is the maximal number of persons in $S$ a person in $S$ can know, $y_{1}$ knows exactly $n$ persons in $S$. By precisely the same reasoning we find that each person in $N_{i}$, $i=1,2, \\ldots, n$, knows exactly $n$ persons in $S$.\nLetting $y_{1}$ take the role of $x$ in our argument, we see that also each $x_{i}$ knows exactly $n$ persons. Note that, by (c), every person in $S$ other than $x, x_{1}, \\ldots, x_{n}$, must be in some $N_{j}$. Therefore every person in $S$ knows exactly $n$ persons in $S$ and thus has the same number of acquaintances in $S$.\n(ii) To maximize the number of subsets, we have to minimize the size of each group. The smallest possible subset is one in which every person knows exactly two persons, and hence there must be exactly five persons in the subset, forming a cycle where two persons stand side by side only if they know each other. Therefore the maximum possible number of subsets is 1990/5 $=398$.", "problem_match": "# Question 4", "solution_match": "# SOLUTION:"}
+{"year": "1990", "problem_label": "5", "tier": 1, "problem": "Show that for every integer $n \\geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.", "solution": "(provided by the Canadian Problems Committee).\nThe basic building blocks will be right angled triangles with sides $p, q$ (which are positive integers) adjacent to the right angle.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=153&width=185&top_left_y=623&top_left_x=902)\n\nIn the first instance, we take $p=q=1$ and construct five basic building blocks: $L_{1}, L_{2}, M, R_{1}$ and $R_{2}$ 。\n\n## [3)\n\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=186&width=115&top_left_y=956&top_left_x=411)\n$L_{1}$\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=186&width=112&top_left_y=956&top_left_x=649)\n$L_{2}$\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=183&width=107&top_left_y=955&top_left_x=933)\nM\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=179&width=107&top_left_y=954&top_left_x=1215)\n$R_{1}$\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=177&width=110&top_left_y=952&top_left_x=1452)\n$R_{2}$\n\nWe shall now build convex hexagons by taking, on the left, one of the blocks $L_{i}$, attaching $n$ copies of the block $M$, and finally attaching one of the blocks $R_{j}$. We must therefore exclude the case when $(i, j)=(2,1)$ for this does not generate a hexagon. Further, for $(i, j)=(1,1)$ or $(i, j)=(1,2)$, we require that $n \\geq 1$, whereas for $(i, j)=$ $(2,2)$, we only need require that $n \\geq 0$.\n\nThus, with the obvious interpretation:\n$L_{1}+n M+R_{1}$ gives a convex hexagon containing $2+4 n+2=4 n+4 \\quad(n \\geq 1)$ congruent triangles;\n$L_{1}+n M+R_{2}$ gives a convex hexagon containing $2+4 n+3=4 n+5(n \\geq 1)$ congruent triangles; and\n$L_{2}+n M+R_{2}$ gives a convex hexagon containing $3+4 n+3=4 n+6 \\quad(n \\geq 0)$ congruent triangles, or $4 n+2(n \\geq 1)$ congruent triangles.\n\nWe shall now modify the lengths of the sides of the right triangle to obtain the case of $4 n+3 \\quad(n \\geq 1)$ congruent triangles.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=438&width=1180&top_left_y=2101&top_left_x=407)\n\nSo we have $2 n+1$ triangles in the top part and $2 n+2$ triangles in the bottom part. In order to match, we need\n\n$$\n(n+1) p=(n+2) q\n$$\n\nso we take\n\n$$\nq=n+1 \\quad \\text { and } \\quad p=n+2\n$$\n\nThis completes the solution.", "problem_match": "# Question 5", "solution_match": "\nFIRST SOLUTION "}
+{"year": "1990", "problem_label": "5", "tier": 1, "problem": "Show that for every integer $n \\geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.", "solution": "(provided by the Canadian Problems Committee):\nThe basic building blocks will be right angled triangles with sides $m, n$ (which are positive integers) adjacent to the right angle.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-11.jpg?height=251&width=180&top_left_y=758&top_left_x=910)\n\nWe construct an \"UPPER CONFIGURATION\", being a rectangle consisting of $m$ building block units of pairs of triangles with the side of length $n$ as base. This gives a base length of $n m$ across the configuration.\n\nWe further construct a \"LOWER CONFIGURATION\", being a triangle with base up, consisting along the base of $n$ building block units. Again, we have a base length of $m n$ across the configuration.\n\nTwo triangles in the upper configuration are shaded horizontally. One triangle in the lower configuration is also shaded horizontally. Another triangle in the lower configuration is shaded vertically.\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-12.jpg?height=1212&width=1356&top_left_y=260&top_left_x=341)\n\nNow consider the figure obtained by joining the two configurations along the base line of common length nm. To create the classes of hexagons defined below, it is necessary that both $n \\geq 3$ and $m \\geq 3$.\n\nWe create a class of convex hexagons (class 1 ) by omitting the three triangles that are shaded horizontally. The other class of convex hexagons (class 2) is obtained by omitting all shaded triangles.\n\nNow count the total number of triangles in the full configuration.\nThe upper configuration gives $2 m$ triangles. The lower configuration gives\n\n$$\n\\sum_{k=1}^{n}(2 k-1)=n^{2} \\quad \\text { triangles. }\n$$\n\nThus the total number of triangles in a hexagon in class 1 is\n\n$$\n2 m-2+n^{2}-1\n$$\n\nand the total number of triangle in a hexagon in class 2 is\n\n$$\n2 m-2+n^{2}-2\n$$\n\nThese, together with the restrictions on $n$ and $m$, generate all positive integers greater than or equal to 11.\n\nFor the integers $6,7,8,9$ and 10 , we give specific examples:\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=394&width=283&top_left_y=513&top_left_x=365)\n\n6\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=329&width=402&top_left_y=584&top_left_x=742)\n\n7\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=394&width=404&top_left_y=513&top_left_x=1240)\n\n8\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=388&width=410&top_left_y=1015&top_left_x=559)\n\n9\n![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=392&width=413&top_left_y=1013&top_left_x=1059)\n\n10\n\nThis completes the solution.\n\nThere are $\\binom{n}{k}$ products of the $a_{i}$ taken $k$ at a time. Amongst these products any given $a_{i}$ will appear $\\binom{n-1}{k-1}$ times, since $\\binom{n-1}{k-1}$ is the number of ways of choosing the other factors of the product. So the $\\mathrm{AM} / \\mathrm{GM}$ inequality gives\n\n## ④\n\n$$\n\\frac{S_{k}}{\\binom{n}{k}} \\geq\\left[\\prod_{i=1}^{n} a_{i}^{\\binom{n-1}{k-1}}\\right]^{\\frac{1}{n}\\binom{n}{n}}\n$$\n\nBut $\\binom{n}{k}=\\frac{n}{k}\\binom{n-1}{k-1}$, leading to\n6 S $\\quad S_{k} \\geq\\binom{ n}{k}\\left(\\prod_{i=1}^{n} a_{i}\\right)^{\\frac{k}{n}}$.\nHence\n田\n\n$$\nS_{k} S_{n-k} \\geq\\binom{ n}{k}\\left(\\prod_{i=1}^{n} a_{i}\\right)^{\\frac{k}{n}}\\binom{n}{n-k}\\left(\\prod_{i=1}^{n} a_{i}\\right)^{\\frac{n-k}{n}}=\\binom{n}{k}^{2}\\left(\\prod_{1}^{n} a_{i}\\right) .\n$$", "problem_match": "# Question 5", "solution_match": "\nSECOND SOLUTION "}

APMO/segmented/en-apmo1991_sol.jsonl

@@ -1,7 +1,7 @@
 {"year": "1991", "problem_label": "1", "tier": 1, "problem": "Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.", "solution": "Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=663&top_left_x=673)\n\nIt is well known that $\\frac{A G}{A M}=\\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\\frac{2}{3}$. Hence $G X=\\frac{1}{2} X Y=\\frac{1}{3} B C$.\nNow look at the similarity between triangles $Q B C$ and $Q G X$ :\n\n$$\n\\frac{Q G}{Q B}=\\frac{G X}{B C}=\\frac{1}{3} \\Longrightarrow Q B=3 Q G \\Longrightarrow Q B=\\frac{3}{4} B G=\\frac{3}{4} \\cdot \\frac{2}{3} B R=\\frac{1}{2} B R .\n$$\n\nFinally, since $\\frac{B M}{B C}=\\frac{B Q}{B R}, M Q$ is a midline in $B C R$. Therefore $M Q=\\frac{1}{2} C R=\\frac{1}{4} A C$ and $M Q \\| A C$. Similarly, $M P=\\frac{1}{4} A B$ and $M P \\| A B$. This is sufficient to establish that $M P Q$ and $A B C$ are similar (with similarity ratio $\\frac{1}{4}$ ).", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}
 {"year": "1991", "problem_label": "1", "tier": 1, "problem": "Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.", "solution": "Let $S$ and $R$ be the midpoints of $A B$ and $A C$, respectively. Since $G$ is the centroid, it lies in the medians $B R$ and $C S$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=1971&top_left_x=676)\n\nDue to the similarity between triangles $Q B C$ and $Q G X$ (which is true because $G X \\| B C$ ), there is an inverse homothety with center $Q$ and ratio $-\\frac{X G}{B C}=\\frac{X Y}{2 B C}$ that takes $B$ to $G$ and $C$ to $X$. This homothety takes the midpoint $M$ of $B C$ to the midpoint $K$ of $G X$.\n\nNow consider the homothety that takes $B$ to $X$ and $C$ to $G$. This new homothety, with ratio $\\frac{X Y}{2 B C}$, also takes $M$ to $K$. Hence lines $B X$ (which contains side $A B$ ), $C G$ (which contains the median $C S$ ), and $M K$ have a common point, which is $S$. Thus $Q$ lies on midline $M S$.\nThe same reasoning proves that $P$ lies on midline $M R$. Since all homothety ratios are the same, $\\frac{M Q}{M S}=\\frac{M P}{M R}$, which shows that $M P Q$ is similar to $M R S$, which in turn is similar to $A B C$, and we are done.", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}
-{"year": "1991", "problem_label": "2", "tier": 1, "problem": "Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points?", "solution": "Embed the points in the cartesian plane such that no two points have the same $y$-coordinate. Let $P_{1}, P_{2}, \\ldots, P_{997}$ be the points and $y_{1}<y_{2}<\\ldots<y_{997}$ be their respective $y$-coordinates. Then the $y$-coordinate of the midpoint of $P_{i} P_{i+1}, i=1,2, \\ldots, 996$ is $\\frac{y_{i}+y_{i+1}}{2}$ and the $y$-coordinate of the midpoint of $P_{i} P_{i+2}, i=1,2, \\ldots, 995$ is $\\frac{y_{i}+y_{i+2}}{2}$. Since\n\n$$\n\\frac{y_{1}+y_{2}}{2}<\\frac{y_{1}+y_{3}}{2}<\\frac{y_{2}+y_{3}}{2}<\\frac{y_{2}+y_{4}}{2}<\\cdots<\\frac{y_{995}+y_{997}}{2}<\\frac{y_{996}+y_{997}}{2}\n$$\n\nthere are at least $996+995=1991$ distinct midpoints, and therefore at least 1991 red points. The equality case happens if we take $P_{i}=(0,2 i), i=1,2, \\ldots, 997$. The midpoints are $(0, i+j)$, $1 \\leq i<j \\leq 997$, which are the points $(0, k)$ with $1+2=3 \\leq k \\leq 996+997=1993$, a total of $1993-3+1=1991$ red points.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
-{"year": "1991", "problem_label": "3", "tier": 1, "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ be positive real numbers such that $a_{1}+a_{2}+\\cdots+a_{n}=b_{1}+b_{2}+$ $\\cdots+b_{n}$. Show that\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "solution": "By the Cauchy-Schwartz inequality,\n\n$$\n\\left(\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}}\\right)\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right) \\geq\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2} .\n$$\n\nSince $\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right)=2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)$,\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}}{2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
+{"year": "1991", "problem_label": "2", "tier": 1, "problem": "Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points?", "solution": "Embed the points in the cartesian plane such that no two points have the same $y$-coordinate. Let $P_{1}, P_{2}, \\ldots, P_{997}$ be the points and $y_{1}<y_{2}<\\ldots<y_{997}$ be their respective $y$-coordinates. Then the $y$-coordinate of the midpoint of $P_{i} P_{i+1}, i=1,2, \\ldots, 996$ is $\\frac{y_{i}+y_{i+1}}{2}$ and the $y$-coordinate of the midpoint of $P_{i} P_{i+2}, i=1,2, \\ldots, 995$ is $\\frac{y_{i}+y_{i+2}}{2}$. Since\n\n$$\n\\frac{y_{1}+y_{2}}{2}<\\frac{y_{1}+y_{3}}{2}<\\frac{y_{2}+y_{3}}{2}<\\frac{y_{2}+y_{4}}{2}<\\cdots<\\frac{y_{995}+y_{997}}{2}<\\frac{y_{996}+y_{997}}{2}\n$$\n\nthere are at least $996+995=1991$ distinct midpoints, and therefore at least 1991 red points. The equality case happens if we take $P_{i}=(0,2 i), i=1,2, \\ldots, 997$. The midpoints are $(0, i+j)$, $1 \\leq i<j \\leq 997$, which are the points $(0, k)$ with $1+2=3 \\leq k \\leq 996+997=1993$, a total of $1993-3+1=1991$ red points.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
+{"year": "1991", "problem_label": "3", "tier": 1, "problem": "Let $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ be positive real numbers such that $a_{1}+a_{2}+\\cdots+a_{n}=b_{1}+b_{2}+$ $\\cdots+b_{n}$. Show that\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "solution": "By the Cauchy-Schwartz inequality,\n\n$$\n\\left(\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}}\\right)\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right) \\geq\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2} .\n$$\n\nSince $\\left(\\left(a_{1}+b_{1}\\right)+\\left(a_{2}+b_{2}\\right)+\\cdots+\\left(a_{n}+b_{n}\\right)\\right)=2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)$,\n\n$$\n\\frac{a_{1}^{2}}{a_{1}+b_{1}}+\\frac{a_{2}^{2}}{a_{2}+b_{2}}+\\cdots+\\frac{a_{n}^{2}}{a_{n}+b_{n}} \\geq \\frac{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}}{2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)}=\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{2}\n$$", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
 {"year": "1991", "problem_label": "4", "tier": 1, "problem": "During a break, $n$ children at school sit in a circle around their teacher to play a game. The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and soon. Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.\n\nAnswer: All powers of 2 .", "solution": "Number the children from 0 to $n-1$. Then the teacher hands candy to children in positions $f(x)=1+2+\\cdots+x \\bmod n=\\frac{x(x+1)}{2} \\bmod n$. Our task is to find the range of $f: \\mathbb{Z}_{n} \\rightarrow \\mathbb{Z}_{n}$, and to verify whether the range is $\\mathbb{Z}_{n}$, that is, whether $f$ is a bijection.\nIf $n=2^{a} m, m>1$ odd, look at $f(x)$ modulo $m$. Since $m$ is odd, $m|f(x) \\Longleftrightarrow m| x(x+1)$. Then, for instance, $f(x) \\equiv 0(\\bmod m)$ for $x=0$ and $x=m-1$. This means that $f(x)$ is not a bijection modulo $m$, and there exists $t$ such that $f(x) \\not \\equiv t(\\bmod m)$ for all $x$. By the Chinese Remainder Theorem,\n\n$$\nf(x) \\equiv t \\quad(\\bmod n) \\Longleftrightarrow \\begin{cases}f(x) \\equiv t & \\left(\\bmod 2^{a}\\right) \\\\ f(x) \\equiv t & (\\bmod m)\\end{cases}\n$$\n\nTherefore, $f$ is not a bijection modulo $n$.\nIf $n=2^{a}$, then\n\n$$\nf(x)-f(y)=\\frac{1}{2}(x(x+1)-y(y+1))=\\frac{1}{2}\\left(x^{2}-y^{2}+x-y\\right)=\\frac{(x-y)(x+y+1)}{2} .\n$$\n\nand\n\n$$\nf(x) \\equiv f(y) \\quad\\left(\\bmod 2^{a}\\right) \\Longleftrightarrow(x-y)(x+y+1) \\equiv 0 \\quad\\left(\\bmod 2^{a+1}\\right)\n$$\n\nIf $x$ and $y$ have the same parity, $x+y+1$ is odd and $(*)$ is equivalent to $x \\equiv y\\left(\\bmod 2^{a+1}\\right)$. If $x$ and $y$ have different parity,\n\n$$\n(*) \\Longleftrightarrow x+y+1 \\equiv 0 \\quad\\left(\\bmod 2^{a+1}\\right)\n$$\n\nHowever, $1 \\leq x+y+1 \\leq 2\\left(2^{a}-1\\right)+1=2^{a+1}-1$, so $x+y+1$ is not a multiple of $2^{a+1}$. Therefore $f$ is a bijection if $n$ is a power of 2 .", "problem_match": "# Problem 4", "solution_match": "# Solution 1"}
 {"year": "1991", "problem_label": "4", "tier": 1, "problem": "During a break, $n$ children at school sit in a circle around their teacher to play a game. The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and soon. Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.\n\nAnswer: All powers of 2 .", "solution": "We give a full description of $a_{n}$, the size of the range of $f$.\nSince congruences modulo $n$ are defined, via Chinese Remainder Theorem, by congruences modulo $p^{\\alpha}$ for all prime divisors $p$ of $n$ and $\\alpha$ being the number of factors $p$ in the factorization of $n, a_{n}=\\prod_{p^{\\alpha} \\| n} a_{p^{\\alpha}}$.\nRefer to the first solution to check the case $p=2: a_{2^{\\alpha}}=2^{\\alpha}$.\nFor an odd prime $p$,\n\n$$\nf(x)=\\frac{x(x+1)}{2}=\\frac{(2 x+1)^{2}-1}{8}\n$$\n\nand since $p$ is odd, there is a bijection between the range of $f$ and the quadratic residues modulo $p^{\\alpha}$, namely $t \\mapsto 8 t+1$. So $a_{p^{\\alpha}}$ is the number of quadratic residues modulo $p^{\\alpha}$.\nLet $g$ be a primitive root of $p^{\\alpha}$. Then there are $\\frac{1}{2} \\phi\\left(p^{\\alpha}\\right)=\\frac{p-1}{2} \\cdot p^{\\alpha-1}$ quadratic residues that are coprime with $p: 1, g^{2}, g^{4}, \\ldots, g^{\\phi\\left(p^{n}\\right)-2}$. If $p$ divides a quadratic residue $k p$, that is, $x^{2} \\equiv k p$ $\\left(\\bmod p^{\\alpha}\\right), \\alpha \\geq 2$, then $p$ divides $x$ and, therefore, also $k$. Hence $p^{2}$ divides this quadratic residue, and these quadratic residues are $p^{2}$ times each quadratic residue of $p^{\\alpha-2}$. Thus\n\n$$\na_{p^{\\alpha}}=\\frac{p-1}{2} \\cdot p^{\\alpha-1}+a_{p^{\\alpha}-2} .\n$$\n\nSince $a_{p}=\\frac{p-1}{2}+1$ and $a_{p^{2}}=\\frac{p-1}{2} \\cdot p+1$, telescoping yields\n\n$$\na_{p^{2 t}}=\\frac{p-1}{2}\\left(p^{2 t-1}+p^{2 t-3}+\\cdots+p\\right)+1=\\frac{p\\left(p^{2 t}-1\\right)}{2(p+1)}+1\n$$\n\nand\n\n$$\na_{p^{2 t-1}}=\\frac{p-1}{2}\\left(p^{2 t-2}+p^{2 t-4}+\\cdots+1\\right)+1=\\frac{p^{2 t}-1}{2(p+1)}+1\n$$\n\nNow the problem is immediate: if $n$ is divisible by an odd prime $p, a_{p^{\\alpha}}<p^{\\alpha}$ for all $\\alpha$, and since $a_{t} \\leq t$ for all $t, a_{n}<n$.", "problem_match": "# Problem 4", "solution_match": "# Solution 2"}
-{"year": "1991", "problem_label": "5", "tier": 1, "problem": "Given are two tangent circles and a point $P$ on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point $P$.", "solution": "Throughout this problem, we will assume that the given circles are externally tangent, since the problem does not have a solution otherwise.\nLet $\\Gamma_{1}$ and $\\Gamma_{2}$ be the given circles and $T$ be their tangency point. Suppose $\\omega$ is a circle that is tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ and passes through $P$.\nNow invert about point $P$, with radius $P T$. Let any line through $P$ that cuts $\\Gamma_{1}$ do so at points $X$ and $Y$. The power of $P$ with respect to $\\Gamma_{1}$ is $P T^{2}=P X \\cdot P Y$, so $X$ and $Y$ are swapped by this inversion. Therefore $\\Gamma_{1}$ is mapped to itself in this inversion. The same applies to $\\Gamma_{2}$. Since circle $\\omega$ passes through $P$, it is mapped to a line tangent to the images of $\\Gamma_{1}$ (itself) and $\\Gamma_{2}$ (also itself), that is, a common tangent line. This common tangent cannot be $P T$, as $P T$ is also mapped to itself. Since $\\Gamma_{1}$ and $\\Gamma_{2}$ have exactly other two common tangent lines, there are two solutions: the inverses of the tangent lines.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-7.jpg?height=918&width=729&top_left_y=997&top_left_x=635)\n\nWe proceed with the construction with the aid of some macro constructions that will be detailed later.\n\nStep 1. Draw the common tangents to $\\Gamma_{1}$ and $\\Gamma_{2}$.\nStep 2. For each common tangent $t$, draw the projection $P_{t}$ of $P$ onto $t$.\nStep 3. Find the inverse $P_{1}$ of $P_{t}$ with respect to the circle with center $P$ and radius $P T$.\nStep 4. $\\omega_{t}$ is the circle with diameter $P P_{1}$.\nLet's work out the details for steps 1 and 3 . Steps 2 and 4 are immediate.\nStep 1. In this particular case in which $\\Gamma_{1}$ and $\\Gamma_{2}$ are externally tangent, there is a small shortcut:\n\n- Draw the circle with diameter on the two centers $O_{1}$ of $\\Gamma_{1}$ and $O_{2}$ of $\\Gamma_{2}$, and find its center $O$.\n- Let this circle meet common tangent line $O P$ at points $Q, R$. The required lines are the perpendicular to $O Q$ at $Q$ and the perpendicular to $O R$ at $R$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=572&width=729&top_left_y=282&top_left_x=635)\n\nLet's show why this construction works. Let $R_{i}$ be the radius of circle $\\Gamma_{i}$ and suppose without loss of generality that $R_{1} \\leq R_{2}$. Note that $O Q=\\frac{1}{2} O_{1} O_{2}=\\frac{R_{1}+R_{2}}{2}, O T=O O_{1}-R_{1}=\\frac{R_{2}-R_{1}}{2}$, so\n\n$$\n\\sin \\angle T Q O=\\frac{O T}{O Q}=\\frac{R_{2}-R_{1}}{R_{1}+R_{2}}\n$$\n\nwhich is also the sine of the angle between $O_{1} O_{2}$ and the common tangent lines.\nLet $t$ be the perpendicular to $O Q$ through $Q$. Then $\\angle\\left(t, O_{1} O_{2}\\right)=\\angle(O Q, Q T)=\\angle T Q O$, and $t$ is parallel to a common tangent line. Since\n\n$$\nd(O, t)=O Q=\\frac{R_{1}+R_{2}}{2}=\\frac{d\\left(O_{1}, t\\right)+d\\left(O_{2}, t\\right)}{2}\n$$\n\nand $O$ is the midpoint of $O_{1} O_{2}, O$ is also at the same distance from $t$ and the common tangent line, so these two lines coincide.\nStep 3. Finding the inverse of a point $X$ given the inversion circle $\\Omega$ with center $O$ is a well known procedure, but we describe it here for the sake of completeness.\n\n- If $X$ lies in $\\Omega$, then its inverse is $X$.\n- If $X$ lies in the interior of $\\Omega$, draw ray $O X$, then the perpendicular line $\\ell$ to $O X$ at $X$. Let $\\ell$ meet $\\Omega$ at a point $Y$. The inverse of $X$ is the intersection $X^{\\prime}$ of $O X$ and the line perpendicular to $O Y$ at $Y$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n- If $X$ is in the exterior of $\\Omega$, draw ray $O X$ and one of the tangent lines $\\ell$ from $X$ to $\\Omega$ (just connect $X$ to one of the intersections of $\\Omega$ and the circle with diameter $O X$ ). Let $\\ell$ touch $\\Omega$ at a point $Y$. The inverse of $X$ is the projection $X^{\\prime}$ of $Y$ onto $O X$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X^{\\prime}$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=347&width=481&top_left_y=2265&top_left_x=408)\n$X$ is inside $\\Omega$\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=398&width=469&top_left_y=2274&top_left_x=1096)", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "1991", "problem_label": "5", "tier": 1, "problem": "Given are two tangent circles and a point $P$ on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point $P$.", "solution": "Throughout this problem, we will assume that the given circles are externally tangent, since the problem does not have a solution otherwise.\nLet $\\Gamma_{1}$ and $\\Gamma_{2}$ be the given circles and $T$ be their tangency point. Suppose $\\omega$ is a circle that is tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ and passes through $P$.\nNow invert about point $P$, with radius $P T$. Let any line through $P$ that cuts $\\Gamma_{1}$ do so at points $X$ and $Y$. The power of $P$ with respect to $\\Gamma_{1}$ is $P T^{2}=P X \\cdot P Y$, so $X$ and $Y$ are swapped by this inversion. Therefore $\\Gamma_{1}$ is mapped to itself in this inversion. The same applies to $\\Gamma_{2}$. Since circle $\\omega$ passes through $P$, it is mapped to a line tangent to the images of $\\Gamma_{1}$ (itself) and $\\Gamma_{2}$ (also itself), that is, a common tangent line. This common tangent cannot be $P T$, as $P T$ is also mapped to itself. Since $\\Gamma_{1}$ and $\\Gamma_{2}$ have exactly other two common tangent lines, there are two solutions: the inverses of the tangent lines.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-7.jpg?height=918&width=729&top_left_y=997&top_left_x=635)\n\nWe proceed with the construction with the aid of some macro constructions that will be detailed later.\n\nStep 1. Draw the common tangents to $\\Gamma_{1}$ and $\\Gamma_{2}$.\nStep 2. For each common tangent $t$, draw the projection $P_{t}$ of $P$ onto $t$.\nStep 3. Find the inverse $P_{1}$ of $P_{t}$ with respect to the circle with center $P$ and radius $P T$.\nStep 4. $\\omega_{t}$ is the circle with diameter $P P_{1}$.\nLet's work out the details for steps 1 and 3 . Steps 2 and 4 are immediate.\nStep 1. In this particular case in which $\\Gamma_{1}$ and $\\Gamma_{2}$ are externally tangent, there is a small shortcut:\n\n- Draw the circle with diameter on the two centers $O_{1}$ of $\\Gamma_{1}$ and $O_{2}$ of $\\Gamma_{2}$, and find its center $O$.\n- Let this circle meet common tangent line $O P$ at points $Q, R$. The required lines are the perpendicular to $O Q$ at $Q$ and the perpendicular to $O R$ at $R$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=572&width=729&top_left_y=282&top_left_x=635)\n\nLet's show why this construction works. Let $R_{i}$ be the radius of circle $\\Gamma_{i}$ and suppose without loss of generality that $R_{1} \\leq R_{2}$. Note that $O Q=\\frac{1}{2} O_{1} O_{2}=\\frac{R_{1}+R_{2}}{2}, O T=O O_{1}-R_{1}=\\frac{R_{2}-R_{1}}{2}$, so\n\n$$\n\\sin \\angle T Q O=\\frac{O T}{O Q}=\\frac{R_{2}-R_{1}}{R_{1}+R_{2}}\n$$\n\nwhich is also the sine of the angle between $O_{1} O_{2}$ and the common tangent lines.\nLet $t$ be the perpendicular to $O Q$ through $Q$. Then $\\angle\\left(t, O_{1} O_{2}\\right)=\\angle(O Q, Q T)=\\angle T Q O$, and $t$ is parallel to a common tangent line. Since\n\n$$\nd(O, t)=O Q=\\frac{R_{1}+R_{2}}{2}=\\frac{d\\left(O_{1}, t\\right)+d\\left(O_{2}, t\\right)}{2}\n$$\n\nand $O$ is the midpoint of $O_{1} O_{2}, O$ is also at the same distance from $t$ and the common tangent line, so these two lines coincide.\nStep 3. Finding the inverse of a point $X$ given the inversion circle $\\Omega$ with center $O$ is a well known procedure, but we describe it here for the sake of completeness.\n\n- If $X$ lies in $\\Omega$, then its inverse is $X$.\n- If $X$ lies in the interior of $\\Omega$, draw ray $O X$, then the perpendicular line $\\ell$ to $O X$ at $X$. Let $\\ell$ meet $\\Omega$ at a point $Y$. The inverse of $X$ is the intersection $X^{\\prime}$ of $O X$ and the line perpendicular to $O Y$ at $Y$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n- If $X$ is in the exterior of $\\Omega$, draw ray $O X$ and one of the tangent lines $\\ell$ from $X$ to $\\Omega$ (just connect $X$ to one of the intersections of $\\Omega$ and the circle with diameter $O X$ ). Let $\\ell$ touch $\\Omega$ at a point $Y$. The inverse of $X$ is the projection $X^{\\prime}$ of $Y$ onto $O X$. This is because $O Y X^{\\prime}$ is a right triangle with altitude $Y X^{\\prime}$, and therefore $O X \\cdot O X^{\\prime}=O Y^{2}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=347&width=481&top_left_y=2265&top_left_x=408)\n$X$ is inside $\\Omega$\n![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=398&width=469&top_left_y=2274&top_left_x=1096)", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo1992_sol.jsonl

@@ -1,5 +1,5 @@
-{"year": "1992", "problem_label": "1", "tier": 1, "problem": "A triangle with sides $a, b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s=(a+b+c) / 2$. Construct a triangle with sides $s-a, s-b$, and $s-c$. This process is repeated until a triangle can no longer be constructed with the sidelengths given.\nFor which original triangles can this process be repeated indefinitely?\nAnswer: Only equilateral triangles.", "solution": "The perimeter of each new triangle constructed by the process is $(s-a)+(s-b)+(s-c)=$ $3 s-(a+b+c)=3 s-2 s=s$, that is, it is halved. Consider a new equivalent process in which a similar triangle with sidelengths $2(s-a), 2(s-b), 2(s-c)$ is constructed, so the perimeter is kept invariant.\nSuppose without loss of generality that $a \\leq b \\leq c$. Then $2(s-c) \\leq 2(s-b) \\leq 2(s-a)$, and the difference between the largest side and the smallest side changes from $c-a$ to $2(s-a)-2(s-c)=$ $2(c-a)$, that is, it doubles. Therefore, if $c-a>0$ then eventually this difference becomes larger than $a+b+c$, and it's immediate that a triangle cannot be constructed with the sidelengths. Hence the only possibility is $c-a=0 \\Longrightarrow a=b=c$, and it is clear that equilateral triangles can yield an infinite process, because all generated triangles are equilateral.", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}
-{"year": "1992", "problem_label": "2", "tier": 1, "problem": "In a circle $C$ with centre $O$ and radius $r$, let $C_{1}, C_{2}$ be two circles with centres $O_{1}, O_{2}$ and radii $r_{1}, r_{2}$ respectively, so that each circle $C_{i}$ is internally tangent to $C$ at $A_{i}$ and so that $C_{1}, C_{2}$ are externally tangent to each other at $A$.\nProve that the three lines $O A, O_{1} A_{2}$, and $O_{2} A_{1}$ are concurrent.", "solution": "Because of the tangencies, the following triples of points (two centers and a tangency point) are collinear:\n\n$$\nO_{1} ; O_{2} ; A, \\quad O ; O_{1} ; A_{1}, \\quad O ; O_{2} ; A_{2}\n$$\n\nBecause of that we can ignore the circles and only draw their centers and tangency points.\n![](https://cdn.mathpix.com/cropped/2024_11_22_c7e8c80a7518426c71e5g-2.jpg?height=1018&width=1095&top_left_y=713&top_left_x=432)\n\nNow the problem is immediate from Ceva's theorem in triangle $O O_{1} O_{2}$, because\n\n$$\n\\frac{O A_{1}}{A_{1} O_{1}} \\cdot \\frac{O_{1} A}{A O_{2}} \\cdot \\frac{O_{2} A_{2}}{A_{2} O}=\\frac{r}{r_{1}} \\cdot \\frac{r_{1}}{r_{2}} \\cdot \\frac{r_{2}}{r}=1\n$$", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
-{"year": "1992", "problem_label": "3", "tier": 1, "problem": "Let $n$ be an integer such that $n>3$. Suppose that we choose three numbers from the set $\\{1,2, \\ldots, n\\}$. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations.\n(a) Show that if we choose all three numbers greater than $n / 2$, then the values of these combinations are all distinct.\n(b) Let $p$ be a prime number such that $p \\leq \\sqrt{n}$. Show that the number of ways of choosing three numbers so that the smallest one is $p$ and the values of the combinations are not all distinct is precisely the number of positive divisors of $p-1$.", "solution": "In both items, the smallest chosen number is at least 2: in part (a), $n / 2>1$ and in part (b), $p$ is a prime. So let $1<x<y<z$ be the chosen numbers. Then all possible combinations are\n\n$$\nx+y+z, \\quad x+y z, \\quad x y+z, \\quad y+z x, \\quad(x+y) z, \\quad(z+x) y, \\quad(x+y) z, \\quad x y z\n$$\n\nSince, for $1<m<n$ and $t>1,(m-1)(n-1) \\geq 1 \\cdot 2 \\Longrightarrow m n>m+n, t n+m-(t m+n)=$ $(t-1)(n-m)>0 \\Longrightarrow t n+m>t m+n$, and $(t+m) n-(t+n) m=t(n-m)>0$,\n\n$$\nx+y+z<z+x y<y+z x<x+y z\n$$\n\nand\n\n$$\n(y+z) x<(x+z) y<(x+y) z<x y z .\n$$\n\nAlso, $(y+z) x-(y+z x)=(x-1) y>0 \\Longrightarrow(y+z) x>y+z x$ and $(x+z) y-(x+y z)=$ $(y-1) x>0 \\Longrightarrow(x+z) y>x+y z$. Therefore the only numbers that can be equal are $x+y z$ and $(y+z) x$. In this case,\n\n$$\nx+y z=(y+z) x \\Longleftrightarrow(y-x)(z-x)=x(x-1)\n$$\n\nNow we can solve the items.\n(a) if $n / 2<x<y<z$ then $z-x<n / 2$, and since $y-x<z-x, y-x<n / 2-1$; then\n\n$$\n(y-x)(z-x)<\\frac{n}{2}\\left(\\frac{n}{2}-1\\right)<x(x-1)\n$$\n\nand therefore $x+y z<(y+z) x$.\n(b) if $x=p$, then $(y-p)(z-p)=p(p-1)$. Since $y-p<z-p,(y-p)^{2}<(y-p)(z-p)=$ $p(p-1) \\Longrightarrow y-p<p$, that is, $p$ does not divide $y-p$. Then $y-p$ is a divisor $d$ of $p-1$ and $z-p=\\frac{p(p-1)}{d}$. Therefore,\n\n$$\nx=p, \\quad, y=p+d, \\quad z=p+\\frac{p(p-1)}{d}\n$$\n\nwhich is a solution for every divisor $d$ of $p-1$ because\n\n$$\nx=p<y=p+d<2 p \\leq p+p \\cdot \\frac{p-1}{d}=z .\n$$\n\nComment: If $x=1$ was allowed, then any choice $1, y, z$ would have repeated numbers in the combination, as $1 \\cdot y+z=y+1 \\cdot z$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
-{"year": "1992", "problem_label": "4", "tier": 1, "problem": "Determine all pairs $(h, s)$ of positive integers with the following property: If one draws $h$ horizontal lines and another $s$ lines which satisfy\n(i) they are not horizontal,\n(ii) no two of them are parallel,\n(iii) no three of the $h+s$ lines are concurrent,\nthen the number of regions formed by these $h+s$ lines is 1992 .\nAnswer: $(995,1),(176,10)$, and $(80,21)$.", "solution": "Let $a_{h, s}$ the number of regions formed by $h$ horizontal lines and $s$ another lines as described in the problem. Let $\\mathcal{F}_{h, s}$ be the union of the $h+s$ lines and pick any line $\\ell$. If it intersects the other lines in $n$ (distinct!) points then $\\ell$ is partitioned into $n-1$ line segments and 2 rays, which delimit regions. Therefore if we remove $\\ell$ the number of regions decreases by exactly $n-1+2=n+1$.\nThen $a_{0,0}=1$ (no lines means there is only one region), and since every one of $s$ lines intersects the other $s-1$ lines, $a_{0, s}=a_{0, s-1}+s$ for $s \\geq 0$. Summing yields\n\n$$\na_{0, s}=s+(s-1)+\\cdots+1+a_{0,0}=\\frac{s^{2}+s+2}{2} .\n$$\n\nEach horizontal line only intersects the $s$ non-horizontal lines, so $a_{h, s}=a_{h-1, s}+s+1$, which implies\n\n$$\na_{h, s}=a_{0, s}+h(s+1)=\\frac{s^{2}+s+2}{2}+h(s+1) .\n$$\n\nOur final task is solving\n\n$$\na_{h, s}=1992 \\Longleftrightarrow \\frac{s^{2}+s+2}{2}+h(s+1)=1992 \\Longleftrightarrow(s+1)(s+2 h)=2 \\cdot 1991=2 \\cdot 11 \\cdot 181\n$$\n\nThe divisors of $2 \\cdot 1991$ are $1,2,11,22,181,362,1991,3982$. Since $s, h>0,2 \\leq s+1<s+2 h$, so the possibilities for $s+1$ can only be 2,11 and 22 , yielding the following possibilities for $(h, s)$ :\n\n$$\n(995,1), \\quad(176,10), \\quad \\text { and }(80,21) .\n$$", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
-{"year": "1992", "problem_label": "5", "tier": 1, "problem": "Find a sequence of maximal length consisting of non-zero integers in which the sum of any seven consecutive terms is positive and that of any eleven consecutive terms is negative.\n\nAnswer: The maximum length is 16 . There are several possible sequences with this length; one such sequence is $(-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7)$.", "solution": "Suppose it is possible to have more than 16 terms in the sequence. Let $a_{1}, a_{2}, \\ldots, a_{17}$ be the first 17 terms of the sequence. Consider the following array of terms in the sequence:\n\n| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |\n| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ |\n| $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ |\n| $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ |\n| $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ |\n| $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |\n\nLet $S$ the sum of the numbers in the array. If we sum by rows we obtain negative sums in each row, so $S<0$; however, it we sum by columns we obtain positive sums in each column, so $S>0$, a contradiction. This implies that the sequence cannot have more than 16 terms. One idea to find a suitable sequence with 16 terms is considering cycles of 7 numbers. For instance, one can try\n\n$$\n-a,-a, b,-a,-a,-a, b,-a,-a, b,-a,-a,-a, b,-a,-a .\n$$\n\nThe sum of every seven consecutive numbers is $-5 a+2 b$ and the sum of every eleven consecutive numbers is $-8 a+3 b$, so $-5 a+2 b>0$ and $-8 a+3 b<0$, that is,\n\n$$\n\\frac{5 a}{2}<b<\\frac{8 a}{3} \\Longleftrightarrow 15 a<6 b<16 a\n$$\n\nThen we can choose, say, $a=7$ and $105<6 b<112 \\Longleftrightarrow b=18$. A valid sequence is then\n\n$$\n-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7\n$$", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "1992", "problem_label": "1", "tier": 1, "problem": "A triangle with sides $a, b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s=(a+b+c) / 2$. Construct a triangle with sides $s-a, s-b$, and $s-c$. This process is repeated until a triangle can no longer be constructed with the sidelengths given.\nFor which original triangles can this process be repeated indefinitely?\nAnswer: Only equilateral triangles.", "solution": "The perimeter of each new triangle constructed by the process is $(s-a)+(s-b)+(s-c)=$ $3 s-(a+b+c)=3 s-2 s=s$, that is, it is halved. Consider a new equivalent process in which a similar triangle with sidelengths $2(s-a), 2(s-b), 2(s-c)$ is constructed, so the perimeter is kept invariant.\nSuppose without loss of generality that $a \\leq b \\leq c$. Then $2(s-c) \\leq 2(s-b) \\leq 2(s-a)$, and the difference between the largest side and the smallest side changes from $c-a$ to $2(s-a)-2(s-c)=$ $2(c-a)$, that is, it doubles. Therefore, if $c-a>0$ then eventually this difference becomes larger than $a+b+c$, and it's immediate that a triangle cannot be constructed with the sidelengths. Hence the only possibility is $c-a=0 \\Longrightarrow a=b=c$, and it is clear that equilateral triangles can yield an infinite process, because all generated triangles are equilateral.", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}
+{"year": "1992", "problem_label": "2", "tier": 1, "problem": "In a circle $C$ with centre $O$ and radius $r$, let $C_{1}, C_{2}$ be two circles with centres $O_{1}, O_{2}$ and radii $r_{1}, r_{2}$ respectively, so that each circle $C_{i}$ is internally tangent to $C$ at $A_{i}$ and so that $C_{1}, C_{2}$ are externally tangent to each other at $A$.\nProve that the three lines $O A, O_{1} A_{2}$, and $O_{2} A_{1}$ are concurrent.", "solution": "Because of the tangencies, the following triples of points (two centers and a tangency point) are collinear:\n\n$$\nO_{1} ; O_{2} ; A, \\quad O ; O_{1} ; A_{1}, \\quad O ; O_{2} ; A_{2}\n$$\n\nBecause of that we can ignore the circles and only draw their centers and tangency points.\n![](https://cdn.mathpix.com/cropped/2024_11_22_c7e8c80a7518426c71e5g-2.jpg?height=1018&width=1095&top_left_y=713&top_left_x=432)\n\nNow the problem is immediate from Ceva's theorem in triangle $O O_{1} O_{2}$, because\n\n$$\n\\frac{O A_{1}}{A_{1} O_{1}} \\cdot \\frac{O_{1} A}{A O_{2}} \\cdot \\frac{O_{2} A_{2}}{A_{2} O}=\\frac{r}{r_{1}} \\cdot \\frac{r_{1}}{r_{2}} \\cdot \\frac{r_{2}}{r}=1\n$$", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
+{"year": "1992", "problem_label": "3", "tier": 1, "problem": "Let $n$ be an integer such that $n>3$. Suppose that we choose three numbers from the set $\\{1,2, \\ldots, n\\}$. Using each of these three numbers only once and using addition, multiplication, and parenthesis, let us form all possible combinations.\n(a) Show that if we choose all three numbers greater than $n / 2$, then the values of these combinations are all distinct.\n(b) Let $p$ be a prime number such that $p \\leq \\sqrt{n}$. Show that the number of ways of choosing three numbers so that the smallest one is $p$ and the values of the combinations are not all distinct is precisely the number of positive divisors of $p-1$.", "solution": "In both items, the smallest chosen number is at least 2: in part (a), $n / 2>1$ and in part (b), $p$ is a prime. So let $1<x<y<z$ be the chosen numbers. Then all possible combinations are\n\n$$\nx+y+z, \\quad x+y z, \\quad x y+z, \\quad y+z x, \\quad(x+y) z, \\quad(z+x) y, \\quad(x+y) z, \\quad x y z\n$$\n\nSince, for $1<m<n$ and $t>1,(m-1)(n-1) \\geq 1 \\cdot 2 \\Longrightarrow m n>m+n, t n+m-(t m+n)=$ $(t-1)(n-m)>0 \\Longrightarrow t n+m>t m+n$, and $(t+m) n-(t+n) m=t(n-m)>0$,\n\n$$\nx+y+z<z+x y<y+z x<x+y z\n$$\n\nand\n\n$$\n(y+z) x<(x+z) y<(x+y) z<x y z .\n$$\n\nAlso, $(y+z) x-(y+z x)=(x-1) y>0 \\Longrightarrow(y+z) x>y+z x$ and $(x+z) y-(x+y z)=$ $(y-1) x>0 \\Longrightarrow(x+z) y>x+y z$. Therefore the only numbers that can be equal are $x+y z$ and $(y+z) x$. In this case,\n\n$$\nx+y z=(y+z) x \\Longleftrightarrow(y-x)(z-x)=x(x-1)\n$$\n\nNow we can solve the items.\n(a) if $n / 2<x<y<z$ then $z-x<n / 2$, and since $y-x<z-x, y-x<n / 2-1$; then\n\n$$\n(y-x)(z-x)<\\frac{n}{2}\\left(\\frac{n}{2}-1\\right)<x(x-1)\n$$\n\nand therefore $x+y z<(y+z) x$.\n(b) if $x=p$, then $(y-p)(z-p)=p(p-1)$. Since $y-p<z-p,(y-p)^{2}<(y-p)(z-p)=$ $p(p-1) \\Longrightarrow y-p<p$, that is, $p$ does not divide $y-p$. Then $y-p$ is a divisor $d$ of $p-1$ and $z-p=\\frac{p(p-1)}{d}$. Therefore,\n\n$$\nx=p, \\quad, y=p+d, \\quad z=p+\\frac{p(p-1)}{d}\n$$\n\nwhich is a solution for every divisor $d$ of $p-1$ because\n\n$$\nx=p<y=p+d<2 p \\leq p+p \\cdot \\frac{p-1}{d}=z .\n$$\n\nComment: If $x=1$ was allowed, then any choice $1, y, z$ would have repeated numbers in the combination, as $1 \\cdot y+z=y+1 \\cdot z$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
+{"year": "1992", "problem_label": "4", "tier": 1, "problem": "Determine all pairs $(h, s)$ of positive integers with the following property: If one draws $h$ horizontal lines and another $s$ lines which satisfy\n(i) they are not horizontal,\n(ii) no two of them are parallel,\n(iii) no three of the $h+s$ lines are concurrent,\nthen the number of regions formed by these $h+s$ lines is 1992 .\nAnswer: $(995,1),(176,10)$, and $(80,21)$.", "solution": "Let $a_{h, s}$ the number of regions formed by $h$ horizontal lines and $s$ another lines as described in the problem. Let $\\mathcal{F}_{h, s}$ be the union of the $h+s$ lines and pick any line $\\ell$. If it intersects the other lines in $n$ (distinct!) points then $\\ell$ is partitioned into $n-1$ line segments and 2 rays, which delimit regions. Therefore if we remove $\\ell$ the number of regions decreases by exactly $n-1+2=n+1$.\nThen $a_{0,0}=1$ (no lines means there is only one region), and since every one of $s$ lines intersects the other $s-1$ lines, $a_{0, s}=a_{0, s-1}+s$ for $s \\geq 0$. Summing yields\n\n$$\na_{0, s}=s+(s-1)+\\cdots+1+a_{0,0}=\\frac{s^{2}+s+2}{2} .\n$$\n\nEach horizontal line only intersects the $s$ non-horizontal lines, so $a_{h, s}=a_{h-1, s}+s+1$, which implies\n\n$$\na_{h, s}=a_{0, s}+h(s+1)=\\frac{s^{2}+s+2}{2}+h(s+1) .\n$$\n\nOur final task is solving\n\n$$\na_{h, s}=1992 \\Longleftrightarrow \\frac{s^{2}+s+2}{2}+h(s+1)=1992 \\Longleftrightarrow(s+1)(s+2 h)=2 \\cdot 1991=2 \\cdot 11 \\cdot 181\n$$\n\nThe divisors of $2 \\cdot 1991$ are $1,2,11,22,181,362,1991,3982$. Since $s, h>0,2 \\leq s+1<s+2 h$, so the possibilities for $s+1$ can only be 2,11 and 22 , yielding the following possibilities for $(h, s)$ :\n\n$$\n(995,1), \\quad(176,10), \\quad \\text { and }(80,21) .\n$$", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
+{"year": "1992", "problem_label": "5", "tier": 1, "problem": "Find a sequence of maximal length consisting of non-zero integers in which the sum of any seven consecutive terms is positive and that of any eleven consecutive terms is negative.\n\nAnswer: The maximum length is 16 . There are several possible sequences with this length; one such sequence is $(-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7)$.", "solution": "Suppose it is possible to have more than 16 terms in the sequence. Let $a_{1}, a_{2}, \\ldots, a_{17}$ be the first 17 terms of the sequence. Consider the following array of terms in the sequence:\n\n| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |\n| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ |\n| $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ |\n| $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ |\n| $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ |\n| $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |\n\nLet $S$ the sum of the numbers in the array. If we sum by rows we obtain negative sums in each row, so $S<0$; however, it we sum by columns we obtain positive sums in each column, so $S>0$, a contradiction. This implies that the sequence cannot have more than 16 terms. One idea to find a suitable sequence with 16 terms is considering cycles of 7 numbers. For instance, one can try\n\n$$\n-a,-a, b,-a,-a,-a, b,-a,-a, b,-a,-a,-a, b,-a,-a .\n$$\n\nThe sum of every seven consecutive numbers is $-5 a+2 b$ and the sum of every eleven consecutive numbers is $-8 a+3 b$, so $-5 a+2 b>0$ and $-8 a+3 b<0$, that is,\n\n$$\n\\frac{5 a}{2}<b<\\frac{8 a}{3} \\Longleftrightarrow 15 a<6 b<16 a\n$$\n\nThen we can choose, say, $a=7$ and $105<6 b<112 \\Longleftrightarrow b=18$. A valid sequence is then\n\n$$\n-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7\n$$", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo1993_sol.jsonl

