Add data for asian pacific mathematics olympiad (APMO) (#2)

- add APMO 1991-1992 problem and solution data files (56c623505457537a2f4993f556c17745197b41fb)
- docs: add script documentation and author info (be2703fd3ff5015cf3f69a7fdade79adae26795d)
- add segmented APMO 2024 problems and solutions (6998e48f28ae3b9d3f985425ebf1bbbc541f3f13)


Co-authored-by: Javi Lau <LxYxvv@users.noreply.huggingface.co>

APMO/download_script/download.py

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+# -----------------------------------------------------------------------------
+# Author: Jiawei Liu
+# Date: 2024-11-21
+# -----------------------------------------------------------------------------
+'''
+Download script for APMO
+To run: 
+`python APMO\download_script\download.py`
+'''
+
+import requests
+from bs4 import BeautifulSoup
+from tqdm import tqdm
+from pathlib import Path
+from requests.adapters import HTTPAdapter
+from urllib3.util.retry import Retry
+from urllib.parse import urljoin
+
+
+def build_session(
+    max_retries: int = 3,
+    backoff_factor: int = 2,
+    session: requests.Session = None
+) -> requests.Session:
+    """
+    Build a requests session with retries
+
+    Args:
+        max_retries (int, optional): Number of retries. Defaults to 3.
+        backoff_factor (int, optional): Backoff factor. Defaults to 2.
+        session (requests.Session, optional): Session object. Defaults to None.
+    """
+    session = session or requests.Session()
+    adapter = HTTPAdapter(max_retries=Retry(total=max_retries, backoff_factor=backoff_factor))
+    session.mount("http://", adapter)
+    session.mount("https://", adapter)
+    session.headers.update({
+        "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3"
+    })
+
+    return session
+
+
+def main():
+    base_url = "https://www.apmo-official.org/problems"
+    req_session = build_session()
+
+    output_dir = Path(__file__).parent.parent / "raw"
+    output_dir.mkdir(parents=True, exist_ok=True)
+
+    resp = req_session.get(base_url)
+    soup = BeautifulSoup(resp.text, 'html.parser')
+
+    # Get the uri list of years
+    year_list_ele = soup.select_one("body > div.container.text-center > div > div:nth-child(2)")
+    year_list = year_list_ele.find_all('a')
+    year_uris = [ele['href'] for ele in year_list]
+
+    for year_uri in tqdm(year_uris):
+        output_file = output_dir / f"en-{Path(year_uri).name}"
+
+        # Check if the file already exists
+        if output_file.exists():
+            continue
+
+        pdf_resp = req_session.get(urljoin(base_url, year_uri))
+
+        if pdf_resp.status_code != 200:
+            print(f"Failed to download {year_uri}")
+            continue
+
+        output_file.write_bytes(pdf_resp.content)
+
+
+if __name__ == "__main__":
+    main()