@@ -1,5 +1,5 @@
-{"year": "1993", "problem_label": "1", "tier": 1, "problem": "Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\\angle A B C$ is 60 degrees. Let $\\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$.\nProve that $C A^{2}=C M \\times C E$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_ef2d48e645c25b0f6b7ag-1.jpg?height=567&width=664&top_left_y=696&top_left_x=656)\n\nTriangles $A E D$ and $C D F$ are similar, because $A D \\| C F$ and $A E \\| C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,\n\n$$\n\\frac{A E}{C D}=\\frac{A D}{C F} \\Longleftrightarrow \\frac{A E}{A C}=\\frac{A C}{C F}\n$$\n\nThe last equality combined with\n\n$$\n\\angle E A C=180^{\\circ}-\\angle B A C=120^{\\circ}=\\angle A C F\n$$\n\nshows that triangles $E A C$ and $A C F$ are also similar. Therefore $\\angle C A M=\\angle C A F=\\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \\cdot C E$, and we are done.", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}
-{"year": "1993", "problem_label": "2", "tier": 1, "problem": "Find the total number of different integer values the function\n\n$$\nf(x)=[x]+[2 x]+\\left[\\frac{5 x}{3}\\right]+[3 x]+[4 x]\n$$\n\ntakes for real numbers $x$ with $0 \\leq x \\leq 100$.\nNote: $[t]$ is the largest integer that does not exceed $t$.\nAnswer: 734.", "solution": "Note that, since $[x+n]=[x]+n$ for any integer $n$,\n\n$$\nf(x+3)=[x+3]+[2(x+3)]+\\left[\\frac{5(x+3)}{3}\\right]+[3(x+3)]+[4(x+3)]=f(x)+35,\n$$\n\none only needs to investigate the interval $[0,3)$.\nThe numbers in this interval at which at least one of the real numbers $x, 2 x, \\frac{5 x}{3}, 3 x, 4 x$ is an integer are\n\n- $0,1,2$ for $x$;\n- $\\frac{n}{2}, 0 \\leq n \\leq 5$ for $2 x$;\n- $\\frac{3 n}{5}, 0 \\leq n \\leq 4$ for $\\frac{5 x}{3}$;\n- $\\frac{n}{3}, 0 \\leq n \\leq 8$ for $3 x$;\n- $\\frac{n}{4}, 0 \\leq n \\leq 11$ for $4 x$.\n\nOf these numbers there are\n\n- 3 integers $(0,1,2)$;\n- 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );\n- 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );\n- 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );\n- 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).\n\nTherefore $f(x)$ increases 22 times per interval. Since $100=33 \\cdot 3+1$, there are $33 \\cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \\frac{1}{2}, 99 \\frac{1}{3}, 99 \\frac{2}{3}, 99 \\frac{1}{4}$, $99 \\frac{3}{4}, 99 \\frac{3}{5}$.\nThe total is then $33 \\cdot 22+8=734$.\nComment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of\n\n$$\n0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30\n$$\n\nin the interval $[0, f(100)]=[0,1166]$. Since $1166 \\equiv 11(\\bmod 35)$, this comprises 33 cycles plus the 8 numbers in the previous list.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
-{"year": "1993", "problem_label": "3", "tier": 1, "problem": "Let\n\n$$\nf(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{0} \\quad \\text { and } \\quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\\cdots+c_{0}\n$$\n\nbe non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\\max \\left(\\left|a_{n}\\right|, \\ldots,\\left|a_{0}\\right|\\right)$ and $c=\\max \\left(\\left|c_{n+1}\\right|, \\ldots,\\left|c_{0}\\right|\\right)$, prove that $\\frac{a}{c} \\leq n+1$.", "solution": "Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \\ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:\n\n- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \\ldots, n$, and $a=c \\Longrightarrow \\frac{a}{c}=1 \\leq n+1$.\n- $|r| \\geq 1$. Then\n\n$$\n\\begin{gathered}\n\\left|a_{0}\\right|=\\left|\\frac{c_{0}}{r}\\right| \\leq c \\\\\n\\left|a_{1}\\right|=\\left|\\frac{c_{1}-a_{0}}{r}\\right| \\leq\\left|c_{1}\\right|+\\left|a_{0}\\right| \\leq 2 c\n\\end{gathered}\n$$\n\nand inductively if $\\left|a_{k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{k+1}\\right|=\\left|\\frac{c_{k+1}-a_{k}}{r}\\right| \\leq\\left|c_{k+1}\\right|+\\left|a_{k}\\right| \\leq c+(k+1) c=(k+2) c\n$$\n\nTherefore, $\\left|a_{k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c \\Longleftrightarrow \\frac{a}{c} \\leq n+1$.\n\n- $0<|r|<1$. Now work backwards: $\\left|a_{n}\\right|=\\left|c_{n+1}\\right| \\leq c$,\n\n$$\n\\left|a_{n-1}\\right|=\\left|c_{n}-r a_{n}\\right| \\leq\\left|c_{n}\\right|+\\left|r a_{n}\\right|<c+c=2 c,\n$$\n\nand inductively if $\\left|a_{n-k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{n-k-1}\\right|=\\left|c_{n-k}-r a_{n-k}\\right| \\leq\\left|c_{n-k}\\right|+\\left|r a_{n-k}\\right|<c+(k+1) c=(k+2) c .\n$$\n\nTherefore, $\\left|a_{n-k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c$ again.", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
-{"year": "1993", "problem_label": "4", "tier": 1, "problem": "Determine all positive integers $n$ for which the equation\n\n$$\nx^{n}+(2+x)^{n}+(2-x)^{n}=0\n$$\n\nhas an integer as a solution.\n\n## Answer: $n=1$.", "solution": "If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.\nFor $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.\nFor $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to\n\n$$\ny^{n}+(1+y)^{n}+(1-y)^{n}=0 .\n$$\n\nLooking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization\n\n$$\na^{n}+b^{n}=(a+b)\\left(a^{n-1}-a^{n-2} b+\\cdots+b^{n-1}\\right) \\quad \\text { for } n \\text { odd, }\n$$\n\nwhich has a sum of $n$ terms as the second factor, the equation is now equivalent to\n\n$$\ny^{n}+(1+y+1-y)\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right)=0\n$$\n\nor\n\n$$\ny^{n}=-2\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right) .\n$$\n\nEach of the $n$ terms in the second factor is odd, and $n$ is odd, so the second factor is odd. Therefore, $y^{n}$ has only one factor 2 , which is a contradiction to the fact that, $y$ being even, $y^{n}$ has at least $n>1$ factors 2 . Hence there are no solutions if $n>1$.", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
-{"year": "1993", "problem_label": "5", "tier": 1, "problem": "Let $P_{1}, P_{2}, \\ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:\n(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \\ldots, 1993$;\n(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \\ldots, 1992$.\n\nProve that for some $i, 0 \\leq i \\leq 1992$, there exists a point $Q$ with coordinates $\\left(q_{x}, q_{y}\\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.", "solution": "Call a point $(x, y) \\in \\mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \\leq i \\leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$.\nIn fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\\left(\\frac{a+c}{2}, \\frac{b+d}{2}\\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "1993", "problem_label": "1", "tier": 1, "problem": "Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\\angle A B C$ is 60 degrees. Let $\\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$.\nProve that $C A^{2}=C M \\times C E$.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_ef2d48e645c25b0f6b7ag-1.jpg?height=567&width=664&top_left_y=696&top_left_x=656)\n\nTriangles $A E D$ and $C D F$ are similar, because $A D \\| C F$ and $A E \\| C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,\n\n$$\n\\frac{A E}{C D}=\\frac{A D}{C F} \\Longleftrightarrow \\frac{A E}{A C}=\\frac{A C}{C F}\n$$\n\nThe last equality combined with\n\n$$\n\\angle E A C=180^{\\circ}-\\angle B A C=120^{\\circ}=\\angle A C F\n$$\n\nshows that triangles $E A C$ and $A C F$ are also similar. Therefore $\\angle C A M=\\angle C A F=\\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \\cdot C E$, and we are done.", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}
+{"year": "1993", "problem_label": "2", "tier": 1, "problem": "Find the total number of different integer values the function\n\n$$\nf(x)=[x]+[2 x]+\\left[\\frac{5 x}{3}\\right]+[3 x]+[4 x]\n$$\n\ntakes for real numbers $x$ with $0 \\leq x \\leq 100$.\nNote: $[t]$ is the largest integer that does not exceed $t$.\nAnswer: 734.", "solution": "Note that, since $[x+n]=[x]+n$ for any integer $n$,\n\n$$\nf(x+3)=[x+3]+[2(x+3)]+\\left[\\frac{5(x+3)}{3}\\right]+[3(x+3)]+[4(x+3)]=f(x)+35,\n$$\n\none only needs to investigate the interval $[0,3)$.\nThe numbers in this interval at which at least one of the real numbers $x, 2 x, \\frac{5 x}{3}, 3 x, 4 x$ is an integer are\n\n- $0,1,2$ for $x$;\n- $\\frac{n}{2}, 0 \\leq n \\leq 5$ for $2 x$;\n- $\\frac{3 n}{5}, 0 \\leq n \\leq 4$ for $\\frac{5 x}{3}$;\n- $\\frac{n}{3}, 0 \\leq n \\leq 8$ for $3 x$;\n- $\\frac{n}{4}, 0 \\leq n \\leq 11$ for $4 x$.\n\nOf these numbers there are\n\n- 3 integers $(0,1,2)$;\n- 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );\n- 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );\n- 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );\n- 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).\n\nTherefore $f(x)$ increases 22 times per interval. Since $100=33 \\cdot 3+1$, there are $33 \\cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \\frac{1}{2}, 99 \\frac{1}{3}, 99 \\frac{2}{3}, 99 \\frac{1}{4}$, $99 \\frac{3}{4}, 99 \\frac{3}{5}$.\nThe total is then $33 \\cdot 22+8=734$.\nComment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of\n\n$$\n0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30\n$$\n\nin the interval $[0, f(100)]=[0,1166]$. Since $1166 \\equiv 11(\\bmod 35)$, this comprises 33 cycles plus the 8 numbers in the previous list.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
+{"year": "1993", "problem_label": "3", "tier": 1, "problem": "Let\n\n$$\nf(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{0} \\quad \\text { and } \\quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\\cdots+c_{0}\n$$\n\nbe non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\\max \\left(\\left|a_{n}\\right|, \\ldots,\\left|a_{0}\\right|\\right)$ and $c=\\max \\left(\\left|c_{n+1}\\right|, \\ldots,\\left|c_{0}\\right|\\right)$, prove that $\\frac{a}{c} \\leq n+1$.", "solution": "Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \\ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:\n\n- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \\ldots, n$, and $a=c \\Longrightarrow \\frac{a}{c}=1 \\leq n+1$.\n- $|r| \\geq 1$. Then\n\n$$\n\\begin{gathered}\n\\left|a_{0}\\right|=\\left|\\frac{c_{0}}{r}\\right| \\leq c \\\\\n\\left|a_{1}\\right|=\\left|\\frac{c_{1}-a_{0}}{r}\\right| \\leq\\left|c_{1}\\right|+\\left|a_{0}\\right| \\leq 2 c\n\\end{gathered}\n$$\n\nand inductively if $\\left|a_{k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{k+1}\\right|=\\left|\\frac{c_{k+1}-a_{k}}{r}\\right| \\leq\\left|c_{k+1}\\right|+\\left|a_{k}\\right| \\leq c+(k+1) c=(k+2) c\n$$\n\nTherefore, $\\left|a_{k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c \\Longleftrightarrow \\frac{a}{c} \\leq n+1$.\n\n- $0<|r|<1$. Now work backwards: $\\left|a_{n}\\right|=\\left|c_{n+1}\\right| \\leq c$,\n\n$$\n\\left|a_{n-1}\\right|=\\left|c_{n}-r a_{n}\\right| \\leq\\left|c_{n}\\right|+\\left|r a_{n}\\right|<c+c=2 c,\n$$\n\nand inductively if $\\left|a_{n-k}\\right| \\leq(k+1) c$\n\n$$\n\\left|a_{n-k-1}\\right|=\\left|c_{n-k}-r a_{n-k}\\right| \\leq\\left|c_{n-k}\\right|+\\left|r a_{n-k}\\right|<c+(k+1) c=(k+2) c .\n$$\n\nTherefore, $\\left|a_{n-k}\\right| \\leq(k+1) c \\leq(n+1) c$ for all $k$, and $a \\leq(n+1) c$ again.", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
+{"year": "1993", "problem_label": "4", "tier": 1, "problem": "Determine all positive integers $n$ for which the equation\n\n$$\nx^{n}+(2+x)^{n}+(2-x)^{n}=0\n$$\n\nhas an integer as a solution.\n\n## Answer: $n=1$.", "solution": "If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.\nFor $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.\nFor $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to\n\n$$\ny^{n}+(1+y)^{n}+(1-y)^{n}=0 .\n$$\n\nLooking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization\n\n$$\na^{n}+b^{n}=(a+b)\\left(a^{n-1}-a^{n-2} b+\\cdots+b^{n-1}\\right) \\quad \\text { for } n \\text { odd, }\n$$\n\nwhich has a sum of $n$ terms as the second factor, the equation is now equivalent to\n\n$$\ny^{n}+(1+y+1-y)\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right)=0\n$$\n\nor\n\n$$\ny^{n}=-2\\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\\cdots+(1-y)^{n-1}\\right) .\n$$\n\nEach of the $n$ terms in the second factor is odd, and $n$ is odd, so the second factor is odd. Therefore, $y^{n}$ has only one factor 2 , which is a contradiction to the fact that, $y$ being even, $y^{n}$ has at least $n>1$ factors 2 . Hence there are no solutions if $n>1$.", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
+{"year": "1993", "problem_label": "5", "tier": 1, "problem": "Let $P_{1}, P_{2}, \\ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:\n(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \\ldots, 1993$;\n(ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \\ldots, 1992$.\n\nProve that for some $i, 0 \\leq i \\leq 1992$, there exists a point $Q$ with coordinates $\\left(q_{x}, q_{y}\\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.", "solution": "Call a point $(x, y) \\in \\mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \\leq i \\leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$.\nIn fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\\left(\\frac{a+c}{2}, \\frac{b+d}{2}\\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo1994_sol.jsonl

@@ -1,7 +1,7 @@
-{"year": "1994", "problem_label": "1", "tier": 1, "problem": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function such that\n(i) For all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x)+f(y)+1 \\geq f(x+y) \\geq f(x)+f(y)\n$$\n\n(ii) For all $x \\in[0,1), f(0) \\geq f(x)$,\n(iii) $-f(-1)=f(1)=1$.\n\nFind all such functions $f$.\nAnswer: $f(x)=\\lfloor x\\rfloor$, the largest integer that does not exceed $x$, is the only function.", "solution": "Plug $y \\rightarrow 1$ in (i):\n\n$$\nf(x)+f(1)+1 \\geq f(x+1) \\geq f(x)+f(1) \\Longleftrightarrow f(x)+1 \\leq f(x+1) \\leq f(x)+2\n$$\n\nNow plug $y \\rightarrow-1$ and $x \\rightarrow x+1$ in (i):\n\n$$\nf(x+1)+f(-1)+1 \\geq f(x) \\geq f(x+1)+f(-1) \\Longleftrightarrow f(x) \\leq f(x+1) \\leq f(x)+1\n$$\n\nHence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \\Longrightarrow f(0)=0$.\nCondition (ii) states that $f(x) \\leq 0$ in $[0,1)$.\nNow plug $y \\rightarrow 1-x$ in (i):\n\n$$\nf(x)+f(1-x)+1 \\leq f(x+(1-x)) \\leq f(x)+f(1-x) \\Longrightarrow f(x)+f(1-x) \\geq 0\n$$\n\nIf $x \\in(0,1)$ then $1-x \\in(0,1)$ as well, so $f(x) \\leq 0$ and $f(1-x) \\leq 0$, which implies $f(x)+f(1-x) \\leq 0$. Thus, $f(x)=f(1-x)=0$ for $x \\in(0,1)$. This combined with $f(0)=0$ and $f(x+1)=f(x)+1$ proves that $f(x)=\\lfloor x\\rfloor$, which satisfies the problem conditions, as since\n$x+y=\\lfloor x\\rfloor+\\lfloor y\\rfloor+\\{x\\}+\\{y\\}$ and $0 \\leq\\{x\\}+\\{y\\}<2 \\Longrightarrow\\lfloor x\\rfloor+\\lfloor y\\rfloor \\leq x+y<\\lfloor x\\rfloor+\\lfloor y\\rfloor+2$ implies\n\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor+1 \\geq\\lfloor x+y\\rfloor \\geq\\lfloor x\\rfloor+\\lfloor y\\rfloor .\n$$", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}
+{"year": "1994", "problem_label": "1", "tier": 1, "problem": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be a function such that\n(i) For all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x)+f(y)+1 \\geq f(x+y) \\geq f(x)+f(y)\n$$\n\n(ii) For all $x \\in[0,1), f(0) \\geq f(x)$,\n(iii) $-f(-1)=f(1)=1$.\n\nFind all such functions $f$.\nAnswer: $f(x)=\\lfloor x\\rfloor$, the largest integer that does not exceed $x$, is the only function.", "solution": "Plug $y \\rightarrow 1$ in (i):\n\n$$\nf(x)+f(1)+1 \\geq f(x+1) \\geq f(x)+f(1) \\Longleftrightarrow f(x)+1 \\leq f(x+1) \\leq f(x)+2\n$$\n\nNow plug $y \\rightarrow-1$ and $x \\rightarrow x+1$ in (i):\n\n$$\nf(x+1)+f(-1)+1 \\geq f(x) \\geq f(x+1)+f(-1) \\Longleftrightarrow f(x) \\leq f(x+1) \\leq f(x)+1\n$$\n\nHence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \\Longrightarrow f(0)=0$.\nCondition (ii) states that $f(x) \\leq 0$ in $[0,1)$.\nNow plug $y \\rightarrow 1-x$ in (i):\n\n$$\nf(x)+f(1-x)+1 \\leq f(x+(1-x)) \\leq f(x)+f(1-x) \\Longrightarrow f(x)+f(1-x) \\geq 0\n$$\n\nIf $x \\in(0,1)$ then $1-x \\in(0,1)$ as well, so $f(x) \\leq 0$ and $f(1-x) \\leq 0$, which implies $f(x)+f(1-x) \\leq 0$. Thus, $f(x)=f(1-x)=0$ for $x \\in(0,1)$. This combined with $f(0)=0$ and $f(x+1)=f(x)+1$ proves that $f(x)=\\lfloor x\\rfloor$, which satisfies the problem conditions, as since\n$x+y=\\lfloor x\\rfloor+\\lfloor y\\rfloor+\\{x\\}+\\{y\\}$ and $0 \\leq\\{x\\}+\\{y\\}<2 \\Longrightarrow\\lfloor x\\rfloor+\\lfloor y\\rfloor \\leq x+y<\\lfloor x\\rfloor+\\lfloor y\\rfloor+2$ implies\n\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor+1 \\geq\\lfloor x+y\\rfloor \\geq\\lfloor x\\rfloor+\\lfloor y\\rfloor .\n$$", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}
 {"year": "1994", "problem_label": "2", "tier": 1, "problem": "Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.", "solution": "Embed $A B C$ in the complex plane, with $A, B$ and $C$ in the circle $|z|=R$, so $O$ is the origin. Represent each point by its lowercase letter. It is well known that $h=a+b+c$, so\n\n$$\nO H=|a+b+c| \\leq|a|+|b|+|c|=3 R .\n$$\n\nThe equality cannot occur because $a, b$, and $c$ are not collinear, so $O H<3 R$.", "problem_match": "# Problem 2", "solution_match": "# Solution 1"}
 {"year": "1994", "problem_label": "2", "tier": 1, "problem": "Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.", "solution": "Suppose with loss of generality that $\\angle A<90^{\\circ}$. Let $B D$ be an altitude. Then\n\n$$\nA H=\\frac{A D}{\\cos \\left(90^{\\circ}-C\\right)}=\\frac{A B \\cos A}{\\sin C}=2 R \\cos A\n$$\n\nBy the triangle inequality,\n\n$$\nO H<A O+A H=R+2 R \\cos A<3 R\n$$\n\nComment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that\n\n$$\nO H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} .\n$$\n\nIn fact, using vectors in a coordinate system with $O$ as origin, by the Euler line\n\n$$\n\\overrightarrow{O H}=3 \\overrightarrow{O G}=3 \\cdot \\frac{\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}}{3}=\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}\n$$\n\nso\n\n$$\nO H^{2}=\\overrightarrow{O H} \\cdot \\overrightarrow{O H}=(\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C}) \\cdot(\\overrightarrow{O A}+\\overrightarrow{O B}+\\overrightarrow{O C})\n$$\n\nExpanding and using the fact that $\\overrightarrow{O X} \\cdot \\overrightarrow{O X}=O X^{2}=R^{2}$ for $X \\in\\{A, B, C\\}$, as well as\n$\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=O A \\cdot O B \\cdot \\cos \\angle A O B=R^{2} \\cos 2 C=R^{2}\\left(1-2 \\sin ^{2} C\\right)=R^{2}\\left(1-2\\left(\\frac{c}{2 R}\\right)^{2}\\right)=R^{2}-\\frac{c^{2}}{2}$, we find that\n\n$$\n\\begin{aligned}\nO H^{2} & =\\overrightarrow{O A} \\cdot \\overrightarrow{O A}+\\overrightarrow{O B} \\cdot \\overrightarrow{O B}+\\overrightarrow{O C} \\cdot \\overrightarrow{O C}+2 \\overrightarrow{O A} \\cdot \\overrightarrow{O B}+2 \\overrightarrow{O A} \\cdot \\overrightarrow{O C}+2 \\overrightarrow{O B} \\cdot \\overrightarrow{O C} \\\\\n& =3 R^{2}+\\left(2 R^{2}-c^{2}\\right)+\\left(2 R^{2}-b^{2}\\right)+\\left(2 R^{2}-a^{2}\\right) \\\\\n& =9 R^{2}-a^{2}-b^{2}-c^{2}\n\\end{aligned}\n$$\n\nas required.\nThis proves that $O H^{2}<9 R^{2} \\Longrightarrow O H<3 R$, and since $a, b, c$ can be arbitrarily small (fix the circumcircle and choose $A, B, C$ arbitrarily close in this circle), the bound is sharp.", "problem_match": "# Problem 2", "solution_match": "# Solution 2"}
-{"year": "1994", "problem_label": "3", "tier": 1, "problem": "Let $n$ be an integer of the form $a^{2}+b^{2}$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \\leq \\sqrt{n}$, then $p$ divides $a b$. Determine all such $n$.\n\nAnswer: $n=2,5,13$.", "solution": "A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \\leq \\sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime.\nSuppose without loss of generality that $a \\geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.\n\n- If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.\n- If $a-b=1$ then $a$ and $b$ are coprime and $a^{2}+b^{2}=(a-b)^{2}+2 a b=2 a b+1=2 b(b+1)+1=$ $2 b^{2}+2 b+1$. So any prime factor of any number smaller than $\\sqrt{2 b^{2}+2 b+1}$ is a divisor of $a b=b(b+1)$.\nOne can check that $b=1$ and $b=2$ yields the solutions $n=1^{2}+2^{2}=5$ (the only prime $p$ is 2 ) and $n=2^{2}+3^{2}=13$ (the only primes $p$ are 2 and 3 ). Suppose that $b>2$.\nConsider, for instance, the prime factors of $b-1 \\leq \\sqrt{2 b^{2}+2 b+1}$, which is coprime with $b$. Any prime must then divide $a=b+1$. Then it divides $(b+1)-(b-1)=2$, that is, $b-1$ can only have 2 as a prime factor, that is, $b-1$ is a power of 2 , and since $b-1 \\geq 2$, $b$ is odd.\nSince $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \\geq 0$, we can also consider any prime divisor of $b+2$. Since $b$ is odd, $b$ and $b+2$ are also coprime, so any prime divisor of $b+2$ must divide $a=b+1$. But $b+1$ and $b+2$ are also coprime, so there can be no such primes. This is a contradiction, and $b \\geq 3$ does not yield any solutions.\n- If $a-b>1$, consider a prime divisor $p$ of $a-b=\\sqrt{a^{2}-2 a b+b^{2}}<\\sqrt{a^{2}+b^{2}}$. Since $p$ divides one of $a$ and $b, p$ divides both numbers (just add or subtract $a-b$ accordingly.) This is a contradiction.\n\nHence the only solutions are $n=2,5,13$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
+{"year": "1994", "problem_label": "3", "tier": 1, "problem": "Let $n$ be an integer of the form $a^{2}+b^{2}$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \\leq \\sqrt{n}$, then $p$ divides $a b$. Determine all such $n$.\n\nAnswer: $n=2,5,13$.", "solution": "A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \\leq \\sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime.\nSuppose without loss of generality that $a \\geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.\n\n- If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.\n- If $a-b=1$ then $a$ and $b$ are coprime and $a^{2}+b^{2}=(a-b)^{2}+2 a b=2 a b+1=2 b(b+1)+1=$ $2 b^{2}+2 b+1$. So any prime factor of any number smaller than $\\sqrt{2 b^{2}+2 b+1}$ is a divisor of $a b=b(b+1)$.\nOne can check that $b=1$ and $b=2$ yields the solutions $n=1^{2}+2^{2}=5$ (the only prime $p$ is 2 ) and $n=2^{2}+3^{2}=13$ (the only primes $p$ are 2 and 3 ). Suppose that $b>2$.\nConsider, for instance, the prime factors of $b-1 \\leq \\sqrt{2 b^{2}+2 b+1}$, which is coprime with $b$. Any prime must then divide $a=b+1$. Then it divides $(b+1)-(b-1)=2$, that is, $b-1$ can only have 2 as a prime factor, that is, $b-1$ is a power of 2 , and since $b-1 \\geq 2$, $b$ is odd.\nSince $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \\geq 0$, we can also consider any prime divisor of $b+2$. Since $b$ is odd, $b$ and $b+2$ are also coprime, so any prime divisor of $b+2$ must divide $a=b+1$. But $b+1$ and $b+2$ are also coprime, so there can be no such primes. This is a contradiction, and $b \\geq 3$ does not yield any solutions.\n- If $a-b>1$, consider a prime divisor $p$ of $a-b=\\sqrt{a^{2}-2 a b+b^{2}}<\\sqrt{a^{2}+b^{2}}$. Since $p$ divides one of $a$ and $b, p$ divides both numbers (just add or subtract $a-b$ accordingly.) This is a contradiction.\n\nHence the only solutions are $n=2,5,13$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
 {"year": "1994", "problem_label": "4", "tier": 1, "problem": "Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?\nAnswer: Yes.", "solution": "The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\\cos (2 n \\theta), \\sin (2 n \\theta))$ for an appropriate $\\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \\theta \\bmod \\pi$, that is, $2|\\sin ((m-n) \\theta)|$. Our task is then finding $\\theta$ such that (i) $\\sin (k \\theta)$ is rational for all $k \\in \\mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\\theta \\in(0, \\pi / 2)$ such that $\\cos \\theta=\\frac{3}{5}$ and therefore $\\sin \\theta=\\frac{4}{5}$ does the job. Proof of (i): We know that $\\sin ((n+1) \\theta)+\\sin ((n-1) \\theta)=2 \\sin (n \\theta) \\cos \\theta$, so if $\\sin ((n-1) \\theta$ and $\\sin (n \\theta)$ are both rational then $\\sin ((n+1) \\theta)$ also is. Since $\\sin (0 \\theta)=0$ and $\\sin \\theta$ are rational, an induction shows that $\\sin (n \\theta)$ is rational for $n \\in \\mathbb{Z}_{>0}$; the result is also true if $n$ is negative because $\\sin$ is an odd function.\nProof of (ii): $P_{m}=P_{n} \\Longleftrightarrow 2 n \\theta=2 m \\theta+2 k \\pi$ for some $k \\in \\mathbb{Z}$, which implies $\\sin ((n-m) \\theta)=$ $\\sin (k \\pi)=0$. We show that $\\sin (k \\theta) \\neq 0$ for all $k \\neq 0$.\nWe prove a stronger result: let $\\sin (k \\theta)=\\frac{a_{k}}{5^{k}}$. Then\n\n$$\n\\begin{aligned}\n\\sin ((k+1) \\theta)+\\sin ((k-1) \\theta)=2 \\sin (k \\theta) \\cos \\theta & \\Longleftrightarrow \\frac{a_{k+1}}{5^{k+1}}+\\frac{a_{k-1}}{5^{k-1}}=2 \\cdot \\frac{a_{k}}{5^{k}} \\cdot \\frac{3}{5} \\\\\n& \\Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1}\n\\end{aligned}\n$$\n\nSince $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \\geq 0$, and $a_{k+1} \\equiv a_{k}(\\bmod 5)$ for $k \\geq 1$ (note that $a_{-1}=-\\frac{4}{25}$ is not an integer!). Thus $a_{k} \\equiv 4(\\bmod 5)$ for all $k \\geq 1$, and $\\sin (k \\theta)=\\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \\equiv 4(\\bmod 5)$. This proves (ii) and we are done.", "problem_match": "# Problem 4", "solution_match": "# Solution 1"}
 {"year": "1994", "problem_label": "4", "tier": 1, "problem": "Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational?\nAnswer: Yes.", "solution": "We present a different construction. Consider the (collinear) points\n\n$$\nP_{k}=\\left(1, \\frac{x_{k}}{y_{k}}\\right),\n$$\n\nsuch that the distance $O P_{k}$ from the origin $O$,\n\n$$\nO P_{k}=\\frac{\\sqrt{x_{k}^{2}+y_{k}^{2}}}{y_{k}}\n$$\n\nis rational, and $x_{k}$ and $y_{k}$ are integers. Clearly, $P_{i} P_{j}=\\left|\\frac{x_{i}}{y_{i}}-\\frac{x_{j}}{y_{j}}\\right|$ is rational.\nPerform an inversion with center $O$ and unit radius. It maps the line $x=1$, which contains all points $P_{k}$, to a circle (minus the origin). Let $Q_{k}$ be the image of $P_{k}$ under this inversion. Then\n\n$$\nQ_{i} Q_{j}=\\frac{1^{2} P_{i} P_{j}}{O P_{i} \\cdot O P_{j}}\n$$\n\nis rational and we are done if we choose $x_{k}$ and $y_{k}$ accordingly. But this is not hard, as we can choose the legs of a Pythagorean triple, say\n\n$$\nx_{k}=k^{2}-1, \\quad y_{k}=2 k\n$$\n\nThis implies $O P_{k}=\\frac{k^{2}+1}{2 k}$, and then\n\n$$\nQ_{i} Q_{j}=\\frac{\\left|\\frac{i^{2}-1}{i}-\\frac{j^{2}-1}{j}\\right|}{\\frac{i^{2}+1}{2 i} \\cdot \\frac{j^{2}+1}{2 j}}=\\frac{|4(i-j)(i j+1)|}{\\left(i^{2}+1\\right)\\left(j^{2}+1\\right)}\n$$", "problem_match": "# Problem 4", "solution_match": "# Solution 2"}
-{"year": "1994", "problem_label": "5", "tier": 1, "problem": "You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:\n\n| $A$ | $B$ | $C$ |\n| :--- | :--- | :--- |\n| 10 | 1010 | 20 |\n| 100 | 1100100 | 400 |\n| 1000 | 1111101000 | 13000 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n\nProve that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.", "solution": "Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then\n\n$$\n2^{b_{k}-1} \\leq 10^{k}<2^{b_{k}} \\Longleftrightarrow \\log _{2} 10^{k}<b_{k} \\leq \\log _{2} 10^{k}+1 \\Longleftrightarrow b_{k}=\\left\\lfloor k \\cdot \\log _{2} 10\\right\\rfloor+1\n$$\n\nand, similarly\n\n$$\nc_{k}=\\left\\lfloor k \\cdot \\log _{5} 10\\right\\rfloor+1\n$$\n\nBeatty's theorem states that if $\\alpha$ and $\\beta$ are irrational positive numbers such that\n\n$$\n\\frac{1}{\\alpha}+\\frac{1}{\\beta}=1\n$$\n\nthen the sequences $\\lfloor k \\alpha\\rfloor$ and $\\lfloor k \\beta\\rfloor, k=1,2, \\ldots$, partition the positive integers.\nThen, since\n\n$$\n\\frac{1}{\\log _{2} 10}+\\frac{1}{\\log _{5} 10}=\\log _{10} 2+\\log _{10} 5=\\log _{10}(2 \\cdot 5)=1\n$$\n\nthe sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done.\nComment: For the sake of completeness, a proof of Beatty's theorem follows.\nLet $x_{n}=\\alpha n$ and $y_{n}=\\beta n, n \\geq 1$ integer. Note that, since $\\alpha m=\\beta n$ implies that $\\frac{\\alpha}{\\beta}$ is rational but\n\n$$\n\\frac{\\alpha}{\\beta}=\\alpha \\cdot \\frac{1}{\\beta}=\\alpha\\left(1-\\frac{1}{\\alpha}\\right)=\\alpha-1\n$$\n\nis irrational, the sequences have no common terms, and all terms in both sequences are irrational.\nThe theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \\alpha<N \\Longleftrightarrow n<\\frac{N}{\\alpha}$, there are $\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is\n\n$$\nT(N)=\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor+\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor\n$$\n\nHowever, $x-1<\\lfloor x\\rfloor<x$ for nonintegers $x$, so\n\n$$\n\\begin{aligned}\n\\frac{N}{\\alpha}-1+\\frac{N}{\\beta}-1<T(N)<\\frac{N}{\\alpha}+\\frac{N}{\\beta} & \\Longleftrightarrow N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right)-2<T(N)<N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right) \\\\\n& \\Longleftrightarrow N-2<T(N)<N,\n\\end{aligned}\n$$\n\nthat is, $T(N)=N-1$.\nTherefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "1994", "problem_label": "5", "tier": 1, "problem": "You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:\n\n| $A$ | $B$ | $C$ |\n| :--- | :--- | :--- |\n| 10 | 1010 | 20 |\n| 100 | 1100100 | 400 |\n| 1000 | 1111101000 | 13000 |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n\nProve that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.", "solution": "Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then\n\n$$\n2^{b_{k}-1} \\leq 10^{k}<2^{b_{k}} \\Longleftrightarrow \\log _{2} 10^{k}<b_{k} \\leq \\log _{2} 10^{k}+1 \\Longleftrightarrow b_{k}=\\left\\lfloor k \\cdot \\log _{2} 10\\right\\rfloor+1\n$$\n\nand, similarly\n\n$$\nc_{k}=\\left\\lfloor k \\cdot \\log _{5} 10\\right\\rfloor+1\n$$\n\nBeatty's theorem states that if $\\alpha$ and $\\beta$ are irrational positive numbers such that\n\n$$\n\\frac{1}{\\alpha}+\\frac{1}{\\beta}=1\n$$\n\nthen the sequences $\\lfloor k \\alpha\\rfloor$ and $\\lfloor k \\beta\\rfloor, k=1,2, \\ldots$, partition the positive integers.\nThen, since\n\n$$\n\\frac{1}{\\log _{2} 10}+\\frac{1}{\\log _{5} 10}=\\log _{10} 2+\\log _{10} 5=\\log _{10}(2 \\cdot 5)=1\n$$\n\nthe sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done.\nComment: For the sake of completeness, a proof of Beatty's theorem follows.\nLet $x_{n}=\\alpha n$ and $y_{n}=\\beta n, n \\geq 1$ integer. Note that, since $\\alpha m=\\beta n$ implies that $\\frac{\\alpha}{\\beta}$ is rational but\n\n$$\n\\frac{\\alpha}{\\beta}=\\alpha \\cdot \\frac{1}{\\beta}=\\alpha\\left(1-\\frac{1}{\\alpha}\\right)=\\alpha-1\n$$\n\nis irrational, the sequences have no common terms, and all terms in both sequences are irrational.\nThe theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \\alpha<N \\Longleftrightarrow n<\\frac{N}{\\alpha}$, there are $\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is\n\n$$\nT(N)=\\left\\lfloor\\frac{N}{\\alpha}\\right\\rfloor+\\left\\lfloor\\frac{N}{\\beta}\\right\\rfloor\n$$\n\nHowever, $x-1<\\lfloor x\\rfloor<x$ for nonintegers $x$, so\n\n$$\n\\begin{aligned}\n\\frac{N}{\\alpha}-1+\\frac{N}{\\beta}-1<T(N)<\\frac{N}{\\alpha}+\\frac{N}{\\beta} & \\Longleftrightarrow N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right)-2<T(N)<N\\left(\\frac{1}{\\alpha}+\\frac{1}{\\beta}\\right) \\\\\n& \\Longleftrightarrow N-2<T(N)<N,\n\\end{aligned}\n$$\n\nthat is, $T(N)=N-1$.\nTherefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo1999_sol.jsonl

@@ -0,0 +1,6 @@
+{"year": "1999", "problem_label": "1", "tier": 1, "problem": "Find the smallest positive integer $n$ with the following property : There does not exist an arithmetic progression of 1993 terms of real numbers containing exactly $n$ integers.", "solution": "and Marking Scheme:\nWe first note that the integer terms of any arithmetic progression are \"equally spaced\", because if the $i$ th term $a_{i}$ and the $(i+j)$ th term $a_{i+j}$ of an arithmetic progression are both integers, then so is the $(i+2 j)$ th term $a_{i+2 j}=a_{i+j}+\\left(a_{i+j}-a_{i}\\right)$.\n\n1 POINT for realizing that the integers must be \"equally spaced\".\nThus, by scaling and translation, we can assume that the integer terms of the arithmetic progression are $1,2, \\cdots, n$ and we need only to consider arithmetic progression of the form\n\n$$\n1,1+\\frac{1}{k}, 1+\\frac{2}{k}, \\cdots, 1+\\frac{k-1}{k}, 2,2+\\frac{1}{k}, \\cdots, n-1, \\cdots, n-1+\\frac{k-1}{k}, n\n$$\n\nThis has $k n-k+1$ terms of which exactly $n$ are integers. Moreover we can add up to $k-1$ terms on either end and get another arithmetic progression without changing the number of integer terms.\n\n2 POINTS for noticing that the maximal sequence has an equal number of terms on either side of the integers appearing in the sequence (this includes the 1 POINT above). In other words, 2 POINTS for the scaled and translated form of the progression including the $k-1$ terms on either side.\n\nThus there are arithmetic progressions with $n$ integers whose length is any integer lying in the interval $[k n-k+1, k n+k-1]$, where $k$ is any positive integer. Thus we want to find the smallest $n>0$ so that, if $k$ is the largest integer satisfying $k n+k-1 \\leq 1998$, then $(k+1) n-(k+1)+1 \\geq 2000$.\n\n4 POINTS for clarifying the nature of the number $n$ in this way, which includes counting the terms of the maximal and minimal sequences containing $n$ integers and bounding them accordingly (this includes the 2 POINTS above).\n\nThat is, putting $k=\\lfloor 1999 /(n+1)\\rfloor$, we want the smallest integer $n$ so that\n\n$$\n\\left\\lfloor\\frac{1999}{n+1}\\right\\rfloor(n-1)+n \\geq 2000\n$$\n\nThis inequality does not hold if\n\n$$\n\\frac{1999}{n+1} \\cdot(n-1)+n<2000\n$$\n\n2 POINTS for setting up an inequality for $n$.\nThis simplifies to $n^{2}<3999$, that is, $n \\leq 63$. Now we check integers from $n=64$ on:\n\n$$\n\\begin{aligned}\n& \\text { for } n=64,\\left\\lfloor\\frac{1999}{65}\\right\\rfloor \\cdot 63+64=30 \\cdot 63+64=1954<2000 ; \\\\\n& \\text { for } n=65,\\left\\lfloor\\frac{1999}{66}\\right\\rfloor \\cdot 64+65=30 \\cdot 64+65=1985<2000 ; \\\\\n& \\text { for } n=66,\\left\\lfloor\\frac{1999}{67}\\right\\rfloor \\cdot 65+66=29 \\cdot 65+66=1951<2000 ; \\\\\n& \\text { for } n=67,\\left\\lfloor\\left\\lfloor\\frac{1999}{68}\\right\\rfloor \\cdot 66+67=29 \\cdot 66+67=1981<2000 ;\\right. \\\\\n& \\text { for } n=68,\\left\\lfloor\\frac{1999}{69}\\right\\rfloor \\cdot 67+68=28 \\cdot 67+68=1944<2000 ; \\\\\n& \\text { for } n=69,\\left\\lfloor\\frac{1999}{70}\\right\\rfloor \\cdot 68+69=28 \\cdot 68+69=1973<2000 ; \\\\\n& \\text { for } n=70,\\left\\lfloor\\frac{1999}{71}\\right\\rfloor \\cdot 69+70=28 \\cdot 69+70=2002 \\geq 2000 .\n\\end{aligned}\n$$\n\nThus the answer is $n=70$.\n1.POINT for checking these rumbers and finding that $n=70$.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution "}
+{"year": "1999", "problem_label": "2", "tier": 1, "problem": "Let $a_{1}, a_{2}, \\cdots$ be a sequence of real numbers satisfying $a_{i+j} \\leq a_{i}+a_{j}$ for all $i, j=1,2, \\cdots$. Prove that\n\n$$\na_{1}+\\frac{a_{2}}{2}+\\frac{a_{3}}{3}+\\cdots+\\frac{a_{n}}{n} \\geq a_{n}\n$$\n\nfor each positive integer $n$.", "solution": "and Marking Scheme:\n\nLetting $b_{i}=a_{i} / i,(i=1,2, \\cdots)$, we prove that\n\n$$\nb_{1}+\\cdots+b_{n} \\geq a_{n} \\quad(n=1,2, \\cdots)\n$$\n\nby induction on $n$. For $n=1, b_{1}=a_{1} \\geq a_{1}$, and the induction starts. Assume that\n\n$$\nb_{1}+\\cdots+b_{k} \\geq a_{k}\n$$\n\nfor all $k=1,2, \\cdots, n-1$. It suffices to prove that $b_{1}+\\cdots+b_{n} \\geq a_{n}$ or equivalently that\n\n$$\n\\begin{aligned}\n& n b_{1}+\\cdots+n b_{n-1} \\geq(n-1) a_{n} . \\\\\n& 3 \\text { POINTS for separating } a_{n} \\text { from } b_{1}, \\cdots, b_{n-1} \\text {. } \\\\\n& n b_{1}+\\cdots+n b_{n-1}=(n-1) b_{1}+(n-2) b_{2}+\\cdots+b_{n-1}+b_{1}+2 b_{2}+\\cdots+(n-1) b_{n-1} \\\\\n& =b_{1}+\\left(b_{1}+b_{2}\\right)+\\cdots+\\left(b_{1}+b_{2}+\\cdots+b_{n-1}\\right)+\\left(a_{1}+a_{2}+\\cdots+a_{n-1}\\right) \\\\\n& \\geq 2\\left(a_{1}+a_{2}+\\cdots+a_{n-1}\\right)=\\sum_{i=1}^{n-1}\\left(a_{i}+a_{n-i}\\right) \\geq(n-1) a_{n} .\n\\end{aligned}\n$$\n\n3 POINTS for the first inequaliny and 1 POINT for the rest.", "problem_match": "\nProblem 2.", "solution_match": "# Solution "}
+{"year": "1999", "problem_label": "3", "tier": 1, "problem": "Let $\\Gamma_{1}$ and $\\Gamma_{2}$ be two circles intersecting at $P$ and $Q$. The common tangent, closer to $P$, of $\\Gamma_{1}$ and $\\Gamma_{2}$ touches $\\Gamma_{1}$ at $A$ and $\\Gamma_{2}$ at $B$. The tangent of $\\Gamma_{1}$ at $P$ meets $\\Gamma_{2}$ at $C$, which is different from $P$ and the extension of $A P$ meets $B C$ at $R$. Prove that the circumcircle of triangle $P Q R$ is tangent to $B P$ and $B R$.", "solution": "and Marking Scheme:\n\nLet $\\alpha=\\angle P A B, \\beta=\\angle A B P$ y $\\gamma=\\angle Q A P$. Then, since $P C$ is tangent to $\\Gamma_{1}$, we have $\\angle Q P C=$ $\\angle Q B C=\\gamma$. Thus $A, B, R, Q$ are concyclic.\n\n3 POINTS for proving that $A, B, R, Q$ are concyclic.\nSince $A B$ is a common tangent to $\\Gamma_{1}$ and $\\Gamma_{2}$ then $\\angle A Q P=\\alpha$ and $\\angle P Q B=\\angle P C B=\\beta$. Therefore, since $A, B, R, Q$ are concyclic, $\\angle A R B=\\angle A Q B=\\alpha+\\beta$ and $\\angle B Q R=\\alpha$. Thus $\\angle P Q R=\\angle P Q B+$ $\\angle B Q R=\\alpha+\\beta$.\n\n$$\n2 \\text { POINTS for proving that } \\angle P Q R=\\angle P R B=\\alpha+\\beta\n$$\n\nSince $\\angle B P R$ is an exterior angle of triangle $A B P, \\angle B P R=\\alpha+\\beta$. We have\n\n$$\n\\angle P Q R=\\angle B P R=\\angle B R P\n$$\n\n1 POINT for proving $\\angle B P R=\\alpha+\\beta$.\nSo circumcircle of $P Q R$ is tangent to $B P$ and $B R$.\n1 POINT for concluding.\nRemark. 2POINTS can be given for proving that $\\angle P R B=\\angle R P B$ and 1 more POINT for attempting to prove (unsuccessfully) that $\\angle P R B=\\angle R P B=\\angle P Q R$.", "problem_match": "\nProblem 3.", "solution_match": "# Solution "}
+{"year": "1999", "problem_label": "4", "tier": 1, "problem": "Determine all pairs $(a, b)$ of integers with the property that the numbers $a^{2}+4 b$ and $b^{2}+4 a$ are both perfect squares.", "solution": "and Marking Scheme:\n\nWithout loss of gencrality, assume that $|b| \\leq|a|$. If $b=0$, then $a$ must be a perfect square. So ( $a=k^{2}, b=0$ ) for each $k \\in Z$ is a solution.\nNow we consider the case $b \\neq 0$. Because $a^{2}+4 b$ is a perfect square, the quadratic equation\n\n$$\nx^{2}+a x-b=0\n$$\n\nhas two non-zero integral roots $x_{1}, x_{2}$.\n2 POINTS for noticing that this equation has integral roots.\nThen $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=-b$, and from this it follows that\n\n$$\n\\frac{1}{\\left|x_{1}\\right|}+\\frac{1}{\\left|x_{2}\\right|} \\geq\\left|\\frac{1}{x_{1}}+\\frac{1}{x_{2}}\\right|=\\frac{|a|}{|b|} \\geq 1\n$$\n\nHence there is at least one root, say $x_{1}$, such that $\\left|x_{1}\\right| \\leq 2$.\n3 POINTS for finding that $\\left|x_{1}\\right| \\leq 2$.\n\nThere are the following possibilities.\n(1) $x_{1}=2$. Substituting $x_{1}=2$ into ( $*$ ) we get $b=2 a+4$. So we have $b^{2}+4 a=(2 a+4)^{2}+4 a=$ $4 a^{2}+20 a+16=(2 a+5)^{2}-9$. It is casy to see that the solution in non-negative integers of the equation $x^{2}-9=y^{2}$ is $(3,0)$. Hence $2 a+5= \\pm 3$. From this we obtain $a=-4, b=-4$ and $a=-1, b=2$. The latter should be rejected because of the assumption $|a| \\geq|b|$.\n(2) $x_{1}=-2$. Substituting $x_{1}=-2$ into (*) we get $b=4-2 a$. Hence $b^{2}+4 a=4 a^{2}-12 a+16=$ $(2 a-3)^{2}+7$. It is easy to show that the solution in non-negative integers of the equation $x^{2}+7=y^{2}$ is $(3,4)$. Hence $2 a-3= \\pm 3$. From this we obtain $a=3, b=-2$.\n(3) $x_{1}=1$. Substituting $x_{1}=1$ into $(*)$ we get $b=a+1$. Hence $b^{2}+4 a=a^{2}+6 a+1=(a+3)^{2}-8$. It is easy to show that the solution in non-negative integers of the equation $x^{2}-8=y^{2}$ is $(3,1)$. Hence $a+3= \\pm 3$. From this we obtain $a=-6, b=-5$.\n(4) $x_{1}=-1$. Substituting $x_{1}=-1$ into (*) we get $b=1-a$. Then $a^{2}+4 b=(a-2)^{2}, b^{2}+4 a=(a+1)^{2}$. Consequently, $a=k, b=1-k(k \\in Z)$ is a solution.\n\nTesting these solutions and by symmetry we obtain the following solutions\n\n$$\n(-4,-4),(-5,-6),(-6,-5),\\left(0, k^{2}\\right),\\left(k^{2}, 0\\right),(k, 1-k)\n$$\n\nwhere $k$ is an arbitrary integer. (Observe that the solution $(3,-2)$ obtained in the second possibility is included in the last solution as a special case.)\n\n1 POINT for writing up the correct answer.", "problem_match": "\nProblem 4.", "solution_match": "# First Solution "}
+{"year": "1999", "problem_label": "4", "tier": 1, "problem": "Determine all pairs $(a, b)$ of integers with the property that the numbers $a^{2}+4 b$ and $b^{2}+4 a$ are both perfect squares.", "solution": "and Marking Scheme:\n\nWithout loss of generality assume that $|b| \\leq|a|$. Then $a^{2}+4 b \\leq a^{2}+4|a|<a^{2}+4|a|+4=(|a|+2)^{2}$. Given that $a^{2}+4 b$ is a perfect square and since $a^{2}+4 b$ y $a^{2}$ have same parity then $a^{2}+4 b \\neq(|a|+1)^{2}$, so\n\n$$\na^{2}+4 b \\leq a^{2}\n$$\n\n2 POINTS for proving (1).\nLet us consider three cases.\nCase 1. $a^{2}+4 b=a^{2}$. Then $b=0$ and $a$ must be a perfect square. So $a=k^{2}, b=0(k \\in Z)$ is a solution.\nCase 2. $a^{2}+4 b=(|a|-2)^{2}$. Then $b=1-|a|$, therefore $b^{2}+4 a=a^{2}-2|a|+4 a+1$ must be a perfect square.\nIf $a>0$ then $b^{2}+4 a=(a+1)^{2}$ is a perfect square for each $a \\in Z$. Consequently $a=k$ and $b=1-k$ ( $k \\in Z^{+}$) is a solution.\nIf $a=0$ then $b=1$, but from (1) $b$ must be non-positive.\nIf $a<0$ then $b^{2}+4 a=m^{2}-6 m+1$ must be a perfect square, where $m=-a>0$. For $m \\geq 8$\n\n$$\n(m-3)^{2}>m^{2}-6 m+1>(m-4)^{2}\n$$\n\ntherefore $m<8$. If $m=1,2,3,4,5$ then $m^{2}-6 m+1<0$. If $m=6, m^{2}-6 m+1=1$ is a perfect square thus $a=-6$ and $b=-5$ is a solution. If $m=7, m^{2}-6 m+1=8$ is not a perfect square.\n\n2 POINTS for case 1 and case 2.\nCase 3. $a^{2}+4 b \\leq(|a|-4)^{2}$. Since $|b| \\leq|a|$ then $b \\geq-|a|$, thus $a^{2}-4|a| \\leq a^{2}+4 b \\leq(|a|-4)^{2}$. It follows that $|a| \\leq 4$. We have following posibilities:\n\n## 1 POINT for finding that $|a| \\leq 4$ in this case.\n\n(a) $|a|=4$. Then $16+4 b=0$ or $b=-4$. Thus $b^{2}+4 a=16 \\pm 16$ must be a perfect square. So $a=-4$ y $b=-4$.\n(b) $|a|=3$. In this case $a^{2}+4 b=9+4 b \\leq 1$, then $9+4 b=0$ or $9+4 b=1$. The equation $9+4 b=0$ does not have integer solutions. The solution of the second equation is $b=-2$. Then $b^{2}+4 a=4 \\pm 12$ must be a perfect square, thus $a=3$.\n(c) $|a|=2 \\cdot a^{2}+4 b=4+4 b \\leq 4$. Since $4+4 b$ is cven and must be a perfect square then $4+4 b=4$ or $4+4 b=0$. Therefore $b=0$ or $b=-1$. If $b=0, b^{2}+4 a= \\pm 8$ is not a perfect square. If $b=-1$ then $b^{2}+4 a=1 \\pm 8$ is a perfect square if $a=2$. Thus $a=2$ and $b=-1$ is a solution.\n(d) $|a|=1$. Then $a^{2}+4 b=4 b+1 \\leq 9$. Since $4 b+1$ must be an odd perfect square then $4 b+1=1$ or $4 b+1=9$. So $b=0$ or $b=2$. If $b=0, b^{2}+4 a= \\pm 4$, then $a=1$. If $b=2$ then $a=-1$, but this is not possible because $|b| \\leq|a|$. Thus $a=1$ y $b=0$ is a solution in this case.\n(c) $|a|=0$. Since $|b| \\leq|a|$ then $b=0$.\n\n1 POINT for concluding case 3.\n\nTesting these solutions and by symmetry we obtain the following solutions:\n\n$$\n\\left(k^{2}, 0\\right),\\left(0, k^{2}\\right),(k, 1-k),(-6,-5),(-5,-6),(-4,-4)\n$$\n\nwhere $k$ is an arbitrary integer. Note that if $(k, 1-k)$ is a solution with $k>0$, then taking $t=1-k$, $k=1-t$, so $(1-t, t)$ is solution. Thus by symetry $(k, 1-k)$ is a solution for any integer.\n\n1 POINT for writing up the correct answer.\nRemark: I POINT can be given for checking that $(k, 1-k)$ is a solution. However NO POINT is given for finding any other particular solution.", "problem_match": "\nProblem 4.", "solution_match": "# Second Solution "}
+{"year": "1999", "problem_label": "5", "tier": 1, "problem": "Let $S$ be a set of $2 n+1$ points in the plane such that no three are collinear and no four concyclic. A circle will be called good if it has 3 points of $S$ on its circumference, $n-1$ points in its interior and $n-1$ in its exterior. Prove that the number of good circles has the same parity as $n$.", "solution": "and Marking Scheme:\n\nLemma 1. Let $P$ and $Q$ be two points of $S$. The number of good circles that contain $P$ and $Q$ on their circumference is odd.\n\n## Proof of Lemma 1.\n\n![](https://cdn.mathpix.com/cropped/2024_11_22_808e310d9512bffbfcbfg-07.jpg?height=521&width=939&top_left_y=520&top_left_x=639)\n\nLet $N$ be the number of good circles that pass through $P$ and $Q$. Number the points on one side of the line $P Q$ by $A_{1}, A_{2}, \\ldots, A_{k}$ and those on the other side by $B_{1}, B_{2}, \\ldots, B_{m}$ in such a way that if $\\angle P A_{i} Q=\\alpha_{i}, \\angle P B_{j} Q=180-\\beta_{j}$ then $\\alpha_{1}>\\alpha_{2}>\\ldots>\\alpha_{k}$ and $\\beta_{1}>\\beta_{2}>\\ldots>\\beta_{m}$.\nNote that the angles $\\alpha_{1}, \\alpha_{2}, \\ldots, \\alpha_{k}, \\beta_{1}, \\beta_{2}, \\ldots, \\beta_{m}$ are all distinct since there are no four points in $S$ that are concyclic.\nObserve that the circle that passes through $P, Q$ and $A_{i}$ has $A_{j}$ in its interior when $\\alpha_{j}>\\alpha_{i}$ that is, when $i>j$; and it contains $B_{j}$ in its interior when $\\alpha_{i}+180-\\beta_{j}>180$, that is, when $\\alpha_{i}>\\beta_{j}$. Similar conditions apply to the circle that contains $P, Q$ and $B_{j}$.\n\n1 POINT for characterizing the points that lie inside a given circle ins terms of these angles, or for similar considerations.\nOrder the angles $\\alpha_{1}, \\alpha_{2}, \\ldots, \\alpha_{k}, \\beta_{1}, \\beta_{2}, \\ldots, \\beta_{m}$ from the greatest to least. Now transform $S$ as follows. Consider a $\\beta_{j}$ that bas an $\\alpha_{i}$ immediately to its left in such an ordering ( $\\ldots>\\alpha_{i}>\\beta_{j} \\ldots$ ). Consider a new set $S^{\\prime}$ that contains the same points as $S$ except for $A_{i}$ and $B_{j}$. These two points will be replaced by $A_{i}^{\\prime}$ and $B_{j}^{\\prime}$ that satisfy $\\angle P A_{i}^{\\prime} Q=\\beta_{j}=\\alpha_{i}^{\\prime}$ and $\\angle P B_{j}^{\\prime} Q=180-\\alpha_{i}^{\\prime \\prime}=180-\\beta_{j}^{\\prime}$. Thus $\\beta_{j}$ and $\\alpha_{i}$ have been interchanged and the ordering of the $\\alpha$ 's and $\\beta$ 's has only changed with respect to the relative order of $\\alpha_{i}$ and $\\beta_{j}$; we continue to have\n\n$$\n\\alpha_{1}>\\alpha_{2}>\\ldots>\\alpha_{i-1}>\\alpha_{i}^{\\prime}>\\alpha_{i+1}>\\ldots>\\alpha_{k}\n$$\n\nand\n\n$$\n\\beta_{1}>\\beta_{2}>\\ldots>\\beta_{j-1}>\\beta_{j}^{\\prime}>\\beta_{j+1}>\\ldots>\\beta_{m}\n$$\n\n1 POINT for this or another useful transformation of the set $S$.\nAnalyze the good circles in this new set $S^{\\prime}$. Clearly, a circle through $P, Q, A_{r}(r \\neq i)$ or through $P, Q, B_{s}(s \\neq j)$ that was good in $S$ will also be good in $S^{\\prime}$, because the order of $A_{r}$ (or $B_{s}$ ) relative to the rest of the points has not changed, and therefore the number of points in the interior or exterior of this circle has not changed. The only changes that could have taken place are:\na) If the circle $P, Q, A_{i}$ was good in $S$, the circle $P, Q, A_{i}^{\\prime}$ may not be good in $S^{\\prime}$.\nb) If the circle $P, Q, B_{j}$ was good in $S$, the circle $P, Q, B_{j}^{\\prime}$ may not be good in $S^{\\prime}$.\nc) If the circle $P, Q, A_{i}$ was not good in $S$, the circle $P, Q, A_{i}^{\\prime}$ may be good in $S^{\\prime}$.\nd) If the circle $P, Q, B_{j}$ was not good in $S$, the circle $P, Q, B_{j}^{\\prime}$ may be good in $S^{\\prime}$.\n\n1 POINT for realizing that the transformation can only change the \"goodness\" of these circles. But observe that the circle $P, Q, A_{i}$ contains the points $A_{1}, A_{2}, \\ldots, A_{i-1}, B_{j}, B_{j+1}, \\ldots, B_{m}$ and does not contain the points $A_{i+1}, A_{i+2}, \\ldots, A_{k}, B_{1}, B_{2}, \\ldots, B_{j-1}$ in its interior. Then this circle is good if and only if $i+m-j=k-i+j-1$, which we rewrite as $j-i=\\frac{1}{2}(m-k+1)$. On the other hand, the circle $P, Q, B_{j}$ contains the points $B_{j+1}, B_{j+2}, \\ldots, B_{m}, A_{1}, A_{2}, \\ldots, A_{i}$ and does not contain the points $B_{1}, B_{2}, \\ldots, B_{j-1}, A_{i+1}, A_{i+2}, \\ldots, A_{k}$ in its interior. Hence this circle is good if and only if $m-j+i=j-1+k-i$, which we rewrite as $j-i=\\frac{1}{2}(m-k+1)$.\nTherefore, the circle $P, Q, A_{i}$ is good if and only if the circle $P, Q, B_{j}$ is good. Similarly, the circle $P, Q, A_{i}^{\\prime}$ is good if and only if the circle $P, Q, B_{j}^{\\prime}$ is good. That is to say, transforming $S$ into $S^{\\prime}$ we lose either 0 or 2 good circles of $S$ and we gain either 0 or 2 good circles in $S^{\\prime}$.\n\n1 POINT for realizing that the \"goodness\" of these circles is changed in pairs.\nContinuing in this way, we may continue to transform $S$ until we obtain a new set $S_{0}$ such that the angles $\\alpha_{1}^{\\prime}, \\alpha_{2}^{\\prime}, \\ldots, \\alpha_{k}^{\\prime}, \\beta_{1}^{\\prime}, \\beta_{2}^{\\prime}, \\ldots, \\beta_{m}^{\\prime}$ satisfy\n\n$$\n\\beta_{1}^{\\prime}>\\beta_{2}^{\\prime}>\\ldots>\\beta_{n}^{\\prime}>\\alpha_{1}^{\\prime}>\\alpha_{2}^{\\prime}>\\ldots>\\alpha_{k}^{\\prime}\n$$\n\nand such that the number of good circles in $S_{0}$ has the same parity as $N$. We claim that $S_{0}$ has exactly one good circle. In this configuration, the circle $P, Q, A_{i}$ does not contain any $B_{j}$ and the circle $P, Q, B_{r}$ does not contain any $A_{s}$ (for all $\\left.i, j\\right)$, because $\\alpha_{a}+\\left(180-\\beta_{b}\\right)<180$ for all $a, b$. Hence, the only possible good circles are $P, Q, B_{m-n+1}$ (which contains the $n-1$ points $B_{m-n+2}, B_{m-n+3}, \\ldots, B_{m}$ ), if $m-n+1>0$, and the circle $P, Q, A_{n}$ (which contains the $n-1$ points $A_{1}, A_{2}, \\ldots, A_{n}$ ), if $n \\leq k$. But, since $m+k=2 n-1$, which we rewrite as $m-n+1=n-k$, exactly one of the inequalitites $m-n+1>0$ and $n \\leq k$ is satisfied. It follows that one of the points $B_{m-n+1}$ and $A_{n}$ corresponds to a good circle and the other does not. Hence, $S_{0}$ has exactly one good circle, and $N$ is odd.\n\n1 POINT for showing that this configuration has exactly one good circle.\n\nNow consider the $\\binom{2 n+1}{2}$ pairs of points in $S$. Let $a_{2 k+1}$ be the number of pairs of points through which exactly $2 k+1$ good circles pass. Then\n\n$$\na_{1}+a_{3}+a_{5}+\\ldots=\\binom{2 n+1}{2}\n$$\n\nBut then the number of good circles in $S$ is\n\n$$\n\\begin{aligned}\n\\frac{1}{3}\\left(a_{1}+3 a_{3}+5 a_{5}+7 a_{7}+\\ldots\\right) & \\equiv a_{1}+3 a_{3}+5 a_{5}+7 a_{7}+\\ldots \\\\\n& \\equiv a_{1}+a_{3}+a_{5}+a_{7}+\\ldots \\\\\n& \\equiv\\binom{2 n+1}{2} \\\\\n& \\equiv n(2 n+1) \\\\\n& \\equiv n(\\bmod 2) .\n\\end{aligned}\n$$\n\nHere we have taken into account that each good circle is counted 3 times in the expression $a_{1}+3 a_{3}+$ $5 a_{5}+7 a_{7}+\\ldots$ The desired result follows.\n\n2 POINTS for this computation.\n\n## Alteraative Proof of Lemma 1.\n\nLet, $A_{1}, A_{2}, \\ldots, A_{2 n-1}$ be the $2 n-1$ given points other than $P$ and $Q$.\nInvert the plane with respect to point $P$. Let $O, B_{1}, B_{2}, \\ldots, B_{2 n-1}$ be the images of points $Q, A_{1}, A_{2}, \\ldots, A_{2 n-1}$, respectively, under this inversion. Call point $B_{i}$ \"good\" if the line $O B_{i}$ splits the points $B_{1}, B_{2} \\ldots, B_{i-1}, B_{i+1}, \\ldots, B_{2 n-1}$ evenly, leaving $n-1$ of them to each side of it. (Notice that no other $B_{j}$ can lie on the line $O B_{i}$, or else the points $P, Q, A_{i}$ and $A_{j}$ would be concyclic.) Then it is clear that the circle through $P, Q$ and $A_{i}$ is good if and only if point $B_{i}$ is good. Therefore, it suffices to prove that the number of good points is odd.\n\n1 POINT for inverting and realizing the equivalence between good circles and good points. Notice that the good points depend only on the relative positions of rays $O B_{1}, O B_{2}, \\ldots, O B_{2 n-1}$, and not on the exact positions of points $B_{1}, B_{2}, \\ldots, B_{2 n-1}$. Therefore we may assume, for simplicity, that $B_{1}, B_{2}, \\ldots, B_{2 n-1}$ lie on the unit circle $\\Gamma$ with center $O$.\n\n1 POINT for this or a similar simplification.\nLet $C_{1}, C_{2}, \\ldots, C_{2 n-1}$ be the points diametrically opposite to $B_{1}, B_{2}, \\ldots, B_{2 n-1}$ in $\\Gamma$. As remarked earlier, no $C_{i}$ can coincide with one of the $B_{j}$ 's. We will call the $B_{i}$ 's \"white points\", and the $C_{i}$ 's \"black points\". We will refer to these $4 n-2$ points as the \"colored points\".\nNow we prove that the number of good points is odd, which will complete the proof of the lemma. We proceed by induction on $n$. If $n=1$, the result is trivial. Now assume that the result is true for $n=k$, and consider $2 k+1$ white points $B_{1}, B_{2}, \\ldots, B_{2 k+1}$ on the circle $\\Gamma$ (no two of which are diametrically opposite), and their diametrically opposite black points $C_{1}, C_{2}, \\ldots, C_{2 k+1}$. Call this configuration of points \"configuration 1 \". It is clear that we must have two consecutive colored points on $\\Gamma$ which have different colors, say $B_{i}$ and $C_{j}$. Now remove points $B_{i}, B_{j}, C_{i}$ and $C_{j}$ from $\\Gamma$, to obtain \"configuration 2 \", a configuration with $2 k-1$ points of each color.\n\n1 POINT for this or a similar transformation of the set.\nIt is easy to verify the following two claims:\n\n1. Point $B_{i}$ is good in configuration 1 if and only if point $B_{j}$ is good in configuration 1.\n2. Let $k \\neq i, j$. Then point $B_{k}$ is good in configuration 1 if and only if it is good in configuration 2.\n\nIt follows that, by removing points $B_{i}, B_{j}, C_{i}$ and $C_{j}$, the number of good points can either stay the same, or decreases by two. In any case, its parity remains unchanged. Since we know, by the induction hypothesis, that the number of good points in configuration 2 is odd, it follows that the number of good points in configuration 1 is also odd. This completes the proof.\n\n## Another Approach to Lemma 1.\n\nOne can give another inductive proof of lemma. 1, which combines the ideas of the two proofs that we have given. The idea is to start as in the first proof, with the characterization of the points inside a given circle.\n\n1 POINT\nThen we transform the set $S$ by removing the points $A_{i}$ and $B_{j}$ instead of replacing then by $A_{i}^{p}$ and $B_{j}^{\\prime}$.\n\n1 POINT\nIt can be shown that every one of the romaining circles going through $P$ and $Q$, contained exactly one of $A_{i}$ and $B_{j}$. Therefore, the only good circles we could have gained or lost are $P, Q, A_{i}$ and $P, Q, B_{j}$.\n\n2 POINTS\nFinally, we show that either both or none of these circles were good, so the parity of the number of good circles isn't changed by this transformation.\n\n1 POINT\n\nRemark: 2 POINTS can be given if the result has been fully proved for a particular case with $n>1$. (If more than one particular case has been analyzed completly, only 2 POINTS.) These points are awarded only if no progress has been made in the general solution of the problem.", "problem_match": "\nProblem 5.", "solution_match": "# Solution "}

APMO/segmented/en-apmo2000_sol.jsonl

@@ -1,6 +1,6 @@
-{"year": "2000", "problem_label": "1", "tier": 1, "problem": "Compute the sum $S=\\sum_{i=0}^{101} \\frac{x_{i}^{3}}{1-3 x_{i}+3 x_{i}^{2}}$ for $x_{i}=\\frac{i}{101}$.\nAnswer: $S=51$.", "solution": "Since $x_{101-i}=\\frac{101-i}{101}=1-\\frac{i}{101}=1-x_{i}$ and\n\n$$\n1-3 x_{i}+3 x_{i}^{2}=\\left(1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}\\right)+x_{i}^{3}=\\left(1-x_{i}\\right)^{3}+x_{i}^{3}=x_{101-i}^{3}+x_{i}^{3},\n$$\n\nwe have, by replacing $i$ by $101-i$ in the second sum,\n\n$$\n2 S=S+S=\\sum_{i=0}^{101} \\frac{x_{i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}+\\sum_{i=0}^{101} \\frac{x_{101-i}^{3}}{x_{i}^{3}+x_{101-i}^{3}}=\\sum_{i=0}^{101} \\frac{x_{i}^{3}+x_{101-i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}=102,\n$$\n\nso $S=51$.", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}
-{"year": "2000", "problem_label": "2", "tier": 1, "problem": "Given the following arrangement of circles:\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=358&width=401&top_left_y=264&top_left_x=793)\n\nEach of the numbers $1,2, \\ldots, 9$ is to be written into one of these circles, so that each circle contains exactly one of these numbers and\n(i) the sums of the four numbers on each side of the triangle are equal;\n(ii) the sums of squares of the four numbers on each side of the triangle are equal.\n\nFind all ways in which this can be done.\nAnswer: The only solutions are\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=361&width=404&top_left_y=1076&top_left_x=792)\nand the ones generated by permuting the vertices, adjusting sides and exchanging the two middle numbers on each side.", "solution": "Let $a, b$, and $c$ be the numbers in the vertices of the triangular arrangement. Let $s$ be the sum of the numbers on each side and $t$ be the sum of the squares of the numbers on each side. Summing the numbers (or their squares) on the three sides repeats each once the numbers on the vertices (or their squares):\n\n$$\n\\begin{gathered}\n3 s=a+b+c+(1+2+\\cdots+9)=a+b+c+45 \\\\\n3 t=a^{2}+b^{2}+c^{2}+\\left(1^{2}+2^{2}+\\cdots+9^{2}\\right)=a^{2}+b^{2}+c^{2}+285\n\\end{gathered}\n$$\n\nAt any rate, $a+b+c$ and $a^{2}+b^{2}+c^{2}$ are both multiples of 3 . Since $x^{2} \\equiv 0,1(\\bmod 3)$, either $a, b, c$ are all multiples of 3 or none is a multiple of 3 . If two of them are $1,2 \\bmod 3$ then $a+b+c \\equiv 0(\\bmod 3)$ implies that the other should be a multiple of 3 , which is not possible. Thus $a, b, c$ are all congruent modulo 3 , that is,\n\n$$\n\\{a, b, c\\}=\\{3,6,9\\}, \\quad\\{1,4,7\\}, \\quad \\text { or } \\quad\\{2,5,8\\}\n$$\n\nCase 1: $\\{a, b, c\\}=\\{3,6,9\\}$. Then $3 t=3^{2}+6^{2}+9^{2}+285 \\Longleftrightarrow t=137$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=372&width=409&top_left_y=2395&top_left_x=789)\n\nIn this case $x^{2}+y^{2}+3^{2}+9^{2}=137 \\Longleftrightarrow x^{2}+y^{2}=47$. However, 47 cannot be written as the sum of two squares. One can check manually, or realize that $47 \\equiv 3(\\bmod 4)$, and since $x^{2}, y^{2} \\equiv 0,1(\\bmod 4), x^{2}+y^{2} \\equiv 0,1,2(\\bmod 4)$ cannot be 47.\nHence there are no solutions in this case.\nCase 2: $\\{a, b, c\\}=\\{1,4,7\\}$. Then $3 t=1^{2}+4^{2}+7^{2}+285 \\Longleftrightarrow t=117$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=360&width=398&top_left_y=434&top_left_x=795)\n\nIn this case $x^{2}+y^{2}+1^{2}+7^{2}=117 \\Longleftrightarrow x^{2}+y^{2}=67 \\equiv 3(\\bmod 4)$, and as in the previous case there are no solutions.\nCase 3: $\\{a, b, c\\}=\\{2,5,8\\}$. Then $3 t=2^{2}+5^{2}+8^{2}+285 \\Longleftrightarrow t=126$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=355&width=401&top_left_y=1025&top_left_x=793)\n\nThen\n\n$$\n\\left\\{\\begin{array} { c } \n{ x ^ { 2 } + y ^ { 2 } + 2 ^ { 2 } + 8 ^ { 2 } = 1 2 6 } \\\\\n{ t ^ { 2 } + u ^ { 2 } + 2 ^ { 2 } + 5 ^ { 2 } = 1 2 6 } \\\\\n{ m ^ { 2 } + n ^ { 2 } + 5 ^ { 2 } + 8 ^ { 2 } = 1 2 6 }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array}{c}\nx^{2}+y^{2}=58 \\\\\nt^{2}+u^{2}=97 \\\\\nm^{2}+n^{2}=37\n\\end{array}\\right.\\right.\n$$\n\nThe only solutions to $t^{2}+u^{2}=97$ and $m^{2}+n^{2}=37$ are $\\{t, u\\}=\\{4,9\\}$ and $\\{m, n\\}=\\{1,6\\}$, respectively (again, one can check manually.) Then $\\{x, y\\}=\\{3,7\\}$, and the solutions are\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=353&width=392&top_left_y=1751&top_left_x=798)\nand the ones generated by permuting the vertices, adjusting sides and exchanging the two middle numbers on each side. There are $3!\\cdot 2^{3}=48$ such solutions.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
+{"year": "2000", "problem_label": "1", "tier": 1, "problem": "Compute the sum $S=\\sum_{i=0}^{101} \\frac{x_{i}^{3}}{1-3 x_{i}+3 x_{i}^{2}}$ for $x_{i}=\\frac{i}{101}$.\nAnswer: $S=51$.", "solution": "Since $x_{101-i}=\\frac{101-i}{101}=1-\\frac{i}{101}=1-x_{i}$ and\n\n$$\n1-3 x_{i}+3 x_{i}^{2}=\\left(1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}\\right)+x_{i}^{3}=\\left(1-x_{i}\\right)^{3}+x_{i}^{3}=x_{101-i}^{3}+x_{i}^{3},\n$$\n\nwe have, by replacing $i$ by $101-i$ in the second sum,\n\n$$\n2 S=S+S=\\sum_{i=0}^{101} \\frac{x_{i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}+\\sum_{i=0}^{101} \\frac{x_{101-i}^{3}}{x_{i}^{3}+x_{101-i}^{3}}=\\sum_{i=0}^{101} \\frac{x_{i}^{3}+x_{101-i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}=102,\n$$\n\nso $S=51$.", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}
+{"year": "2000", "problem_label": "2", "tier": 1, "problem": "Given the following arrangement of circles:\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=358&width=401&top_left_y=264&top_left_x=793)\n\nEach of the numbers $1,2, \\ldots, 9$ is to be written into one of these circles, so that each circle contains exactly one of these numbers and\n(i) the sums of the four numbers on each side of the triangle are equal;\n(ii) the sums of squares of the four numbers on each side of the triangle are equal.\n\nFind all ways in which this can be done.\nAnswer: The only solutions are\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=361&width=404&top_left_y=1076&top_left_x=792)\nand the ones generated by permuting the vertices, adjusting sides and exchanging the two middle numbers on each side.", "solution": "Let $a, b$, and $c$ be the numbers in the vertices of the triangular arrangement. Let $s$ be the sum of the numbers on each side and $t$ be the sum of the squares of the numbers on each side. Summing the numbers (or their squares) on the three sides repeats each once the numbers on the vertices (or their squares):\n\n$$\n\\begin{gathered}\n3 s=a+b+c+(1+2+\\cdots+9)=a+b+c+45 \\\\\n3 t=a^{2}+b^{2}+c^{2}+\\left(1^{2}+2^{2}+\\cdots+9^{2}\\right)=a^{2}+b^{2}+c^{2}+285\n\\end{gathered}\n$$\n\nAt any rate, $a+b+c$ and $a^{2}+b^{2}+c^{2}$ are both multiples of 3 . Since $x^{2} \\equiv 0,1(\\bmod 3)$, either $a, b, c$ are all multiples of 3 or none is a multiple of 3 . If two of them are $1,2 \\bmod 3$ then $a+b+c \\equiv 0(\\bmod 3)$ implies that the other should be a multiple of 3 , which is not possible. Thus $a, b, c$ are all congruent modulo 3 , that is,\n\n$$\n\\{a, b, c\\}=\\{3,6,9\\}, \\quad\\{1,4,7\\}, \\quad \\text { or } \\quad\\{2,5,8\\}\n$$\n\nCase 1: $\\{a, b, c\\}=\\{3,6,9\\}$. Then $3 t=3^{2}+6^{2}+9^{2}+285 \\Longleftrightarrow t=137$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=372&width=409&top_left_y=2395&top_left_x=789)\n\nIn this case $x^{2}+y^{2}+3^{2}+9^{2}=137 \\Longleftrightarrow x^{2}+y^{2}=47$. However, 47 cannot be written as the sum of two squares. One can check manually, or realize that $47 \\equiv 3(\\bmod 4)$, and since $x^{2}, y^{2} \\equiv 0,1(\\bmod 4), x^{2}+y^{2} \\equiv 0,1,2(\\bmod 4)$ cannot be 47.\nHence there are no solutions in this case.\nCase 2: $\\{a, b, c\\}=\\{1,4,7\\}$. Then $3 t=1^{2}+4^{2}+7^{2}+285 \\Longleftrightarrow t=117$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=360&width=398&top_left_y=434&top_left_x=795)\n\nIn this case $x^{2}+y^{2}+1^{2}+7^{2}=117 \\Longleftrightarrow x^{2}+y^{2}=67 \\equiv 3(\\bmod 4)$, and as in the previous case there are no solutions.\nCase 3: $\\{a, b, c\\}=\\{2,5,8\\}$. Then $3 t=2^{2}+5^{2}+8^{2}+285 \\Longleftrightarrow t=126$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=355&width=401&top_left_y=1025&top_left_x=793)\n\nThen\n\n$$\n\\left\\{\\begin{array} { c } \n{ x ^ { 2 } + y ^ { 2 } + 2 ^ { 2 } + 8 ^ { 2 } = 1 2 6 } \\\\\n{ t ^ { 2 } + u ^ { 2 } + 2 ^ { 2 } + 5 ^ { 2 } = 1 2 6 } \\\\\n{ m ^ { 2 } + n ^ { 2 } + 5 ^ { 2 } + 8 ^ { 2 } = 1 2 6 }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array}{c}\nx^{2}+y^{2}=58 \\\\\nt^{2}+u^{2}=97 \\\\\nm^{2}+n^{2}=37\n\\end{array}\\right.\\right.\n$$\n\nThe only solutions to $t^{2}+u^{2}=97$ and $m^{2}+n^{2}=37$ are $\\{t, u\\}=\\{4,9\\}$ and $\\{m, n\\}=\\{1,6\\}$, respectively (again, one can check manually.) Then $\\{x, y\\}=\\{3,7\\}$, and the solutions are\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=353&width=392&top_left_y=1751&top_left_x=798)\nand the ones generated by permuting the vertices, adjusting sides and exchanging the two middle numbers on each side. There are $3!\\cdot 2^{3}=48$ such solutions.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
 {"year": "2000", "problem_label": "3", "tier": 1, "problem": "Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.", "solution": "Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-4.jpg?height=775&width=869&top_left_y=572&top_left_x=562)\n\nThe orthogonal projection of $K$ onto side $B C$ is $M$. Let $R$ and $S$ be the orthogonal projections of $K$ onto lines $A B$ and $A C$, respectively. Points $R, M$, and $S$ lie in the Simson line of $K$ with respect to $A B C$. Since $K$ is in the bisector of $\\angle B A C, A R K S$ is a kite, and the Simson line $R M S$ is perpendicular to $A N$, and therefore parallel to $P Q$.\nNow consider the homothety with center $A$ that takes $O$ to $K$. Since $O P \\perp A B$ and $K R \\perp A B$, $O P$ and $K R$ are parallel, which means that $P$ is taken to $R$. Finally, line $P Q$ is parallel to line $R S$, so line $P Q$ is taken to line $R S$ by the homothety. Then $Q$ is taken to $M$, and since $O$ is taken to $K$, line $O Q$ is taken to line $M K$. We are done now: this means that $O Q$ is parallel to $M K$, which is perpendicular to $B C$ (it is its perpendicular bisector, as $M B=M C$ and $K B=K C$.)", "problem_match": "# Problem 3", "solution_match": "# Solution 1"}
 {"year": "2000", "problem_label": "3", "tier": 1, "problem": "Let $A B C$ be a triangle. Let $M$ and $N$ be the points in which the median and angle bisector, respectively, at $A$ meet the side $B C$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $N A$ meets $M A$ and $B A$, respectively, and $O$ be the point in which the perpendicular at $P$ to $B A$ meets $A N$ produced. Prove that $Q O$ is perpendicular to $B C$.", "solution": "Consider a cartesian plane with $A=(0,0)$ as the origin and the bisector $A N$ as $x$-axis. Thus $A B$ has equation $y=m x$ and $A C$ has equation $y=-m x$. Let $B=(b, m b)$ and $C=(c,-m c)$. By symmetry, the problem is immediate if $A B=A C$, that is, if $b=c$. Suppose that $b \\neq c$ from now on. Line $B C$ has slope $\\frac{m b-(-m c)}{b-c}=\\frac{m(b+c)}{b-c}$. Let $N=(n, 0)$.\nPoint $M$ is the midpoint $\\left(\\frac{b+c}{2}, \\frac{m b-m c}{2}\\right)$ of $B C$, so $A M$ has slope $\\frac{m(b-c)}{b+c}$.\nThe line through $N$ that is perpendicular to the $x$-axis $A N$ is $x=n$. Therefore\n\n$$\nP=(n, m n) \\quad \\text { and } \\quad Q=\\left(n, \\frac{m(b-c) n}{b+c}\\right) .\n$$\n\nIn the right triangle $A P O$, with altitude $A N, A N \\cdot A O=A P^{2}$. Thus\n\n$$\nn \\cdot A O=(0-n)^{2}+(0-m n)^{2} \\Longleftrightarrow A O=n\\left(m^{2}+1\\right) \\Longrightarrow O=\\left(n\\left(m^{2}+1\\right), 0\\right)\n$$\n\nFinally, the slope of $O Q$ is\n\n$$\n\\frac{\\frac{m(b-c) n}{b+c}-0}{n-n\\left(m^{2}+1\\right)}=-\\frac{b-c}{(b+c) m}\n$$\n\nSince the product of the slopes of $O Q$ and $B C$ is\n\n$$\n-\\frac{b-c}{(b+c) m} \\cdot \\frac{m(b+c)}{b-c}=-1\n$$\n\n$O Q$ and $B C$ are perpendicular, and we are done.\nComment: The second solution shows that $N$ can be any point in the bisector of $\\angle A$. In fact, if we move $N$ in the bisector and construct $O, P$ and $Q$ accordingly, then all lines $O Q$ obtained are parallel: just consider a homothety with center $A$ and variable ratios.", "problem_match": "# Problem 3", "solution_match": "# Solution 2"}
-{"year": "2000", "problem_label": "4", "tier": 1, "problem": "Let $n, k$ be given positive integers with $n>k$. Prove that\n\n$$\n\\frac{1}{n+1} \\cdot \\frac{n^{n}}{k^{k}(n-k)^{n-k}}<\\frac{n!}{k!(n-k)!}<\\frac{n^{n}}{k^{k}(n-k)^{n-k}} .\n$$", "solution": "The inequality is equivalent to\n\n$$\n\\frac{n^{n}}{n+1}<\\binom{n}{k} k^{k}(n-k)^{n-k}<n^{n}\n$$\n\nwhich suggests investigating the binomial expansion of\n\n$$\nn^{n}=((n-k)+k)^{n}=\\sum_{i=0}^{n}\\binom{n}{i}(n-k)^{n-i} k^{i}\n$$\n\nThe $(k+1)$ th term $T_{k+1}$ of the expansion is $\\binom{n}{k} k^{k}(n-k)^{n-k}$, and all terms in the expansion are positive, which implies the right inequality.\nNow, for $1 \\leq i \\leq n$,\n\n$$\n\\frac{T_{i+1}}{T_{i}}=\\frac{\\binom{n}{i}(n-k)^{n-i} k^{i}}{\\binom{n}{i-1}(n-k)^{n-i+1} k^{i-1}}=\\frac{(n-i+1) k}{i(n-k)}\n$$\n\nand\n\n$$\n\\frac{T_{i+1}}{T_{i}}>1 \\Longleftrightarrow(n-i+1) k>i(n-k) \\Longleftrightarrow i<k+\\frac{k}{n} \\Longleftrightarrow i \\leq k\n$$\n\nThis means that\n\n$$\nT_{1}<T_{2}<\\cdots<T_{k+1}>T_{k+2}>\\cdots>T_{n+1}\n$$\n\nthat is, $T_{k+1}=\\binom{n}{k} k^{k}(n-k)^{n-k}$ is the largest term in the expansion. The maximum term is greater that the average, which is the sum $n^{n}$ divided by the quantity $n+1$, therefore\n\n$$\n\\binom{n}{k} k^{k}(n-k)^{n-k}>\\frac{n^{n}}{n+1}\n$$\n\nas required.\nComment: If we divide further by $n^{n}$ one finds\n\n$$\n\\frac{1}{n+1}<\\binom{n}{k}\\left(\\frac{k}{n}\\right)^{k}\\left(1-\\frac{k}{n}\\right)^{n-k}<1\n$$\n\nThe middle term is the probability $P(X=k)$ of $k$ successes in a binomial distribution with $n$ trials and success probability $p=\\frac{k}{n}$. The right inequality is immediate from the fact that $P(X=k)$ is not the only possible event in this distribution, and the left inequality comes from the fact that the mode of the binomial distribution are given by $\\lfloor(n+1) p\\rfloor=\\left\\lfloor(n+1) \\frac{k}{n}\\right\\rfloor=k$ and $\\lceil(n+1) p-1\\rceil=k$. However, the proof of this fact is identical to the above solution.", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
-{"year": "2000", "problem_label": "5", "tier": 1, "problem": "Given a permutation $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$ of the sequence $0,1, \\ldots, n$. A transposition of $a_{i}$ with $a_{j}$ is called legal if $a_{i}=0$ for $i>0$, and $a_{i-1}+1=a_{j}$. The permutation $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$ is called regular if after a number of legal transpositions it becomes $(1,2, \\ldots, n, 0)$. For which numbers $n$ is the permutation $(1, n, n-1, \\ldots, 3,2,0)$ regular?\n\nAnswer: $n=2$ and $n=2^{k}-1, k$ positive integer.", "solution": "A legal transposition consists of looking at the number immediately before 0 and exchanging 0 and its successor; therefore, we can perform at most one legal transposition to any permutation, and a legal transposition is not possible only and if only 0 is preceded by $n$.\nIf $n=1$ or $n=2$ there is nothing to do, so $n=1=2^{1}-1$ and $n=2$ are solutions. Suppose that $n>3$ in the following.\nCall a pass a maximal sequence of legal transpositions that move 0 to the left. We first illustrate what happens in the case $n=15$, which is large enough to visualize what is going on. The first pass is\n\n$$\n\\begin{aligned}\n& (1,15,14,13,12,11,10,9,8,7,6,5,4,3,2,0) \\\\\n& (1,15,14,13,12,11,10,9,8,7,6,5,4,0,2,3) \\\\\n& (1,15,14,13,12,11,10,9,8,7,6,0,4,5,2,3) \\\\\n& (1,15,14,13,12,11,10,9,8,0,6,7,4,5,2,3) \\\\\n& (1,15,14,13,12,11,10,0,8,9,6,7,4,5,2,3) \\\\\n& (1,15,14,13,12,0,10,11,8,9,6,7,4,5,2,3) \\\\\n& (1,15,14,0,12,13,10,11,8,9,6,7,4,5,2,3) \\\\\n& (1,0,14,15,12,13,10,11,8,9,6,7,4,5,2,3)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 2, the second pass is\n\n$$\n\\begin{aligned}\n& (1,2,14,15,12,13,10,11,8,9,6,7,4,5,0,3) \\\\\n& (1,2,14,15,12,13, \\mathbf{1 0}, 11,8,9,0,7,4,5,6,3) \\\\\n& (1,2, \\mathbf{1 4}, 15,12,13,0,11,8,9,10,7,4,5,6,3) \\\\\n& (1,2,0,15,12,13,14,11,8,9,10,7,4,5,6,3)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 3 , the third pass is\n\n$$\n\\begin{aligned}\n& (1,2,3,15,12,13,14,11,8,9,10,7,4,5,6,0) \\\\\n& (1,2,3,15,12,13,14,11,8,9,10,0,4,5,6,7) \\\\\n& (1,2,3,15,12,13,14,0,8,9,10,11,4,5,6,7) \\\\\n& (1,2,3,0,12,13,14,15,8,9,10,11,4,5,6,7)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 4, the fourth pass is\n\n$$\n\\begin{aligned}\n& (1,2,3,4,12,13,14,15,8,9,10,11,0,5,6,7) \\\\\n& (1,2,3,4,0,13,14,15,8,9,10,11,12,5,6,7)\n\\end{aligned}\n$$\n\nAnd then one can successively perform the operations to eventually find\n\n$$\n(1,2,3,4,5,6,7,0,8,9,10,11,12,13,14,15)\n$$\n\nafter which 0 will move one unit to the right with each transposition, and $n=15$ is a solution. The general case follows.\n\nCase 1: $n>2$ even: After the first pass, in which 0 is transposed successively with $3,5, \\ldots, n-1$, after which 0 is right after $n$, and no other legal transposition can be performed. So $n$ is not a solution in this case.\nCase 2: $n=2^{k}-1$ : Denote $N=n+1, R=2^{r},[a: b]=(a, a+1, a+2, \\ldots, b)$, and concatenation by a comma. Let $P_{r}$ be the permutation\n\n$$\n[1: R-1],(0),[N-R: N-1],[N-2 R: N-R-1], \\ldots,[2 R: 3 R-1],[R: 2 R-1]\n$$\n\n$P_{r}$ is formed by the blocks $[1: R-1],(0)$, and other $2^{k-r}-1$ blocks of size $R=2^{r}$ with consecutive numbers, beginning with $t R$ and finishing with $(t+1) R-1$, in decreasing order of $t$. Also define $P_{0}$ as the initial permutation.\nThen it can be verified that $P_{r+1}$ is obtained from $P_{r}$ after a number of legal transpositions: it can be easily verified that $P_{0}$ leads to $P_{1}$, as 0 is transposed successively with $3,5, \\ldots, n-1$, shifting cyclically all numbers with even indices; this is $P_{1}$.\nStarting from $P_{r}, r>0$, 0 is successively transposed with $R, 3 R, \\ldots, N-R$. The numbers $0, N-R, N-3 R, \\ldots, 3 R, R$ are cyclically shifted. This means that $R$ precedes 0 , and the blocks become\n\n$$\n\\begin{gathered}\n{[1: R],(0),[N-R+1: N-1],[N-2 R: N-R],[N-3 R+1: N-2 R-1], \\ldots,} \\\\\n{[3 R+1: 4 R-1],[2 R: 3 R],[R+1: 2 R-1]}\n\\end{gathered}\n$$\n\nNote that the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number.\nNow $0, N-R+1, N-3 R+1, \\ldots, 3 R+1, R+1$ are shifted. Note that, for every $i$ th block, $i$ odd greater than 1 , the first number is cyclically shifted, and the blocks become\n\n$$\n\\begin{gathered}\n{[1: R+1],(0),[N-R+2: N-1],[N-2 R: N-R+1],[N-3 R+2: N-2 R-1], \\ldots,} \\\\\n{[3 R+1: 4 R-1],[2 R: 3 R+1],[R+2: 2 R-1]}\n\\end{gathered}\n$$\n\nThe same phenomenom happened: the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number. This pattern continues: $0, N-R+u, N-3 R+u, \\ldots, R+u$ are shifted, $u=0,1,2, \\ldots, R-1$, the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number, until they vanish. We finish with\n\n$$\n[1: 2 R-1],(0),[N-2 R: N-1], \\ldots,[2 R: 4 R-1]\n$$\n\nwhich is precisely $P_{r+1}$.\nSince $P_{k}=[1: N-1],(0), n=2^{k}-1$ is a solution.\nCase 3: $n$ is odd, but is not of the form $2^{k}-1$. Write $n+1$ as $n+1=2^{a}(2 b+1), b \\geq 1$, and define $P_{0}, \\ldots, P_{a}$ as in the previous case. Since $2^{a}$ divides $N=n+1$, the same rules apply, and we obtain $P_{a}$ :\n\n$$\n\\left[1: 2^{a}-1\\right],(0),\\left[N-2^{a}: N-1\\right],\\left[N-2^{a+1}: N-2^{a}-1\\right], \\ldots,\\left[2^{a+1}: 3 \\cdot 2^{a}-1\\right],\\left[2^{a}: 2^{a+1}-1\\right]\n$$\n\nBut then 0 is transposed with $2^{a}, 3 \\cdot 2^{a}, \\ldots,(2 b-1) \\cdot 2^{a}=N-2^{a+1}$, after which 0 is put immediately after $N-1=n$, and cannot be transposed again. Therefore, $n$ is not a solution. All cases were studied, and so we are done.\n\nComment: The general problem of finding the number of regular permutations for any $n$ seems to be difficult. A computer finds the first few values\n\n$$\n1,2,5,14,47,189,891,4815,29547\n$$\n\nwhich is not catalogued at oeis.org.", "problem_match": "# Problem 5", "solution_match": "# Solution\n"}
+{"year": "2000", "problem_label": "4", "tier": 1, "problem": "Let $n, k$ be given positive integers with $n>k$. Prove that\n\n$$\n\\frac{1}{n+1} \\cdot \\frac{n^{n}}{k^{k}(n-k)^{n-k}}<\\frac{n!}{k!(n-k)!}<\\frac{n^{n}}{k^{k}(n-k)^{n-k}} .\n$$", "solution": "The inequality is equivalent to\n\n$$\n\\frac{n^{n}}{n+1}<\\binom{n}{k} k^{k}(n-k)^{n-k}<n^{n}\n$$\n\nwhich suggests investigating the binomial expansion of\n\n$$\nn^{n}=((n-k)+k)^{n}=\\sum_{i=0}^{n}\\binom{n}{i}(n-k)^{n-i} k^{i}\n$$\n\nThe $(k+1)$ th term $T_{k+1}$ of the expansion is $\\binom{n}{k} k^{k}(n-k)^{n-k}$, and all terms in the expansion are positive, which implies the right inequality.\nNow, for $1 \\leq i \\leq n$,\n\n$$\n\\frac{T_{i+1}}{T_{i}}=\\frac{\\binom{n}{i}(n-k)^{n-i} k^{i}}{\\binom{n}{i-1}(n-k)^{n-i+1} k^{i-1}}=\\frac{(n-i+1) k}{i(n-k)}\n$$\n\nand\n\n$$\n\\frac{T_{i+1}}{T_{i}}>1 \\Longleftrightarrow(n-i+1) k>i(n-k) \\Longleftrightarrow i<k+\\frac{k}{n} \\Longleftrightarrow i \\leq k\n$$\n\nThis means that\n\n$$\nT_{1}<T_{2}<\\cdots<T_{k+1}>T_{k+2}>\\cdots>T_{n+1}\n$$\n\nthat is, $T_{k+1}=\\binom{n}{k} k^{k}(n-k)^{n-k}$ is the largest term in the expansion. The maximum term is greater that the average, which is the sum $n^{n}$ divided by the quantity $n+1$, therefore\n\n$$\n\\binom{n}{k} k^{k}(n-k)^{n-k}>\\frac{n^{n}}{n+1}\n$$\n\nas required.\nComment: If we divide further by $n^{n}$ one finds\n\n$$\n\\frac{1}{n+1}<\\binom{n}{k}\\left(\\frac{k}{n}\\right)^{k}\\left(1-\\frac{k}{n}\\right)^{n-k}<1\n$$\n\nThe middle term is the probability $P(X=k)$ of $k$ successes in a binomial distribution with $n$ trials and success probability $p=\\frac{k}{n}$. The right inequality is immediate from the fact that $P(X=k)$ is not the only possible event in this distribution, and the left inequality comes from the fact that the mode of the binomial distribution are given by $\\lfloor(n+1) p\\rfloor=\\left\\lfloor(n+1) \\frac{k}{n}\\right\\rfloor=k$ and $\\lceil(n+1) p-1\\rceil=k$. However, the proof of this fact is identical to the above solution.", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
+{"year": "2000", "problem_label": "5", "tier": 1, "problem": "Given a permutation $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$ of the sequence $0,1, \\ldots, n$. A transposition of $a_{i}$ with $a_{j}$ is called legal if $a_{i}=0$ for $i>0$, and $a_{i-1}+1=a_{j}$. The permutation $\\left(a_{0}, a_{1}, \\ldots, a_{n}\\right)$ is called regular if after a number of legal transpositions it becomes $(1,2, \\ldots, n, 0)$. For which numbers $n$ is the permutation $(1, n, n-1, \\ldots, 3,2,0)$ regular?\n\nAnswer: $n=2$ and $n=2^{k}-1, k$ positive integer.", "solution": "A legal transposition consists of looking at the number immediately before 0 and exchanging 0 and its successor; therefore, we can perform at most one legal transposition to any permutation, and a legal transposition is not possible only and if only 0 is preceded by $n$.\nIf $n=1$ or $n=2$ there is nothing to do, so $n=1=2^{1}-1$ and $n=2$ are solutions. Suppose that $n>3$ in the following.\nCall a pass a maximal sequence of legal transpositions that move 0 to the left. We first illustrate what happens in the case $n=15$, which is large enough to visualize what is going on. The first pass is\n\n$$\n\\begin{aligned}\n& (1,15,14,13,12,11,10,9,8,7,6,5,4,3,2,0) \\\\\n& (1,15,14,13,12,11,10,9,8,7,6,5,4,0,2,3) \\\\\n& (1,15,14,13,12,11,10,9,8,7,6,0,4,5,2,3) \\\\\n& (1,15,14,13,12,11,10,9,8,0,6,7,4,5,2,3) \\\\\n& (1,15,14,13,12,11,10,0,8,9,6,7,4,5,2,3) \\\\\n& (1,15,14,13,12,0,10,11,8,9,6,7,4,5,2,3) \\\\\n& (1,15,14,0,12,13,10,11,8,9,6,7,4,5,2,3) \\\\\n& (1,0,14,15,12,13,10,11,8,9,6,7,4,5,2,3)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 2, the second pass is\n\n$$\n\\begin{aligned}\n& (1,2,14,15,12,13,10,11,8,9,6,7,4,5,0,3) \\\\\n& (1,2,14,15,12,13, \\mathbf{1 0}, 11,8,9,0,7,4,5,6,3) \\\\\n& (1,2, \\mathbf{1 4}, 15,12,13,0,11,8,9,10,7,4,5,6,3) \\\\\n& (1,2,0,15,12,13,14,11,8,9,10,7,4,5,6,3)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 3 , the third pass is\n\n$$\n\\begin{aligned}\n& (1,2,3,15,12,13,14,11,8,9,10,7,4,5,6,0) \\\\\n& (1,2,3,15,12,13,14,11,8,9,10,0,4,5,6,7) \\\\\n& (1,2,3,15,12,13,14,0,8,9,10,11,4,5,6,7) \\\\\n& (1,2,3,0,12,13,14,15,8,9,10,11,4,5,6,7)\n\\end{aligned}\n$$\n\nAfter exchanging 0 and 4, the fourth pass is\n\n$$\n\\begin{aligned}\n& (1,2,3,4,12,13,14,15,8,9,10,11,0,5,6,7) \\\\\n& (1,2,3,4,0,13,14,15,8,9,10,11,12,5,6,7)\n\\end{aligned}\n$$\n\nAnd then one can successively perform the operations to eventually find\n\n$$\n(1,2,3,4,5,6,7,0,8,9,10,11,12,13,14,15)\n$$\n\nafter which 0 will move one unit to the right with each transposition, and $n=15$ is a solution. The general case follows.\n\nCase 1: $n>2$ even: After the first pass, in which 0 is transposed successively with $3,5, \\ldots, n-1$, after which 0 is right after $n$, and no other legal transposition can be performed. So $n$ is not a solution in this case.\nCase 2: $n=2^{k}-1$ : Denote $N=n+1, R=2^{r},[a: b]=(a, a+1, a+2, \\ldots, b)$, and concatenation by a comma. Let $P_{r}$ be the permutation\n\n$$\n[1: R-1],(0),[N-R: N-1],[N-2 R: N-R-1], \\ldots,[2 R: 3 R-1],[R: 2 R-1]\n$$\n\n$P_{r}$ is formed by the blocks $[1: R-1],(0)$, and other $2^{k-r}-1$ blocks of size $R=2^{r}$ with consecutive numbers, beginning with $t R$ and finishing with $(t+1) R-1$, in decreasing order of $t$. Also define $P_{0}$ as the initial permutation.\nThen it can be verified that $P_{r+1}$ is obtained from $P_{r}$ after a number of legal transpositions: it can be easily verified that $P_{0}$ leads to $P_{1}$, as 0 is transposed successively with $3,5, \\ldots, n-1$, shifting cyclically all numbers with even indices; this is $P_{1}$.\nStarting from $P_{r}, r>0$, 0 is successively transposed with $R, 3 R, \\ldots, N-R$. The numbers $0, N-R, N-3 R, \\ldots, 3 R, R$ are cyclically shifted. This means that $R$ precedes 0 , and the blocks become\n\n$$\n\\begin{gathered}\n{[1: R],(0),[N-R+1: N-1],[N-2 R: N-R],[N-3 R+1: N-2 R-1], \\ldots,} \\\\\n{[3 R+1: 4 R-1],[2 R: 3 R],[R+1: 2 R-1]}\n\\end{gathered}\n$$\n\nNote that the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number.\nNow $0, N-R+1, N-3 R+1, \\ldots, 3 R+1, R+1$ are shifted. Note that, for every $i$ th block, $i$ odd greater than 1 , the first number is cyclically shifted, and the blocks become\n\n$$\n\\begin{gathered}\n{[1: R+1],(0),[N-R+2: N-1],[N-2 R: N-R+1],[N-3 R+2: N-2 R-1], \\ldots,} \\\\\n{[3 R+1: 4 R-1],[2 R: 3 R+1],[R+2: 2 R-1]}\n\\end{gathered}\n$$\n\nThe same phenomenom happened: the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number. This pattern continues: $0, N-R+u, N-3 R+u, \\ldots, R+u$ are shifted, $u=0,1,2, \\ldots, R-1$, the first block and the ones on even positions greater than 2 have one more number and the other blocks have one less number, until they vanish. We finish with\n\n$$\n[1: 2 R-1],(0),[N-2 R: N-1], \\ldots,[2 R: 4 R-1]\n$$\n\nwhich is precisely $P_{r+1}$.\nSince $P_{k}=[1: N-1],(0), n=2^{k}-1$ is a solution.\nCase 3: $n$ is odd, but is not of the form $2^{k}-1$. Write $n+1$ as $n+1=2^{a}(2 b+1), b \\geq 1$, and define $P_{0}, \\ldots, P_{a}$ as in the previous case. Since $2^{a}$ divides $N=n+1$, the same rules apply, and we obtain $P_{a}$ :\n\n$$\n\\left[1: 2^{a}-1\\right],(0),\\left[N-2^{a}: N-1\\right],\\left[N-2^{a+1}: N-2^{a}-1\\right], \\ldots,\\left[2^{a+1}: 3 \\cdot 2^{a}-1\\right],\\left[2^{a}: 2^{a+1}-1\\right]\n$$\n\nBut then 0 is transposed with $2^{a}, 3 \\cdot 2^{a}, \\ldots,(2 b-1) \\cdot 2^{a}=N-2^{a+1}$, after which 0 is put immediately after $N-1=n$, and cannot be transposed again. Therefore, $n$ is not a solution. All cases were studied, and so we are done.\n\nComment: The general problem of finding the number of regular permutations for any $n$ seems to be difficult. A computer finds the first few values\n\n$$\n1,2,5,14,47,189,891,4815,29547\n$$\n\nwhich is not catalogued at oeis.org.", "problem_match": "# Problem 5", "solution_match": "# Solution\n\n"}