APMO/md/en-apmo1989_sol.md

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+# APMO 1989 - Problems and Solutions 
+
+## Problem 1
+
+Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let
+
+$$
+S=x_{1}+x_{2}+\cdots+x_{n}
+$$
+
+Prove that
+
+$$
+\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq 1+S+\frac{S^{2}}{2!}+\frac{S^{3}}{3!}+\cdots+\frac{S^{n}}{n!}
+$$
+
+## Solution 1
+
+Let $\sigma_{k}$ be the $k$ th symmetric polynomial, namely
+
+$$
+\sigma_{k}=\sum_{\substack{|S|=k \\ S \subseteq\{1,2, \ldots, n\}}} \prod_{i \in S} x_{i},
+$$
+
+and more explicitly
+
+$$
+\sigma_{1}=S, \quad \sigma_{2}=x_{1} x_{2}+x_{1} x_{3}+\cdots+x_{n-1} x_{n}, \quad \text { and so on. }
+$$
+
+Then
+
+$$
+\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right)=1+\sigma_{1}+\sigma_{2}+\cdots+\sigma_{n}
+$$
+
+The expansion of
+
+$$
+S^{k}=\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}=\underbrace{\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cdots\left(x_{1}+x_{2}+\cdots+x_{n}\right)}_{k \text { times }}
+$$
+
+has 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
+
+$$
+S^{k} \geq k!\sigma_{k} \Longleftrightarrow \sigma_{k} \leq \frac{S^{k}}{k!}
+$$
+
+Summing the obtained inequalities for $k=1,2, \ldots, n$ yields the result.
+
+## Solution 2
+
+By AM-GM,
+
+$$
+\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \leq\left(\frac{\left(1+x_{1}\right)+\left(1+x_{2}\right)+\cdots+\left(1+x_{n}\right)}{n}\right)^{n}=\left(1+\frac{S}{n}\right)^{n}
+$$
+
+By the binomial theorem,
+
+$$
+\left(1+\frac{S}{n}\right)^{n}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{S}{n}\right)^{k}=\sum_{k=0}^{n} \frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \sum_{k=0}^{n} \frac{S^{k}}{k!}
+$$
+
+and the result follows.
+Comment: Maclaurin's inequality states that
+
+$$
+\frac{\sigma_{1}}{n} \geq \sqrt{\frac{\sigma_{2}}{\binom{n}{2}}} \geq \cdots \geq \sqrt[k]{\frac{\sigma_{k}}{\binom{n}{k}}} \geq \cdots \geq \sqrt[n]{\frac{\sigma_{n}}{\binom{n}{n}}}
+$$
+
+Then $\sigma_{k} \leq\binom{ n}{k} \frac{S^{k}}{n^{k}}=\frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \frac{S^{k}}{k!}$.
+
+## Problem 2
+
+Prove that the equation
+
+$$
+6\left(6 a^{2}+3 b^{2}+c^{2}\right)=5 n^{2}
+$$
+
+has 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
+
+$$
+6 a^{2}+3 b^{2}+c^{2}=30 n_{0}^{2}
+$$
+
+The number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to
+
+$$
+2 a^{2}+b^{2}+3 c_{0}^{2}=10 n_{0}^{2}
+$$
+
+Now look at the equation modulo 8:
+
+$$
+b^{2}+3 c_{0}^{2} \equiv 2\left(n_{0}^{2}-a^{2}\right) \quad(\bmod 8)
+$$
+
+Integers $b$ 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
+
+$$
+b^{2}+3 c_{0}^{2} \equiv 0 \quad(\bmod 8)
+$$
+
+If $b$ 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
+
+$$
+a^{2}+2 b_{0}^{2}+6 c_{1}^{2}=5 n_{0}^{2}
+$$
+
+Look at the last equation modulo 8:
+
+$$
+a^{2}+3 n_{0}^{2} \equiv 2\left(c_{1}^{2}-b_{0}^{2}\right) \quad(\bmod 8)
+$$
+
+A similar argument shows that $a$ and $n_{0}$ are both even.
+We have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find
+
+$$
+6\left(6(a / 2)^{2}+3(b / 2)^{2}+(c / 2)^{2}\right)=5(n / 2)^{2}
+$$
+
+and 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 3
+
+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 $E_{1} E_{2} E_{3}$.
+Answer: $\frac{25}{49}$.
+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 $A_{3} B_{1}$ 。
+![](https://cdn.mathpix.com/cropped/2024_11_22_831b46bbbb7a3a492868g-3.jpg?height=441&width=449&top_left_y=750&top_left_x=769)
+
+By Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,
+
+$$
+\frac{A_{1} B_{1}}{A_{1} A_{2}} \cdot \frac{D_{1} A_{3}}{D_{1} B_{1}} \cdot \frac{C_{2} A_{2}}{C_{2} A_{3}}=1 \Longleftrightarrow \frac{D_{1} A_{3}}{D_{1} B_{1}}=2 \cdot 3=6 \Longleftrightarrow \frac{D_{1} B_{1}}{A_{3} B_{1}}=\frac{1}{7}
+$$
+
+Since $A_{3} G=\frac{2}{3} A_{3} B_{1}$, 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
+
+$$
+\frac{G D_{1}}{G A_{3}}=\frac{4}{14}=\frac{2}{7}
+$$
+
+Similar 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 $-\frac{2}{7}$.
+By Menelao's theorem on triangle $A_{1} A_{2} B_{2}$ and line $C_{1} E_{1} A_{3}$,
+
+$$
+\frac{C_{1} A_{1}}{C_{1} A_{2}} \cdot \frac{E_{1} B_{2}}{E_{1} A_{1}} \cdot \frac{A_{3} A_{2}}{A_{3} B_{2}}=1 \Longleftrightarrow \frac{E_{1} B_{2}}{E_{1} A_{1}}=3 \cdot \frac{1}{2}=\frac{3}{2} \Longleftrightarrow \frac{A_{1} E_{1}}{A_{1} B_{2}}=\frac{2}{5}
+$$
+
+If $A_{1} B_{2}=15 u$, 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
+
+$$
+\frac{G E_{1}}{G A_{1}}=\frac{4}{10}=\frac{2}{5}
+$$
+
+Similar 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 $\frac{2}{5}$.
+Then $D_{1} D_{2} D_{3}$ 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 4
+
+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
+
+$$
+4 m \frac{\left(m-\frac{n^{2}}{4}\right)}{3 n}
+$$
+
+triples $(a, b, c)$ 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 $D_{i} \cup D_{j} \subseteq\{1,2, \ldots, n\}$,
+
+$$
+\left|D_{i} \cap D_{j}\right|=\left|D_{i}\right|+\left|D_{j}\right|-\left|D_{i} \cup D_{j}\right| \leq d_{i}+d_{j}-n
+$$
+
+Summing 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
+
+$$
+T \geq \frac{1}{3} \sum_{(i, j) \in S}\left(d_{i}+d_{j}-n\right)
+$$
+
+Each 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
+
+$$
+T \geq \frac{1}{3}\left(\sum_{i=1}^{n} d_{i}^{2}-m n\right) \geq \frac{1}{3}\left(\frac{\left(\sum_{i=1}^{n} d_{i}\right)^{2}}{n}-m n\right) .
+$$
+
+Finally, 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
+
+$$
+T \geq \frac{1}{3}\left(\frac{(2 m)^{2}}{n}-m n\right)=4 m \frac{\left(m-\frac{n^{2}}{4}\right)}{3 n} .
+$$
+
+Comment: 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 5
+
+Determine all functions $f$ from the reals to the reals for which
+(1) $f(x)$ is strictly increasing,
+(2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$.
+(Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)
+
+Answer: $f(x)=x+c, c \in \mathbb{R}$ constant.
+
+## Solution
+
+Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\underbrace{f(f(\ldots f}_{n \text { times }}(x)))$.
+Plug $x \rightarrow f_{n+1}(x)$ in (2): since $g\left(f_{n+1}(x)\right)=g\left(f\left(f_{n}(x)\right)\right)=f_{n}(x)$,
+
+$$
+f_{n+2}(x)+f_{n}(x)=2 f_{n+1}(x)
+$$
+
+that is,
+
+$$
+f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_{n}(x)
+$$
+
+Therefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find
+
+$$
+f_{n}(x)-x=n(f(x)-x) .
+$$
+
+Since $g$ has the same properties as $f$,
+
+$$
+g_{n}(x)-x=n(g(x)-x)=-n(f(x)-x) .
+$$
+
+Finally, $g$ 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.
+Let $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,
+
+$$
+x+n(f(x)-x)>y+n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]>y-x
+$$
+
+and
+
+$$
+x-n(f(x)-x)>y-n(f(y)-y) \Longleftrightarrow n[(f(x)-x)-(f(y)-y)]<x-y
+$$
+
+Summing it up,
+
+$$
+|n[(f(x)-x)-(f(y)-y)]|<x-y \quad \text { for all } n \in \mathbb{Z}_{>0}
+$$
+
+Suppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,
+
+$$
+|n(a-b)|<x-y
+$$
+
+which 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, $f(x)=x+c$.
+It is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.
+