APMO/segmented/en-apmo2004_sol.jsonl

@@ -1,8 +1,8 @@
-{"year": "2004", "problem_label": "1", "tier": 1, "problem": "Determine all finite nonempty sets $S$ of positive integers satisfying\n\n$$\n\\frac{i+j}{(i, j)} \\text { is an element of } S \\text { for all } i, j \\text { in } S \\text {, }\n$$\n\nwhere $(i, j)$ is the greatest common divisor of $i$ and $j$.\nAnswer: $S=\\{2\\}$.", "solution": "Let $k \\in S$. Then $\\frac{k+k}{(k, k)}=2$ is in $S$ as well.\nSuppose for the sake of contradiction that there is an odd number in $S$, and let $k$ be the largest such odd number. Since $(k, 2)=1, \\frac{k+2}{(k, 2)}=k+2>k$ is in $S$ as well, a contradiction. Hence $S$ has no odd numbers.\nNow suppose that $\\ell>2$ is the second smallest number in $S$. Then $\\ell$ is even and $\\frac{\\ell+2}{(\\ell, 2)}=\\frac{\\ell}{2}+1$ is in $S$. Since $\\ell>2 \\Longrightarrow \\frac{\\ell}{2}+1>2, \\frac{\\ell}{2}+1 \\geq \\ell \\Longleftrightarrow \\ell \\leq 2$, a contradiction again.\nTherefore $S$ can only contain 2 , and $S=\\{2\\}$ is the only solution.", "problem_match": "# Problem 1", "solution_match": "# Solution\n"}
+{"year": "2004", "problem_label": "1", "tier": 1, "problem": "Determine all finite nonempty sets $S$ of positive integers satisfying\n\n$$\n\\frac{i+j}{(i, j)} \\text { is an element of } S \\text { for all } i, j \\text { in } S \\text {, }\n$$\n\nwhere $(i, j)$ is the greatest common divisor of $i$ and $j$.\nAnswer: $S=\\{2\\}$.", "solution": "Let $k \\in S$. Then $\\frac{k+k}{(k, k)}=2$ is in $S$ as well.\nSuppose for the sake of contradiction that there is an odd number in $S$, and let $k$ be the largest such odd number. Since $(k, 2)=1, \\frac{k+2}{(k, 2)}=k+2>k$ is in $S$ as well, a contradiction. Hence $S$ has no odd numbers.\nNow suppose that $\\ell>2$ is the second smallest number in $S$. Then $\\ell$ is even and $\\frac{\\ell+2}{(\\ell, 2)}=\\frac{\\ell}{2}+1$ is in $S$. Since $\\ell>2 \\Longrightarrow \\frac{\\ell}{2}+1>2, \\frac{\\ell}{2}+1 \\geq \\ell \\Longleftrightarrow \\ell \\leq 2$, a contradiction again.\nTherefore $S$ can only contain 2 , and $S=\\{2\\}$ is the only solution.", "problem_match": "# Problem 1", "solution_match": "# Solution\n\n"}
 {"year": "2004", "problem_label": "2", "tier": 1, "problem": "Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\\mathrm{AOH}, \\mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.", "solution": "Suppose, without loss of generality, that $B$ and $C$ lies in the same side of line $O H$. Such line is the Euler line of $A B C$, so the centroid $G$ lies in this line.\n![](https://cdn.mathpix.com/cropped/2024_11_22_7992a4504bd2a9e623e4g-2.jpg?height=444&width=618&top_left_y=523&top_left_x=685)\n\nLet $M$ be the midpoint of $B C$. Then the distance between $M$ and the line $O H$ is the average of the distances from $B$ and C to OH , and the sum of the areas of triangles BOH and COH is\n\n$$\n[B O H]+[C O H]=\\frac{O H \\cdot d(B, O H)}{2}+\\frac{O H \\cdot d(C, O H)}{2}=\\frac{O H \\cdot 2 d(M, O H)}{2} .\n$$\n\nSince $A G=2 G M, d(A, O H)=2 d(M, O H)$. Hence\n\n$$\n[B O H]+[C O H]=\\frac{O H \\cdot d(A, O H)}{2}=[A O H]\n$$\n\nand the result follows.", "problem_match": "# Problem 2", "solution_match": "# Solution 1"}
 {"year": "2004", "problem_label": "2", "tier": 1, "problem": "Let $O$ be the circumcentre and $H$ the orthocentre of an acute triangle $A B C$. Prove that the area of one of the triangles $\\mathrm{AOH}, \\mathrm{BOH}$ and COH is equal to the sum of the areas of the other two.", "solution": "One can use barycentric coordinates: it is well known that\n\n$$\n\\begin{gathered}\nA=(1: 0: 0), \\quad B=(0: 1: 0), \\quad C=(0: 0: 1), \\\\\nO=(\\sin 2 A: \\sin 2 B: \\sin 2 C) \\quad \\text { and } \\quad H=(\\tan A: \\tan B: \\tan C) .\n\\end{gathered}\n$$\n\nThen the (signed) area of $A O H$ is proportional to\n\n$$\n\\left|\\begin{array}{ccc}\n1 & 0 & 0 \\\\\n\\sin 2 A & \\sin 2 B & \\sin 2 C \\\\\n\\tan A & \\tan B & \\tan C\n\\end{array}\\right|\n$$\n\nAdding all three expressions we find that the sum of the signed sums of the areas is a constant times\n\n$$\n\\left|\\begin{array}{ccc}\n1 & 0 & 0 \\\\\n\\sin 2 A & \\sin 2 B & \\sin 2 C \\\\\n\\tan A & \\tan B & \\tan C\n\\end{array}\\right|+\\left|\\begin{array}{ccc}\n0 & 1 & 0 \\\\\n\\sin 2 A & \\sin 2 B & \\sin 2 C \\\\\n\\tan A & \\tan B & \\tan C\n\\end{array}\\right|+\\left|\\begin{array}{ccc}\n0 & 0 & 1 \\\\\n\\sin 2 A & \\sin 2 B & \\sin 2 C \\\\\n\\tan A & \\tan B & \\tan C\n\\end{array}\\right|\n$$\n\nBy multilinearity of the determinant, this sum equals\n\n$$\n\\left|\\begin{array}{ccc}\n1 & 1 & 1 \\\\\n\\sin 2 A & \\sin 2 B & \\sin 2 C \\\\\n\\tan A & \\tan B & \\tan C\n\\end{array}\\right|\n$$\n\nwhich contains, in its rows, the coordinates of the centroid, the circumcenter, and the orthocenter. Since these three points lie in the Euler line of $A B C$, the signed sum of the areas is 0 , which means that one of the areas of $A O H, B O H, C O H$ is the sum of the other two areas.\nComment: Both solutions can be adapted to prove a stronger result: if the centroid $G$ of triangle $A B C$ belongs to line $X Y$ then one of the areas of triangles $A X Y, B X Y$, and $C X Y$ is equal to the sum of the other two.", "problem_match": "# Problem 2", "solution_match": "# Solution 2"}
-{"year": "2004", "problem_label": "3", "tier": 1, "problem": "Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p, q$ of $S$, the number of lines in $\\mathcal{L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour.\nNote: A line $\\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\\ell$ with neither point on $\\ell$.", "solution": "Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even.\nNow it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ is odd for all $q \\neq p$ and $r \\neq p$, which is equivalent to proving that $n(p, q)+n(p, r)+n(q, r)$ is always odd. For this purpose, consider the seven numbered regions defined by lines $p q$, $p r$, and $q r$ :\n![](https://cdn.mathpix.com/cropped/2024_11_22_7992a4504bd2a9e623e4g-3.jpg?height=515&width=809&top_left_y=950&top_left_x=586)\n\nAny line that do not pass through any of points $p, q, r$ meets the sides $p q, q r, p r$ of triangle $p q r$ in an even number of points (two sides or no sides), so these lines do not affect the parity of $n(p, q)+n(p, r)+n(q, r)$. Hence the only lines that need to be considered are the ones that pass through one of vertices $p, q, r$ and cuts the opposite side in the triangle $p q r$.\nLet $n_{i}$ be the number of points in region $i, p, q$, and $r$ excluded, as depicted in the diagram. Then the lines through $p$ that separate $q$ and $r$ are the lines passing through $p$ and points from regions 1,4 , and 7 . The same applies for $p, q$ and regions 2,5 , and 7 ; and $p, r$ and regions 3,6 , and 7. Therefore\n\n$$\n\\begin{aligned}\nn(p, q)+n(q, r)+n(p, r) & \\equiv\\left(n_{2}+n_{5}+n_{7}\\right)+\\left(n_{1}+n_{4}+n_{7}\\right)+\\left(n_{3}+n_{6}+n_{7}\\right) \\\\\n& \\equiv n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}=2004-3 \\equiv 1 \\quad(\\bmod 2)\n\\end{aligned}\n$$\n\nand the result follows.\nComment: The problem statement is also true if 2004 is replaced by any even number and is not true if 2004 is replaced by any odd number greater than 1.", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
-{"year": "2004", "problem_label": "4", "tier": 1, "problem": "For a real number $x$, let $\\lfloor x\\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor\n$$\n\nis even for every positive integer $n$.", "solution": "Consider four cases:\n\n- $n \\leq 5$. Then $\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=0$ is an even number.\n- $n$ and $n+1$ are both composite (in particular, $n \\geq 8$ ). Then $n=a b$ and $n+1=c d$ for $a, b, c, d \\geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and smaller than $n$, and one can choose $a, b, c, d$ such that exactly one of these four numbers is even. Hence $\\frac{(n-1)!}{n(n+1)}$ is an integer. As $n \\geq 8>6,(n-1)!$ has at least three even factors, so $\\frac{(n-1)!}{n(n+1)}$ is an even integer.\n- $n \\geq 7$ is an odd prime. By Wilson's theorem, $(n-1)!\\equiv-1(\\bmod n)$, that is, $\\frac{(n-1)!+1}{n}$ is an integer, as $\\frac{(n-1)!+n+1}{n}=\\frac{(n-1)!+1}{n}+1$ is. As before, $\\frac{(n-1)!}{n+1}$ is an even integer; therefore $\\frac{(n-1)!+n+1}{n+1}=\\frac{(n-1)!}{n+1}+1$ is an odd integer.\nAlso, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\\frac{(n-1)!+n+1}{n+1}$, so $\\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=\\frac{(n-1)!+n+1}{n(n+1)}-1\n$$\n\nis even.\n\n- $n+1 \\geq 7$ is an odd prime. Again, since $n$ is composite, $\\frac{(n-1)!}{n}$ is an even integer, and $\\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\\equiv-1(\\bmod n+1) \\Longleftrightarrow(n-1)!\\equiv 1$ $(\\bmod n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\\frac{(n-1)!+n}{n}$. Then $\\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=\\frac{(n-1)!+n}{n(n+1)}-1\n$$\n\nis even.", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
+{"year": "2004", "problem_label": "3", "tier": 1, "problem": "Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p, q$ of $S$, the number of lines in $\\mathcal{L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour.\nNote: A line $\\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\\ell$ with neither point on $\\ell$.", "solution": "Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even.\nNow it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ is odd for all $q \\neq p$ and $r \\neq p$, which is equivalent to proving that $n(p, q)+n(p, r)+n(q, r)$ is always odd. For this purpose, consider the seven numbered regions defined by lines $p q$, $p r$, and $q r$ :\n![](https://cdn.mathpix.com/cropped/2024_11_22_7992a4504bd2a9e623e4g-3.jpg?height=515&width=809&top_left_y=950&top_left_x=586)\n\nAny line that do not pass through any of points $p, q, r$ meets the sides $p q, q r, p r$ of triangle $p q r$ in an even number of points (two sides or no sides), so these lines do not affect the parity of $n(p, q)+n(p, r)+n(q, r)$. Hence the only lines that need to be considered are the ones that pass through one of vertices $p, q, r$ and cuts the opposite side in the triangle $p q r$.\nLet $n_{i}$ be the number of points in region $i, p, q$, and $r$ excluded, as depicted in the diagram. Then the lines through $p$ that separate $q$ and $r$ are the lines passing through $p$ and points from regions 1,4 , and 7 . The same applies for $p, q$ and regions 2,5 , and 7 ; and $p, r$ and regions 3,6 , and 7. Therefore\n\n$$\n\\begin{aligned}\nn(p, q)+n(q, r)+n(p, r) & \\equiv\\left(n_{2}+n_{5}+n_{7}\\right)+\\left(n_{1}+n_{4}+n_{7}\\right)+\\left(n_{3}+n_{6}+n_{7}\\right) \\\\\n& \\equiv n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}=2004-3 \\equiv 1 \\quad(\\bmod 2)\n\\end{aligned}\n$$\n\nand the result follows.\nComment: The problem statement is also true if 2004 is replaced by any even number and is not true if 2004 is replaced by any odd number greater than 1.", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
+{"year": "2004", "problem_label": "4", "tier": 1, "problem": "For a real number $x$, let $\\lfloor x\\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor\n$$\n\nis even for every positive integer $n$.", "solution": "Consider four cases:\n\n- $n \\leq 5$. Then $\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=0$ is an even number.\n- $n$ and $n+1$ are both composite (in particular, $n \\geq 8$ ). Then $n=a b$ and $n+1=c d$ for $a, b, c, d \\geq 2$. Moreover, since $n$ and $n+1$ are coprime, $a, b, c, d$ are all distinct and smaller than $n$, and one can choose $a, b, c, d$ such that exactly one of these four numbers is even. Hence $\\frac{(n-1)!}{n(n+1)}$ is an integer. As $n \\geq 8>6,(n-1)!$ has at least three even factors, so $\\frac{(n-1)!}{n(n+1)}$ is an even integer.\n- $n \\geq 7$ is an odd prime. By Wilson's theorem, $(n-1)!\\equiv-1(\\bmod n)$, that is, $\\frac{(n-1)!+1}{n}$ is an integer, as $\\frac{(n-1)!+n+1}{n}=\\frac{(n-1)!+1}{n}+1$ is. As before, $\\frac{(n-1)!}{n+1}$ is an even integer; therefore $\\frac{(n-1)!+n+1}{n+1}=\\frac{(n-1)!}{n+1}+1$ is an odd integer.\nAlso, $n$ and $n+1$ are coprime and $n$ divides the odd integer $\\frac{(n-1)!+n+1}{n+1}$, so $\\frac{(n-1)!+n+1}{n(n+1)}$ is also an odd integer. Then\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=\\frac{(n-1)!+n+1}{n(n+1)}-1\n$$\n\nis even.\n\n- $n+1 \\geq 7$ is an odd prime. Again, since $n$ is composite, $\\frac{(n-1)!}{n}$ is an even integer, and $\\frac{(n-1)!+n}{n}$ is an odd integer. By Wilson's theorem, $n!\\equiv-1(\\bmod n+1) \\Longleftrightarrow(n-1)!\\equiv 1$ $(\\bmod n+1)$. This means that $n+1$ divides $(n-1)!+n$, and since $n$ and $n+1$ are coprime, $n+1$ also divides $\\frac{(n-1)!+n}{n}$. Then $\\frac{(n-1)!+n}{n(n+1)}$ is also an odd integer and\n\n$$\n\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor=\\frac{(n-1)!+n}{n(n+1)}-1\n$$\n\nis even.", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
 {"year": "2004", "problem_label": "5", "tier": 1, "problem": "Prove that\n\n$$\n\\left(a^{2}+2\\right)\\left(b^{2}+2\\right)\\left(c^{2}+2\\right) \\geq 9(a b+b c+c a)\n$$\n\nfor all real numbers $a, b, c>0$.", "solution": "Let $p=a+b+c, q=a b+b c+c a$, and $r=a b c$. The inequality simplifies to\n\n$$\na^{2} b^{2} c^{2}+2\\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\\right)+4\\left(a^{2}+b^{2}+c^{2}\\right)+8-9(a b+b c+c a) \\geq 0\n$$\n\nSince $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=q^{2}-2 p r$ and $a^{2}+b^{2}+c^{2}=p^{2}-2 q$,\n\n$$\nr^{2}+2 q^{2}-4 p r+4 p^{2}-8 q+8-9 q \\geq 0\n$$\n\nwhich simplifies to\n\n$$\nr^{2}+2 q^{2}+4 p^{2}-17 q-4 p r+8 \\geq 0\n$$\n\nBearing in mind that equality occurs for $a=b=c=1$, which means that, for instance, $p=3 r$, one can rewrite $(I)$ as\n\n$$\n\\left(r-\\frac{p}{3}\\right)^{2}-\\frac{10}{3} p r+\\frac{35}{9} p^{2}+2 q^{2}-17 q+8 \\geq 0\n$$\n\nSince $(a b-b c)^{2}+(b c-c a)^{2}+(c a-a b)^{2} \\geq 0$ is equivalent to $q^{2} \\geq 3 p r$, rewrite $(I I)$ as\n\n$$\n\\left(r-\\frac{p}{3}\\right)^{2}+\\frac{10}{9}\\left(q^{2}-3 p r\\right)+\\frac{35}{9} p^{2}+\\frac{8}{9} q^{2}-17 q+8 \\geq 0\n$$\n\nFinally, $a=b=c=1$ implies $q=3$; then rewrite (III) as\n\n$$\n\\left(r-\\frac{p}{3}\\right)^{2}+\\frac{10}{9}\\left(q^{2}-3 p r\\right)+\\frac{35}{9}\\left(p^{2}-3 q\\right)+\\frac{8}{9}(q-3)^{2} \\geq 0\n$$\n\nThis final inequality is true because $q^{2} \\geq 3 p r$ and $p^{2}-3 q=\\frac{1}{2}\\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\\right] \\geq 0$.", "problem_match": "# Problem 5", "solution_match": "# Solution 1"}
 {"year": "2004", "problem_label": "5", "tier": 1, "problem": "Prove that\n\n$$\n\\left(a^{2}+2\\right)\\left(b^{2}+2\\right)\\left(c^{2}+2\\right) \\geq 9(a b+b c+c a)\n$$\n\nfor all real numbers $a, b, c>0$.", "solution": "We prove the stronger inequality\n\n$$\n\\left(a^{2}+2\\right)\\left(b^{2}+2\\right)\\left(c^{2}+2\\right) \\geq 3(a+b+c)^{2}\n$$\n\nwhich implies the proposed inequality because $(a+b+c)^{2} \\geq 3(a b+b c+c a)$ is equivalent to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\geq 0$, which is immediate.\nThe inequality $(*)$ is equivalent to\n\n$$\n\\left(\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)-3\\right) a^{2}-6(b+c) a+2\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)-3(b+c)^{2} \\geq 0\n$$\n\nSeeing this inequality as a quadratic inequality in $a$ with positive leading coefficient $\\left(b^{2}+2\\right)\\left(c^{2}+\\right.$ 2) $-3=b^{2} c^{2}+2 b^{2}+2 c^{2}+1$, it suffices to prove that its discriminant is non-positive, which is equivalent to\n\n$$\n(3(b+c))^{2}-\\left(\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)-3\\right)\\left(2\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)-3(b+c)^{2}\\right) \\leq 0\n$$\n\nThis simplifies to\n\n$$\n-2\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)+3(b+c)^{2}+6 \\leq 0\n$$\n\nNow we look $(* *)$ as a quadratic inequality in $b$ with negative leading coefficient $-2 c^{2}-1$ :\n\n$$\n\\left(-2 c^{2}-1\\right) b^{2}+6 c b-c^{2}-2 \\leq 0\n$$\n\nIf suffices to show that the discriminant of $(* *)$ is non-positive, which is equivalent to\n\n$$\n9 c^{2}-\\left(2 c^{2}+1\\right)\\left(c^{2}+2\\right) \\leq 0\n$$\n\nIt simplifies to $-2\\left(c^{2}-1\\right)^{2} \\leq 0$, which is true. The equality occurs for $c^{2}=1$, that is, $c=1$, for which $b=\\frac{6 c}{2\\left(2 c^{2}+1\\right)}=1$, and $a=\\frac{6(b+c)}{2\\left(\\left(b^{2}+2\\right)\\left(c^{2}+2\\right)-3\\right)}=1$.", "problem_match": "# Problem 5", "solution_match": "# Solution 2"}
 {"year": "2004", "problem_label": "5", "tier": 1, "problem": "Prove that\n\n$$\n\\left(a^{2}+2\\right)\\left(b^{2}+2\\right)\\left(c^{2}+2\\right) \\geq 9(a b+b c+c a)\n$$\n\nfor all real numbers $a, b, c>0$.", "solution": "Let $A, B, C$ angles in $(0, \\pi / 2)$ such that $a=\\sqrt{2} \\tan A, b=\\sqrt{2} \\tan B$, and $c=\\sqrt{2} \\tan C$. Then the inequality is equivalent to\n\n$$\n4 \\sec ^{2} A \\sec ^{2} B \\sec ^{2} C \\geq 9(\\tan A \\tan B+\\tan B \\tan C+\\tan C \\tan A)\n$$\n\nSubstituting $\\sec x=\\frac{1}{\\cos x}$ for $x \\in\\{A, B, C\\}$ and clearing denominators, the inequality is equivalent to\n\n$$\n\\cos A \\cos B \\cos C(\\sin A \\sin B \\cos C+\\cos A \\sin B \\sin C+\\sin A \\cos B \\sin C) \\leq \\frac{4}{9}\n$$\n\nSince\n\n$$\n\\begin{aligned}\n& \\cos (A+B+C)=\\cos A \\cos (B+C)-\\sin A \\sin (B+C) \\\\\n= & \\cos A \\cos B \\cos C-\\cos A \\sin B \\sin C-\\sin A \\cos B \\sin C-\\sin A \\sin B \\cos C,\n\\end{aligned}\n$$\n\nwe rewrite our inequality as\n\n$$\n\\cos A \\cos B \\cos C(\\cos A \\cos B \\cos C-\\cos (A+B+C)) \\leq \\frac{4}{9}\n$$\n\nThe cosine function is concave down on $(0, \\pi / 2)$. Therefore, if $\\theta=\\frac{A+B+C}{3}$, by the AM-GM inequality and Jensen's inequality,\n\n$$\n\\cos A \\cos B \\cos C \\leq\\left(\\frac{\\cos A+\\cos B+\\cos C}{3}\\right)^{3} \\leq \\cos ^{3} \\frac{A+B+C}{3}=\\cos ^{3} \\theta\n$$\n\nTherefore, since $\\cos A \\cos B \\cos C-\\cos (A+B+C)=\\sin A \\sin B \\cos C+\\cos A \\sin B \\sin C+$ $\\sin A \\cos B \\sin C>0$, and recalling that $\\cos 3 \\theta=4 \\cos ^{3} \\theta-3 \\cos \\theta$,\n$\\cos A \\cos B \\cos C(\\cos A \\cos B \\cos C-\\cos (A+B+C)) \\leq \\cos ^{3} \\theta\\left(\\cos ^{3} \\theta-\\cos 3 \\theta\\right)=3 \\cos ^{4} \\theta\\left(1-\\cos ^{2} \\theta\\right)$. Finally, by AM-GM (notice that $1-\\cos ^{2} \\theta=\\sin ^{2} \\theta>0$ ),\n$3 \\cos ^{4} \\theta\\left(1-\\cos ^{2} \\theta\\right)=\\frac{3}{2} \\cos ^{2} \\theta \\cdot \\cos ^{2} \\theta\\left(2-2 \\cos ^{2} \\theta\\right) \\leq \\frac{3}{2}\\left(\\frac{\\cos ^{2} \\theta+\\cos ^{2} \\theta+\\left(2-2 \\cos ^{2} \\theta\\right)}{3}\\right)^{3}=\\frac{4}{9}$,\nand the result follows.", "problem_match": "# Problem 5", "solution_match": "# Solution 3"}

APMO/segmented/en-apmo2007_sol.jsonl

@@ -1,5 +1,8 @@
-{"year": "2007", "problem_label": "1", "tier": 1, "problem": "Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.", "solution": "Without loss of generality, we may assume that $S$ contains only positive integers. Let\n\n$$\nS=\\left\\{2^{a_{i}} 3^{b_{i}} \\mid a_{i}, b_{i} \\in \\mathbb{Z}, a_{i}, b_{i} \\geq 0,1 \\leq i \\leq 9\\right\\}\n$$\n\nIt suffices to show that there are $1 \\leq i_{1}, i_{2}, i_{3} \\leq 9$ such that\n\n$$\na_{i_{1}}+a_{i_{2}}+a_{i_{3}} \\equiv b_{i_{1}}+b_{i_{2}}+b_{i_{3}} \\equiv 0 \\quad(\\bmod 3) .\n$$\n\nFor $n=2^{a} 3^{b} \\in S$, let's call $(a(\\bmod 3), b(\\bmod 3))$ the type of $n$. Then there are 9 possible types:\n\n$$\n(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\n$$\n\nLet $N(i, j)$ be the number of integers in $S$ of type $(i, j)$. We obtain 3 distinct integers whose product is a perfect cube when\n(1) $N(i, j) \\geq 3$ for some $i, j$, or\n(2) $N(i, 0) N(i, 1) N(i, 2) \\neq 0$ for some $i=0,1,2$, or\n(3) $N(0, j) N(1, j) N(2, j) \\neq 0$ for some $j=0,1,2$, or\n(4) $N\\left(i_{1}, j_{1}\\right) N\\left(i_{2}, j_{2}\\right) N\\left(i_{3}, j_{3}\\right) \\neq 0$, where $\\left\\{i_{1}, i_{2}, i_{3}\\right\\}=\\left\\{j_{1}, j_{2}, j_{3}\\right\\}=\\{0,1,2\\}$.\n\nAssume that none of the conditions (1) (3) holds. Since $N(i, j) \\leq 2$ for all $(i, j)$, there are at least five $N(i, j)$ 's that are nonzero. Furthermore, among those nonzero $N(i, j)$ 's, no three have the same $i$ nor the same $j$. Using these facts, one may easily conclude that the condition (4) should hold. (For example, if one places each nonzero $N(i, j)$ in the $(i, j)$-th box of a regular $3 \\times 3$ array of boxes whose rows and columns are indexed by 0,1 and 2 , then one can always find three boxes, occupied by at least one nonzero $N(i, j)$, whose rows and columns are all distinct. This implies (4).)\n\nSecond solution. Up to $(\\dagger)$, we do the same as above and get 9 possible types:\n\n$$\n(a(\\bmod 3), b(\\bmod 3))=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\n$$\n\nfor $n=2^{a} 3^{b} \\in S$.\nNote that (i) among any 5 integers, there exist 3 whose sum is $0(\\bmod 3)$, and that (ii) if $i, j, k \\in\\{0,1,2\\}$, then $i+j+k \\equiv 0(\\bmod 3)$ if and only if $i=j=k$ or $\\{i, j, k\\}=\\{0,1,2\\}$.\n\nLet's define\n$T$ : the set of types of the integers in $S$;\n$N(i)$ : the number of integers in $S$ of the type $(i, \\cdot)$;\n$M(i)$ : the number of integers $j \\in\\{0,1,2\\}$ such that $(i, j) \\in T$.\nIf $N(i) \\geq 5$ for some $i$, the result follows from (i). Otherwise, for some permutation $(i, j, k)$ of $(0,1,2)$,\n\n$$\nN(i) \\geq 3, \\quad N(j) \\geq 3, \\quad N(k) \\geq 1\n$$\n\nIf $M(i)$ or $M(j)$ is 1 or 3 , the result follows from (ii). Otherwise $M(i)=M(j)=2$. Then either\n\n$$\n(i, x),(i, y),(j, x),(j, y) \\in T \\quad \\text { or } \\quad(i, x),(i, y),(j, x),(j, z) \\in T\n$$\n\nfor some permutation $(x, y, z)$ of $(0,1,2)$. Since $N(k) \\geq 1$, at least one of $(k, x),(k, y)$ and $(k, z)$ contained in $T$. Therefore, in any case, the result follows from (ii). (For example, if $(k, y) \\in T$, then take $(i, y),(j, y),(k, y)$ or $(i, x),(j, z),(k, y)$ from T.)", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}
-{"year": "2007", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be an acute angled triangle with $\\angle B A C=60^{\\circ}$ and $A B>A C$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $A B C$. Prove that\n\n$$\n2 \\angle A H I=3 \\angle A B C .\n$$", "solution": "Let $D$ be the intersection point of the lines $A H$ and $B C$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $A B C$ and the line $A H$. Let the line through $I$ perpendicular to $B C$ meet $B C$ and the minor arc $B C$ of the circumcircle $O$ at $E$ and $N$, respectively. We have\n$\\angle B I C=180^{\\circ}-(\\angle I B C+\\angle I C B)=180^{\\circ}-\\frac{1}{2}(\\angle A B C+\\angle A C B)=90^{\\circ}+\\frac{1}{2} \\angle B A C=120^{\\circ}$\nand also $\\angle B N C=180^{\\circ}-\\angle B A C=120^{\\circ}=\\angle B I C$. Since $I N \\perp B C$, the quadrilateral $B I C N$ is a kite and thus $I E=E N$.\n\nNow, since $H$ is the orthocenter of the triangle $A B C, H D=D K$. Also because $E D \\perp I N$ and $E D \\perp H K$, we conclude that $I H K N$ is an isosceles trapezoid with $I H=N K$.\n\nHence\n\n$$\n\\angle A H I=180^{\\circ}-\\angle I H K=180^{\\circ}-\\angle A K N=\\angle A B N .\n$$\n\nSince $I E=E N$ and $B E \\perp I N$, the triangles $I B E$ and $N B E$ are congruent. Therefore\n\n$$\n\\angle N B E=\\angle I B E=\\angle I B C=\\angle I B A=\\frac{1}{2} \\angle A B C\n$$\n\nand thus\n\n$$\n\\angle A H I=\\angle A B N=\\frac{3}{2} \\angle A B C .\n$$\n\nSecond solution. Let $P, Q$ and $R$ be the intersection points of $B H, C H$ and $A H$ with $A C, A B$ and $B C$, respectively. Then we have $\\angle I B H=\\angle I C H$. Indeed,\n\n$$\n\\angle I B H=\\angle A B P-\\angle A B I=30^{\\circ}-\\frac{1}{2} \\angle A B C\n$$\n\nand\n\n$$\n\\angle I C H=\\angle A C I-\\angle A C H=\\frac{1}{2} \\angle A C B-30^{\\circ}=30^{\\circ}-\\frac{1}{2} \\angle A B C,\n$$\n\nbecause $\\angle A B H=\\angle A C H=30^{\\circ}$ and $\\angle A C B+\\angle A B C=120^{\\circ}$. (Note that $\\angle A B P>\\angle A B I$ and $\\angle A C I>\\angle A C H$ because $A B$ is the longest side of the triangle $A B C$ under the given conditions.) Therefore BIHC is a cyclic quadrilateral and thus\n\n$$\n\\angle B H I=\\angle B C I=\\frac{1}{2} \\angle A C B .\n$$\n\nOn the other hand,\n\n$$\n\\angle B H R=90^{\\circ}-\\angle H B R=90^{\\circ}-(\\angle A B C-\\angle A B H)=120^{\\circ}-\\angle A B C\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\angle A H I & =180^{\\circ}-\\angle B H I-\\angle B H R=60^{\\circ}-\\frac{1}{2} \\angle A C B+\\angle A B C \\\\\n& =60^{\\circ}-\\frac{1}{2}\\left(120^{\\circ}-\\angle A B C\\right)+\\angle A B C=\\frac{3}{2} \\angle A B C .\n\\end{aligned}\n$$", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}
+{"year": "2007", "problem_label": "1", "tier": 1, "problem": "Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.", "solution": "Without loss of generality, we may assume that $S$ contains only positive integers. Let\n\n$$\nS=\\left\\{2^{a_{i}} 3^{b_{i}} \\mid a_{i}, b_{i} \\in \\mathbb{Z}, a_{i}, b_{i} \\geq 0,1 \\leq i \\leq 9\\right\\}\n$$\n\nIt suffices to show that there are $1 \\leq i_{1}, i_{2}, i_{3} \\leq 9$ such that\n\n$$\na_{i_{1}}+a_{i_{2}}+a_{i_{3}} \\equiv b_{i_{1}}+b_{i_{2}}+b_{i_{3}} \\equiv 0 \\quad(\\bmod 3) .\n$$\n\nFor $n=2^{a} 3^{b} \\in S$, let's call $(a(\\bmod 3), b(\\bmod 3))$ the type of $n$. Then there are 9 possible types:\n\n$$\n(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\n$$\n\nLet $N(i, j)$ be the number of integers in $S$ of type $(i, j)$. We obtain 3 distinct integers whose product is a perfect cube when\n(1) $N(i, j) \\geq 3$ for some $i, j$, or\n(2) $N(i, 0) N(i, 1) N(i, 2) \\neq 0$ for some $i=0,1,2$, or\n(3) $N(0, j) N(1, j) N(2, j) \\neq 0$ for some $j=0,1,2$, or\n(4) $N\\left(i_{1}, j_{1}\\right) N\\left(i_{2}, j_{2}\\right) N\\left(i_{3}, j_{3}\\right) \\neq 0$, where $\\left\\{i_{1}, i_{2}, i_{3}\\right\\}=\\left\\{j_{1}, j_{2}, j_{3}\\right\\}=\\{0,1,2\\}$.\n\nAssume that none of the conditions (1) (3) holds. Since $N(i, j) \\leq 2$ for all $(i, j)$, there are at least five $N(i, j)$ 's that are nonzero. Furthermore, among those nonzero $N(i, j)$ 's, no three have the same $i$ nor the same $j$. Using these facts, one may easily conclude that the condition (4) should hold. (For example, if one places each nonzero $N(i, j)$ in the $(i, j)$-th box of a regular $3 \\times 3$ array of boxes whose rows and columns are indexed by 0,1 and 2 , then one can always find three boxes, occupied by at least one nonzero $N(i, j)$, whose rows and columns are all distinct. This implies (4).)", "problem_match": "\nProblem 1.", "solution_match": "\nSolution."}
+{"year": "2007", "problem_label": "1", "tier": 1, "problem": "Let $S$ be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube.", "solution": "Up to $(\\dagger)$, we do the same as above and get 9 possible types:\n\n$$\n(a(\\bmod 3), b(\\bmod 3))=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\n$$\n\nfor $n=2^{a} 3^{b} \\in S$.\nNote that (i) among any 5 integers, there exist 3 whose sum is $0(\\bmod 3)$, and that (ii) if $i, j, k \\in\\{0,1,2\\}$, then $i+j+k \\equiv 0(\\bmod 3)$ if and only if $i=j=k$ or $\\{i, j, k\\}=\\{0,1,2\\}$.\n\nLet's define\n$T$ : the set of types of the integers in $S$;\n$N(i)$ : the number of integers in $S$ of the type $(i, \\cdot)$;\n$M(i)$ : the number of integers $j \\in\\{0,1,2\\}$ such that $(i, j) \\in T$.\nIf $N(i) \\geq 5$ for some $i$, the result follows from (i). Otherwise, for some permutation $(i, j, k)$ of $(0,1,2)$,\n\n$$\nN(i) \\geq 3, \\quad N(j) \\geq 3, \\quad N(k) \\geq 1\n$$\n\nIf $M(i)$ or $M(j)$ is 1 or 3 , the result follows from (ii). Otherwise $M(i)=M(j)=2$. Then either\n\n$$\n(i, x),(i, y),(j, x),(j, y) \\in T \\quad \\text { or } \\quad(i, x),(i, y),(j, x),(j, z) \\in T\n$$\n\nfor some permutation $(x, y, z)$ of $(0,1,2)$. Since $N(k) \\geq 1$, at least one of $(k, x),(k, y)$ and $(k, z)$ contained in $T$. Therefore, in any case, the result follows from (ii). (For example, if $(k, y) \\in T$, then take $(i, y),(j, y),(k, y)$ or $(i, x),(j, z),(k, y)$ from T.)", "problem_match": "\nProblem 1.", "solution_match": "\nSecond solution."}
+{"year": "2007", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be an acute angled triangle with $\\angle B A C=60^{\\circ}$ and $A B>A C$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $A B C$. Prove that\n\n$$\n2 \\angle A H I=3 \\angle A B C .\n$$", "solution": "Let $D$ be the intersection point of the lines $A H$ and $B C$. Let $K$ be the intersection point of the circumcircle $O$ of the triangle $A B C$ and the line $A H$. Let the line through $I$ perpendicular to $B C$ meet $B C$ and the minor arc $B C$ of the circumcircle $O$ at $E$ and $N$, respectively. We have\n$\\angle B I C=180^{\\circ}-(\\angle I B C+\\angle I C B)=180^{\\circ}-\\frac{1}{2}(\\angle A B C+\\angle A C B)=90^{\\circ}+\\frac{1}{2} \\angle B A C=120^{\\circ}$\nand also $\\angle B N C=180^{\\circ}-\\angle B A C=120^{\\circ}=\\angle B I C$. Since $I N \\perp B C$, the quadrilateral $B I C N$ is a kite and thus $I E=E N$.\n\nNow, since $H$ is the orthocenter of the triangle $A B C, H D=D K$. Also because $E D \\perp I N$ and $E D \\perp H K$, we conclude that $I H K N$ is an isosceles trapezoid with $I H=N K$.\n\nHence\n\n$$\n\\angle A H I=180^{\\circ}-\\angle I H K=180^{\\circ}-\\angle A K N=\\angle A B N .\n$$\n\nSince $I E=E N$ and $B E \\perp I N$, the triangles $I B E$ and $N B E$ are congruent. Therefore\n\n$$\n\\angle N B E=\\angle I B E=\\angle I B C=\\angle I B A=\\frac{1}{2} \\angle A B C\n$$\n\nand thus\n\n$$\n\\angle A H I=\\angle A B N=\\frac{3}{2} \\angle A B C .\n$$", "problem_match": "\nProblem 2.", "solution_match": "\nSolution."}
+{"year": "2007", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be an acute angled triangle with $\\angle B A C=60^{\\circ}$ and $A B>A C$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $A B C$. Prove that\n\n$$\n2 \\angle A H I=3 \\angle A B C .\n$$", "solution": "Let $P, Q$ and $R$ be the intersection points of $B H, C H$ and $A H$ with $A C, A B$ and $B C$, respectively. Then we have $\\angle I B H=\\angle I C H$. Indeed,\n\n$$\n\\angle I B H=\\angle A B P-\\angle A B I=30^{\\circ}-\\frac{1}{2} \\angle A B C\n$$\n\nand\n\n$$\n\\angle I C H=\\angle A C I-\\angle A C H=\\frac{1}{2} \\angle A C B-30^{\\circ}=30^{\\circ}-\\frac{1}{2} \\angle A B C,\n$$\n\nbecause $\\angle A B H=\\angle A C H=30^{\\circ}$ and $\\angle A C B+\\angle A B C=120^{\\circ}$. (Note that $\\angle A B P>\\angle A B I$ and $\\angle A C I>\\angle A C H$ because $A B$ is the longest side of the triangle $A B C$ under the given conditions.) Therefore BIHC is a cyclic quadrilateral and thus\n\n$$\n\\angle B H I=\\angle B C I=\\frac{1}{2} \\angle A C B .\n$$\n\nOn the other hand,\n\n$$\n\\angle B H R=90^{\\circ}-\\angle H B R=90^{\\circ}-(\\angle A B C-\\angle A B H)=120^{\\circ}-\\angle A B C\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\angle A H I & =180^{\\circ}-\\angle B H I-\\angle B H R=60^{\\circ}-\\frac{1}{2} \\angle A C B+\\angle A B C \\\\\n& =60^{\\circ}-\\frac{1}{2}\\left(120^{\\circ}-\\angle A B C\\right)+\\angle A B C=\\frac{3}{2} \\angle A B C .\n\\end{aligned}\n$$", "problem_match": "\nProblem 2.", "solution_match": "\nSecond solution."}
 {"year": "2007", "problem_label": "3", "tier": 1, "problem": "Consider $n$ disks $C_{1}, C_{2}, \\ldots, C_{n}$ in a plane such that for each $1 \\leq i<n$, the center of $C_{i}$ is on the circumference of $C_{i+1}$, and the center of $C_{n}$ is on the circumference of $C_{1}$. Define the score of such an arrangement of $n$ disks to be the number of pairs $(i, j)$ for which $C_{i}$ properly contains $C_{j}$. Determine the maximum possible score.", "solution": "The answer is $(n-1)(n-2) / 2$.\nLet's call a set of $n$ disks satisfying the given conditions an $n$-configuration. For an $n$ configuration $\\mathcal{C}=\\left\\{C_{1}, \\ldots, C_{n}\\right\\}$, let $S_{\\mathcal{C}}=\\left\\{(i, j) \\mid C_{i}\\right.$ properly contains $\\left.C_{j}\\right\\}$. So, the score of an $n$-configuration $\\mathcal{C}$ is $\\left|S_{\\mathcal{C}}\\right|$.\n\nWe'll show that (i) there is an $n$-configuration $\\mathcal{C}$ for which $\\left|S_{\\mathcal{C}}\\right|=(n-1)(n-2) / 2$, and that (ii) $\\left|S_{\\mathcal{C}}\\right| \\leq(n-1)(n-2) / 2$ for any $n$-configuration $\\mathcal{C}$.\n\nLet $C_{1}$ be any disk. Then for $i=2, \\ldots, n-1$, take $C_{i}$ inside $C_{i-1}$ so that the circumference of $C_{i}$ contains the center of $C_{i-1}$. Finally, let $C_{n}$ be a disk whose center is on the circumference of $C_{1}$ and whose circumference contains the center of $C_{n-1}$. This gives $S_{\\mathcal{C}}=\\{(i, j) \\mid 1 \\leq i<j \\leq n-1\\}$ of size $(n-1)(n-2) / 2$, which proves (i).\n\nFor any $n$-configuration $\\mathcal{C}, S_{\\mathcal{C}}$ must satisfy the following properties:\n(1) $(i, i) \\notin S_{\\mathcal{C}}$,\n(2) $(i+1, i) \\notin S_{\\mathcal{C}},(1, n) \\notin S_{\\mathcal{C}}$,\n(3) if $(i, j),(j, k) \\in S_{\\mathcal{C}}$, then $(i, k) \\in S_{\\mathcal{C}}$,\n(4) if $(i, j) \\in S_{\\mathcal{C}}$, then $(j, i) \\notin S_{\\mathcal{C}}$.\n\nNow we show that a set $G$ of ordered pairs of integers between 1 and $n$, satisfying the conditions $(1) \\sim(4)$, can have no more than $(n-1)(n-2) / 2$ elements. Suppose that there exists a set $G$ that satisfies the conditions (1) $\\sim(4)$, and has more than $(n-1)(n-2) / 2$ elements. Let $n$ be the least positive integer with which there exists such a set $G$. Note that $G$ must have $(i, i+1)$ for some $1 \\leq i \\leq n$ or $(n, 1)$, since otherwise $G$ can have at most\n\n$$\n\\binom{n}{2}-n=\\frac{n(n-3)}{2}<\\frac{(n-1)(n-2)}{2}\n$$\n\nelements. Without loss of generality we may assume that $(n, 1) \\in G$. Then $(1, n-1) \\notin G$, since otherwise the condition (3) yields $(n, n-1) \\in G$ contradicting the condition (2). Now let $G^{\\prime}=\\{(i, j) \\in G \\mid 1 \\leq i, j \\leq n-1\\}$, then $G^{\\prime}$ satisfies the conditions (1) (4), with $n-1$.\n\nWe now claim that $\\left|G-G^{\\prime}\\right| \\leq n-2$ :\nSuppose that $\\left|G-G^{\\prime}\\right|>n-2$, then $\\left|G-G^{\\prime}\\right|=n-1$ and hence for each $1 \\leq i \\leq n-1$, either $(i, n)$ or $(n, i)$ must be in $G$. We already know that $(n, 1) \\in G$ and $(n-1, n) \\in G$ (because $(n, n-1) \\notin G$ ) and this implies that $(n, n-2) \\notin G$ and $(n-2, n) \\in G$. If we keep doing this process, we obtain $(1, n) \\in G$, which is a contradiction.\n\nSince $\\left|G-G^{\\prime}\\right| \\leq n-2$, we obtain\n\n$$\n\\left|G^{\\prime}\\right| \\geq \\frac{(n-1)(n-2)}{2}-(n-2)=\\frac{(n-2)(n-3)}{2}\n$$\n\nThis, however, contradicts the minimality of $n$, and hence proves (ii).", "problem_match": "\nProblem 3.", "solution_match": "\nSolution."}
-{"year": "2007", "problem_label": "4", "tier": 1, "problem": "Let $x, y$ and $z$ be positive real numbers such that $\\sqrt{x}+\\sqrt{y}+\\sqrt{z}=1$. Prove that\n\n$$\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\geq 1\n$$", "solution": "We first note that\n\n$$\n\\begin{aligned}\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}} & =\\frac{x^{2}-x(y+z)+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{x(y+z)}{\\sqrt{2 x^{2}(y+z)}} \\\\\n& =\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\sqrt{\\frac{y+z}{2}} \\\\\n& \\geq \\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{\\sqrt{y}+\\sqrt{z}}{2} .\n\\end{aligned}\n$$\n\nSimilarly, we have\n\n$$\n\\begin{aligned}\n& \\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}} \\geq \\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{\\sqrt{z}+\\sqrt{x}}{2}, \\\\\n& \\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\geq \\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\frac{\\sqrt{x}+\\sqrt{y}}{2} .\n\\end{aligned}\n$$\n\nWe now add (1) (3) to get\n\n$$\n\\begin{aligned}\n& \\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\\\\n& \\geq \\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z} \\\\\n& =\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+1 .\n\\end{aligned}\n$$\n\nThus, it suffices to show that\n\n$$\n\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}} \\geq 0 .\n$$\n\nNow, assume without loss of generality, that $x \\geq y \\geq z$. Then we have\n\n$$\n\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}} \\geq 0\n$$\n\nand\n\n$$\n\\begin{aligned}\n& \\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}=\\frac{(y-z)(x-z)}{\\sqrt{2 z^{2}(x+y)}}-\\frac{(y-z)(x-y)}{\\sqrt{2 y^{2}(z+x)}} \\\\\n& \\geq \\frac{(y-z)(x-y)}{\\sqrt{2 z^{2}(x+y)}}-\\frac{(y-z)(x-y)}{\\sqrt{2 y^{2}(z+x)}}=(y-z)(x-y)\\left(\\frac{1}{\\sqrt{2 z^{2}(x+y)}}-\\frac{1}{\\sqrt{2 y^{2}(z+x)}}\\right)\n\\end{aligned}\n$$\n\nThe last quantity is non-negative due to the fact that\n\n$$\ny^{2}(z+x)=y^{2} z+y^{2} x \\geq y z^{2}+z^{2} x=z^{2}(x+y)\n$$\n\nThis completes the proof.\n\nSecond solution. By Cauchy-Schwarz inequality,\n\n$$\n\\begin{aligned}\n& \\left(\\frac{x^{2}}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}) \\geq(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})^{2}=1\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n& \\left(\\frac{y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{x y}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}) \\geq\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right)^{2}\n\\end{aligned}\n$$\n\nWe now combine (5) and (6) to find\n\n$$\n\\begin{aligned}\n& \\left(\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(x+y)}+\\sqrt{2(y+z)}+\\sqrt{2(z+x)}) \\\\\n& \\geq 1+\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right)^{2} \\geq 2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) .\n\\end{aligned}\n$$\n\nThus, it suffices to show that\n\n$$\n2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}\n$$\n\nConsider the following inequality using AM-GM inequality\n\n$$\n\\left[\\sqrt{\\frac{y z}{x}}+\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right)\\right]^{2} \\geq 4 \\sqrt{\\frac{y z}{x}}\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right)=2(y+z)\n$$\n\nor equivalently\n\n$$\n\\sqrt{\\frac{y z}{x}}+\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)} .\n$$\n\nSimilarly, we have\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{z x}{y}}+\\left(\\frac{1}{2} \\sqrt{\\frac{x y}{z}}+\\frac{1}{2} \\sqrt{\\frac{y z}{x}}\\right) \\geq \\sqrt{2(z+x)} \\\\\n& \\sqrt{\\frac{x y}{z}}+\\left(\\frac{1}{2} \\sqrt{\\frac{y z}{x}}+\\frac{1}{2} \\sqrt{\\frac{z x}{y}}\\right) \\geq \\sqrt{2(x+y)}\n\\end{aligned}\n$$\n\nAdding the last three inequalities, we get\n\n$$\n2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)} .\n$$\n\nThis completes the proof.", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}
+{"year": "2007", "problem_label": "4", "tier": 1, "problem": "Let $x, y$ and $z$ be positive real numbers such that $\\sqrt{x}+\\sqrt{y}+\\sqrt{z}=1$. Prove that\n\n$$\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\geq 1\n$$", "solution": "We first note that\n\n$$\n\\begin{aligned}\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}} & =\\frac{x^{2}-x(y+z)+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{x(y+z)}{\\sqrt{2 x^{2}(y+z)}} \\\\\n& =\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\sqrt{\\frac{y+z}{2}} \\\\\n& \\geq \\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{\\sqrt{y}+\\sqrt{z}}{2} .\n\\end{aligned}\n$$\n\nSimilarly, we have\n\n$$\n\\begin{aligned}\n& \\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}} \\geq \\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{\\sqrt{z}+\\sqrt{x}}{2}, \\\\\n& \\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\geq \\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\frac{\\sqrt{x}+\\sqrt{y}}{2} .\n\\end{aligned}\n$$\n\nWe now add (1) (3) to get\n\n$$\n\\begin{aligned}\n& \\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\\\\n& \\geq \\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\sqrt{x}+\\sqrt{y}+\\sqrt{z} \\\\\n& =\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+1 .\n\\end{aligned}\n$$\n\nThus, it suffices to show that\n\n$$\n\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}+\\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}} \\geq 0 .\n$$\n\nNow, assume without loss of generality, that $x \\geq y \\geq z$. Then we have\n\n$$\n\\frac{(x-y)(x-z)}{\\sqrt{2 x^{2}(y+z)}} \\geq 0\n$$\n\nand\n\n$$\n\\begin{aligned}\n& \\frac{(z-x)(z-y)}{\\sqrt{2 z^{2}(x+y)}}+\\frac{(y-z)(y-x)}{\\sqrt{2 y^{2}(z+x)}}=\\frac{(y-z)(x-z)}{\\sqrt{2 z^{2}(x+y)}}-\\frac{(y-z)(x-y)}{\\sqrt{2 y^{2}(z+x)}} \\\\\n& \\geq \\frac{(y-z)(x-y)}{\\sqrt{2 z^{2}(x+y)}}-\\frac{(y-z)(x-y)}{\\sqrt{2 y^{2}(z+x)}}=(y-z)(x-y)\\left(\\frac{1}{\\sqrt{2 z^{2}(x+y)}}-\\frac{1}{\\sqrt{2 y^{2}(z+x)}}\\right)\n\\end{aligned}\n$$\n\nThe last quantity is non-negative due to the fact that\n\n$$\ny^{2}(z+x)=y^{2} z+y^{2} x \\geq y z^{2}+z^{2} x=z^{2}(x+y)\n$$\n\nThis completes the proof.", "problem_match": "\nProblem 4.", "solution_match": "\nSolution."}
+{"year": "2007", "problem_label": "4", "tier": 1, "problem": "Let $x, y$ and $z$ be positive real numbers such that $\\sqrt{x}+\\sqrt{y}+\\sqrt{z}=1$. Prove that\n\n$$\n\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}} \\geq 1\n$$", "solution": "By Cauchy-Schwarz inequality,\n\n$$\n\\begin{aligned}\n& \\left(\\frac{x^{2}}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}) \\geq(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})^{2}=1\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n& \\left(\\frac{y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{x y}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}) \\geq\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right)^{2}\n\\end{aligned}\n$$\n\nWe now combine (5) and (6) to find\n\n$$\n\\begin{aligned}\n& \\left(\\frac{x^{2}+y z}{\\sqrt{2 x^{2}(y+z)}}+\\frac{y^{2}+z x}{\\sqrt{2 y^{2}(z+x)}}+\\frac{z^{2}+x y}{\\sqrt{2 z^{2}(x+y)}}\\right) \\\\\n& \\quad \\times(\\sqrt{2(x+y)}+\\sqrt{2(y+z)}+\\sqrt{2(z+x)}) \\\\\n& \\geq 1+\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right)^{2} \\geq 2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) .\n\\end{aligned}\n$$\n\nThus, it suffices to show that\n\n$$\n2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)}\n$$\n\nConsider the following inequality using AM-GM inequality\n\n$$\n\\left[\\sqrt{\\frac{y z}{x}}+\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right)\\right]^{2} \\geq 4 \\sqrt{\\frac{y z}{x}}\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right)=2(y+z)\n$$\n\nor equivalently\n\n$$\n\\sqrt{\\frac{y z}{x}}+\\left(\\frac{1}{2} \\sqrt{\\frac{z x}{y}}+\\frac{1}{2} \\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)} .\n$$\n\nSimilarly, we have\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{z x}{y}}+\\left(\\frac{1}{2} \\sqrt{\\frac{x y}{z}}+\\frac{1}{2} \\sqrt{\\frac{y z}{x}}\\right) \\geq \\sqrt{2(z+x)} \\\\\n& \\sqrt{\\frac{x y}{z}}+\\left(\\frac{1}{2} \\sqrt{\\frac{y z}{x}}+\\frac{1}{2} \\sqrt{\\frac{z x}{y}}\\right) \\geq \\sqrt{2(x+y)}\n\\end{aligned}\n$$\n\nAdding the last three inequalities, we get\n\n$$\n2\\left(\\sqrt{\\frac{y z}{x}}+\\sqrt{\\frac{z x}{y}}+\\sqrt{\\frac{x y}{z}}\\right) \\geq \\sqrt{2(y+z)}+\\sqrt{2(z+x)}+\\sqrt{2(x+y)} .\n$$\n\nThis completes the proof.", "problem_match": "\nProblem 4.", "solution_match": "\nSecond solution."}
 {"year": "2007", "problem_label": "5", "tier": 1, "problem": "A regular $(5 \\times 5)$-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles, exactly one light is switched on. Find all the possible positions of this light.", "solution": "We assign the following first labels to the 25 positions of the lights:\n\n| 1 | 1 | 0 | 1 | 1 |\n| :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 0 | 0 | 0 |\n| 1 | 1 | 0 | 1 | 1 |\n| 0 | 0 | 0 | 0 | 0 |\n| 1 | 1 | 0 | 1 | 1 |\n\nFor each on-off combination of lights in the array, define its first value to be the sum of the first labels of those positions at which the lights are switched on. It is easy to check that toggling any switch always leads to an on-off combination of lights whose first value has the same parity(the remainder when divided by 2) as that of the previous on-off combination.\n\nThe $90^{\\circ}$ rotation of the first labels gives us another labels (let us call it the second labels) which also makes the parity of the second value(the sum of the second labels of those positions at which the lights are switched on) invariant under toggling.\n\n| 1 | 0 | 1 | 0 | 1 |\n| :--- | :--- | :--- | :--- | :--- |\n| 1 | 0 | 1 | 0 | 1 |\n| 0 | 0 | 0 | 0 | 0 |\n| 1 | 0 | 1 | 0 | 1 |\n| 1 | 0 | 1 | 0 | 1 |\n\nSince the parity of the first and the second values of the initial status is 0 , after certain number of toggles the parity must remain unchanged with respect to the first labels and the second labels as well. Therefore, if exactly one light is on after some number of toggles, the label of that position must be 0 with respect to both labels. Hence according to the above pictures, the possible positions are the ones marked with $*_{i}$ 's in the following picture:\n\n|  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n|  | $*_{2}$ |  | $*_{1}$ |  |\n|  |  | $*_{0}$ |  |  |\n|  | $*_{3}$ |  | $*_{4}$ |  |\n|  |  |  |  |  |\n\nNow we demonstrate that all five positions are possible:\nToggling the positions checked by t (the order of toggling is irrelevant) in the first picture makes the center $\\left(*_{0}\\right)$ the only position with light on and the second picture makes the position $*_{1}$ the only position with light on. The other $*_{i}$ 's can be obtained by rotating the second picture appropriately.\n![](https://cdn.mathpix.com/cropped/2024_11_22_16a4255df619f73cfb01g-11.jpg?height=287&width=326&top_left_y=705&top_left_x=708)\n\n|  | t |  | t |  |\n| :---: | :---: | :---: | :---: | :---: |\n| t | t |  | t | t |\n|  | t |  |  |  |\n|  |  | t | t | t |\n|  |  |  | t |  |", "problem_match": "\nProblem 5.", "solution_match": "\nSolution."}

APMO/segmented/en-apmo2012_sol.jsonl

@@ -1,5 +1,5 @@
 {"year": "2012", "problem_label": "1", "tier": 1, "problem": "", "solution": "Let us denote by $\\triangle X Y Z$ the area of the triangle $X Y Z$. Let $x=\\triangle P A B, y=\\triangle P B C$ and $z=\\triangle P C A$.\n\nFrom\n\n$$\ny: z=\\triangle B C P: \\triangle A C P=B F: A F=\\triangle B P F: \\triangle A P F=(x-1): 1\n$$\n\nfollows that $z(x-1)=y$, which yields $(z+1) x=x+y+z$. Similarly, we get $(x+1) y=x+y+z$ and $(y+1) z=x+y+z$. Thus, we obtain $(x+1) y=(y+1) z=$ $(z+1) x$.\n\nWe may assume without loss of generality that $x \\leq y, z$. If we assume that $y>z$ holds, then we get $(y+1) z>(z+1) x$, which is a contradiction. Similarly, we see that $y<z$ leads to a contradiction $(x+1) y<(y+1) z$. Therefore, we must have $y=z$. Then, we also get from $(y+1) z=(z+1) x$ that $x=z$ must hold. We now obtain from $(x-1): 1=y: z=1: 1$ that $x=y=z=2$ holds. Therefore, we conclude that the area of the triangle $A B C$ equals $x+y+z=6$.", "problem_match": "# Problem 1.", "solution_match": "\nSolution:"}
 {"year": "2012", "problem_label": "2", "tier": 1, "problem": "", "solution": "If we insert numbers as in the figure below ( 0 's are to be inserted in the remaining blank boxes), then we see that the condition of the problem is satisfied and the total number of all the numbers inserted is 5 .\n\n| 0 | 1 | 0 |  |\n| :--- | :--- | :--- | :--- |\n| 1 | 1 | 1 |  |\n| 0 | 1 | 0 |  |\n|  |  |  |  |\n\nWe will show that the sum of all the numbers to be inserted in the boxes of the given grid cannot be more than 5 if the distribution of the numbers has to satisfy the requirement of the problem. Let $n=2012$. Let us say that the row number (the column number) of a box in the given grid is $i(j$, respectively) if the box lies on the $i$-th row and the $j$-th column. For a pair of positive integers $x$ and $y$, denote by $R(x, y)$ the sum of the numbers inserted in all of the boxes whose row number is greater than or equal to $x$ and less than or equal to $y$ (assign the value 0 if $x>y$ ).\n\nFirst let $a$ be the largest integer satisfying $1 \\leq a \\leq n$ and $R(1, a-1) \\leq 1$, and then choose the smallest integer $c$ satisfying $a \\leq c \\leq n$ and $R(c+1, n) \\leq 1$. It is possible to choose such a pair $a, c$ since $R(1,0)=0$ and $R(n+1, n)=0$. If $a<c$, then we have $a<n$ and so, by the maximality of $a$, we must have $R(1, a)>1$, while from the minimality of $c$, we must have $R(a+1, n)>1$. Then by splitting the grid into 2 rectangles by means of the horizontal line bordering the $a$-th row and the $a+1$-th row, we get the splitting contradicting the requirement of the problem. Thus, we must have $a=c$.\n\nSimilarly, if for any pair of integers $x, y$ we define $C(x, y)$ to be the sum of the numbers inserted in all of the boxes whose column number is greater than or equal to $x$ and less than or equal to $y(C(x, y)=0$ if $x>y)$, then we get a number $b$ for which\n\n$$\nC(1, b-1) \\leq 1, \\quad C(b+1, n) \\leq 1, \\quad 1 \\leq b \\leq n\n$$\n\nIf we let $r$ be the number inserted in the box whose row number is $a$ and the column number is $b$, then since $r \\leq 1$, we conclude that the sum of the numbers inserted into all of the boxes is\n\n$$\n\\leq R(1, a-1)+R(a+1, n)+C(1, b-1)+C(b+1, n)+r \\leq 5\n$$", "problem_match": "# Problem 2.", "solution_match": "\nSolution:"}
-{"year": "2012", "problem_label": "3", "tier": 1, "problem": "", "solution": "For integers $a, b$ and a positive integer $m$, let us write $a \\equiv b(\\bmod m)$ if $a-b$ is divisible by $m$. Since $\\frac{n^{p}+1}{p^{n}+1}$ must be a positive integer, we see that $p^{n} \\leq n^{p}$ must hold. This means that if $p=2$, then $2^{n} \\leq n^{2}$ must hold. As it is easy to show by induction that $2^{n}>n^{2}$ holds if $n \\geq 5$, we conclude that if $p=2$, then $n \\leq 4$ must be satisfied. And we can check that $(p, n)=(2,2),(2,4)$ satisfy the condition of the problem, while $(2,3)$ does not.\n\nNext, we consider the case where $p \\geq 3$.\nSuppose $s$ is an integer satisfying $s \\geq p$. If $s^{p} \\leq p^{s}$ for such an $s$, then we have\n\n$$\n\\begin{aligned}\n(s+1)^{p} & =s^{p}\\left(1+\\frac{1}{s}\\right)^{p} \\leq p^{s}\\left(1+\\frac{1}{p}\\right)^{p} \\\\\n& =p^{s} \\sum_{r=0}^{p}{ }_{p} C_{r} \\frac{1}{p^{r}}<p^{s} \\sum_{r=0}^{p} \\frac{1}{r!} \\\\\n& \\leq p^{s}\\left(1+\\sum_{r=1}^{p} \\frac{1}{2^{r-1}}\\right) \\\\\n& <p^{s}(1+2) \\leq p^{s+1}\n\\end{aligned}\n$$\n\nThus we have $(s+1)^{p}<p^{s+1}$, and by induction on $n$, we can conclude that if $n>p$, then $n^{p}<p^{n}$. This implies that we must have $n \\leq p$ in order to satisfy our requirement $p^{n} \\leq n^{p}$.\n\nWe note that since $p^{n}+1$ is even, so is $n^{p}+1$, which, in turn implies that $n$ must be odd and therefore, $p^{n}+1$ is divisible by $p+1$, and $n^{p}+1$ is also divisible by $p+1$. Thus we have $n^{p} \\equiv-1(\\bmod (p+1))$, and therefore, $n^{2 p} \\equiv 1(\\bmod (p+1))$.\n\nNow, let $e$ be the smallest positive integer for which $n^{e} \\equiv 1(\\bmod (p+1))$. Then, we can write $2 p=e x+y$, where $x, y$ are non-negative integers and $0 \\leq y<e$, and we have\n\n$$\n1 \\equiv n^{2 p}=\\left(n^{e}\\right)^{x} \\cdot n^{y} \\equiv n^{y} \\quad(\\bmod (p+1))\n$$\n\nwhich implies, because of the minimality of $e$, that $y=0$ must hold. This means that $2 p$ is an integral multiple of $e$, and therefore, $e$ must equal one of the numbers $1,2, p, 2 p$.\n\nNow, if $e=1, p$, then we get $n^{p} \\equiv 1(\\bmod (p+1))$, which contradicts the fact that $p$ is an odd prime. Since $n$ and $p+1$ are relatively prime, we have by Euler's Theorem that $n^{\\varphi(p+1)} \\equiv 1(\\bmod (p+1))$, where $\\varphi(m)$ denotes the number of integers $j(1 \\leq j \\leq m)$ which are relatively prime with $m$. From $\\varphi(p+1)<p+1<2 p$ and the minimality of $e$, we can then conclude that $e=2$ must hold.\n\nFrom $n^{2} \\equiv 1(\\bmod (p+1))$, we get\n\n$$\n-1 \\equiv n^{p}=n^{\\left(2 \\cdot \\frac{p-1}{2}+1\\right)} \\equiv n \\quad(\\bmod (p+1))\n$$\n\nwhich implies that $p+1$ divides $n+1$. Therefore, we must have $p \\leq n$, which, together with the fact $n \\leq p$, show that $p=n$ must hold.\n\nIt is clear that the pair $(p, p)$ for any prime $p \\geq 3$ satisfies the condition of the problem, and thus, we conclude that the pairs $(p, n)$ which satisfy the condition of the problem must be of the form $(2,4)$ and $(p, p)$ with any prime $p$.\n\nAlternate Solution. Let us consider the case where $p \\geq 3$. As we saw in the preceding solution, $n$ must be odd if the pair $(p, n)$ satisfy the condition of the problem. Now, let $q$ be a prime factor of $p+1$. Then, since $p+1$ divides $p^{n}+1, q$ must be a prime factor of $p^{n}+1$ and of $n^{p}+1$ as well. Suppose $q \\geq 3$. Then, from $n^{p} \\equiv-1(\\bmod q)$, it follows that $n^{2 p} \\equiv 1(\\bmod q)$ holds. If we let $e$ be the smallest positive integer satisfying $n^{e} \\equiv 1(\\bmod q)$, then by using the same argument as we used in the preceding solution, we can conclude that $e$ must equal one of the numbers $1,2, p, 2 p$. If $e=1, p$, then we get $n^{p} \\equiv 1(\\bmod q)$, which contradicts the assumption $q \\geq 3$. Since $n$ is not a multiple of $q$, by Fermat's Little Theorem we get $n^{q-1} \\equiv 1(\\bmod q)$, and therefore, we get by the minimality of $e$ that $e=2$ must hold. From $n^{2} \\equiv 1(\\bmod q)$, we also get\n\n$$\nn^{p}=n^{\\left(2 \\cdot \\frac{p-1}{2}+1\\right)} \\equiv n \\quad(\\bmod q),\n$$\n\nand since $n^{p} \\equiv-1(\\bmod q)$, we have $n \\equiv-1(\\bmod q)$ as well.\nNow, if $q=2$ then since $n$ is odd, we have $n \\equiv-1(\\bmod q)$ as well. Thus, we conclude that for an arbitrary prime factor $q$ of $p+1, n \\equiv-1(\\bmod q)$ must hold.\n\nSuppose, for a prime $q, q^{k}$ for some positive integer $k$ is a factor of $p+1$. Then $q^{k}$ must be a factor of $n^{p}+1$ as well. But since\n\n$$\n\\begin{gathered}\nn^{p}+1=(n+1)\\left(n^{p-1}-n^{p-2}+\\cdots-n+1\\right) \\quad \\text { and } \\\\\nn^{p-1}-n^{p-2}+\\cdots-n+1 \\equiv(-1)^{p-1}-(-1)^{p-2}+\\cdots-(-1)+1 \\not \\equiv 0 \\quad(\\bmod q)\n\\end{gathered}\n$$\n\nwe see that $q^{k}$ must divide $n+1$. By applying the argument above for each prime factor $q$ of $p+1$, we can then conclude that $n+1$ must be divisible by $p+1$, and as we did in the preceding proof, we can conclude that $n=p$ must hold.", "problem_match": "# Problem 3.", "solution_match": "# Solution\n"}
+{"year": "2012", "problem_label": "3", "tier": 1, "problem": "", "solution": "For integers $a, b$ and a positive integer $m$, let us write $a \\equiv b(\\bmod m)$ if $a-b$ is divisible by $m$. Since $\\frac{n^{p}+1}{p^{n}+1}$ must be a positive integer, we see that $p^{n} \\leq n^{p}$ must hold. This means that if $p=2$, then $2^{n} \\leq n^{2}$ must hold. As it is easy to show by induction that $2^{n}>n^{2}$ holds if $n \\geq 5$, we conclude that if $p=2$, then $n \\leq 4$ must be satisfied. And we can check that $(p, n)=(2,2),(2,4)$ satisfy the condition of the problem, while $(2,3)$ does not.\n\nNext, we consider the case where $p \\geq 3$.\nSuppose $s$ is an integer satisfying $s \\geq p$. If $s^{p} \\leq p^{s}$ for such an $s$, then we have\n\n$$\n\\begin{aligned}\n(s+1)^{p} & =s^{p}\\left(1+\\frac{1}{s}\\right)^{p} \\leq p^{s}\\left(1+\\frac{1}{p}\\right)^{p} \\\\\n& =p^{s} \\sum_{r=0}^{p}{ }_{p} C_{r} \\frac{1}{p^{r}}<p^{s} \\sum_{r=0}^{p} \\frac{1}{r!} \\\\\n& \\leq p^{s}\\left(1+\\sum_{r=1}^{p} \\frac{1}{2^{r-1}}\\right) \\\\\n& <p^{s}(1+2) \\leq p^{s+1}\n\\end{aligned}\n$$\n\nThus we have $(s+1)^{p}<p^{s+1}$, and by induction on $n$, we can conclude that if $n>p$, then $n^{p}<p^{n}$. This implies that we must have $n \\leq p$ in order to satisfy our requirement $p^{n} \\leq n^{p}$.\n\nWe note that since $p^{n}+1$ is even, so is $n^{p}+1$, which, in turn implies that $n$ must be odd and therefore, $p^{n}+1$ is divisible by $p+1$, and $n^{p}+1$ is also divisible by $p+1$. Thus we have $n^{p} \\equiv-1(\\bmod (p+1))$, and therefore, $n^{2 p} \\equiv 1(\\bmod (p+1))$.\n\nNow, let $e$ be the smallest positive integer for which $n^{e} \\equiv 1(\\bmod (p+1))$. Then, we can write $2 p=e x+y$, where $x, y$ are non-negative integers and $0 \\leq y<e$, and we have\n\n$$\n1 \\equiv n^{2 p}=\\left(n^{e}\\right)^{x} \\cdot n^{y} \\equiv n^{y} \\quad(\\bmod (p+1))\n$$\n\nwhich implies, because of the minimality of $e$, that $y=0$ must hold. This means that $2 p$ is an integral multiple of $e$, and therefore, $e$ must equal one of the numbers $1,2, p, 2 p$.\n\nNow, if $e=1, p$, then we get $n^{p} \\equiv 1(\\bmod (p+1))$, which contradicts the fact that $p$ is an odd prime. Since $n$ and $p+1$ are relatively prime, we have by Euler's Theorem that $n^{\\varphi(p+1)} \\equiv 1(\\bmod (p+1))$, where $\\varphi(m)$ denotes the number of integers $j(1 \\leq j \\leq m)$ which are relatively prime with $m$. From $\\varphi(p+1)<p+1<2 p$ and the minimality of $e$, we can then conclude that $e=2$ must hold.\n\nFrom $n^{2} \\equiv 1(\\bmod (p+1))$, we get\n\n$$\n-1 \\equiv n^{p}=n^{\\left(2 \\cdot \\frac{p-1}{2}+1\\right)} \\equiv n \\quad(\\bmod (p+1))\n$$\n\nwhich implies that $p+1$ divides $n+1$. Therefore, we must have $p \\leq n$, which, together with the fact $n \\leq p$, show that $p=n$ must hold.\n\nIt is clear that the pair $(p, p)$ for any prime $p \\geq 3$ satisfies the condition of the problem, and thus, we conclude that the pairs $(p, n)$ which satisfy the condition of the problem must be of the form $(2,4)$ and $(p, p)$ with any prime $p$.\n\nAlternate Solution. Let us consider the case where $p \\geq 3$. As we saw in the preceding solution, $n$ must be odd if the pair $(p, n)$ satisfy the condition of the problem. Now, let $q$ be a prime factor of $p+1$. Then, since $p+1$ divides $p^{n}+1, q$ must be a prime factor of $p^{n}+1$ and of $n^{p}+1$ as well. Suppose $q \\geq 3$. Then, from $n^{p} \\equiv-1(\\bmod q)$, it follows that $n^{2 p} \\equiv 1(\\bmod q)$ holds. If we let $e$ be the smallest positive integer satisfying $n^{e} \\equiv 1(\\bmod q)$, then by using the same argument as we used in the preceding solution, we can conclude that $e$ must equal one of the numbers $1,2, p, 2 p$. If $e=1, p$, then we get $n^{p} \\equiv 1(\\bmod q)$, which contradicts the assumption $q \\geq 3$. Since $n$ is not a multiple of $q$, by Fermat's Little Theorem we get $n^{q-1} \\equiv 1(\\bmod q)$, and therefore, we get by the minimality of $e$ that $e=2$ must hold. From $n^{2} \\equiv 1(\\bmod q)$, we also get\n\n$$\nn^{p}=n^{\\left(2 \\cdot \\frac{p-1}{2}+1\\right)} \\equiv n \\quad(\\bmod q),\n$$\n\nand since $n^{p} \\equiv-1(\\bmod q)$, we have $n \\equiv-1(\\bmod q)$ as well.\nNow, if $q=2$ then since $n$ is odd, we have $n \\equiv-1(\\bmod q)$ as well. Thus, we conclude that for an arbitrary prime factor $q$ of $p+1, n \\equiv-1(\\bmod q)$ must hold.\n\nSuppose, for a prime $q, q^{k}$ for some positive integer $k$ is a factor of $p+1$. Then $q^{k}$ must be a factor of $n^{p}+1$ as well. But since\n\n$$\n\\begin{gathered}\nn^{p}+1=(n+1)\\left(n^{p-1}-n^{p-2}+\\cdots-n+1\\right) \\quad \\text { and } \\\\\nn^{p-1}-n^{p-2}+\\cdots-n+1 \\equiv(-1)^{p-1}-(-1)^{p-2}+\\cdots-(-1)+1 \\not \\equiv 0 \\quad(\\bmod q)\n\\end{gathered}\n$$\n\nwe see that $q^{k}$ must divide $n+1$. By applying the argument above for each prime factor $q$ of $p+1$, we can then conclude that $n+1$ must be divisible by $p+1$, and as we did in the preceding proof, we can conclude that $n=p$ must hold.", "problem_match": "# Problem 3.", "solution_match": "# Solution\n\n"}
 {"year": "2012", "problem_label": "4", "tier": 1, "problem": "", "solution": "If $A B=A C$, then we get $B F=C F$ and the conclusion of the problem is clearly satisfied. So, we assume that $A B \\neq A C$ in the sequel.\n\nDue to symmetry, we may suppose without loss of generality that $A B>A C$. Let $K$ be the point on the circle $\\Gamma$ such that $A K$ is a diameter of this circle. Then, we get\n\n$$\n\\angle B C K=\\angle A C K-\\angle A C B=90^{\\circ}-\\angle A C B=\\angle C B H\n$$\n\nand\n\n$$\n\\angle C B K=\\angle A B K-\\angle A B C=90^{\\circ}-\\angle A B C=\\angle B C H,\n$$\n\nfrom which we conclude that the triangles $B C K$ and $C B H$ are congruent. Therefore, the quadrilateral $B K C H$ is a parallelogram, and its diagonal $H K$ passes through the center $M$ of the other diagonal $B C$. Therefore, the 3 points $H, M, K$ lie on the same straight line, and we have $\\angle A E M=\\angle A E K=90^{\\circ}$.\n\nFrom $\\angle A E D=90^{\\circ}=\\angle A D M$, we see that the 4 points $A, E, D, M$ lie on the circumference of the same circle, from which we obtain $\\angle A M B=\\angle A E D=$ $\\angle A E F=\\angle A C F$. Putting this fact together with the fact that $\\angle A B M=\\angle A F C$, we conclude that the triangles $A B M$ and $A F C$ are similar, and we get $\\frac{A M}{B M}=\\frac{A C}{F C}$. By a similar argument, we get that the triangles $A C M$ and $A F B$ are similar, and\ntherefore, that $\\frac{A M}{C M}=\\frac{A B}{F B}$ holds. Noting that $B M=C M$, we also get $\\frac{A C}{F C}=\\frac{A B}{F B}$, from which we can conclude that $\\frac{B F}{C F}=\\frac{A B}{A C}$, proving the assertion of the problem.", "problem_match": "# Problem 4.", "solution_match": "\nSolution:"}
 {"year": "2012", "problem_label": "5", "tier": 1, "problem": "", "solution": "Let us note first that if $i \\neq j$, then since $a_{i} a_{j} \\leq \\frac{a_{i}^{2}+a_{j}^{2}}{2}$, we have\n\n$$\nn-a_{i} a_{j} \\geq n-\\frac{a_{i}^{2}+a_{j}^{2}}{2} \\geq n-\\frac{n}{2}=\\frac{n}{2}>0 .\n$$\n\nIf we set $b_{i}=\\left|a_{i}\\right|(i=1,2, \\ldots, n)$, then we get $b_{1}^{2}+b_{2}^{2}+\\cdots+b_{n}^{2}=n$ and $\\frac{1}{n-a_{i} a_{j}} \\leq$ $\\frac{1}{n-b_{i} b_{j}}$, which shows that it is enough to prove the assertion of the problem in the case where all of $a_{1}, a_{2}, \\cdots, a_{n}$ are non-negative. Hence, we assume from now on that $a_{1}, a_{2}, \\cdots, a_{n}$ are all non-negative.\n\nBy multiplying by $n$ the both sides of the desired inequality we get the inequality:\n\n$$\n\\sum_{1 \\leq i<j \\leq n} \\frac{n}{n-a_{i} a_{j}} \\leq \\frac{n^{2}}{2}\n$$\n\nand since $\\frac{n}{n-a_{i} a_{j}}=1+\\frac{a_{i} a_{j}}{n-a_{i} a_{j}}$, we obtain from the inequality above by subtracting $\\frac{n(n-1)}{2}$ from both sides the following inequality:\n\n$$\n\\sum_{1 \\leq i<j \\leq n} \\frac{a_{i} a_{j}}{n-a_{i} a_{j}} \\leq \\frac{n}{2}\n$$\n\nWe will show that this inequality (i) holds.\nIf for some $i$ the equality $a_{i}^{2}=n$ is valid, then $a_{j}=0$ must hold for all $j \\neq i$ and the inequality (i) is trivially satisfied. So, we assume from now on that $a_{i}^{2}<n$ is valid for each $i$.\n\nLet us assume that $i \\neq j$ from now on. Since $0 \\leq a_{i} a_{j} \\leq\\left(\\frac{a_{i}+a_{j}}{2}\\right)^{2} \\leq \\frac{a_{i}^{2}+a_{j}^{2}}{2}$ holds, we have\n\n$$\n\\frac{a_{i} a_{j}}{n-a_{i} a_{j}} \\leq \\frac{a_{i} a_{j}}{n-\\frac{a_{i}^{2}+a_{j}^{2}}{2}} \\leq \\frac{\\left(\\frac{a_{i}+a_{j}}{2}\\right)^{2}}{n-\\frac{a_{i}^{2}+a_{j}^{2}}{2}}=\\frac{1}{2} \\cdot \\frac{\\left(a_{i}+a_{j}\\right)^{2}}{\\left(n-a_{i}^{2}\\right)+\\left(n-a_{j}^{2}\\right)}\n$$\n\nSince $n-a_{i}^{2}>0, n-a_{j}^{2}>0$, we also get from the Cauchy-Schwarz inequality that\n\n$$\n\\left(\\frac{a_{j}^{2}}{n-a_{i}^{2}}+\\frac{a_{i}^{2}}{n-a_{j}^{2}}\\right)\\left(\\left(n-a_{i}^{2}\\right)+\\left(n-a_{j}^{2}\\right)\\right) \\geq\\left(a_{i}+a_{j}\\right)^{2},\n$$\n\nfrom which it follows that\n\n$$\n\\frac{\\left(a_{i}+a_{j}\\right)^{2}}{\\left(n-a_{i}^{2}\\right)+\\left(n-a_{j}^{2}\\right)} \\leq\\left(\\frac{a_{j}^{2}}{n-a_{i}^{2}}+\\frac{a_{i}^{2}}{n-a_{j}^{2}}\\right)\n$$\n\nholds. Combining the inequalities (ii) and (iii), we get\n\n$$\n\\begin{aligned}\n\\sum_{1 \\leq i<j \\leq n} \\frac{a_{i} a_{j}}{n-a_{i} a_{j}} & \\leq \\frac{1}{2} \\sum_{1 \\leq i<j \\leq n}\\left(\\frac{a_{j}^{2}}{n-a_{i}^{2}}+\\frac{a_{i}^{2}}{n-a_{j}^{2}}\\right) \\\\\n& =\\frac{1}{2} \\sum_{i \\neq j} \\frac{a_{j}^{2}}{n-a_{i}^{2}} \\\\\n& =\\frac{1}{2} \\sum_{i=1}^{n} \\frac{n-a_{i}^{2}}{n-a_{i}^{2}} \\\\\n& =\\frac{n}{2}\n\\end{aligned}\n$$\n\nwhich establishes the desired inequality (i).", "problem_match": "# Problem 5.", "solution_match": "\nSolution:"}

APMO/segmented/en-apmo2020_sol.jsonl

@@ -0,0 +1,8 @@
+{"year": "2020", "problem_label": "1", "tier": 1, "problem": "Let $\\Gamma$ be the circumcircle of $\\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.", "solution": "From the conditions, we have\n![](https://cdn.mathpix.com/cropped/2024_11_22_207f52578b01f443eb43g-1.jpg?height=960&width=743&top_left_y=772&top_left_x=710)\n\nLet $P$ be the intersection of $A C$ and $B F$. Then we have\n\n$$\n\\angle P A E=\\angle C B A=\\angle B A C=\\angle B F C .\n$$\n\nThis implies $A, P, F, E$ are concyclic. It follows that\n\n$$\n\\angle F P E=\\angle F A E=\\angle F B A,\n$$\n\nand hence $A B$ and $E P$ are parallel. So $E, P, D$ are collinear, and the result follows.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution 1"}
+{"year": "2020", "problem_label": "1", "tier": 1, "problem": "Let $\\Gamma$ be the circumcircle of $\\triangle A B C$. Let $D$ be a point on the side $B C$. The tangent to $\\Gamma$ at $A$ intersects the parallel line to $B A$ through $D$ at point $E$. The segment $C E$ intersects $\\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $A C, B F, D E$ are concurrent.", "solution": "Let $E^{\\prime}$ be any point on the extension of $E A$. From $\\angle A E D=\\angle E^{\\prime} A B=\\angle A C D$, points $A, D, C, E$ are concyclic.\n![](https://cdn.mathpix.com/cropped/2024_11_22_207f52578b01f443eb43g-2.jpg?height=714&width=673&top_left_y=215&top_left_x=753)\n\nLet $P$ be the intersection of $B F$ and $D E$. From $\\angle A F P=\\angle A C B=\\angle A E P$, the points $A, P, F, E$ are concyclic. In addition, from $\\angle E P A=\\angle E F A=\\angle D B A$, points $A, B, D, P$ are concyclic.\nBy considering the radical centre of $(B D F E),(A P F E)$ and $(B D P A)$, we find that the lines $B D, A P, E F$ are concurrent at $C$. The result follows.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution 2"}
+{"year": "2020", "problem_label": "2", "tier": 1, "problem": "Show that $r=2$ is the largest real number $r$ which satisfies the following condition:\n\nIf a sequence $a_{1}, a_{2}, \\ldots$ of positive integers fulfills the inequalities\n\n$$\na_{n} \\leq a_{n+2} \\leq \\sqrt{a_{n}^{2}+r a_{n+1}}\n$$\n\nfor every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \\geq M$.", "solution": ". First, let us assume that $r>2$, and take a positive integer $a \\geq 1 /(r-2)$.\nThen, if we let $a_{n}=a+\\lfloor n / 2\\rfloor$ for $n=1,2, \\ldots$, the sequence $a_{n}$ satisfies the inequalities\n\n$$\n\\sqrt{a_{n}^{2}+r a_{n+1}} \\geq \\sqrt{a_{n}^{2}+r a_{n}} \\geq \\sqrt{a_{n}^{2}+\\left(2+\\frac{1}{a}\\right) a_{n}} \\geq a_{n}+1=a_{n+2}\n$$\n\nbut since $a_{n+2}>a_{n}$ for any $n$, we see that $r$ does not satisfy the condition given in the problem.\nNow we show that $r=2$ does satisfy the condition of the problem. Suppose $a_{1}, a_{2}, \\ldots$ is a sequence of positive integers satisfying the inequalities given in the problem, and there exists a positive integer $m$ for which $a_{m+2}>a_{m}$ is satisfied.\nBy induction we prove the following assertion:\n\n$$\na_{m+2 k} \\leq a_{m+2 k-1}=a_{m+1} \\text { holds for every positive integer } k\n$$\n\nThe truth of $(\\dagger)$ for $k=1$ follows from the inequalities below\n\n$$\n2 a_{m+2}-1=a_{m+2}^{2}-\\left(a_{m+2}-1\\right)^{2} \\leq a_{m}^{2}+2 a_{m+1}-\\left(a_{m+2}-1\\right)^{2} \\leq 2 a_{m+1}\n$$\n\nLet us assume that $(\\dagger)$ holds for some positive integer $k$. From\n\n$$\na_{m+1}^{2} \\leq a_{m+2 k+1}^{2} \\leq a_{m+2 k-1}^{2}+2 a_{m+2 k} \\leq a_{m+1}^{2}+2 a_{m+1}<\\left(a_{m+1}+1\\right)^{2}\n$$\n\nit follows that $a_{m+2 k+1}=a_{m+1}$ must hold. Furthermore, since $a_{m+2 k} \\leq a_{m+1}$, we have\n\n$$\na_{m+2 k+2}^{2} \\leq a_{m+2 k}^{2}+2 a_{m+2 k+1} \\leq a_{m+1}^{2}+2 a_{m+1}<\\left(a_{m+1}+1\\right)^{2}\n$$\n\nfrom which it follows that $a_{m+2 k+2} \\leq a_{m+1}$, which proves the assertion $(\\dagger)$.\nWe can conclude that for the value of $m$ with which we started our argument above, $a_{m+2 k+1}=a_{m+1}$ holds for every positive integer $k$. Therefore, in order to finish the proof, it is enough to show that $a_{m+2 k}$ becomes constant after some value of $k$. Since every $a_{m+2 k}$ is a positive integer less than or equal to $a_{m+1}$, there exists $k=K$ for which $a_{m+2 K}$ takes the maximum value. By the monotonicity of $a_{m+2 k}$, it then follows that $a_{m+2 k}=a_{m+2 K}$ for all $k \\geq K$.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 1"}
+{"year": "2020", "problem_label": "2", "tier": 1, "problem": "Show that $r=2$ is the largest real number $r$ which satisfies the following condition:\n\nIf a sequence $a_{1}, a_{2}, \\ldots$ of positive integers fulfills the inequalities\n\n$$\na_{n} \\leq a_{n+2} \\leq \\sqrt{a_{n}^{2}+r a_{n+1}}\n$$\n\nfor every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \\geq M$.", "solution": "We only give an alternative proof of the assertion $(\\dagger)$ in solution 1 . Let $\\left\\{a_{n}\\right\\}$ be a sequence satisfying the inequalities given in the problem. We will use the following key observations:\n(a) If $a_{n+1} \\leq a_{n}$ for some $n \\geq 1$, then\n\n$$\na_{n} \\leq a_{n+2} \\leq \\sqrt{a_{n}^{2}+2 a_{n+1}}<\\sqrt{a_{n}^{2}+2 a_{n}+1}=a_{n}+1,\n$$\n\nhence $a_{n}=a_{n+2}$.\n(b) If $a_{n} \\leq a_{n+1}$ for some $n \\geq 1$, then\n\n$$\na_{n} \\leq a_{n+2} \\leq \\sqrt{a_{n}^{2}+2 a_{n+1}}<\\sqrt{a_{n+1}^{2}+2 a_{n+1}+1}=a_{n+1}+1\n$$\n\nhence $a_{n} \\leq a_{n+2} \\leq a_{n+1}$.\nNow let $m$ be a positive integer such that $a_{m+2}>a_{m}$. By the observations above, we must have $a_{m}<a_{m+2} \\leq a_{m+1}$. Thus the assertion ( $\\dagger$ ) is true for $k=1$. Assume that the assertion holds for some positive integer $k$. Using observation (a), we get $a_{m+2 k+1}=a_{m+2 k-1}=a_{m+1}$. Thus $a_{m+2 k} \\leq a_{m+2 k+1}$, and then using observation (b), we get $a_{m+2 k+2} \\leq a_{m+2 k+1}=a_{m+1}$, which proves the assertion ( $\\dagger$ ).", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}
+{"year": "2020", "problem_label": "3", "tier": 1, "problem": "Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways.", "solution": "We claim that $k=2^{a}$ for all $a \\geq 0$.\nLet $A=\\{1,2,4,8, \\ldots\\}$ and $B=\\mathbb{N} \\backslash A$. For any set $T$, let $s(T)$ denote the sum of the elements of $T$. (If $T$ is empty, we let $s(T)=0$.)\nWe first show that any positive integer $k=2^{a}$ satisfies the desired property. Let $B^{\\prime}$ be a subset of $B$ with $a$ elements, and let $S=A \\cup B^{\\prime}$. Recall that any nonnegative integer has a unique binary representation. Hence, for any integer $t>s\\left(B^{\\prime}\\right)$ and any subset $B^{\\prime \\prime} \\subseteq B^{\\prime}$, the number $t-s\\left(B^{\\prime \\prime}\\right)$ can be written as a sum of distinct elements of $A$ in a unique way. This means that $t$ can be written as a sum of distinct elements of $B^{\\prime}$ in exactly $2^{a}$ ways.\nNext, assume that some positive integer $k$ satisfies the desired property for a positive integer $m \\geq 2$ and a set $S$. Clearly, $S$ is infinite.\nLemma: For all sufficiently large $x \\in S$, the smallest element of $S$ larger than $x$ is $2 x$.\nProof of Lemma: Let $x \\in S$ with $x>3 m$, and let $x<y<2 x$. We will show that $y \\notin S$. Suppose first that $y>x+m$. Then $y-x$ can be written as a sum of distinct elements of $S$ not including $x$ in $k$ ways. If $y \\in S$, then $y$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction. Suppose now that $y \\leq x+m$. We consider $z \\in(2 x-m, 2 x)$. Similarly as before, $z-x$ can be written as a sum of distinct elements of $S$ not including $x$ or $y$ in $k$ ways. If $y \\in S$, then since $m<z-y<x, z-y$ can be written as a sum of distinct elements of $S$ not including $x$ or $y$. This means that $z$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction.\nWe now show that $2 x \\in S$; assume for contradiction that this is not the case. Observe that $2 x$ can be written as a sum of distinct elements of $S$ including $x$ in exactly $k-1$ ways. This means that $2 x$ can also be written as a sum of distinct elements of $S$ not including $x$. If this sum includes any number less than $x-m$, then removing this number, we can write some number $y \\in(x+m, 2 x)$ as a sum of distinct elements of $S$ not including $x$. Now if $y=y^{\\prime}+x$ where $y^{\\prime} \\in(m, x)$ then $y^{\\prime}$ can be written as\na sum of distinct elements of $S$ including $x$ in exactly $k$ ways. Therefore $y$ can be written as a sum of distinct elements of $S$ in at least $k+1$ ways, a contradiction. Hence the sum only includes numbers in the range $[x-m, x)$. Clearly two numbers do not suffice. On the other hand, three such numbers sum to at least $3(x-m)>2 x$, a contradiction.\nFrom the Lemma, we have that $S=T \\cup U$, where $T$ is finite and $U=\\{x, 2 x, 4 x, 8 x, \\ldots\\}$ for some positive integer $x$. Let $y$ be any positive integer greater than $s(T)$. For any subset $T^{\\prime} \\subseteq T$, if $y-s\\left(T^{\\prime}\\right) \\equiv 0(\\bmod x)$, then $y-s\\left(T^{\\prime}\\right)$ can be written as a sum of distinct elements of $U$ in a unique way; otherwise $y-s\\left(T^{\\prime}\\right)$ cannot be written as a sum of distinct elements of $U$. Hence the number of ways to write $y$ as a sum of distinct elements of $S$ is equal to the number of subsets $T^{\\prime} \\subseteq T$ such that $s\\left(T^{\\prime}\\right) \\equiv y(\\bmod x)$. Since this holds for all $y$, for any $0 \\leq a \\leq x-1$ there are exactly $k$ subsets $T^{\\prime} \\subseteq T$ such that $s\\left(T^{\\prime}\\right) \\equiv a(\\bmod x)$. This means that there are $k x$ subsets of $T$ in total. But the number of subsets of $T$ is a power of 2 , and therefore $k$ is a power of 2 , as claimed.", "problem_match": "\nProblem 3.", "solution_match": "# Solution:"}
+{"year": "2020", "problem_label": "3", "tier": 1, "problem": "Determine all positive integers $k$ for which there exist a positive integer $m$ and a set $S$ of positive integers such that any integer $n>m$ can be written as a sum of distinct elements of $S$ in exactly $k$ ways.", "solution": ". We give an alternative proof of the first half of the lemma in the Solution 1 above.\nLet $s_{1}<s_{2}<\\cdots$ be the elements of $S$. For any positive integer $r$, define $A_{r}(x)=\\prod_{n=1}^{r}\\left(1+x^{s_{n}}\\right)$. For each $n$ such that $m \\leq n<s_{r+1}$, all $k$ ways of writing $n$ as a sum of elements of $S$ must only use $s_{1}, \\ldots, s_{r}$, so the coefficient of $x^{n}$ in $A_{r}(x)$ is $k$. Similarly the number of ways of writing $s_{r+1}$ as a sum of elements of $S$ without using $s_{r+1}$ is exactly $k-1$. Hence the coefficient of $x^{s_{r+1}}$ in $A_{r}(x)$ is $k-1$.\nFix a $t$ such that $s_{t}>2(m+1)$. Write\n\n$$\nA_{t-1}(x)=u(x)+k\\left(x^{m+1}+\\cdots+x^{s_{t}-1}\\right)+x^{s_{t}} v(x)\n$$\n\nfor some $u(x), v(x)$ where $u(x)$ is of degree at most $m$.\nNote that\n\n$$\nA_{t+1}(x)=A_{t-1}(x)+x^{s_{t}} A_{t-1}(x)+x^{s_{t+1}} A_{t-1}(x)+x^{s_{t}+s_{t+1}} A_{t-1}(x)\n$$\n\nIf $s_{t+1}+m+1<2 s_{t}$, we can find the term $x^{s_{t+1}+m+1}$ in $x^{s_{t}} A_{t-1}(x)$ and in $x^{s_{t+1}} A_{t-1}(x)$. Hence the coefficient of $x^{s_{t+1}+m+1}$ in $A_{t+1}(x)$ is at least $2 k$, which is impossible. So $s_{t+1} \\geq 2 s_{t}-(m+1)>$ $s_{t}+m+1$.\nNow\n\n$$\nA_{t}(x)=A_{t-1}(x)+x^{s_{t}} u(x)+k\\left(x^{s_{t}+m+1}+\\cdots x^{2 s_{t}-1}\\right)+x^{2 s_{t}} v(x)\n$$\n\nRecall that the coefficent of $x^{s_{t+1}}$ in $A_{t}(x)$ is $k-1$. But if $s_{t}+m+1<s_{t+1}<s_{2 t}$, then the coefficient of $x^{s_{t+1}}$ in $A_{t}(x)$ is at least $k$, which is a contradiction. Therefore $s_{t+1} \\geq 2 s_{t}$.", "problem_match": "\nProblem 3.", "solution_match": "\nSolution 2"}
+{"year": "2020", "problem_label": "4", "tier": 1, "problem": "Let $\\mathbb{Z}$ denote the set of all integers. Find all polynomials $P(x)$ with integer coefficients that satisfy the following property:\nFor any infinite sequence $a_{1}, a_{2}, \\ldots$ of integers in which each integer in $\\mathbb{Z}$ appears exactly once, there exist indices $i<j$ and an integer $k$ such that $a_{i}+a_{i+1}+\\cdots+a_{j}=P(k)$.", "solution": "Part 1: All polynomials with $\\operatorname{deg} P=1$ satisfy the given property.\nSuppose $P(x)=c x+d$, and assume without loss of generality that $c>d \\geq 0$. Denote $s_{i}=a_{1}+a_{2}+$ $\\cdots+a_{i}(\\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \\geq 2$ and $s_{j}-s_{i} \\equiv d$ $(\\bmod c)$.\nConsider $c+1$ indices $e_{1}, e_{2}, \\ldots, e_{c+1}>1$ such that $a_{e_{l}} \\equiv d(\\bmod c)$. By the pigeonhole principle, among the $n+1$ pairs $\\left(s_{e_{1}-1}, s_{e_{1}}\\right),\\left(s_{e_{2}-1}, s_{e_{2}}\\right), \\ldots,\\left(s_{e_{n+1}-1}, s_{e_{n+1}}\\right)$, some two are equal, say $\\left(s_{m-1}, s_{m}\\right)$ and $\\left(s_{n-1}, s_{n}\\right)$. We can then take $i=m-1$ and $j=n$.\nPart 2: All polynomials with $\\operatorname{deg} P \\neq 1$ do not satisfy the given property.\nLemma: If $\\operatorname{deg} P \\neq 1$, then for any positive integers $A, B$, and $C$, there exists an integer $y$ with $|y|>C$ such that no value in the range of $P$ falls within the interval $[y-A, y+B]$.\nProof of Lemma: The claim is immediate when $P$ is constant or when $\\operatorname{deg} P$ is even since $P$ is bounded from below. Let $P(x)=a_{n} x^{n}+\\cdots+a_{1} x+a_{0}$ be of odd degree greater than 1 , and assume without\nloss of generality that $a_{n}>0$. Since $P(x+1)-P(x)=a_{n} n x^{n-1}+\\ldots$, and $n-1>0$, the gap between $P(x)$ and $P(x+1)$ grows arbitrarily for large $x$. The claim follows.\nSuppose $\\operatorname{deg} P \\neq 1$. We will inductively construct a sequence $\\left\\{a_{i}\\right\\}$ such that for any indices $i<j$ and any integer $k$ it holds that $a_{i}+a_{i+1}+\\cdots+a_{j} \\neq P(k)$. Suppose that we have constructed the sequence up to $a_{i}$, and $m$ is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take $a_{i+2}=m$. Consider all the new sums of at least two consecutive terms; each of them contains $a_{i+1}$. Hence all such sums are in the interval $\\left[a_{i+1}-A, a_{i+1}+B\\right]$ for fixed constants $A, B$. The lemma allows us to choose $a_{i+1}$ so that all such sums avoid the range of $P$.\nAlternate Solution for Part 1: Again, suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \\geq 0$. Let $S_{i}=\\left\\{a_{j}+a_{j+1}+\\cdots+a_{i}(\\bmod c) \\mid j=1,2, \\ldots, i\\right\\}$. Then $S_{i+1}=\\left\\{s_{i}+a_{i+1}\\right.$ $\\left.(\\bmod c) \\mid s_{i} \\in S_{i}\\right\\} \\cup\\left\\{a_{i+1}(\\bmod c)\\right\\}$. Hence $\\left|S_{i+1}\\right|=\\left|S_{i}\\right|$ or $S_{i+1}=\\left|S_{i}\\right|+1$, with the former occuring exactly when $0 \\in S_{i}$. Since $\\left|S_{i}\\right| \\leq c$, the latter can only occur finitely many times, so there exists $I$ such that $0 \\in S_{i}$ for all $i \\geq I$. Let $t>I$ be an index with $a_{t} \\equiv d(\\bmod c)$. Then we can find a sum of at least two consecutive terms ending at $a_{t}$ and congruent to $d(\\bmod c)$.\nAlternate Construction when $P(x)$ is constant or of even degree\nIf $P(x)$ is of even degree, then $P$ is bounded from below or from above. In case of $P$ is constant or bounded from above, then there exists a positive integer $c$ such that $P(x)<c$. Let $\\left\\{a_{i}\\right\\}$ be the sequence\n\n$$\n0,1,-1,2,3,-2,4,5,-3, \\cdots\n$$\n\nwhich is given by $a_{3 n+1}=2 n, a_{3 n+2}=2 n+1, a_{3 n+3}=-(n+1)$ for all $n \\geq 0$. Notice that for any $i<j$ we have $a_{i}+\\cdots+a_{j} \\geq 0$. Then for the sequence $\\left\\{b_{n}\\right\\}$ defined by $b_{n}=a_{n}+c$, clearly $b_{i}+\\cdots+b_{j} \\geq\\left(a_{i}+\\cdots+a_{j}\\right)+2 c>c$ which is out side the range of $P(x)$.\n\nNow if $P$ is bounded from below, there exist a positive integer $c$ such that $P(x)>-c$. In this case, take $b_{n}$ to be $b_{n}=-a_{n}-c$. Then for all $i<j$ we have $b_{i}+\\cdots+b_{j} \\leq-\\left(a_{1}+\\cdots a_{n}\\right)-2 c<-c$ which is again out side the range of $P(x)$.", "problem_match": "\nProblem 4.", "solution_match": "# Solution:"}
+{"year": "2020", "problem_label": "5", "tier": 1, "problem": "Let $n \\geq 3$ be a fixed integer. The number 1 is written $n$ times on a blackboard. Below the blackboard, there are two buckets that are initially empty. A move consists of erasing two of the numbers $a$ and $b$, replacing them with the numbers 1 and $a+b$, then adding one stone to the first bucket and $\\operatorname{gcd}(a, b)$ stones to the second bucket. After some finite number of moves, there are $s$ stones in the first bucket and $t$ stones in the second bucket, where $s$ and $t$ are positive integers. Find all possible values of the ratio $\\frac{t}{s}$.", "solution": "The answer is the set of all rational numbers in the interval $[1, n-1)$. First, we show that no other numbers are possible. Clearly the ratio is at least 1, since for every move, at least one stone is added to the second bucket. Note that the number $s$ of stones in the first bucket is always equal to $p-n$, where $p$ is the sum of the numbers on the blackboard. We will assume that the numbers are written in a row, and whenever two numbers $a$ and $b$ are erased, $a+b$ is written in the place of the number on the right. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be the numbers on the blackboard from left to right, and let\n\n$$\nq=0 \\cdot a_{1}+1 \\cdot a_{2}+\\cdots+(n-1) a_{n}\n$$\n\nSince each number $a_{i}$ is at least 1 , we always have\n\n$$\nq \\leq(n-1) p-(1+\\cdots+(n-1))=(n-1) p-\\frac{n(n-1)}{2}=(n-1) s+\\frac{n(n-1)}{2}\n$$\n\nAlso, if a move changes $a_{i}$ and $a_{j}$ with $i<j$, then $t$ changes by $\\operatorname{gcd}\\left(a_{i}, a_{j}\\right) \\leq a_{i}$ and $q$ increases by\n\n$$\n(j-1) a_{i}-(i-1)\\left(a_{i}-1\\right) \\geq i a_{i}-(i-1)\\left(a_{i}-1\\right) \\geq a_{i}\n$$\n\nHence $q-t$ never decreases. We may assume without loss of generality that the first move involves the rightmost 1. Then immediately after this move, $q=0+1+\\cdots+(n-2)+(n-1) \\cdot 2=\\frac{(n+2)(n-1)}{2}$ and\n$t=1$. So after that move, we always have\n\n$$\n\\begin{aligned}\nt & \\leq q+1-\\frac{(n+2)(n-1)}{2} \\\\\n& \\leq(n-1) s+\\frac{n(n-1)}{2}-\\frac{(n+2)(n-1)}{2}+1 \\\\\n& =(n-1) s-(n-2)<(n-1) s\n\\end{aligned}\n$$\n\nHence, $\\frac{t}{s}<n-1$. So $\\frac{t}{s}$ must be a rational number in $[1, n-1)$.\n\nAfter a single move, we have $\\frac{t}{s}=1$, so it remains to prove that $\\frac{t}{s}$ can be any rational number in $(1, n-1)$. We will now show by induction on $n$ that for any positive integer $a$, it is possible to reach a situation where there are $n-1$ occurrences of 1 on the board and the number $a^{n-1}$, with $t$ and $s$ equal to $a^{n-2}(a-1)(n-1)$ and $a^{n-1}-1$, respectively. For $n=2$, this is clear as there is only one possible move at each step, so after $a-1$ moves $s$ and $t$ will both be equal to $a-1$. Now assume that the claim is true for $n-1$, where $n>2$. Call the algorithm which creates this situation using $n-1$ numbers algorithm $A$. Then to reach the situation for size $n$, we apply algorithm $A$, to create the number $a^{n-2}$. Next, apply algorithm $A$ again and then add the two large numbers, repeat until we get the number $a^{n-1}$. Then algorithm $A$ was applied $a$ times and the two larger numbers were added $a-1$ times. Each time the two larger numbers are added, $t$ increases by $a^{n-2}$ and each time algorithm $A$ is applied, $t$ increases by $a^{n-3}(a-1)(n-2)$. Hence, the final value of $t$ is\n\n$$\nt=(a-1) a^{n-2}+a \\cdot a^{n-3}(a-1)(n-2)=a^{n-2}(a-1)(n-1)\n$$\n\nThis completes the induction.\nNow we can choose 1 and the large number $b$ times for any positive integer $b$, and this will add $b$ stones to each bucket. At this point we have\n\n$$\n\\frac{t}{s}=\\frac{a^{n-2}(a-1)(n-1)+b}{a^{n-1}-1+b}\n$$\n\nSo we just need to show that for any rational number $\\frac{p}{q} \\in(1, n-1)$, there exist positive integers $a$ and $b$ such that\n\n$$\n\\frac{p}{q}=\\frac{a^{n-2}(a-1)(n-1)+b}{a^{n-1}-1+b}\n$$\n\nRearranging, we see that this happens if and only if\n\n$$\nb=\\frac{q a^{n-2}(a-1)(n-1)-p\\left(a^{n-1}-1\\right)}{p-q} .\n$$\n\nIf we choose $a \\equiv 1(\\bmod p-q)$, then this will be an integer, so we just need to check that the numerator is positive for sufficiently large $a$.\n\n$$\n\\begin{aligned}\nq a^{n-2}(a-1)(n-1)-p\\left(a^{n-1}-1\\right) & >q a^{n-2}(a-1)(n-1)-p a^{n-1} \\\\\n& =a^{n-2}(a(q(n-1)-p)-(n-1))\n\\end{aligned}\n$$\n\nwhich is positive for sufficiently large $a$ since $q(n-1)-p>0$.\n\nAlternative solution for the upper bound. Rather than starting with $n$ occurrences of 1 , we may start with infinitely many 1 s , but we are restricted to having at most $n-1$ numbers which are not equal to 1 on the board at any time. It is easy to see that this does not change the problem. Note also that we can ignore the 1 we write on the board each move, so the allowed move is to rub off two numbers and write their sum. We define the width and score of a number on the board as follows. Colour that number red, then reverse every move up to that point all the way back to the situation when the numbers are all 1 s . Whenever a red number is split, colour the two replacement numbers\nred. The width of the original number is equal to the maximum number of red integers greater than 1 which appear on the board at the same time. The score of the number is the number of stones which were removed from the second bucket during these splits. Then clearly the width of any number is at most $n-1$. Also, $t$ is equal to the sum of the scores of the final numbers. We claim that if a number $p>1$ has a width of at most $w$, then its score is at most $(p-1) w$. We will prove this by strong induction on $p$. If $p=1$, then clearly $p$ has a score of 0 , so the claim is true. If $p>1$, then $p$ was formed by adding two smaller numbers $a$ and $b$. Clearly $a$ and $b$ both have widths of at most $w$. Moreover, if $a$ has a width of $w$, then at some point in the reversed process there will be $w$ numbers in the set $\\{2,3,4, \\ldots\\}$ that have split from $a$, and hence there can be no such numbers at this point which have split from $b$. Between this point and the final situation, there must always be at least one number in the set $\\{2,3,4, \\ldots\\}$ that split from $a$, so the width of $b$ is at most $w-1$. Therefore, $a$ and $b$ cannot both have widths of $w$, so without loss of generality, $a$ has width at most $w$ and $b$ has width at most $w-1$. Then by the inductive hypothesis, $a$ has score at most $(a-1) w$ and $b$ has score at most $(b-1)(w-1)$. Hence, the score of $p$ is at most\n\n$$\n\\begin{aligned}\n(a-1) w+(b-1)(w-1)+\\operatorname{gcd}(a, b) & \\leq(a-1) w+(b-1)(w-1)+b \\\\\n& =(p-1) w+1-w \\\\\n& \\leq(p-1) w .\n\\end{aligned}\n$$\n\nThis completes the induction.\nNow, since each number $p$ in the final configuration has width at most $(n-1)$, it has score less than $(n-1)(p-1)$. Hence the number $t$ of stones in the second bucket is less than the sum over the values of $(n-1)(p-1)$, and $s$ is equal to the sum of the the values of $(p-1)$. Therefore, $\\frac{t}{s}<n-1$.", "problem_match": "\nProblem 5.", "solution_match": "# Solution:"}

APMO/segmented/en-apmo2021_sol.jsonl

@@ -0,0 +1,8 @@
+{"year": "2021", "problem_label": "1", "tier": 1, "problem": "Prove that for each real number $r>2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^{2}=r\\lfloor x\\rfloor$.\nNote: $\\lfloor x\\rfloor$ denotes the largest integer less than or equal to $x$", "solution": "Let $r>2$ be a real number. Let $x$ be a positive real number such that $x^{2}=r\\lfloor x\\rfloor$ with $\\lfloor x\\rfloor=k$. Since $x>0$ and $x^{2}=r k$, we also have $k>0$. From $k \\leq x<k+1$, we get $k^{2} \\leq x^{2}=r k<$ $k^{2}+2 k+1 \\leq k^{2}+3 k$, hence $k \\leq r<k+3$, or $r-3<k \\leq r$. There are at most three positive integers in the interval $(r-3, r]$. Thus there are at most three possible values for $k$. Consequently, there are at most three positive solutions to the given equation.\nNow suppose that $k$ is a positive integer in the interval $[r-2, r]$. There are at least two such positive integer. Observe that $k \\leq \\sqrt{r k} \\leq \\sqrt{(k+2) k}<k+1$ and so $r k=r\\lfloor\\sqrt{r k}\\rfloor$. We conclude that the equation $x^{2}=r\\lfloor x\\rfloor$ has at least two positive solutions, namely $x=\\sqrt{r k}$ with $k \\in[r-2, r]$.", "problem_match": "\nProblem 1.", "solution_match": "\nSolution "}
+{"year": "2021", "problem_label": "2", "tier": 1, "problem": "For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \\leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.\nDetermine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \\leq 2021$.", "solution": "There are two possible families of solutions:\n\n- $P(x)=x+d$, for some integer $d \\geq-2022$.\n- $P(x)=-x+d$, for some integer $d \\leq 2022$.\n\nSuppose $P$ satisfies the problem conditions. Clearly $P$ cannot be a constant polynomial. Notice that a polynomial $P$ satifies the conditions if and only if $-P$ also satisfies them. Hence, we may assume the leading coefficient of $P$ is positive. Then, there exists positive integer $M$ such that $P(x)>0$ for $x \\geq M$.\n\nLemma 1. For any positive integer $n$, the integers $P(1), P(2), \\ldots, P(n)$ leave pairwise distinct remainders upon division by $n$.\n\nProof. Assume for contradiction that this is not the case. Then, for some $1 \\leq y<z \\leq n$, there exists $0 \\leq r \\leq n-1$ such that $P(y) \\equiv P(z) \\equiv r(\\bmod n)$. Since $P(a n+b) \\equiv P(b)(\\bmod n)$ for all $a, b$ integers, we have $P(a n+y) \\equiv P(a n+z) \\equiv r(\\bmod n)$ for any integer $a$. Let $A$ be a positive integer such that $A n \\geq M$, and let $k$ be a positive integer such that $k>2 A+2021$. Each of the $2(k-A)$ integers $P(A n+y), P(A n+z), P((A+1) n+y), P((A+1) n+z), \\ldots, P((k-1) n+y), P((k-1) n+z)$ leaves one of the $k$ remainders\n\n$$\nr, n+r, 2 n+r, \\ldots,(k-1) n+r\n$$\n\nupon division by $k n$. This implies that at least $2(k-A)-k=k-2 A$ (possibly overlapping) pairs leave the same remainder upon division by $k n$. Since $k-2 A>2021$ and all of the $2(k-A)$ integers are positive, we find more than 2021 pairs $a, b$ with $a<b \\leq k n$ for which $|P(b)|-|P(a)|$ is divisible by $k n$-hence, $P_{k n}>2021$, a contradiction.\n\nNext, we show that $P$ is linear. Assume that this is not the case, i.e., $\\operatorname{deg} P \\geq 2$. Then we can find a positive integer $k$ such that $P(k)-P(1) \\geq k$. This means that among the integers $P(1), P(2), \\ldots, P(P(k)-P(1))$, two of them, namely $P(k)$ and $P(1)$, leave the same remainder upon division by $P(k)-P(1)$ - contradicting the lemma (by taking $n=P(k)-P(1)$ ). Hence, $P$ must be linear.\nWe can now write $P(x)=c x+d$ with $c>0$. We prove that $c=1$ by two ways.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution "}
+{"year": "2021", "problem_label": "2", "tier": 1, "problem": "For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \\leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.\nDetermine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \\leq 2021$.", "solution": "If $c \\geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the Lemma. Hence $c=1$.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 1"}
+{"year": "2021", "problem_label": "2", "tier": 1, "problem": "For a polynomial $P$ and a positive integer $n$, define $P_{n}$ as the number of positive integer pairs $(a, b)$ such that $a<b \\leq n$ and $|P(a)|-|P(b)|$ is divisible by $n$.\nDetermine all polynomial $P$ with integer coefficients such that for all positive integers $n, P_{n} \\leq 2021$.", "solution": "Suppose $c \\geq 2$. Let $n$ be a positive integer such that $n>2 c M, n\\left(1-\\frac{3}{2 c}\\right)>2022$ and $2 c \\mid n$. Notice that for any positive integers $i$ such that $\\frac{3 n}{2 c}+i<n, P\\left(\\frac{3 n}{2 c}+i\\right)-P\\left(\\frac{n}{2 c}+i\\right)=n$. Hence, $\\left(\\frac{n}{2 c}+i, \\frac{3 n}{2 c}+i\\right)$ satifies the condition in the question for all positive integers $i$ such that $\\frac{3 n}{2 c}+i<n$. Hence, $P_{n}>2021$, a contradiction. Then, $c=1$.\n\nIf $d \\leq-2023$, then there are at least 2022 pairs $a<b$ such that $P(a)=P(b)$, namely $(a, b)=$ $(1,-2 d-1),(2,-2 d-2), \\ldots,(-d-1,-d+1)$. This implies that $d \\geq-2022$.\nFinally, we verify that $P(x)=x+d$ satisfies the condition for any $d \\geq-2022$. Fix a positive integer $n$. Note that $\\| P(b)|-|P(a)||<n$ for all positive integers $a<b \\leq n$, so the only pairs $a, b$ for which $|P(b)|-|P(a)|$ could be divisible by $n$ are those for which $|P(a)|=|P(b)|$. When $d \\geq-2022$, there are indeed at most 2021 such pairs.", "problem_match": "\nProblem 2.", "solution_match": "\nSolution 2"}
+{"year": "2021", "problem_label": "3", "tier": 1, "problem": "Let $A B C D$ be a cyclic convex quadrilateral and $\\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.", "solution": "Let $L$ be the intersection of the bisectors of $\\angle A B C$ and $\\angle B C D$. Let $N$ be the $E$-excenter of $\\triangle B C E$. Let $\\angle B A C=\\angle B D C=\\alpha, \\angle D B C=\\beta$ and $\\angle A C B=\\gamma$.\nWe have the following:\n\n$$\n\\begin{array}{r}\n\\angle C B L=\\frac{1}{2} \\angle A B C=90^{\\circ}-\\frac{1}{2} \\alpha-\\frac{1}{2} \\gamma \\text { and } \\angle B C L=90^{\\circ}-\\frac{1}{2} \\alpha-\\frac{1}{2} \\beta, \\\\\n\\angle C B N=90^{\\circ}-\\frac{1}{2} \\beta \\text { and } \\angle B C N=90^{\\circ}-\\frac{1}{2} \\gamma, \\\\\n\\angle M B L=\\angle M B C+\\angle C B L=90^{\\circ}-\\frac{1}{2} \\gamma \\text { and } \\angle M C L=90^{\\circ}-\\frac{1}{2} \\beta, \\\\\n\\angle L C N=\\angle L B N=180^{\\circ}-\\frac{1}{2}(\\alpha+\\beta+\\gamma) .\n\\end{array}\n$$\n\nApplying the sine rule to $\\triangle M B L$ and $\\triangle M C L$ we obtain\n\n$$\n\\frac{M B}{M L}=\\frac{M C}{M L}=\\frac{\\sin \\angle B L M}{\\sin \\angle M B L}=\\frac{\\sin \\angle C L M}{\\sin \\angle M C L}\n$$\n\nIt follows that\n\n$$\n\\frac{\\sin \\angle B L M}{\\sin \\angle C L M}=\\frac{\\sin \\angle M B L}{\\sin \\angle M C L}=\\frac{\\cos (\\gamma / 2)}{\\cos (\\beta / 2)}\n$$\n\nNow\n\n$$\n\\frac{\\sin \\angle B L M}{\\sin \\angle M L C} \\cdot \\frac{\\sin \\angle L C N}{\\sin \\angle N C B} \\cdot \\frac{\\sin \\angle N B C}{\\sin \\angle N B L}=\\frac{\\cos (\\gamma / 2)}{\\cos (\\beta / 2)} \\cdot \\frac{\\sin \\left(90^{\\circ}-\\frac{1}{2} \\beta\\right)}{\\sin \\left(90^{\\circ}-\\frac{1}{2} \\gamma\\right)}=1\n$$\n\nHence $L M, B N, C N$ are concurrent and therefore $L, M, N$ are collinear.\n\n## Alternative proof\n\nWe proceed similarly as above until the equation (1).\nWe use the following lemma.\nLemma: If $\\pi>\\alpha, \\beta, \\gamma, \\delta>0, \\alpha+\\beta=\\gamma+\\delta<\\pi$, and $\\frac{\\sin \\alpha}{\\sin \\beta}=\\frac{\\sin \\gamma}{\\sin \\delta}$, then $\\alpha=\\gamma$ and $\\beta=\\delta$.\nProof of Lemma: Let $\\theta=\\alpha+\\beta=\\gamma+\\delta$. Then $\\frac{\\sin (\\theta-\\beta)}{\\sin \\beta}=\\frac{\\sin (\\theta-\\delta)}{\\sin \\delta}$.\n\n$$\n\\begin{gathered}\n\\Longleftrightarrow \\sin (\\theta-\\beta) \\sin \\delta=\\sin (\\theta-\\delta) \\sin \\beta \\\\\n\\Longleftrightarrow(\\sin \\theta \\cos \\beta-\\sin \\beta \\cos \\theta) \\sin \\delta=(\\sin \\theta \\cos \\delta-\\sin \\delta \\cos \\theta) \\sin \\beta \\\\\n\\Longleftrightarrow \\sin \\theta \\cos \\beta \\sin \\delta=\\sin \\theta \\cos \\delta \\sin \\beta \\\\\n\\Longleftrightarrow \\sin \\theta \\sin (\\beta-\\delta)=0\n\\end{gathered}\n$$\n\nSince $0<\\theta<\\pi$, then $\\sin \\theta \\neq 0$. Therefore, $\\sin (\\beta-\\delta)=0$, and we must have $\\beta=\\delta$.\nApplying the sine rule to $\\triangle N B L$ and $\\triangle N C L$ we obtain\n\n$$\n\\begin{aligned}\n& \\frac{N B}{N L}=\\frac{\\sin \\angle B L N}{\\sin \\angle L B N} \\\\\n& \\frac{N C}{N L}=\\frac{\\sin \\angle C L N}{\\sin \\angle L C N}\n\\end{aligned}\n$$\n\nSince $\\angle L B N=\\angle L C N$, it follows that\n\n$$\n\\frac{\\sin \\angle B L N}{\\sin \\angle C L N}=\\frac{N B}{N C}=\\frac{\\sin \\angle B C N}{\\sin \\angle C B N}=\\frac{\\cos (\\gamma / 2)}{\\cos (\\beta / 2)}=\\frac{\\sin \\angle B L M}{\\sin \\angle C L M}\n$$\n\nBy the lemma, it is concluded that $\\angle B L M=\\angle B L N$ and $\\angle C L M=\\angle C L N$. Therefore, $L, M, N$ are collinear.", "problem_match": "\nProblem 3.", "solution_match": "# Solution 1"}
+{"year": "2021", "problem_label": "3", "tier": 1, "problem": "Let $A B C D$ be a cyclic convex quadrilateral and $\\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals $A C$ and $B D$, let $L$ be the center of the circle tangent to sides $A B, B C$, and $C D$, and let $M$ be the midpoint of the arc $B C$ of $\\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $B C E$ opposite $E$ lies on the line $L M$.", "solution": "Denote by $N$ the excenter of triangle $B C E$ opposite $E$. Since $B L$ bisects $\\angle A B C$, we have $\\angle C B L=$ $\\frac{\\angle A B C}{2}$. Since $M$ is the midpoint of arc $B C$, we have $\\angle M B C=\\frac{1}{2}(\\angle M B C+\\angle M C B)$ It follows by angle chasing that\n\n$$\n\\begin{aligned}\n\\angle M B L & =\\angle M B C+\\angle C B L=\\frac{1}{2}(\\angle M B C+\\angle M C B+\\angle A B C) \\\\\n& =\\frac{1}{2}(\\angle M B A+\\angle M C B)=90^{\\circ}-\\frac{\\angle B C E}{2}=\\angle B C N .\n\\end{aligned}\n$$\n\nDenote by $X$ and $Y$ the second intersections of lines $B M$ and $C M$ with the circumcircle of $B C L$, respectively. Since $\\angle M B C=\\angle M C B$, we have $B C \\| X Y$. It suffices to show that $B N \\| X L$ and $C N \\| Y L$. Indeed, from this it follows that $\\triangle B C N \\sim \\triangle X Y L$, and therefore a homothety with center $M$ that maps $B$ to $X$ and $C$ to $Y$ also maps $N$ to $L$, implying that $N$ lies on the line $L M$.\nBy symmetry, it suffices to show that $C N \\| Y L$, which is equivalent to showing that $\\angle B C N=\\angle X Y L$. But we have $\\angle B C N=\\angle M B L=\\angle X B L=\\angle X Y L$, completing the proof.", "problem_match": "\nProblem 3.", "solution_match": "# Solution 2"}
+{"year": "2021", "problem_label": "4", "tier": 1, "problem": "Given a $32 \\times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is good if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:\n(a) No good subset consists of 888 cells.\n(b) There exists a good subset consisting of at least 666 cells.", "solution": "(a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those cheese-cells and the other ones gap-cells. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th element is $C$ if the $i$-th step of the mouse was onto a cheese-cell, and $G$ if it was onto a gap-cell. By assumption, $s$ contains $888 C$ 's. Note that $s$ does not contain a contiguous block of 4 (or more) $C$ 's. Hence $s$ contains at least $888 / 3=296$ such $C$-blocks and thus at least $295 G^{\\prime}$ 's. But since each gap-cell is traversed at most twice, this implies there are at least $\\lceil 295 / 2\\rceil=148$ gap-cells, for a total of $888+148=1036>32^{2}$ cells, a contradiction.\n(b) Let $L_{i}, X_{i}$ be two $2^{i} \\times 2^{i}$ tiles that allow the mouse to \"turn left\" and \"cross\", respectively. In detail, the \"turn left\" tiles allow the mouse to enter at its bottom left cell facing up and to leave at its bottom left cell facing left. The \"cross\" tiles allow the mouse to enter at its top right facing down and leave at its bottom left facing left, while also to enter at its bottom left facing up and leave at its top right facing right.\n(a) Basic tiles\n(b) Inductive construction\n(c) $16 \\times 16$\n![](https://cdn.mathpix.com/cropped/2024_11_22_5e137eeb9eefa8d28dceg-4.jpg?height=373&width=1633&top_left_y=1078&top_left_x=249)\n\nNote that given two $2^{i} \\times 2^{i}$ tiles $L_{i}, X_{i}$ we can construct larger $2^{i+1} \\times 2^{i+1}$ tiles $L_{i+1}, X_{i+1}$ inductively as shown on in (b). The construction works because the path intersects itself (or the other path) only inside the smaller $X$-tiles where it works by induction.\nFor a tile $T$, let $|T|$ be the number of pieces of cheese in it. By straightforward induction, $\\left|L_{i}\\right|=\\left|X_{i}\\right|+1$ and $\\left|L_{i+1}\\right|=4 \\cdot\\left|L_{i}\\right|-1$. From the initial condition $\\left|L_{1}\\right|=3$. We now easily compute $\\left|L_{2}\\right|=11,\\left|L_{3}\\right|=43,\\left|L_{4}\\right|=171$, and $\\left|L_{5}\\right|=683$. Hence we get the desired subset.\n\n## Another proof of (a).\n\nLet $X_{N}$ be the largest possible density of cheese-cells in a good subset on an $N \\times N$ table. We will show that $X_{N} \\leq 4 / 5+o(1)$. Specifically, this gives $X_{32} \\leq 817 / 1024$. We look at the (discrete analogue) of the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a different empty square, hence $X_{N} \\leq 4 / 5$.", "problem_match": "\nProblem 4.", "solution_match": "# Solution."}
+{"year": "2021", "problem_label": "5", "tier": 1, "problem": "Determine all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that $f(f(a)-b)+b f(2 a)$ is a perfect square for all integers $a$ and $b$.", "solution": ".\n\nThere are two families of functions which satisfy the condition:\n(1) $f(n)= \\begin{cases}0 & \\text { if } n \\text { is even, and } \\\\ \\text { any perfect square } & \\text { if } n \\text { is odd }\\end{cases}$\n(2) $f(n)=n^{2}$, for every integer $n$.\n\nIt is straightforward to verify that the two families of functions are indeed solutions. Now, suppose that f is any function which satisfies the condition that $f(f(a)-b)+b f(2 a)$ is a perfect square for every pair $(a, b)$ of integers. We denote this condition by $\\left(^{*}\\right)$. We will show that $f$ must belong to either Family (1) or Family (2).\nClaim 1. $f(0)=0$ and $f(n)$ is a perfect square for every integer $n$.\nProof. Plugging $(a, b) \\rightarrow(0, f(0))$ in $\\left(^{*}\\right)$ shows that $f(0)(f(0)+1)=z^{2}$ for some integer $z$. Thus, $(2 f(0)+1-2 z)(2 f(0)+1+2 z)=1$. Therefore, $f(0)$ is either -1 or 0.\nSuppose, for sake of contradiction, that $f(0)=-1$. For any integer $a$, plugging $(a, b) \\rightarrow(a, f(a))$ implies that $f(a) f(2 a)-1$ is a square. Thus, for each $a \\in \\mathbb{Z}$, there exists $x \\in \\mathbb{Z}$ such that $f(a) f(2 a)=x^{2}+1$ This implies that any prime divisor of $f(a)$ is either 2 or is congruent to $1(\\bmod 4)$, and that $4 \\nmid f(a)$, for every $a \\in \\mathbb{Z}$.\nPlugging $(a, b) \\rightarrow(0,3)$ in $\\left(^{*}\\right)$ shows that $f(-4)-3$ is a square. Thus, there is $y \\in \\mathbb{Z}$ such that $f(-4)=y^{2}+3$. Since $4 \\nmid f(-4)$, we note that $f(-4)$ is a positive integer congruent to $3(\\bmod 4)$, but any prime dividing $f(-4)$ is either 2 or is congruent to $1(\\bmod 4)$. This gives a contradiction. Therefore, $f(0)$ must be 0 .\nFor every integer $n$, plugging $(a, b) \\rightarrow(0,-n)$ in $\\left(^{*}\\right)$ shows that $f(n)$ is a square.\nReplacing $b$ with $f(a)-b$, we find that for all integers $a$ and $b$,\n\n$$\nf(b)+(f(a)-b) f(2 a) \\text { is a square. }\n$$\n\nNow, let $S$ be the set of all integers $n$ such that $f(n)=0$. We have two cases:\n\n- Case 1: $S$ is unbounded from above.\n\nWe claim that $f(2 n)=0$ for any integer $n$. Fix some integer $n$, and let $k \\in S$ with $k>f(n)$. Then, plugging $(a, b) \\mapsto(n, k)$ in $\\left({ }^{* *}\\right)$ gives us that $f(k)+(f(n)-k) f(2 n)=(f(n)-k) f(2 n)$ is a square. But $f(n)-k<0$ and $f(2 n)$ is a square by Claim 1. This is possible only if $f(2 n)=0$. In summary, $f(n)=0$ whenever $n$ is even and Claim 1 shows that $f(n)$ is a square whenever $n$ is odd.\n\n- Case 2: $S$ is bounded from above.\n\nLet $T$ be the set of all integers $n$ such that $f(n)=n^{2}$. We show that $T$ is unbounded from above. In fact, we show that $\\frac{p+1}{2} \\in T$ for all primes $p$ big enough.\nFix a prime number $p$ big enough, and let $n=\\frac{p+1}{2}$. Plugging $(a, b) \\mapsto(n, 2 n)$ in ( $\\left.{ }^{* *}\\right)$ shows us that $f(2 n)(f(n)-2 n+1)$ is a square for any integer $n$. For $p$ big enough, we have $2 n \\notin S$, so $f(2 n)$ is a non-zero square. As a result, when $p$ is big enough, $f(n)$ and $f(n)-2 n+1=f(n)-p$ are both squares. Writing $f(n)=k^{2}$ and $f(n)-p=m^{2}$ for some $k, m \\geq 0$, we have\n\n$$\n(k+m)(k-m)=k^{2}-m^{2}=p \\Longrightarrow k+m=p, k-m=1 \\Longrightarrow k=n, m=n-1\n$$\n\nThus, $f(n)=k^{2}=n^{2}$, giving us $n=\\frac{p+1}{2} \\in T$.\nNext, for all $k \\in T$ and $n \\in \\mathbb{Z}$, plugging $(a, b) \\mapsto(n, k)$ in $(* *)$ shows us that $k^{2}+(f(n)-k) f(2 n)$ is a square. But that means $(2 k-f(2 n))^{2}-\\left(f(2 n)^{2}-4 f(n) f(2 n)\\right)=4\\left(k^{2}+(f(n)-k) f(2 n)\\right)$ is also a square. When $k$ is large enough, we have $\\left|f(2 n)^{2}-4 f(n) f(2 n)\\right|+1<|2 k-f(2 n)|$. As a result, we must have $f(2 n)^{2}=4 f(n) f(2 n)$ and thus $f(2 n) \\in\\{0,4 f(n)\\}$ for all integers $n$.\nFinally, we prove that $f(n)=n^{2}$ for all integers $n$. Fix $n$, and take $k \\in T$ big enough such that $2 k \\notin S$. Then, we have $f(k)=k^{2}$ and $f(2 k)=4 f(k)=4 k^{2}$. Plugging $(a, b) \\mapsto(k, n)$ to $(* *)$ shows us that $f(n)+\\left(k^{2}-n\\right) 4 k^{2}=\\left(2 k^{2}-n\\right)^{2}+\\left(f(n)-n^{2}\\right)$ is a square. Since $T$ is unbounded from above, we can take $k \\in T$ such that $2 k \\notin S$ and also $\\left|2 k^{2}-n\\right|>\\left|f(n)-n^{2}\\right|$. This forces $f(n)=n^{2}$, giving us the second family of solution.\n\n## Another approach of Case 1.\n\nClaim 2. One of the following is true.\n(i) For every integer $n, f(2 n)=0$.\n(ii) There exists an integer $K>0$ such that for every integer $n \\geq K, f(n)>0$.\n\nProof. Suppose that there exists an integer $\\alpha \\neq 0$ such that $f(2 \\alpha)>0$. We claim that for every integer $n \\geq f(\\alpha)+1$, we have $f(n)>0$.\nFor every $n \\geq f(\\alpha)+1$, plugging $(a, b) \\rightarrow(\\alpha, f(\\alpha)-n)$ in $\\left(^{*}\\right)$ shows that $f(n)+(f(\\alpha)-n) f(2 \\alpha)$ is a square, and in particular, is non-negative. Hence, $f(n) \\geq(n-f(\\alpha)) f(2 \\alpha)>0$, as desired.\nIf $f$ belongs to Case (i), Claim 1 shows that $f$ belongs to Family (1).\nIf $f$ belongs to Case (ii), then $S$ is bounded from above. From Case 2 we get $f(n)=n^{2}$.", "problem_match": "\nProblem 5.", "solution_match": "# Solution 1"}

APMO/segmented/en-apmo2022_sol.jsonl

@@ -0,0 +1,10 @@
+{"year": "2022", "problem_label": "1", "tier": 1, "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".1\n\nBy inspection, we see that the pairs $(a, b)$ with $a=b$ are solutions, and so too are the pairs $(a, 1)$. We will see that these are the only solutions.\n\n- Case 1. Consider the case $b<a$. Since $b-1$ is a multiple of $a-1$, it follows that $b=1$. This yields the second set of solutions described above.\n- Case 2. This leaves the case $b \\geq a$. Since the positive integer $a^{3}$ is a multiple of $b^{2}$, there is a positive integer $c$ such that $a^{3}=b^{2} c$.\nNote that $a \\equiv b \\equiv 1$ modulo $a-1$. So we have\n\n$$\n1 \\equiv a^{3}=b^{2} c \\equiv c \\quad(\\bmod a-1) .\n$$\n\nIf $c<a$, then we must have $c=1$, hence, $a^{3}=b^{2}$. So there is a positive integer $d$ such that $a=d^{2}$ and $b=d^{3}$. Now $a-1 \\mid b-1$ yields $d^{2}-1 \\mid d^{3}-1$. This implies that $d+1 \\mid d(d+1)+1$, which is impossible.\nIf $c \\geq a$, then $b^{2} c \\geq b^{2} a \\geq a^{3}=b^{2} c$. So there's equality throughout, implying $a=c=b$. This yields the first set of solutions described above.\nTherefore, the solutions described above are the only solutions.", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}
+{"year": "2022", "problem_label": "1", "tier": 1, "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".2\n\nWe will start by showing that there are positive integers $x, c, d$ such that $a=x^{2} c d$ and $b=x^{3} c$. Let $g=\\operatorname{gcd}(a, b)$ so that $a=g d$ and $b=g x$ for some coprime $d$ and $x$. Then, $b^{2} \\mid a^{3}$ is equivalent to $g^{2} x^{2} \\mid g^{3} d^{3}$, which is equivalent to $x^{2} \\mid g d^{3}$. Since $x$ and $d$ are coprime, this implies $x^{2} \\mid g$. Hence, $g=x^{2} c$ for some $c$, giving $a=x^{2} c d$ and $b=x^{3} c$ as required.\nNow, it remains to find all positive integers $x, c, d$ satisfying\n\n$$\nx^{2} c d-1 \\mid x^{3} c-1\n$$\n\nThat is, $x^{3} c \\equiv 1\\left(\\bmod x^{2} c d-1\\right)$. Assuming that this congruence holds, it follows that $d \\equiv x^{3} c d \\equiv x$ $\\left(\\bmod x^{2} c d-1\\right)$. Then, either $x=d$ or $x-d \\geq x^{2} c d-1$ or $d-x \\geq x^{2} c d-1$.\n\n- If $x=d$ then $b=a$.\n- If $x-d \\geq x^{2} c d-1$, then $x-d \\geq x^{2} c d-1 \\geq x-1 \\geq x-d$. Hence, each of these inequalities must in fact be an equality. This implies that $x=c=d=1$, which implies that $a=b=1$.\n- If $d-x \\geq x^{2} c d-1$, then $d-x \\geq x^{2} c d-1 \\geq d-1 \\geq d-x$. Hence, each of these inequalities must in fact be an equality. This implies that $x=c=1$, which implies that $b=1$.\nHence the only solutions are the pairs $(a, b)$ such that $a=b$ or $b=1$. These pairs can be checked to satisfy the given conditions.", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}
+{"year": "2022", "problem_label": "1", "tier": 1, "problem": "Find all pairs $(a, b)$ of positive integers such that $a^{3}$ is a multiple of $b^{2}$ and $b-1$ is a multiple of $a-1$. Note: An integer $n$ is said to be a multiple of an integer $m$ if there is an integer $k$ such that $n=k m$.", "solution": ".3\n\nAll answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.\nTo show that these are all solutions, note that we can easily eliminate the case $a=1$ or $b=1$. Thus, assume that $a, b \\neq 1$ and $a \\neq b$. By the second divisibility, we see that $a-1 \\mid b-a$. However, $\\operatorname{gcd}(a, b) \\mid b-a$ and $a-1$ is relatively prime to $\\operatorname{gcd}(a, b)$. This implies that $(a-1) \\operatorname{gcd}(a, b) \\mid b-a$, which implies $\\operatorname{gcd}(a, b) \\left\\lvert\\, \\frac{b-1}{a-1}-1\\right.$.\nThe last relation implies that $\\operatorname{gcd}(a, b)<\\frac{b-1}{a-1}$, since the right-hand side are positive. However, due to the first divisibility,\n\n$$\n\\operatorname{gcd}(a, b)^{3}=\\operatorname{gcd}\\left(a^{3}, b^{3}\\right) \\geq \\operatorname{gcd}\\left(b^{2}, b^{3}\\right)=b^{2} .\n$$\n\nCombining these two inequalities, we get that\n\n$$\nb^{\\frac{2}{3}}<\\frac{b-1}{a-1}<2 \\frac{b}{a}\n$$\n\nThis implies $a<2 b^{\\frac{1}{3}}$. However, $b^{2} \\mid a^{3}$ gives $b \\leq a^{\\frac{3}{2}}$. This forces\n\n$$\na<2\\left(a^{\\frac{3}{2}}\\right)^{\\frac{1}{3}}=2 \\sqrt{a} \\Longrightarrow a<4 .\n$$\n\nExtracting $a=2,3$ by hand yields no additional solution.", "problem_match": "\nProblem 1.", "solution_match": "# Solution 1"}
+{"year": "2022", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be a right triangle with $\\angle B=90^{\\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\\triangle A C D$ and the circumcircle of $\\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.\n![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)", "solution": ".1\n\nLet the line $E F$ intersect the line $B C$ at $P$ and the circumcircle of $\\triangle A C D$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.\nFirst, notice that $\\triangle B E D$ is isosceles with $E B=E D$. This implies $\\angle E B C=\\angle E D P$.\nThen, $\\angle D A G=\\angle D F G=\\angle E B C=\\angle E D P$ which implies $A G \\| D C$. Hence, $A G C D$ is an isosceles trapezoid.\nAlso, $A G \\| D C$ and $A E=E D$. This implies $\\triangle A E G \\cong \\triangle D E P$ and $A G=D P$.\nSince $B$ is the foot of the perpendicular from $A$ onto the side $C D$ of the isosceles trapezoid $A G C D$, we have $P B=P D+D B=A G+D B=B C$, which does not depend on the choice of $D$. Hence, the initial statement is proven.", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}
+{"year": "2022", "problem_label": "2", "tier": 1, "problem": "Let $A B C$ be a right triangle with $\\angle B=90^{\\circ}$. Point $D$ lies on the line $C B$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $A D$ and let $F$ be the second intersection point of the circumcircle of $\\triangle A C D$ and the circumcircle of $\\triangle B D E$. Prove that as $D$ varies, the line $E F$ passes through a fixed point.\n![](https://cdn.mathpix.com/cropped/2024_11_22_59f4362abe92b4f1e49cg-2.jpg?height=909&width=1292&top_left_y=1492&top_left_x=449)", "solution": ".2\n\nSet up a coordinate system where $B C$ is along the positive $x$-axis, $B A$ is along the positive $y$-axis, and $B$ is the origin. Take $A=(0, a), B=(0,0), C=(c, 0), D=(-d, 0)$ where $a, b, c, d>0$. Then $E=\\left(-\\frac{d}{2}, \\frac{a}{2}\\right)$. The general equation of a circle is\n\n$$\nx^{2}+y^{2}+2 f x+2 g y+h=0\n$$\n\nSubstituting the coordinates of $A, D, C$ into (1) and solving for $f, g, h$, we find that the equation of the circumcircle of $\\triangle A D C$ is\n\n$$\nx^{2}+y^{2}+(d-c) x+\\left(\\frac{c d}{a}-a\\right) y-c d=0\n$$\n\nSimilarly, the equation of the circumcircle of $\\triangle B D E$ is\n\n$$\nx^{2}+y^{2}+d x+\\left(\\frac{d^{2}}{2 a}-\\frac{a}{2}\\right) y=0\n$$\n\nThen (3)-(2) gives the equation of the line $D F$ which is\n\n$$\nc x+\\frac{a^{2}+d^{2}-2 c d}{2 a} y+c d=0\n$$\n\nSolving (3) and (4) simultaneously, we get\n\n$$\nF=\\left(\\frac{c\\left(d^{2}-a^{2}-2 c d\\right)}{a^{2}+(d-2 c)^{2}}, \\frac{2 a c(c-d)}{a^{2}+(d-2 c)^{2}}\\right)\n$$\n\nand the other solution $D=(-d, 0)$.\nFrom this we obtain the equation of the line $E F$ which is $a x+(d-2 c) y+a c=0$. It passes through $P(-c, 0)$ which is independent of $d$.", "problem_match": "\nProblem 2.", "solution_match": "# Solution 2"}
+{"year": "2022", "problem_label": "3", "tier": 1, "problem": "Find all positive integers $k<202$ for which there exists a positive integer $n$ such that\n\n$$\n\\left\\{\\frac{n}{202}\\right\\}+\\left\\{\\frac{2 n}{202}\\right\\}+\\cdots+\\left\\{\\frac{k n}{202}\\right\\}=\\frac{k}{2}\n$$\n\nwhere $\\{x\\}$ denote the fractional part of $x$.\nNote: $\\{x\\}$ denotes the real number $k$ with $0 \\leq k<1$ such that $x-k$ is an integer.", "solution": "Denote the equation in the problem statement as $\\left(^{*}\\right)$, and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \\ldots, k n$ by 202 is 101 . Since $\\left\\{\\frac{i n}{202}\\right\\}$ is invariant in each residue class modulo 202 for each $1 \\leq i \\leq k$, it suffices to consider $0 \\leq n<202$.\n\nIf $n=0$, so is $\\left\\{\\frac{i n}{202}\\right\\}$, meaning that $(*)$ does not hold for any $k$. If $n=101$, then it can be checked that $\\left(^{*}\\right)$ is satisfied if and only if $k=1$. From now on, we will assume that $101 \\nmid n$.\nFor each $1 \\leq i \\leq k$, let $a_{i}=\\left\\lfloor\\frac{i n}{202}\\right\\rfloor=\\frac{i n}{202}-\\left\\{\\frac{i n}{202}\\right\\}$. Rewriting $\\left(^{*}\\right)$ and multiplying the equation by 202, we find that\n\n$$\nn(1+2+\\ldots+k)-202\\left(a_{1}+a_{2}+\\ldots+a_{k}\\right)=101 k\n$$\n\nEquivalently, letting $z=a_{1}+a_{2}+\\ldots+a_{k}$,\n\n$$\nn k(k+1)-404 z=202 k\n$$\n\nSince $n$ is not divisible by 101 , which is prime, it follows that $101 \\mid k(k+1)$. In particular, $101 \\mid k$ or $101 \\mid k+1$. This means that $k \\in\\{100,101,201\\}$. We claim that all these values of $k$ work.\n\n- If $k=201$, we may choose $n=1$. The remainders when dividing $n, 2 n, \\ldots, k n$ by 202 are 1,2 , ..., 201, which have an average of 101 .\n- If $k=100$, we may choose $n=2$. The remainders when dividing $n, 2 n, \\ldots, k n$ by 202 are 2,4 , ..., 200, which have an average of 101.\n- If $k=101$, we may choose $n=51$. To see this, note that the first four remainders are $51,102,153$, 2 , which have an average of 77 . The next four remainders $(53,104,155,4)$ are shifted upwards from the first four remainders by 2 each, and so on, until the 25 th set of the remainders ( 99 , $150,201,50)$ which have an average of 125 . Hence, the first 100 remainders have an average of $\\frac{77+125}{2}=101$. The 101th remainder is also 101 , meaning that the average of all 101 remainders is 101 .\n\nIn conclusion, all values $k \\in\\{1,100,101,201\\}$ satisfy the initial condition.", "problem_match": "\nProblem 3.", "solution_match": "# Solution\n\n"}
+{"year": "2022", "problem_label": "4", "tier": 1, "problem": "Let $n$ and $k$ be positive integers. Cathy is playing the following game. There are $n$ marbles and $k$ boxes, with the marbles labelled 1 to $n$. Initially, all marbles are placed inside one box. Each turn, Cathy chooses a box and then moves the marbles with the smallest label, say $i$, to either any empty box or the box containing marble $i+1$. Cathy wins if at any point there is a box containing only marble $n$.\nDetermine all pairs of integers $(n, k)$ such that Cathy can win this game.", "solution": "We claim Cathy can win if and only if $n \\leq 2^{k-1}$.\n\nFirst, note that each non-empty box always contains a consecutive sequence of labeled marbles. This is true since Cathy is always either removing from or placing in the lowest marble in a box. As a consequence, every move made is reversible.\n\nNext, we prove by induction that Cathy can win if $n=2^{k-1}$. The base case of $n=k=1$ is trivial. Assume a victory can be obtained for $m$ boxes and $2^{m-1}$ marbles. Consider the case of $m+1$ boxes and $2^{m}$ marbles. Cathy can first perform a sequence of moves so that only marbles $2^{m-1}, \\ldots, 2^{m}$ are left in the starting box, while keeping one box, say $B$, empty. Now move the marble $2^{m-1}$ to box $B$, then reverse all of the initial moves while treating $B$ as the starting box. At the end of that, we will have marbles $2^{m-1}+1, \\ldots, 2^{m}$ in the starting box, marbles $1,2, \\ldots, 2^{m-1}$ in box $B$, and $m-1$ empty boxes. By repeating the original sequence of moves on marbles $2^{m-1}+1, \\ldots, 2^{m}$, using the $m$ boxes that are not box $B$, we can reach a state where only marble $2^{m}$ remains in the starting box. Therefore\na victory is possible if $n=2^{k-1}$ or smaller.\n\nWe now prove by induction that Cathy loses if $n=2^{k-1}+1$. The base case of $n=2$ and $k=1$ is trivial. Assume a victory is impossible for $m$ boxes and $2^{m-1}+1$ marbles. For the sake of contradiction, suppose that victory is possible for $m+1$ boxes and $2^{m}+1$ marbles. In a winning sequence of moves, consider the last time a marble $2^{m-1}+1$ leaves the starting box, call this move $X$. After $X$, there cannot be a time when marbles $1, \\ldots, 2^{m-1}+1$ are all in the same box. Otherwise, by reversing these moves after $X$ and deleting marbles greater than $2^{m-1}+1$, it gives us a winning sequence of moves for $2^{m-1}+1$ marbles and $m$ boxes (as the original starting box is not used here), contradicting the inductive hypothesis. Hence starting from $X$, marbles 1 will never be in the same box as any marbles greater than or equal to $2^{m-1}+1$.\n\nNow delete marbles $2, \\ldots, 2^{m-1}$ and consider the winning moves starting from $X$. Marble 1 would only move from one empty box to another, while blocking other marbles from entering its box. Thus we effectively have a sequence of moves for $2^{m-1}+1$ marbles, while only able to use $m$ boxes. This again contradicts the inductive hypothesis. Therefore, a victory is not possible if $n=2^{k-1}+1$ or greater.", "problem_match": "\nProblem 4.", "solution_match": "# Solution\n\n"}
+{"year": "2022", "problem_label": "5", "tier": 1, "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": "The minimum value is $-\\frac{1}{8}$. There are eight equality cases in total. The first one is\n\n$$\n\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right) .\n$$\n\nCyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \\rightarrow$ $(-a,-b,-c,-d))$ all four entries in each of the four quadruples give four more equality cases.", "problem_match": "\nProblem 5.", "solution_match": "# Solution\n\n"}
+{"year": "2022", "problem_label": "5", "tier": 1, "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": ".1\n\nSince the expression is cyclic, we could WLOG $a=\\max \\{a, b, c, d\\}$. Let\n\n$$\nS(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)\n$$\n\nNote that we have given $(a, b, c, d)$ such that $S(a, b, c, d)=-\\frac{1}{8}$. Therefore, to prove that $S(a, b, c, d) \\geq$ $-\\frac{1}{8}$, we just need to consider the case where $S(a, b, c, d)<0$.\n\n- Exactly 1 of $a-b, b-c, c-d, d-a$ is negative.\n\nSince $a=\\max \\{a, b, c, d\\}$, then we must have $d-a<0$. This forces $a>b>c>d$. Now, let us write\n\n$$\nS(a, b, c, d)=-(a-b)(b-c)(c-d)(a-d)\n$$\n\nWrite $a-b=y, b-c=x, c-d=w$ for some positive reals $w, x, y>0$. Plugging to the original condition, we have\n\n$$\n(d+w+x+y)^{2}+(d+w+x)^{2}+(d+w)^{2}+d^{2}-1=0(*)\n$$\n\nand we want to prove that $w x y(w+x+y) \\leq \\frac{1}{8}$. Consider the expression $(*)$ as a quadratic in $d$ :\n\n$$\n4 d^{2}+d(6 w+4 x+2 y)+\\left((w+x+y)^{2}+(w+x)^{2}+w^{2}-1\\right)=0\n$$\n\nSince $d$ is a real number, then the discriminant of the given equation has to be non-negative, i.e. we must have\n\n$$\n\\begin{aligned}\n4 & \\geq 4\\left((w+x+y)^{2}+(w+x)^{2}+w^{2}\\right)-(3 w+2 x+y)^{2} \\\\\n& =\\left(3 w^{2}+2 w y+3 y^{2}\\right)+4 x(w+x+y) \\\\\n& \\geq 8 w y+4 x(w+x+y) \\\\\n& =4(x(w+x+y)+2 w y)\n\\end{aligned}\n$$\n\nHowever, AM-GM gives us\n\n$$\nw x y(w+x+y) \\leq \\frac{1}{2}\\left(\\frac{x(w+x+y)+2 w y}{2}\\right)^{2} \\leq \\frac{1}{8}\n$$\n\nThis proves $S(a, b, c, d) \\geq-\\frac{1}{8}$ for any $a, b, c, d \\in \\mathbb{R}$ such that $a>b>c>d$. Equality holds if and only if $w=y, x(w+x+y)=2 w y$ and $w x y(w+x+y)=\\frac{1}{8}$. Solving these equations gives us $w^{4}=\\frac{1}{16}$ which forces $w=\\frac{1}{2}$ since $w>0$. Solving for $x$ gives us $x(x+1)=\\frac{1}{2}$, and we will get $x=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}$ as $x>0$. Plugging back gives us $d=-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}$, and this gives us\n\n$$\n(a, b, c, d)=\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}\\right)\n$$\n\nThus, any cyclic permutation of the above solution will achieve the minimum equality.\n\n- Exactly 3 of $a-b, b-c, c-d, d-a$ are negative Since $a=\\max \\{a, b, c, d\\}$, then $a-b$ has to be positive. So we must have $b<c<d<a$. Now, note that\n\n$$\n\\begin{aligned}\nS(a, b, c, d) & =(a-b)(b-c)(c-d)(d-a) \\\\\n& =(a-d)(d-c)(c-b)(b-a) \\\\\n& =S(a, d, c, b)\n\\end{aligned}\n$$\n\nNow, note that $a>d>c>b$. By the previous case, $S(a, d, c, b) \\geq-\\frac{1}{8}$, which implies that\n\n$$\nS(a, b, c, d)=S(a, d, c, b) \\geq-\\frac{1}{8}\n$$\n\nas well. Equality holds if and only if\n\n$$\n(a, b, c, d)=\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right)\n$$\n\nand its cyclic permutation.", "problem_match": "\nProblem 5.", "solution_match": "# Solution 5"}
+{"year": "2022", "problem_label": "5", "tier": 1, "problem": "Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Determine the minimum value of $(a-b)(b-c)(c-d)(d-a)$ and determine all values of $(a, b, c, d)$ such that the minimum value is achieved.", "solution": ".2\n\nThe minimum value is $-\\frac{1}{8}$. There are eight equality cases in total. The first one is\n\n$$\n\\left(\\frac{1}{4}+\\frac{\\sqrt{3}}{4},-\\frac{1}{4}-\\frac{\\sqrt{3}}{4}, \\frac{1}{4}-\\frac{\\sqrt{3}}{4},-\\frac{1}{4}+\\frac{\\sqrt{3}}{4}\\right) .\n$$\n\nCyclic shifting all the entries give three more quadruples. Moreover, flipping the sign $((a, b, c, d) \\rightarrow$ $(-a,-b,-c,-d)$ ) all four entries in each of the four quadruples give four more equality cases. We then begin the proof by the following optimization:\n\nClaim 1. In order to get the minimum value, we must have $a+b+c+d=0$.\n\nProof. Assume not, let $\\delta=\\frac{a+b+c+d}{4}$ and note that\n\n$$\n(a-\\delta)^{2}+(b-\\delta)^{2}+(c-\\delta)^{2}+(d-\\delta)^{2}<a^{2}+b^{2}+c^{2}+d^{2}\n$$\n\nso by shifting by $\\delta$ and scaling, we get an even smaller value of $(a-b)(b-c)(c-d)(d-a)$.\n\nThe key idea is to substitute the variables\n\n$$\n\\begin{aligned}\n& x=a c+b d \\\\\n& y=a b+c d \\\\\n& z=a d+b c\n\\end{aligned}\n$$\n\nso that the original expression is just $(x-y)(x-z)$. We also have the conditions $x, y, z \\geq-0.5$ because of:\n\n$$\n2 x+\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=(a+c)^{2}+(b+d)^{2} \\geq 0\n$$\n\nMoreover, notice that\n\n$$\n0=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(x+y+z) \\Longrightarrow x+y+z=\\frac{-1}{2}\n$$\n\nNow, we reduce to the following optimization problem.\nClaim 2. Let $x, y, z \\geq-0.5$ such that $x+y+z=-0.5$. Then, the minimum value of\n\n$$\n(x-y)(x-z)\n$$\n\nis $-1 / 8$. Moreover, the equality case occurs when $x=-1 / 4$ and $\\{y, z\\}=\\{1 / 4,-1 / 2\\}$.\nProof. We notice that\n\n$$\n\\begin{aligned}\n(x-y)(x-z)+\\frac{1}{8} & =\\left(2 y+z+\\frac{1}{2}\\right)\\left(2 z+y+\\frac{1}{2}\\right)+\\frac{1}{8} \\\\\n& =\\frac{1}{8}(4 y+4 z+1)^{2}+\\left(y+\\frac{1}{2}\\right)\\left(z+\\frac{1}{2}\\right) \\geq 0\n\\end{aligned}\n$$\n\nThe last inequality is true since both $y+\\frac{1}{2}$ and $z+\\frac{1}{2}$ are not less than zero.\nThe equality in the last inequality is attained when either $y+\\frac{1}{2}=0$ or $z+\\frac{1}{2}=0$, and $4 y+4 z+1=0$. This system of equations give $(y, z)=(1 / 4,-1 / 2)$ or $(y, z)=(-1 / 2,1 / 4)$ as the desired equality cases.\n\nNote: We can also prove (the weakened) Claim 2 by using Lagrange Multiplier, as follows. We first prove that, in fact, $x, y, z \\in[-0.5,0.5]$. This can be proved by considering that\n\n$$\n-2 x+\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=(a-c)^{2}+(b-d)^{2} \\geq 0\n$$\n\nWe will prove the Claim 2, only that in this case, $x, y, z \\in[-0.5,0.5]$. This is already sufficient to prove the original question. We already have the bounded domain $[-0.5,0.5]^{3}$, so the global minimum must occur somewhere. Thus, it suffices to consider two cases:\n\n- If the global minimum lies on the boundary of $[-0.5,0.5]^{3}$. Then, one of $x, y, z$ must be -0.5 or 0.5 . By symmetry between $y$ and $z$, we split to a few more cases.\n- If $x=0.5$, then $y=z=-0.5$, so $(x-y)(x-z)=1$, not the minimum.\n- If $x=-0.5$, then both $y$ and $z$ must be greater or equal to $x$, so $(x-y)(x-z) \\geq 0$, not the minimum.\n- If $y=0.5$, then $x=z=-0.5$, so $(x-y)(x-z)=0$, not the minimum.\n- If $y=-0.5$, then $z=-x$, so\n\n$$\n(x-y)(x-z)=2 x(x+0.5)\n$$\n\nwhich obtain the minimum at $x=-1 / 4$. This gives the desired equality case.\n\n- If the global minimum lies in the interior $(-0.5,0.5)^{3}$, then we apply Lagrange multiplier:\n\n$$\n\\begin{aligned}\n& \\frac{\\partial}{\\partial x}(x-y)(x-z)=\\lambda \\frac{\\partial}{\\partial x}(x+y+z) \\\\\n& \\frac{\\partial}{\\partial y}(x-y)(x-z)=\\lambda \\frac{\\partial}{\\partial y}(x+y+z) \\\\\n& \\frac{\\partial}{\\partial z}(x-y)(x-z) \\Longrightarrow z-x=\\lambda \\frac{\\partial}{\\partial z}(x+y+z) \\\\\n& \\Longrightarrow y-x=\\lambda .\n\\end{aligned}\n$$\n\nAdding the last two equations gives $\\lambda=0$, or $x=y=z$. This gives $(x-y)(x-z)=0$, not the minimum.\n\nHaving exhausted all cases, we are done.", "problem_match": "\nProblem 5.", "solution_match": "# Solution 5"}

APMO/segmented/en-apmo2023_sol.jsonl

@@ -1,7 +1,7 @@
 {"year": "2023", "problem_label": "1", "tier": 1, "problem": "Let $n \\geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \\ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.\nShow that it is possible to arrange these squares in a way such that every square touches exactly two other squares.", "solution": "Set aside the squares with sidelengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$.\nString the squares of each set $A, B$ along two parallel diagonals, one for each diagonal. Now use the four largest squares along two perpendicular diagonals to finish the construction: one will have sidelengths $n$ and $n-3$, and the other, sidelengths $n-1$ and $n-2$. If the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-1$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration, in which the $A$ and $B$-diagonals are represented by unit squares, and the sidelengths $a_{i}$ of squares from $A$ and $b_{j}$ of squares from $B$ are indicated within each square, follows:\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-01.jpg?height=592&width=1346&top_left_y=1195&top_left_x=321)\n\nSince $\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\sqrt{2}+\\frac{((n-3)+(n-2)) \\sqrt{2}}{2}=\\left(b_{1}+b_{2}+\\cdots+b_{\\ell}+2\\right) \\sqrt{2}+\\frac{(n+(n-1)) \\sqrt{2}}{2}$, this case is done.\nIf the sum of the sidelengths of the squares in $A$ is 1 unit larger than the sum of the sidelengths of the squares in $B$, attach the squares with sidelengths $n-3$ and $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-01.jpg?height=621&width=1337&top_left_y=2094&top_left_x=328)\n\nSince $\\left(a_{1}+a_{2}+\\cdots+a_{k}\\right) \\sqrt{2}+\\frac{((n-3)+(n-1)) \\sqrt{2}}{2}=\\left(b_{1}+b_{2}+\\cdots+b_{\\ell}+1\\right) \\sqrt{2}+\\frac{(n+(n-2)) \\sqrt{2}}{2}$, this case is also done.\nIn both cases, the distance between the $A$-diagonal and the $B$-diagonal is $\\frac{((n-3)+n) \\sqrt{2}}{2}=\\frac{(2 n-3) \\sqrt{2}}{2}$. Since $a_{i}, b_{j} \\leq n-4, \\frac{\\left(a_{i}+b_{j}\\right) \\sqrt{2}}{2}<\\frac{(2 n-4) \\sqrt{2}}{2}<\\frac{(2 n-3) \\sqrt{2}}{2}$, and therefore the $A$ - and $B$-diagonals do not overlap.\nFinally, we prove that it is possible to split the squares of sidelengths 1 to $n-4$ into two sets $A$ and $B$ such that the sum of the sidelengths of the squares in $A$ is 1 or 2 units larger than the sum of the sidelengths of the squares in $B$. One can do that in several ways; we present two possibilities:\n\n- Direct construction: Split the numbers from 1 to $n-4$ into several sets of four consecutive numbers $\\{t, t+1, t+2, t+3\\}$, beginning with the largest numbers; put squares of sidelengths $t$ and $t+3$ in $A$ and squares of sidelengths $t+1$ and $t+2$ in $B$. Notice that $t+(t+3)=$ $(t+1)+(t+2)$. In the end, at most four numbers remain.\n- If only 1 remains, put the corresponding square in $A$, so the sum of the sidelengths of the squares in $A$ is one unit larger that those in $B$;\n- If 1 and 2 remains, put the square of sidelength 2 in $A$ and the square of sidelength 1 in $B$ (the difference is 1 );\n- If 1,2 , and 3 remains, put the squares of sidelengths 1 and 3 in $A$, and the square of sidelength 2 in $B$ (the difference is 2 );\n- If $1,2,3$, and 4 remains, put the squares of sidelengths 2 and 4 in $A$, and the squares of sidelengths 1 and 3 in $B$ (the difference is 2 ).\n- Indirect construction: Starting with $A$ and $B$ as empty sets, add the squares of sidelengths $n-4, n-3, \\ldots, 2$ to either $A$ or $B$ in that order such that at each stage the difference between the sum of the sidelengths in $A$ and the sum of the sidelengths of B is minimized. By induction it is clear that after adding an integer $j$ to one of the sets, this difference is at most $j$. In particular, the difference is 0,1 or 2 at the end. Finally adding the final 1 to one of the sets can ensure that the final difference is 1 or 2 . If necessary, flip $A$ and $B$.", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}
 {"year": "2023", "problem_label": "1", "tier": 1, "problem": "Let $n \\geq 5$ be an integer. Consider $n$ squares with side lengths $1,2, \\ldots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices.\nShow that it is possible to arrange these squares in a way such that every square touches exactly two other squares.", "solution": "Solve the problem by induction in $n$. Construct examples for $n=5,6,7,8,9,10$ (one can use the constructions from the previous solution, for instance). For $n>10$, set aside the six larger squares and arrange them in the following fashion:\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-02.jpg?height=815&width=1051&top_left_y=1948&top_left_x=471)\n\nBy the induction hypothesis, one can arrange the remaining $n-6$ squares away from the six larger squares, so we are done.", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}
-{"year": "2023", "problem_label": "2", "tier": 1, "problem": "Find all integers $n$ satisfying $n \\geq 2$ and $\\frac{\\sigma(n)}{p(n)-1}=n$, in which $\\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.\n\nAnswer: $n=6$.", "solution": "Let $n=p_{1}^{\\alpha_{1}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\\sigma(n)=\\left(1+p_{1}+\\cdots+p_{1}^{\\alpha_{1}}\\right) \\ldots\\left(1+p_{k}+\\cdots+p_{k}^{\\alpha_{k}}\\right)$. Hence\n$p_{k}-1=\\frac{\\sigma(n)}{n}=\\prod_{i=1}^{k}\\left(1+\\frac{1}{p_{i}}+\\cdots+\\frac{1}{p_{i}^{\\alpha_{i}}}\\right)<\\prod_{i=1}^{k} \\frac{1}{1-\\frac{1}{p_{i}}}=\\prod_{i=1}^{k}\\left(1+\\frac{1}{p_{i}-1}\\right) \\leq \\prod_{i=1}^{k}\\left(1+\\frac{1}{i}\\right)=k+1$,\nthat is, $p_{k}-1<k+1$, which is impossible for $k \\geq 3$, because in this case $p_{k}-1 \\geq 2 k-2 \\geq k+1$. Then $k \\leq 2$ and $p_{k}<k+2 \\leq 4$, which implies $p_{k} \\leq 3$.\nIf $k=1$ then $n=p^{\\alpha}$ and $\\sigma(n)=1+p+\\cdots+p^{\\alpha}$, and in this case $n \\nmid \\sigma(n)$, which is not possible. Thus $k=2$, and $n=2^{\\alpha} 3^{\\beta}$ with $\\alpha, \\beta>0$. If $\\alpha>1$ or $\\beta>1$,\n\n$$\n\\frac{\\sigma(n)}{n}>\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right)=2 .\n$$\n\nTherefore $\\alpha=\\beta=1$ and the only answer is $n=6$.\nComment: There are other ways to deal with the case $n=2^{\\alpha} 3^{\\beta}$. For instance, we have $2^{\\alpha+2} 3^{\\beta}=\\left(2^{\\alpha+1}-1\\right)\\left(3^{\\beta+1}-1\\right)$. Since $2^{\\alpha+1}-1$ is not divisible by 2 , and $3^{\\beta+1}-1$ is not divisible by 3 , we have\n\n$$\n\\left\\{\\begin{array} { l } \n{ 2 ^ { \\alpha + 1 } - 1 = 3 ^ { \\beta } } \\\\\n{ 3 ^ { \\beta + 1 } - 1 = 2 ^ { \\alpha + 2 } }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array} { c } \n{ 2 ^ { \\alpha + 1 } - 1 = 3 ^ { \\beta } } \\\\\n{ 3 \\cdot ( 2 ^ { \\alpha + 1 } - 1 ) - 1 = 2 \\cdot 2 ^ { \\alpha + 1 } }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array}{r}\n2^{\\alpha+1}=4 \\\\\n3^{\\beta}=3\n\\end{array}\\right.\\right.\\right.\n$$\n\nand $n=2^{\\alpha} 3^{\\beta}=6$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
-{"year": "2023", "problem_label": "3", "tier": 1, "problem": "Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram.", "solution": "Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $I_{3} I_{4}$ and $C D$ ) be $\\theta$. Then\n\n$$\nr_{2}-r_{1}=I_{1} I_{2} \\sin \\theta=I_{3} I_{4} \\sin \\theta=r_{4}-r_{3}\n$$\n\nwhich implies $r_{1}+r_{4}=r_{2}+r_{3}$. Similar arguments show that $r_{1}+r_{2}=r_{3}+r_{4}$. Thus we obtain $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-05.jpg?height=509&width=812&top_left_y=842&top_left_x=591)\n\nNow let's consider the possible positions of $W, X, Y, Z$. Suppose $A Z \\neq C X$. Without loss of generality assume $A Z>C X$. Since the incircles of $A W Z$ and $C Y X$ are symmetric about the centre of the parallelogram $A B C D$, this implies $C Y>A W$. Using similar arguments, we have\n\n$$\nC Y>A W \\Longrightarrow B W>D Y \\Longrightarrow D Z>B X \\Longrightarrow C X>A Z\n$$\n\nwhich is a contradiction. Therefore $A Z=C X \\Longrightarrow A W=C Y$ and $W X Y Z$ is a parallelogram.\nComment: There are several ways to prove that $r_{1}=r_{3}$ and $r_{2}=r_{4}$. The proposer shows the following three alternative approaches:\nUsing parallel lines: Let $O$ be the centre of parallelogram $A B C D$ and $P$ be the centre of parallelogram $I_{1} I_{2} I_{3} I_{4}$. Since $A I_{1}$ and $C I_{3}$ are angle bisectors, we must have $A I_{1} \\| C I_{3}$. Let $\\ell_{1}$ be the line through $O$ parallel to $A I_{1}$. Since $A O=O C, \\ell_{1}$ is halfway between $A I_{1}$ and $C I_{3}$. Hence $P$ must lie on $\\ell_{1}$.\nSimilarly, $P$ must also lie on $\\ell_{2}$, the line through $O$ parallel to $B I_{2}$. Thus $P$ is the intersection of $\\ell_{1}$ and $\\ell_{2}$, which must be $O$. So the four incentres and hence the four incircles must be symmetric about $O$, which implies $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\nUsing a rotation: Let the bisectors of $\\angle D A B$ and $\\angle A B C$ meet at $X$ and the bisectors of $\\angle B C D$ and $\\angle C D A$ meet at $Y$. Then $I_{1}$ is on $A X, I_{2}$ is on $B X, I_{3}$ is on $C Y$, and $I_{4}$ is on $D Y$. Let $O$ be the centre of $A B C D$. Then a 180 degree rotation about $O$ takes $\\triangle A X B$ to $\\triangle C Y D$. Under the same transformation $I_{1} I_{2}$ is mapped to a parallel segment $I_{1}^{\\prime} I_{2}^{\\prime}$ with $I_{1}^{\\prime}$ on $C Y$ and $I_{2}^{\\prime}$ on $D Y$. Since $I_{1} I_{2} I_{3} I_{4}$ is a parallelogram, $I_{3} I_{4}=I_{1} I_{2}$ and $I_{3} I_{4} \\| I_{1} I_{2}$. Hence $I_{1}^{\\prime} I_{2}^{\\prime}$ and $I_{3} I_{4}$ are parallel, equal length segments on sides $C Y, D Y$ and we conclude that $I_{1}^{\\prime}=I_{3}, I_{2}^{\\prime}=I_{4}$. Hence the centre of $I_{1} I_{2} I_{3} I_{4}$ is also $O$ and we establish that by rotational symmetry that $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\nUsing congruent triangles: Let $A I_{1}$ and $B I_{2}$ intersect at $E$ and let $C I_{3}$ and $D I_{4}$ intersect at $F$. Note that $\\triangle A B E$ and $\\triangle C D F$ are congruent, since $A B=C D$ and corresponding pairs of angles are equal (equal opposite angles parallelogram $A B C D$ are each bisected).\n\nSince $A I_{1} \\| C I_{3}$ and $I_{1} I_{2} \\| I_{4} I_{3}, \\angle I_{2} I_{1} E=\\angle I_{4} I_{3} F$. Similarly $\\angle I_{1} I_{2} E=\\angle I_{3} I_{4} F$. Furthermore $I_{1} I_{2}=I_{3} I_{4}$. Hence triangles $I_{2} I_{1} E$ and $I 4 I_{3} F$ are also congruent.\nHence $A B E I_{1} I_{2}$ and $D C F I_{3} I_{4}$ are congruent. Therefore, the perpendicular distance from $I_{1}$ to $A B$ equals the perpendicular distance from $I_{3}$ to $C D$, that is, $r_{1}=r_{3}$. Similarly $r_{2}=r_{4}$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
+{"year": "2023", "problem_label": "2", "tier": 1, "problem": "Find all integers $n$ satisfying $n \\geq 2$ and $\\frac{\\sigma(n)}{p(n)-1}=n$, in which $\\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.\n\nAnswer: $n=6$.", "solution": "Let $n=p_{1}^{\\alpha_{1}} \\cdot \\ldots \\cdot p_{k}^{\\alpha_{k}}$ be the prime factorization of $n$ with $p_{1}<\\ldots<p_{k}$, so that $p(n)=p_{k}$ and $\\sigma(n)=\\left(1+p_{1}+\\cdots+p_{1}^{\\alpha_{1}}\\right) \\ldots\\left(1+p_{k}+\\cdots+p_{k}^{\\alpha_{k}}\\right)$. Hence\n$p_{k}-1=\\frac{\\sigma(n)}{n}=\\prod_{i=1}^{k}\\left(1+\\frac{1}{p_{i}}+\\cdots+\\frac{1}{p_{i}^{\\alpha_{i}}}\\right)<\\prod_{i=1}^{k} \\frac{1}{1-\\frac{1}{p_{i}}}=\\prod_{i=1}^{k}\\left(1+\\frac{1}{p_{i}-1}\\right) \\leq \\prod_{i=1}^{k}\\left(1+\\frac{1}{i}\\right)=k+1$,\nthat is, $p_{k}-1<k+1$, which is impossible for $k \\geq 3$, because in this case $p_{k}-1 \\geq 2 k-2 \\geq k+1$. Then $k \\leq 2$ and $p_{k}<k+2 \\leq 4$, which implies $p_{k} \\leq 3$.\nIf $k=1$ then $n=p^{\\alpha}$ and $\\sigma(n)=1+p+\\cdots+p^{\\alpha}$, and in this case $n \\nmid \\sigma(n)$, which is not possible. Thus $k=2$, and $n=2^{\\alpha} 3^{\\beta}$ with $\\alpha, \\beta>0$. If $\\alpha>1$ or $\\beta>1$,\n\n$$\n\\frac{\\sigma(n)}{n}>\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right)=2 .\n$$\n\nTherefore $\\alpha=\\beta=1$ and the only answer is $n=6$.\nComment: There are other ways to deal with the case $n=2^{\\alpha} 3^{\\beta}$. For instance, we have $2^{\\alpha+2} 3^{\\beta}=\\left(2^{\\alpha+1}-1\\right)\\left(3^{\\beta+1}-1\\right)$. Since $2^{\\alpha+1}-1$ is not divisible by 2 , and $3^{\\beta+1}-1$ is not divisible by 3 , we have\n\n$$\n\\left\\{\\begin{array} { l } \n{ 2 ^ { \\alpha + 1 } - 1 = 3 ^ { \\beta } } \\\\\n{ 3 ^ { \\beta + 1 } - 1 = 2 ^ { \\alpha + 2 } }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array} { c } \n{ 2 ^ { \\alpha + 1 } - 1 = 3 ^ { \\beta } } \\\\\n{ 3 \\cdot ( 2 ^ { \\alpha + 1 } - 1 ) - 1 = 2 \\cdot 2 ^ { \\alpha + 1 } }\n\\end{array} \\Longleftrightarrow \\left\\{\\begin{array}{r}\n2^{\\alpha+1}=4 \\\\\n3^{\\beta}=3\n\\end{array}\\right.\\right.\\right.\n$$\n\nand $n=2^{\\alpha} 3^{\\beta}=6$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
+{"year": "2023", "problem_label": "3", "tier": 1, "problem": "Let $A B C D$ be a parallelogram. Let $W, X, Y$, and $Z$ be points on sides $A B, B C, C D$, and $D A$, respectively, such that the incenters of triangles $A W Z, B X W, C Y X$ and $D Z Y$ form a parallelogram. Prove that $W X Y Z$ is a parallelogram.", "solution": "Let the four incenters be $I_{1}, I_{2}, I_{3}$, and $I_{4}$ with inradii $r_{1}, r_{2}, r_{3}$, and $r_{4}$ respectively (in the order given in the question). Without loss of generality, let $I_{1}$ be closer to $A B$ than $I_{2}$. Let the acute angle between $I_{1} I_{2}$ and $A B$ (and hence also the angle between $I_{3} I_{4}$ and $C D$ ) be $\\theta$. Then\n\n$$\nr_{2}-r_{1}=I_{1} I_{2} \\sin \\theta=I_{3} I_{4} \\sin \\theta=r_{4}-r_{3}\n$$\n\nwhich implies $r_{1}+r_{4}=r_{2}+r_{3}$. Similar arguments show that $r_{1}+r_{2}=r_{3}+r_{4}$. Thus we obtain $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-05.jpg?height=509&width=812&top_left_y=842&top_left_x=591)\n\nNow let's consider the possible positions of $W, X, Y, Z$. Suppose $A Z \\neq C X$. Without loss of generality assume $A Z>C X$. Since the incircles of $A W Z$ and $C Y X$ are symmetric about the centre of the parallelogram $A B C D$, this implies $C Y>A W$. Using similar arguments, we have\n\n$$\nC Y>A W \\Longrightarrow B W>D Y \\Longrightarrow D Z>B X \\Longrightarrow C X>A Z\n$$\n\nwhich is a contradiction. Therefore $A Z=C X \\Longrightarrow A W=C Y$ and $W X Y Z$ is a parallelogram.\nComment: There are several ways to prove that $r_{1}=r_{3}$ and $r_{2}=r_{4}$. The proposer shows the following three alternative approaches:\nUsing parallel lines: Let $O$ be the centre of parallelogram $A B C D$ and $P$ be the centre of parallelogram $I_{1} I_{2} I_{3} I_{4}$. Since $A I_{1}$ and $C I_{3}$ are angle bisectors, we must have $A I_{1} \\| C I_{3}$. Let $\\ell_{1}$ be the line through $O$ parallel to $A I_{1}$. Since $A O=O C, \\ell_{1}$ is halfway between $A I_{1}$ and $C I_{3}$. Hence $P$ must lie on $\\ell_{1}$.\nSimilarly, $P$ must also lie on $\\ell_{2}$, the line through $O$ parallel to $B I_{2}$. Thus $P$ is the intersection of $\\ell_{1}$ and $\\ell_{2}$, which must be $O$. So the four incentres and hence the four incircles must be symmetric about $O$, which implies $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\nUsing a rotation: Let the bisectors of $\\angle D A B$ and $\\angle A B C$ meet at $X$ and the bisectors of $\\angle B C D$ and $\\angle C D A$ meet at $Y$. Then $I_{1}$ is on $A X, I_{2}$ is on $B X, I_{3}$ is on $C Y$, and $I_{4}$ is on $D Y$. Let $O$ be the centre of $A B C D$. Then a 180 degree rotation about $O$ takes $\\triangle A X B$ to $\\triangle C Y D$. Under the same transformation $I_{1} I_{2}$ is mapped to a parallel segment $I_{1}^{\\prime} I_{2}^{\\prime}$ with $I_{1}^{\\prime}$ on $C Y$ and $I_{2}^{\\prime}$ on $D Y$. Since $I_{1} I_{2} I_{3} I_{4}$ is a parallelogram, $I_{3} I_{4}=I_{1} I_{2}$ and $I_{3} I_{4} \\| I_{1} I_{2}$. Hence $I_{1}^{\\prime} I_{2}^{\\prime}$ and $I_{3} I_{4}$ are parallel, equal length segments on sides $C Y, D Y$ and we conclude that $I_{1}^{\\prime}=I_{3}, I_{2}^{\\prime}=I_{4}$. Hence the centre of $I_{1} I_{2} I_{3} I_{4}$ is also $O$ and we establish that by rotational symmetry that $r_{1}=r_{3}$ and $r_{2}=r_{4}$.\nUsing congruent triangles: Let $A I_{1}$ and $B I_{2}$ intersect at $E$ and let $C I_{3}$ and $D I_{4}$ intersect at $F$. Note that $\\triangle A B E$ and $\\triangle C D F$ are congruent, since $A B=C D$ and corresponding pairs of angles are equal (equal opposite angles parallelogram $A B C D$ are each bisected).\n\nSince $A I_{1} \\| C I_{3}$ and $I_{1} I_{2} \\| I_{4} I_{3}, \\angle I_{2} I_{1} E=\\angle I_{4} I_{3} F$. Similarly $\\angle I_{1} I_{2} E=\\angle I_{3} I_{4} F$. Furthermore $I_{1} I_{2}=I_{3} I_{4}$. Hence triangles $I_{2} I_{1} E$ and $I 4 I_{3} F$ are also congruent.\nHence $A B E I_{1} I_{2}$ and $D C F I_{3} I_{4}$ are congruent. Therefore, the perpendicular distance from $I_{1}$ to $A B$ equals the perpendicular distance from $I_{3}$ to $C D$, that is, $r_{1}=r_{3}$. Similarly $r_{2}=r_{4}$.", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
 {"year": "2023", "problem_label": "4", "tier": 1, "problem": "Let $c>0$ be a given positive real and $\\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ such that\n\n$$\nf((c+1) x+f(y))=f(x+2 y)+2 c x \\quad \\text { for all } x, y \\in \\mathbb{R}_{>0}\n$$\n\nAnswer: $f(x)=2 x$ for all $x>0$.", "solution": "We first prove that $f(x) \\geq 2 x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2 y$ for some positive $y$. Choose $x$ such that $f((c+1) x+f(y))$ and $f(x+2 y)$ cancel out, that is,\n\n$$\n(c+1) x+f(y)=x+2 y \\Longleftrightarrow x=\\frac{2 y-f(y)}{c}\n$$\n\nNotice that $x>0$ because $2 y-f(y)>0$. Then $2 c x=0$, which is not possible. This contradiction yields $f(y) \\geq 2 y$ for all $y>0$.\nNow suppose, again for the sake of contradiction, that $f(y)>2 y$ for some $y>0$. Define the following sequence: $a_{0}$ is an arbitrary real greater than $2 y$, and $f\\left(a_{n}\\right)=f\\left(a_{n-1}\\right)+2 c x$, so that\n\n$$\n\\left\\{\\begin{array}{r}\n(c+1) x+f(y)=a_{n} \\\\\nx+2 y=a_{n-1}\n\\end{array} \\Longleftrightarrow x=a_{n-1}-2 y \\quad \\text { and } \\quad a_{n}=(c+1)\\left(a_{n-1}-2 y\\right)+f(y) .\\right.\n$$\n\nIf $x=a_{n-1}-2 y>0$ then $a_{n}>f(y)>2 y$, so inductively all the substitutions make sense.\nFor the sake of simplicity, let $b_{n}=a_{n}-2 y$, so $b_{n}=(c+1) b_{n-1}+f(y)-2 y \\quad(*)$. Notice that $x=b_{n-1}$ in the former equation, so $f\\left(a_{n}\\right)=f\\left(a_{n-1}\\right)+2 c b_{n-1}$. Telescoping yields\n\n$$\nf\\left(a_{n}\\right)=f\\left(a_{0}\\right)+2 c \\sum_{i=0}^{n-1} b_{i} .\n$$\n\nOne can find $b_{n}$ from the recurrence equation $(*): b_{n}=\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)(c+1)^{n}-\\frac{f(y)-2 y}{c}$, and then\n\n$$\n\\begin{aligned}\nf\\left(a_{n}\\right) & =f\\left(a_{0}\\right)+2 c \\sum_{i=0}^{n-1}\\left(\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)(c+1)^{i}-\\frac{f(y)-2 y}{c}\\right) \\\\\n& =f\\left(a_{0}\\right)+2\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)\\left((c+1)^{n}-1\\right)-2 n(f(y)-2 y)\n\\end{aligned}\n$$\n\nSince $f\\left(a_{n}\\right) \\geq 2 a_{n}=2 b_{n}+4 y$,\n\n$$\n\\begin{aligned}\n& f\\left(a_{0}\\right)+2\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)\\left((c+1)^{n}-1\\right)-2 n(f(y)-2 y) \\geq 2 b_{n}+4 y \\\\\n= & 2\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)(c+1)^{n}-2 \\frac{f(y)-2 y}{c}\n\\end{aligned}\n$$\n\nwhich implies\n\n$$\nf\\left(a_{0}\\right)+2 \\frac{f(y)-2 y}{c} \\geq 2\\left(b_{0}+\\frac{f(y)-2 y}{c}\\right)+2 n(f(y)-2 y)\n$$\n\nwhich is not true for sufficiently large $n$.\nA contradiction is reached, and thus $f(y)=2 y$ for all $y>0$. It is immediate that this function satisfies the functional equation.", "problem_match": "# Problem 4", "solution_match": "# Solution 1"}
 {"year": "2023", "problem_label": "4", "tier": 1, "problem": "Let $c>0$ be a given positive real and $\\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ such that\n\n$$\nf((c+1) x+f(y))=f(x+2 y)+2 c x \\quad \\text { for all } x, y \\in \\mathbb{R}_{>0}\n$$\n\nAnswer: $f(x)=2 x$ for all $x>0$.", "solution": "After proving that $f(y) \\geq 2 y$ for all $y>0$, one can define $g(x)=f(x)-2 x, g: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{\\geq 0}$, and our goal is proving that $g(x)=0$ for all $x>0$. The problem is now rewritten as\n\n$$\n\\begin{aligned}\n& g((c+1) x+g(y)+2 y)+2((c+1) x+g(y)+2 y)=g(x+2 y)+2(x+2 y)+2 c x \\\\\n\\Longleftrightarrow & g((c+1) x+g(y)+2 y)+2 g(y)=g(x+2 y) .\n\\end{aligned}\n$$\n\nThis readily implies that $g(x+2 y) \\geq 2 g(y)$, which can be interpreted as $z>2 y \\Longrightarrow g(z) \\geq$ $2 g(y)$, by plugging $z=x+2 y$.\nNow we prove by induction that $z>2 y \\Longrightarrow g(z) \\geq 2 m \\cdot g(y)$ for any positive integer $2 m$. In fact, since $(c+1) x+g(y)+2 y>2 y, g((c+1) x+g(y)+2 y) \\geq 2 m \\cdot g(y)$, and by (??),\n\n$$\ng(x+2 y) \\geq 2 m \\cdot g(y)+2 g(y)=2(m+1) g(y)\n$$\n\nand we are done by plugging $z=x+2 y$ again.\nThe problem now is done: if $g(y)>0$ for some $y>0$, choose a fixed $z>2 y$ arbitrarily and and integer $m$ such that $m>\\frac{g(z)}{2 g(y)}$. Then $g(z)<2 m \\cdot g(y)$, contradiction.", "problem_match": "# Problem 4", "solution_match": "# Solution 2"}
 {"year": "2023", "problem_label": "5", "tier": 1, "problem": "There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows:\nFirst, he chooses an endpoint of each segment as a \"sink\". Then he places the present at the endpoint of the segment he is at. The present moves as follows:\n\n- If it is on a line segment, it moves towards the sink.\n- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.\n\nIf the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2 n-1$ friends.", "solution": "Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will prove that the friends eligible to receive presents are the ones on even-numbered intersection points.\nFirst part: at most $n$ friends can receive a present.\nThe solution relies on a well-known result: the $n$ lines determine regions inside the circle; then it is possible to paint the regions with two colors such that no regions with a common (line) boundary have the same color. The proof is an induction on $n$ : the fact immediately holds for $n=0$, and the induction step consists on taking away one line $\\ell$, painting the regions obtained with $n-1$ lines, drawing $\\ell$ again and flipping all colors on exactly one half plane determined by $\\ell$.\nNow consider the line starting on point 1. Color the regions in red and blue such that neighboring regions have different colors, and such that the two regions that have point 1 as a vertex are red on the right and blue on the left, from Tony's point of view. Finally, assign to each red region the clockwise direction and to each blue region the anticlockwise direction. Because of the coloring, every boundary will have two directions assigned, but the directions are the same since every boundary divides regions of different colors. Then the present will follow the directions assigned to the regions: it certainly does for both regions in the beginning, and when the present reaches an intersection it will keep bordering one of the two regions it was dividing. To finish this part of the problem, consider the regions that share a boundary with the circle. The directions alternate between outcoming and incoming, starting from 1 (outcoming), so all even-numbered vertices are directed as incoming and are the only ones able to receive presents. Second part: all even-numbered vertices can receive a present.\nFirst notice that, since every two chords intersect, every chord separates the endpoints of each of the other $n-1$ chords. Therefore, there are $n-1$ vertices on each side of every chord, and each chord connects vertices $k$ and $k+n, 1 \\leq k \\leq n$.\nWe prove a stronger result by induction in $n$ : let $k$ be an integer, $1 \\leq k \\leq n$. Direct each chord from $i$ to $i+n$ if $1 \\leq i \\leq k$ and from $i+n$ to $i$ otherwise; in other words, the sinks are $k+1, k+2, \\ldots, k+n$. Now suppose that each chord sends a present, starting from the vertex opposite to each sink, and all presents move with the same rules. Then $k-i$ sends a present to $k+i+1, i=0,1, \\ldots, n-1$ (indices taken modulo $2 n$ ). In particular, for $i=k-1$, Tony, in vertex 1 , send a present to vertex $2 k$. Also, the $n$ paths the presents make do not cross (but they may touch.) More formally, for all $i, 1 \\leq i \\leq n$, if one path takes a present from $k-i$ to $k+i+1$, separating the circle into two regions, all paths taking a present from $k-j$ to $k+j+1, j<i$, are completely contained in one region, and all paths taking a present from\n$k-j$ to $k+j+1, j>i$, are completely contained in the other region. For instance, possible ${ }^{1}$ paths for $k=3$ and $n=5$ follow:\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=541&width=552&top_left_y=289&top_left_x=706)\n\nThe result is true for $n=1$. Let $n>1$ and assume the result is true for less chords. Consider the chord that takes $k$ to $k+n$ and remove it. Apply the induction hypothesis to the remaining $n-1$ lines: after relabeling, presents would go from $k-i$ to $k+i+2,1 \\leq i \\leq n-1$ if the chord were not there.\nReintroduce the chord that takes $k$ to $k+n$. From the induction hypothesis, the chord intersects the paths of the presents in the following order: the $i$-th path the chord intersects is the the one that takes $k-i$ to $k+i, i=1,2, \\ldots, n-1$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=612&width=732&top_left_y=1259&top_left_x=248)\n\nPaths without chord $k \\rightarrow k+n$\n![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=612&width=732&top_left_y=1256&top_left_x=1025)\n\nCorrected paths with chord $k \\rightarrow k+n$\n\nThen the presents cover the following new paths: the present from $k$ will leave its chord and take the path towards $k+1$; then, for $i=1,2, \\ldots, n-1$, the present from $k-i$ will meet the chord from $k$ to $k+n$, move towards the intersection with the path towards $k+i+1$ and go to $k+i+1$, as desired. Notice that the paths still do not cross. The induction (and the solution) is now complete.", "problem_match": "# Problem 5", "solution_match": "# Solution 1"}

APMO/segmented/en-apmo2024_sol.jsonl

@@ -1,8 +1,8 @@
 {"year": "2024", "problem_label": "1", "tier": 1, "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=723&width=1003&top_left_y=678&top_left_x=492)\n\nLet $\\ell$ be the radical axis of circles $B Q X$ and $C P X$. Since $X$ and $Y$ are on $\\ell$, it is sufficient to show that $A$ is on $\\ell$. Let line $A X$ intersect segments $B C$ and $D E$ at $Z$ and $Z^{\\prime}$, respectively. Then it is sufficient to show that $Z$ is on $\\ell$. By $B C \\| D E$, we obtain\n\n$$\n\\frac{B Z}{Z C}=\\frac{D Z^{\\prime}}{Z^{\\prime} E}=\\frac{P Z}{Z Q},\n$$\n\nthus $B Z \\cdot Q Z=C Z \\cdot P Z$, which implies that $Z$ is on $\\ell$.", "problem_match": "# Problem 1", "solution_match": "# Solution 1"}
 {"year": "2024", "problem_label": "1", "tier": 1, "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=721&width=998&top_left_y=1995&top_left_x=492)\n\nLet circle $B Q X$ intersect line $A B$ at a point $S$ which is different from $B$. Then $\\angle D E X=$ $\\angle X Q C=\\angle B S X$, thus $S$ is on circle $D E X$. Similarly, let circle $C P X$ intersect line $A C$ at a point $T$ which is different from $C$. Then $T$ is on circle $D E X$. The power of $A$ with respect to the circle $D E X$ is $A S \\cdot A D=A T \\cdot A E$. Since $\\frac{A D}{A B}=\\frac{A E}{A C}, A S \\cdot A B=A T \\cdot A C$. Then $A$ is in the radical axis of circles $B Q X$ and $C P X$, which implies that three points $A, X$ and $Y$ are collinear.", "problem_match": "# Problem 1", "solution_match": "# Solution 2"}
 {"year": "2024", "problem_label": "1", "tier": 1, "problem": "Let $A B C$ be an acute triangle. Let $D$ be a point on side $A B$ and $E$ be a point on side $A C$ such that lines $B C$ and $D E$ are parallel. Let $X$ be an interior point of $B C E D$. Suppose rays $D X$ and $E X$ meet side $B C$ at points $P$ and $Q$, respectively such that both $P$ and $Q$ lie between $B$ and $C$. Suppose that the circumcircles of triangles $B Q X$ and $C P X$ intersect at a point $Y \\neq X$. Prove that points $A, X$, and $Y$ are collinear.", "solution": "Consider the (direct) homothety that takes triangle $A D E$ to triangle $A B C$, and let $Y^{\\prime}$ be the image of $Y$ under this homothety; in other words, let $Y^{\\prime}$ be the intersection of the line parallel to $B Y$ through $D$ and the line parallel to $C Y$ through $E$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-02.jpg?height=732&width=1008&top_left_y=682&top_left_x=484)\n\nThe homothety implies that $A, Y$, and $Y^{\\prime}$ are collinear, and that $\\angle D Y^{\\prime} E=\\angle B Y C$. Since $B Q X Y$ and $C P X Y$ are cyclic,\n$\\angle D Y^{\\prime} E=\\angle B Y C=\\angle B Y X+\\angle X Y C=\\angle X Q P+\\angle X P Q=180^{\\circ}-\\angle P X Q=180^{\\circ}-\\angle D X E$,\nwhich implies that $D Y^{\\prime} E X$ is cyclic. Therefore\n\n$$\n\\angle D Y^{\\prime} X=\\angle D E X=\\angle P Q X=\\angle B Y X\n$$\n\nwhich, combined with $D Y^{\\prime} \\| B Y$, implies $Y^{\\prime} X \\| Y X$. This proves that $X, Y$, and $Y^{\\prime}$ are collinear, which in turn shows that $A, X$, and $Y$ are collinear.", "problem_match": "# Problem 1", "solution_match": "# Solution 3"}
-{"year": "2024", "problem_label": "2", "tier": 1, "problem": "Consider a $100 \\times 100$ table, and identify the cell in row $a$ and column $b, 1 \\leq a, b \\leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \\leq k \\leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves from $(a, b)$ to $(c, d)$ such that $(|a-c|,|b-d|)$ is either $(1, k)$ or $(k, 1)$. The $k$-knight starts at cell $(1,1)$, and performs several moves. A sequence of moves is a sequence of cells $\\left(x_{0}, y_{0}\\right)=(1,1)$, $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right), \\ldots,\\left(x_{n}, y_{n}\\right)$ such that, for all $i=1,2, \\ldots, n, 1 \\leq x_{i}, y_{i} \\leq 100$ and the $k$-knight can move from $\\left(x_{i-1}, y_{i-1}\\right)$ to $\\left(x_{i}, y_{i}\\right)$. In this case, each cell $\\left(x_{i}, y_{i}\\right)$ is said to be reachable. For each $k$, find $L(k)$, the number of reachable cells.\n\nAnswer: $L(k)=\\left\\{\\begin{array}{ll}100^{2}-(2 k-100)^{2} & \\text { if } k \\text { is even } \\\\ \\frac{100^{2}-(2 k-100)^{2}}{2} & \\text { if } k \\text { is odd }\\end{array}\\right.$.", "solution": "Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \\geq 1$ or $x+k \\leq 100$ or $y-k \\geq 1$ or $y+k \\leq 100$, that is, $x \\geq k+1$ or $x \\leq 100-k$ or $y \\geq k+1$ or $y \\leq 100-k(*)$. Therefore the cells $(x, y)$ for which $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$ are unreachable. Let $S$ be this set of unreachable cells in this square, namely the square of cells $(x, y), 101-k \\leq x, y \\leq k$. If condition $(*)$ is valid for both $(x, y)$ and $(x \\pm 2, y \\pm 2)$ then one can move from $(x, y)$ to $(x \\pm 2, y \\pm 2)$, if they are both in the table, with two moves: either $x \\leq 50$ or $x \\geq 51$; the same is true for $y$. In the first case, move $(x, y) \\rightarrow(x+k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow$ $(x \\pm 1, y+k) \\rightarrow(x \\pm 2, y)$. In the second case, move $(x, y) \\rightarrow(x-k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow(x \\pm 1, y-k) \\rightarrow(x \\pm 2, y)$.\nHence if the table is colored in two colors like a chessboard, if $k \\leq 50$, cells with the same color as $(1,1)$ are reachable. Moreover, if $k$ is even, every other move changes the color of the occupied cell, and all cells are potentially reachable; otherwise, only cells with the same color as $(1,1)$ can be visited. Therefore, if $k$ is even then the reachable cells consists of all cells except the center square defined by $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$, that is, $L(k)=100^{2}-(2 k-100)^{2}$; if $k$ is odd, then only half of the cells are reachable: the ones with the same color as $(1,1)$, and $L(k)=\\frac{1}{2}\\left(100^{2}-(2 k-100)^{2}\\right)$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n"}
-{"year": "2024", "problem_label": "3", "tier": 1, "problem": "Let $n$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers. Prove that\n\n$$\n\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}} \\geq \\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}-\\frac{1}{2^{n}}\n$$", "solution": "We first prove the following lemma:\nLemma 1. For $k$ positive integer and $x, y>0$,\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$\n\nThe proof goes by induction. For $k=1$, we have\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2}+\\left(\\frac{2}{1+y}\\right)^{2} \\geq 2\\left(\\frac{2}{1+x y}\\right)\n$$\n\nwhich reduces to\n\n$$\nx y(x-y)^{2}+(x y-1)^{2} \\geq 0 .\n$$\n\nFor $k>1$, by the inequality $2\\left(A^{2}+B^{2}\\right) \\geq(A+B)^{2}$ applied at $A=\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}$ and $B=\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}$ followed by the induction hypothesis\n\n$$\n\\begin{aligned}\n2\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}}\\right) & \\geq\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}+\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}\\right)^{2} \\\\\n& \\geq\\left(2\\left(\\frac{2}{1+x y}\\right)^{2^{k-2}}\\right)^{2}=4\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n\\end{aligned}\n$$\n\nfrom which the lemma follows.\nThe problem now can be deduced from summing the following applications of the lemma, multiplied by the appropriate factor:\n\n$$\n\\begin{aligned}\n\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}} & \\geq \\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n} \\cdot 1}\\right)^{2^{n-1}} \\\\\n\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n-1}} & \\geq \\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} \\\\\n\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-2}}\\right)^{2^{n-2}}+\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} & \\geq \\frac{1}{2^{n-3}}\\left(\\frac{2}{1+a_{n-2} a_{n-1} a_{n}}\\right)^{2^{n-3}} \\\\\n\\ldots & )^{2^{k}} \\\\\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k}}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k+1} \\ldots a_{n-1} a_{n}}\\right)^{2^{k-1}} & \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+a_{k} \\ldots a_{n-2} a_{n-1} a_{n}}\\right)^{2} \\\\\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\frac{1}{2}\\left(\\frac{2}{1+a_{2} \\ldots a_{n-1} a_{n}}\\right)^{2} & \\geq \\frac{2}{1+a_{1} \\ldots a_{n-2} a_{n-1} a_{n}}\n\\end{aligned}\n$$\n\nComment: Equality occurs if and only if $a_{1}=a_{2}=\\cdots=a_{n}=1$.\n\nComment: The main motivation for the lemma is trying to \"telescope\" the sum\n\n$$\n\\frac{1}{2^{n}}+\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}}\n$$\n\nthat is,\n\n$$\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\cdots+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}}\n$$\n\nto obtain an expression larger than or equal to\n\n$$\n\\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}\n$$\n\nIt seems reasonable to obtain a inequality that can be applied from right to left, decreases the exponent of the factor $1 / 2^{k}$ by 1 , and multiplies the variables in the denominator. Given that, the lemma is quite natural:\n\n$$\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+x y}\\right)^{2^{i-1}}\n$$\n\nor\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$", "problem_match": "# Problem 3", "solution_match": "# Solution\n"}
-{"year": "2024", "problem_label": "4", "tier": 1, "problem": "Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \\ldots, a_{t-1}$ of $0,1, \\ldots, t-$ 1 such that, for every $0 \\leq i \\leq t-1$, the binomial coefficient $\\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \\neq t+i$.", "solution": "We constantly make use of Kummer's theorem which, in particular, implies that $\\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $i),\\binom{n}{k}$ is odd if and only if $S(k) \\subseteq S(n)$. Moreover, if we set $k<n, S(k)$ is a proper subset of $S(n)$, that is, $|S(k)|<|S(n)|$.\nWe start with a lemma that guides us how the permutation should be set.\n\n## Lemma 1.\n\n$$\n\\sum_{i=0}^{t-1}|S(t+i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n$$\n\nThe proof is just realizing that $S(2 i)=\\{1+x, x \\in S(i)\\}$ and $S(2 i+1)=\\{0\\} \\cup\\{1+x, x \\in S(i)\\}$, because $2 i$ in binary is $i$ followed by a zero and $2 i+1$ in binary is $i$ followed by a one. Therefore\n\n$$\n\\begin{aligned}\n\\sum_{i=0}^{t-1}|S(t+i)| & =\\sum_{i=0}^{2 t-1}|S(i)|-\\sum_{i=0}^{t-1}|S(i)|=\\sum_{i=0}^{t-1}|S(2 i)|+\\sum_{i=0}^{t-1}|S(2 i+1)|-\\sum_{i=0}^{t-1}|S(i)| \\\\\n& =\\sum_{i=0}^{t-1}|S(i)|+\\sum_{i=0}^{t-1}(1+|S(i)|)-\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n\\end{aligned}\n$$\n\nThe lemma has an immediate corollary: since $t+i>2 a_{i}$ and $\\binom{t+i}{2 a_{i}}$ is odd for all $i, 0 \\leq i \\leq t-1$, $S\\left(2 a_{i}\\right) \\subset S(t+i)$ with $\\left|S\\left(2 a_{i}\\right)\\right| \\leq|S(t+i)|-1$. Since the sum of $\\left|S\\left(2 a_{i}\\right)\\right|$ is $t$ less than the sum of $|S(t+i)|$, and there are $t$ values of $i$, equality must occur, that is, $\\left|S\\left(2 a_{i}\\right)\\right|=|S(t+i)|-1$, which in conjunction with $S\\left(2 a_{i}\\right) \\subset S(t+i)$ means that $t+i-2 a_{i}=2^{k_{i}}$ for every $i, 0 \\leq i \\leq t-1$, $k_{i} \\in S(t+i)$ (more precisely, $\\left\\{k_{i}\\right\\}=S(t+i) \\backslash S\\left(2 a_{i}\\right)$.)\nIn particular, for $t+i$ odd, this means that $t+i-2 a_{i}=1$, because the only odd power of 2 is 1. Then $a_{i}=\\frac{t+i-1}{2}$ for $t+i$ odd, which takes up all the numbers greater than or equal to $\\frac{t-1}{2}$. Now we need to distribute the numbers that are smaller than $\\frac{t-1}{2}$ (call these numbers small). If $t+i$ is even then by Lucas' Theorem $\\binom{t+i}{2 a_{i}} \\equiv\\binom{\\frac{t+i}{2}}{a_{i}}(\\bmod 2)$, so we pair numbers from $\\lceil t / 2\\rceil$ to $t-1$ (call these numbers big) with the small numbers.\nSay that a set $A$ is paired with another set $B$ whenever $|A|=|B|$ and there exists a bijection $\\pi: A \\rightarrow B$ such that $S(a) \\subset S(\\pi(a))$ and $|S(a)|=|S(\\pi(a))|-1$; we also say that $a$ and $\\pi(a)$ are paired. We prove by induction in $t$ that $A_{t}=\\{0,1,2, \\ldots,\\lfloor t / 2\\rfloor-1\\}$ (the set of small numbers) and $B_{t}=\\{\\lceil t / 2\\rceil, \\ldots, t-2, t-1\\}$ (the set of big numbers) can be uniquely paired.\nThe claim is immediate for $t=1$ and $t=2$. For $t>2$, there is exactly one power of two in $B_{t}$, since $t / 2 \\leq 2^{a}<t \\Longleftrightarrow a=\\left\\lceil\\log _{2}(t / 2)\\right\\rceil$. Let $2^{a}$ be this power of two. Then, since $2^{a} \\geq t / 2$, no number in $A_{t}$ has a one in position $a$ in binary. Since for every number $x, 2^{a} \\leq x<t, a \\in S(x)$ and $a \\notin S(y)$ for all $y \\in A_{t}, x$ can only be paired with $x-2^{a}$, since $S(x)$ needs to be stripped of exactly one position. This takes cares of $x \\in B_{t}, 2^{a} \\leq x<t$, and $y \\in A_{t}, 0 \\leq y<t-2^{a}$.\nNow we need to pair the numbers from $A^{\\prime}=\\left\\{t-2^{a}, t-2^{a}+1, \\ldots,\\lfloor t / 2\\rfloor-1\\right\\} \\subset A$ with the numbers from $B^{\\prime}=\\left\\{\\lceil t / 2\\rceil,\\lceil t / 2\\rceil+1, \\ldots, 2^{a}-1\\right\\} \\subset B$. In order to pair these $t-2\\left(t-2^{a}\\right)=$ $2^{a+1}-t<t$ numbers, we use the induction hypothesis and a bijection between $A^{\\prime} \\cup B^{\\prime}$ and $B_{2^{a+1}-t} \\cup A_{2^{a+1}-t}$. Let $S=S\\left(2^{a}-1\\right)=\\{0,1,2, \\ldots, a-1\\}$. Then take a pair $x, y, x \\in A_{2^{a+1}-t}$ and $y \\in B_{2^{a+1}-t}$ and biject it with $2^{a}-1-x \\in B^{\\prime}$ and $2^{a}-1-y \\in A^{\\prime}$. In fact,\n\n$$\n0 \\leq x \\leq\\left\\lfloor\\frac{2^{a+1}-t}{2}\\right\\rfloor-1=2^{a}-\\left\\lceil\\frac{t}{2}\\right\\rceil-1 \\Longleftrightarrow\\left\\lceil\\frac{t}{2}\\right\\rceil \\leq 2^{a}-1-x \\leq 2^{a}-1\n$$\n\nand\n\n$$\n\\left\\lceil\\frac{2^{a+1}-t}{2}\\right\\rceil=2^{a}-\\left\\lfloor\\frac{t}{2}\\right\\rfloor \\leq y \\leq 2^{a+1}-t-1 \\Longleftrightarrow t-2^{a} \\leq 2^{a}-1-y \\leq\\left\\lfloor\\frac{t}{2}\\right\\rfloor-1\n$$\n\nMoreover, $S\\left(2^{a}-1-x\\right)=S \\backslash S(x)$ and $S\\left(2^{a}-1-y\\right)=S \\backslash S(y)$ are complements with respect to $S$, and $S(x) \\subset S(y)$ and $|S(x)|=|S(y)|-1$ implies $S\\left(2^{a}-1-y\\right) \\subset S\\left(2^{a}-1-x\\right)$ and $\\left|S\\left(2^{a}-1-y\\right)\\right|=\\left|S\\left(2^{a}-1-x\\right)\\right|-1$. Therefore a pairing between $A^{\\prime}$ and $B^{\\prime}$ corresponds to a pairing between $A_{2^{a+1}-t}$ and $B_{2^{a+1}-t}$. Since the latter pairing is unique, the former pairing is also unique, and the result follows.\nWe illustrate the bijection by showing the case $t=23$ :\n\n$$\nA_{23}=\\{0,1,2, \\ldots, 10\\}, \\quad B_{23}=\\{12,13,14, \\ldots, 22\\}\n$$\n\nThe pairing is\n\n$$\n\\left(\\begin{array}{ccccccccccc}\n12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\\\\n8 & 9 & 10 & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6\n\\end{array}\\right)\n$$\n\nin which the bijection is between\n\n$$\n\\left(\\begin{array}{cccc}\n12 & 13 & 14 & 15 \\\\\n8 & 9 & 10 & 7\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(\\begin{array}{llll}\n3 & 2 & 1 & 0 \\\\\n7 & 6 & 5 & 8\n\\end{array}\\right) \\rightarrow\\left(\\begin{array}{llll}\n5 & 6 & 7 & 8 \\\\\n1 & 2 & 3 & 0\n\\end{array}\\right) .\n$$", "problem_match": "# Problem 4", "solution_match": "# Solution\n"}
+{"year": "2024", "problem_label": "2", "tier": 1, "problem": "Consider a $100 \\times 100$ table, and identify the cell in row $a$ and column $b, 1 \\leq a, b \\leq 100$, with the ordered pair $(a, b)$. Let $k$ be an integer such that $51 \\leq k \\leq 99$. A $k$-knight is a piece that moves one cell vertically or horizontally and $k$ cells to the other direction; that is, it moves from $(a, b)$ to $(c, d)$ such that $(|a-c|,|b-d|)$ is either $(1, k)$ or $(k, 1)$. The $k$-knight starts at cell $(1,1)$, and performs several moves. A sequence of moves is a sequence of cells $\\left(x_{0}, y_{0}\\right)=(1,1)$, $\\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right), \\ldots,\\left(x_{n}, y_{n}\\right)$ such that, for all $i=1,2, \\ldots, n, 1 \\leq x_{i}, y_{i} \\leq 100$ and the $k$-knight can move from $\\left(x_{i-1}, y_{i-1}\\right)$ to $\\left(x_{i}, y_{i}\\right)$. In this case, each cell $\\left(x_{i}, y_{i}\\right)$ is said to be reachable. For each $k$, find $L(k)$, the number of reachable cells.\n\nAnswer: $L(k)=\\left\\{\\begin{array}{ll}100^{2}-(2 k-100)^{2} & \\text { if } k \\text { is even } \\\\ \\frac{100^{2}-(2 k-100)^{2}}{2} & \\text { if } k \\text { is odd }\\end{array}\\right.$.", "solution": "Cell $(x, y)$ is directly reachable from another cell if and only if $x-k \\geq 1$ or $x+k \\leq 100$ or $y-k \\geq 1$ or $y+k \\leq 100$, that is, $x \\geq k+1$ or $x \\leq 100-k$ or $y \\geq k+1$ or $y \\leq 100-k(*)$. Therefore the cells $(x, y)$ for which $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$ are unreachable. Let $S$ be this set of unreachable cells in this square, namely the square of cells $(x, y), 101-k \\leq x, y \\leq k$. If condition $(*)$ is valid for both $(x, y)$ and $(x \\pm 2, y \\pm 2)$ then one can move from $(x, y)$ to $(x \\pm 2, y \\pm 2)$, if they are both in the table, with two moves: either $x \\leq 50$ or $x \\geq 51$; the same is true for $y$. In the first case, move $(x, y) \\rightarrow(x+k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow$ $(x \\pm 1, y+k) \\rightarrow(x \\pm 2, y)$. In the second case, move $(x, y) \\rightarrow(x-k, y \\pm 1) \\rightarrow(x, y \\pm 2)$ or $(x, y) \\rightarrow(x \\pm 1, y-k) \\rightarrow(x \\pm 2, y)$.\nHence if the table is colored in two colors like a chessboard, if $k \\leq 50$, cells with the same color as $(1,1)$ are reachable. Moreover, if $k$ is even, every other move changes the color of the occupied cell, and all cells are potentially reachable; otherwise, only cells with the same color as $(1,1)$ can be visited. Therefore, if $k$ is even then the reachable cells consists of all cells except the center square defined by $101-k \\leq x \\leq k$ and $101-k \\leq y \\leq k$, that is, $L(k)=100^{2}-(2 k-100)^{2}$; if $k$ is odd, then only half of the cells are reachable: the ones with the same color as $(1,1)$, and $L(k)=\\frac{1}{2}\\left(100^{2}-(2 k-100)^{2}\\right)$.", "problem_match": "# Problem 2", "solution_match": "# Solution\n\n"}
+{"year": "2024", "problem_label": "3", "tier": 1, "problem": "Let $n$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers. Prove that\n\n$$\n\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}} \\geq \\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}-\\frac{1}{2^{n}}\n$$", "solution": "We first prove the following lemma:\nLemma 1. For $k$ positive integer and $x, y>0$,\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$\n\nThe proof goes by induction. For $k=1$, we have\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2}+\\left(\\frac{2}{1+y}\\right)^{2} \\geq 2\\left(\\frac{2}{1+x y}\\right)\n$$\n\nwhich reduces to\n\n$$\nx y(x-y)^{2}+(x y-1)^{2} \\geq 0 .\n$$\n\nFor $k>1$, by the inequality $2\\left(A^{2}+B^{2}\\right) \\geq(A+B)^{2}$ applied at $A=\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}$ and $B=\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}$ followed by the induction hypothesis\n\n$$\n\\begin{aligned}\n2\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}}\\right) & \\geq\\left(\\left(\\frac{2}{1+x}\\right)^{2^{k-1}}+\\left(\\frac{2}{1+y}\\right)^{2^{k-1}}\\right)^{2} \\\\\n& \\geq\\left(2\\left(\\frac{2}{1+x y}\\right)^{2^{k-2}}\\right)^{2}=4\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n\\end{aligned}\n$$\n\nfrom which the lemma follows.\nThe problem now can be deduced from summing the following applications of the lemma, multiplied by the appropriate factor:\n\n$$\n\\begin{aligned}\n\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}} & \\geq \\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n} \\cdot 1}\\right)^{2^{n-1}} \\\\\n\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n-1}} & \\geq \\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} \\\\\n\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-2}}\\right)^{2^{n-2}}+\\frac{1}{2^{n-2}}\\left(\\frac{2}{1+a_{n-1} a_{n}}\\right)^{2^{n-2}} & \\geq \\frac{1}{2^{n-3}}\\left(\\frac{2}{1+a_{n-2} a_{n-1} a_{n}}\\right)^{2^{n-3}} \\\\\n\\ldots & )^{2^{k}} \\\\\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k}}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+a_{k+1} \\ldots a_{n-1} a_{n}}\\right)^{2^{k-1}} & \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+a_{k} \\ldots a_{n-2} a_{n-1} a_{n}}\\right)^{2} \\\\\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\frac{1}{2}\\left(\\frac{2}{1+a_{2} \\ldots a_{n-1} a_{n}}\\right)^{2} & \\geq \\frac{2}{1+a_{1} \\ldots a_{n-2} a_{n-1} a_{n}}\n\\end{aligned}\n$$\n\nComment: Equality occurs if and only if $a_{1}=a_{2}=\\cdots=a_{n}=1$.\n\nComment: The main motivation for the lemma is trying to \"telescope\" the sum\n\n$$\n\\frac{1}{2^{n}}+\\sum_{i=1}^{n} \\frac{1}{2^{i}}\\left(\\frac{2}{1+a_{i}}\\right)^{2^{i}}\n$$\n\nthat is,\n\n$$\n\\frac{1}{2}\\left(\\frac{2}{1+a_{1}}\\right)^{2}+\\cdots+\\frac{1}{2^{n-1}}\\left(\\frac{2}{1+a_{n-1}}\\right)^{2^{n-1}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+a_{n}}\\right)^{2^{n}}+\\frac{1}{2^{n}}\\left(\\frac{2}{1+1}\\right)^{2^{n}}\n$$\n\nto obtain an expression larger than or equal to\n\n$$\n\\frac{2}{1+a_{1} a_{2} \\ldots a_{n}}\n$$\n\nIt seems reasonable to obtain a inequality that can be applied from right to left, decreases the exponent of the factor $1 / 2^{k}$ by 1 , and multiplies the variables in the denominator. Given that, the lemma is quite natural:\n\n$$\n\\frac{1}{2^{k}}\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\frac{1}{2^{k}}\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq \\frac{1}{2^{k-1}}\\left(\\frac{2}{1+x y}\\right)^{2^{i-1}}\n$$\n\nor\n\n$$\n\\left(\\frac{2}{1+x}\\right)^{2^{k}}+\\left(\\frac{2}{1+y}\\right)^{2^{k}} \\geq 2\\left(\\frac{2}{1+x y}\\right)^{2^{k-1}}\n$$", "problem_match": "# Problem 3", "solution_match": "# Solution\n\n"}
+{"year": "2024", "problem_label": "4", "tier": 1, "problem": "Prove that for every positive integer $t$ there is a unique permutation $a_{0}, a_{1}, \\ldots, a_{t-1}$ of $0,1, \\ldots, t-$ 1 such that, for every $0 \\leq i \\leq t-1$, the binomial coefficient $\\binom{t+i}{2 a_{i}}$ is odd and $2 a_{i} \\neq t+i$.", "solution": "We constantly make use of Kummer's theorem which, in particular, implies that $\\binom{n}{k}$ is odd if and only if $k$ and $n-k$ have ones in different positions in binary. In other words, if $S(x)$ is the set of positions of the digits 1 of $x$ in binary (in which the digit multiplied by $2^{i}$ is in position $i),\\binom{n}{k}$ is odd if and only if $S(k) \\subseteq S(n)$. Moreover, if we set $k<n, S(k)$ is a proper subset of $S(n)$, that is, $|S(k)|<|S(n)|$.\nWe start with a lemma that guides us how the permutation should be set.\n\n## Lemma 1.\n\n$$\n\\sum_{i=0}^{t-1}|S(t+i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n$$\n\nThe proof is just realizing that $S(2 i)=\\{1+x, x \\in S(i)\\}$ and $S(2 i+1)=\\{0\\} \\cup\\{1+x, x \\in S(i)\\}$, because $2 i$ in binary is $i$ followed by a zero and $2 i+1$ in binary is $i$ followed by a one. Therefore\n\n$$\n\\begin{aligned}\n\\sum_{i=0}^{t-1}|S(t+i)| & =\\sum_{i=0}^{2 t-1}|S(i)|-\\sum_{i=0}^{t-1}|S(i)|=\\sum_{i=0}^{t-1}|S(2 i)|+\\sum_{i=0}^{t-1}|S(2 i+1)|-\\sum_{i=0}^{t-1}|S(i)| \\\\\n& =\\sum_{i=0}^{t-1}|S(i)|+\\sum_{i=0}^{t-1}(1+|S(i)|)-\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(i)|=t+\\sum_{i=0}^{t-1}|S(2 i)|\n\\end{aligned}\n$$\n\nThe lemma has an immediate corollary: since $t+i>2 a_{i}$ and $\\binom{t+i}{2 a_{i}}$ is odd for all $i, 0 \\leq i \\leq t-1$, $S\\left(2 a_{i}\\right) \\subset S(t+i)$ with $\\left|S\\left(2 a_{i}\\right)\\right| \\leq|S(t+i)|-1$. Since the sum of $\\left|S\\left(2 a_{i}\\right)\\right|$ is $t$ less than the sum of $|S(t+i)|$, and there are $t$ values of $i$, equality must occur, that is, $\\left|S\\left(2 a_{i}\\right)\\right|=|S(t+i)|-1$, which in conjunction with $S\\left(2 a_{i}\\right) \\subset S(t+i)$ means that $t+i-2 a_{i}=2^{k_{i}}$ for every $i, 0 \\leq i \\leq t-1$, $k_{i} \\in S(t+i)$ (more precisely, $\\left\\{k_{i}\\right\\}=S(t+i) \\backslash S\\left(2 a_{i}\\right)$.)\nIn particular, for $t+i$ odd, this means that $t+i-2 a_{i}=1$, because the only odd power of 2 is 1. Then $a_{i}=\\frac{t+i-1}{2}$ for $t+i$ odd, which takes up all the numbers greater than or equal to $\\frac{t-1}{2}$. Now we need to distribute the numbers that are smaller than $\\frac{t-1}{2}$ (call these numbers small). If $t+i$ is even then by Lucas' Theorem $\\binom{t+i}{2 a_{i}} \\equiv\\binom{\\frac{t+i}{2}}{a_{i}}(\\bmod 2)$, so we pair numbers from $\\lceil t / 2\\rceil$ to $t-1$ (call these numbers big) with the small numbers.\nSay that a set $A$ is paired with another set $B$ whenever $|A|=|B|$ and there exists a bijection $\\pi: A \\rightarrow B$ such that $S(a) \\subset S(\\pi(a))$ and $|S(a)|=|S(\\pi(a))|-1$; we also say that $a$ and $\\pi(a)$ are paired. We prove by induction in $t$ that $A_{t}=\\{0,1,2, \\ldots,\\lfloor t / 2\\rfloor-1\\}$ (the set of small numbers) and $B_{t}=\\{\\lceil t / 2\\rceil, \\ldots, t-2, t-1\\}$ (the set of big numbers) can be uniquely paired.\nThe claim is immediate for $t=1$ and $t=2$. For $t>2$, there is exactly one power of two in $B_{t}$, since $t / 2 \\leq 2^{a}<t \\Longleftrightarrow a=\\left\\lceil\\log _{2}(t / 2)\\right\\rceil$. Let $2^{a}$ be this power of two. Then, since $2^{a} \\geq t / 2$, no number in $A_{t}$ has a one in position $a$ in binary. Since for every number $x, 2^{a} \\leq x<t, a \\in S(x)$ and $a \\notin S(y)$ for all $y \\in A_{t}, x$ can only be paired with $x-2^{a}$, since $S(x)$ needs to be stripped of exactly one position. This takes cares of $x \\in B_{t}, 2^{a} \\leq x<t$, and $y \\in A_{t}, 0 \\leq y<t-2^{a}$.\nNow we need to pair the numbers from $A^{\\prime}=\\left\\{t-2^{a}, t-2^{a}+1, \\ldots,\\lfloor t / 2\\rfloor-1\\right\\} \\subset A$ with the numbers from $B^{\\prime}=\\left\\{\\lceil t / 2\\rceil,\\lceil t / 2\\rceil+1, \\ldots, 2^{a}-1\\right\\} \\subset B$. In order to pair these $t-2\\left(t-2^{a}\\right)=$ $2^{a+1}-t<t$ numbers, we use the induction hypothesis and a bijection between $A^{\\prime} \\cup B^{\\prime}$ and $B_{2^{a+1}-t} \\cup A_{2^{a+1}-t}$. Let $S=S\\left(2^{a}-1\\right)=\\{0,1,2, \\ldots, a-1\\}$. Then take a pair $x, y, x \\in A_{2^{a+1}-t}$ and $y \\in B_{2^{a+1}-t}$ and biject it with $2^{a}-1-x \\in B^{\\prime}$ and $2^{a}-1-y \\in A^{\\prime}$. In fact,\n\n$$\n0 \\leq x \\leq\\left\\lfloor\\frac{2^{a+1}-t}{2}\\right\\rfloor-1=2^{a}-\\left\\lceil\\frac{t}{2}\\right\\rceil-1 \\Longleftrightarrow\\left\\lceil\\frac{t}{2}\\right\\rceil \\leq 2^{a}-1-x \\leq 2^{a}-1\n$$\n\nand\n\n$$\n\\left\\lceil\\frac{2^{a+1}-t}{2}\\right\\rceil=2^{a}-\\left\\lfloor\\frac{t}{2}\\right\\rfloor \\leq y \\leq 2^{a+1}-t-1 \\Longleftrightarrow t-2^{a} \\leq 2^{a}-1-y \\leq\\left\\lfloor\\frac{t}{2}\\right\\rfloor-1\n$$\n\nMoreover, $S\\left(2^{a}-1-x\\right)=S \\backslash S(x)$ and $S\\left(2^{a}-1-y\\right)=S \\backslash S(y)$ are complements with respect to $S$, and $S(x) \\subset S(y)$ and $|S(x)|=|S(y)|-1$ implies $S\\left(2^{a}-1-y\\right) \\subset S\\left(2^{a}-1-x\\right)$ and $\\left|S\\left(2^{a}-1-y\\right)\\right|=\\left|S\\left(2^{a}-1-x\\right)\\right|-1$. Therefore a pairing between $A^{\\prime}$ and $B^{\\prime}$ corresponds to a pairing between $A_{2^{a+1}-t}$ and $B_{2^{a+1}-t}$. Since the latter pairing is unique, the former pairing is also unique, and the result follows.\nWe illustrate the bijection by showing the case $t=23$ :\n\n$$\nA_{23}=\\{0,1,2, \\ldots, 10\\}, \\quad B_{23}=\\{12,13,14, \\ldots, 22\\}\n$$\n\nThe pairing is\n\n$$\n\\left(\\begin{array}{ccccccccccc}\n12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\\\\n8 & 9 & 10 & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6\n\\end{array}\\right)\n$$\n\nin which the bijection is between\n\n$$\n\\left(\\begin{array}{cccc}\n12 & 13 & 14 & 15 \\\\\n8 & 9 & 10 & 7\n\\end{array}\\right) \\quad \\text { and } \\quad\\left(\\begin{array}{llll}\n3 & 2 & 1 & 0 \\\\\n7 & 6 & 5 & 8\n\\end{array}\\right) \\rightarrow\\left(\\begin{array}{llll}\n5 & 6 & 7 & 8 \\\\\n1 & 2 & 3 & 0\n\\end{array}\\right) .\n$$", "problem_match": "# Problem 4", "solution_match": "# Solution\n\n"}
 {"year": "2024", "problem_label": "5", "tier": 1, "problem": "Line $\\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \\neq Q$, while circumcircles of the triangles $P A S$ and $P B R$ intersect at $M \\neq P$. Let lines $M P$ and $N Q$ meet at point $X$, lines $A B$ and $C D$ meet at point $K$ and lines $B C$ and $A D$ meet at point $L$. Prove that point $X$ lies on line $K L$.", "solution": "We start with the following lemma.\nLemma 1. Points $M, N, P, Q$ are concyclic.\nPoint $M$ is the Miquel point of lines $A P=A B, P S=\\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the triangle determined by the common lines $A D, \\ell$, and $B C$, which is $L R S$.\nThen, since quadrilaterals $Q N R C, P M A S$, and $A B C D$ are all cyclic, using directed angles (modulo $180^{\\circ}$ )\n\n$$\n\\begin{aligned}\n\\measuredangle N M P & =\\measuredangle N M S+\\measuredangle S M P=\\measuredangle N R S+\\measuredangle S A P=\\measuredangle N R Q+\\measuredangle D A B=\\measuredangle N R Q+\\measuredangle D C B \\\\\n& =\\measuredangle N R Q+\\measuredangle Q C R=\\measuredangle N R Q+\\measuredangle Q N R=\\measuredangle N Q R=\\measuredangle N Q P,\n\\end{aligned}\n$$\n\nwhich implies that $M N Q P$ is a cyclic quadrilateral.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-08.jpg?height=975&width=1115&top_left_y=1306&top_left_x=439)\n\nLet $E$ be the Miquel point of $A B C D$ (that is, of lines $A B, B C, C D, D A$ ). It is well known that $E$ lies in the line $t$ connecting the intersections of the opposite lines of $A B C D$. Let lines $N Q$ and $t$ meet at $T$. If $T \\neq E$, using directed angles, looking at the circumcircles of $L A B$ (which contains, by definition, $E$ and $M$ ), $A P S$ (which also contains $M$ ), and $M N Q P$,\n\n$$\n\\measuredangle T E M=\\measuredangle L E M=\\measuredangle L A M=\\measuredangle S A M=\\measuredangle S P M=\\measuredangle Q P M=\\measuredangle Q N M=\\measuredangle T N M,\n$$\n\nthat is, $T$ lies in the circumcircle $\\omega$ of $E M N$. If $T=E$, the same computation shows that $\\measuredangle L E M=\\measuredangle E N M$, which means that $t$ is tangent to $\\omega$.\n\nNow let lines $M P$ and $t$ meet at $V$. An analogous computation shows, by looking at the circumcircles of $L C D$ (which contains $E$ and $N$ ), $C Q R$, and $M N Q P$, that $V$ lies in $\\omega$ as well, and that if $V=E$ then $t$ is tangent to $\\omega$.\nTherefore, since $\\omega$ meet $t$ at $T, V$, and $E$, either $T=V$ if both $T \\neq E$ and $V \\neq E$ or $T=V=E$. At any rate, the intersection of lines $M P$ and $N Q$ lies in $t$.", "problem_match": "# Problem 5", "solution_match": "# Solution 1"}
 {"year": "2024", "problem_label": "5", "tier": 1, "problem": "Line $\\ell$ intersects sides $B C$ and $A D$ of cyclic quadrilateral $A B C D$ in its interior points $R$ and $S$ respectively, and intersects ray $D C$ beyond point $C$ at $Q$, and ray $B A$ beyond point $A$ at $P$. Circumcircles of the triangles $Q C R$ and $Q D S$ intersect at $N \\neq Q$, while circumcircles of the triangles $P A S$ and $P B R$ intersect at $M \\neq P$. Let lines $M P$ and $N Q$ meet at point $X$, lines $A B$ and $C D$ meet at point $K$ and lines $B C$ and $A D$ meet at point $L$. Prove that point $X$ lies on line $K L$.", "solution": "Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution.\n\nLemma 2. Denote by $\\operatorname{pow}_{\\omega} X$ the power of point $X$ with respect to circle $\\omega$. Let $\\Gamma_{1}$ and $\\Gamma_{2}$ be circles with different centers. Considering $A B C$ as the reference triangle in barycentric coordinates, the radical axis of $\\Gamma_{1}$ and $\\Gamma_{2}$ is given by\n\n$$\n\\left(\\operatorname{pow}_{\\Gamma_{1}} A-\\operatorname{pow}_{\\Gamma_{2}} A\\right) x+\\left(\\operatorname{pow}_{\\Gamma_{1}} B-\\operatorname{pow}_{\\Gamma_{2}} B\\right) y+\\left(\\operatorname{pow}_{\\Gamma_{1}} C-\\operatorname{pow}_{\\Gamma_{2}} C\\right) z=0\n$$\n\nProof: Let $\\Gamma_{i}$ have the equation $\\Gamma_{i}(x, y, z)=-a^{2} y z-b^{2} z x-c^{2} x y+(x+y+z)\\left(r_{i} x+s_{i} y+t_{i} z\\right)$. Then $\\operatorname{pow}_{\\Gamma_{i}} P=\\Gamma_{i}(P)$. In particular, $\\operatorname{pow}_{\\Gamma_{i}} A=\\Gamma_{i}(1,0,0)=r_{i}$ and, similarly, $\\operatorname{pow}_{\\Gamma_{i}} B=s_{i}$ and $\\operatorname{pow}_{\\Gamma_{i}} C=t_{i}$.\nFinally, the radical axis is\n\n$$\n\\begin{aligned}\n& \\operatorname{pow}_{\\Gamma_{1}} P=\\operatorname{pow}_{\\Gamma_{2}} P \\\\\n\\Longleftrightarrow & \\Gamma_{1}(x, y, z)=\\Gamma_{2}(x, y, z) \\\\\n\\Longleftrightarrow & r_{1} x+s_{1} y+t_{1} z=r_{2} x+s_{2} y+t_{2} z \\\\\n\\Longleftrightarrow & \\left(\\operatorname{pow}_{\\Gamma_{1}} A-\\operatorname{pow}_{\\Gamma_{2}} A\\right) x+\\left(\\operatorname{pow}_{\\Gamma_{1}} B-\\operatorname{pow}_{\\Gamma_{2}} B\\right) y+\\left(\\operatorname{pow}_{\\Gamma_{1}} C-\\operatorname{pow}_{\\Gamma_{2}} C\\right) z=0 .\n\\end{aligned}\n$$\n\nWe still use the Miquel point $E$ of $A B C D$. Notice that the problem is equivalent to proving that lines $M P, N Q$, and $E K$ are concurrent. The main idea is writing these three lines as radical axes. In fact, by definition of points $M, N$, and $E$ :\n\n- $M P$ is the radical axis of the circumcircles of $P A S$ and $P B R$;\n- $N Q$ is the radical axis of the circumcircles of $Q C R$ and $Q D S$;\n- $E K$ is the radical axis of the circumcircles of $K B C$ and $K A D$.\n\nLooking at these facts and the diagram, it makes sense to take triangle $K Q P$ the reference triangle. Because of that, we do not really need to draw circles nor even points $M$ and $N$, as all powers can be computed directly from points in lines $K P, K Q$, and $P Q$.\n![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-09.jpg?height=515&width=1043&top_left_y=2050&top_left_x=478)\n\nAssociate $P$ with the $x$-coordinate, $Q$ with the $y$-coordinate, and $K$ with the $z$-coordinate. Applying the lemma, the equations of lines $P M, Q N$, and $E K$ are\n\n- MP: $(K A \\cdot K P-K B \\cdot K P) x+(Q S \\cdot Q P-Q R \\cdot Q P) y=0$\n- $N Q:(K C \\cdot K Q-K D \\cdot K Q) x+(P R \\cdot P Q-P S \\cdot P Q) z=0$\n- MP: $(-Q C \\cdot Q K+Q D \\cdot Q K) y+(P B \\cdot P K-P A \\cdot P K) z=0$\n\nThese equations simplify to\n\n- $M P:(A B \\cdot K P) x+(P Q \\cdot R S) y=0$\n- $N Q:(-C D \\cdot K Q) x+(P Q \\cdot R S) z=0$\n- $M P:(C D \\cdot K Q) y+(A B \\cdot K P) z=0$\n\nNow, if $u=A B \\cdot K P, v=P Q \\cdot R S$, and $w=C D \\cdot K Q$, it suffices to show that\n\n$$\n\\left|\\begin{array}{ccc}\nu & v & 0 \\\\\n-w & 0 & v \\\\\n0 & w & u\n\\end{array}\\right|=0\n$$\n\nwhich is a straightforward computation.", "problem_match": "# Problem 5", "solution_match": "# Solution 2"}