APMO/md/en-apmo1990_sol.md

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+# Question 1 
+
+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.
+For each value of $\angle B A C$, how many non-similar triangles are there in which $A E G F$ is a cyclic quadrilateral?
+
+## FIRST SOLUTION
+
+Let $I$ be the intersection of $A G$ and $E F$.
+Let $\delta=A I . I G-F I$ IE. Then
+
+$$
+A I=A D / 2, \quad I G=A D / 6, \quad F I=B C / 4=I E
+$$
+
+Further, applying the cosine rule to triangles $A B D, A C D$ we get
+
+$$
+\begin{aligned}
+A B^{2} & =B C^{2} / 4+A D^{2}-A D \cdot B C \cdot \cos \angle B D A, \\
+A C^{2} & =B C^{2} / 4+A D^{2}+A D \cdot B C \cdot \cos \angle B D A, \\
+\text { so } \quad A D^{2} & =\left(A B^{2}+A C^{2}-B C^{2} / 2\right) / 2
+\end{aligned}
+$$
+
+Hence
+
+$$
+\begin{aligned}
+\delta & =\left(A B^{2}+A C^{2}-2 B C^{2}\right) / 24 \\
+& =\left(4 A B \cdot A C \cdot \cos \angle B A C-A B^{2}-A C^{2}\right)
+\end{aligned}
+$$
+
+Now $A E F G$ is a cyclic quadrilateral if and only if $\delta=0$, i.e. if and only if
+
+$$
+\begin{aligned}
+\cos \angle B A C & =\left(A B^{2}+A B^{2}\right) /(4 \cdot A B \cdot A C) \\
+& =(A B / A C+A C / A B) / 4
+\end{aligned}
+$$
+
+5
+Now $A B / A C+A C / A B \geq 2$. Hence $\cos \angle B A C \geq 1 / 2$ and so $\angle B A C \leq 60^{\circ}$.
+For $\angle B A C>60^{\circ}$ there is no triangle with the required property.
+For $\angle B A C=60^{\circ}$ there exists, within similarity, precisely one triangle (which is equilateral) having the required property.
+For $\angle B A C<60^{\circ}$ 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).
+
+SECOND SOLUTION (Mr Marcus Brazil, La Trobe University, Bundoora, Melbourne, Australia):
+We require, as above,
+
+$$
+A I \cdot I G=E I \cdot I F
+$$
+
+(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.
+![](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
+
+$$
+\begin{aligned}
+\angle E G F & =180^{\circ}-\left(B_{1}+C_{1}\right) \\
+& \geq 180^{\circ}-(B+C) \\
+& =A .
+\end{aligned}
+$$
+
+(1)
+
+Thus, if $A E G F$ is cyclic, we would have $\angle E G F+\angle A=180^{\circ}$. Therefore it is necessary that $0<\angle A \leq 90^{\circ}$.
+
+## Continuation "A"
+
+Let $O$ be the circumcentre of $\triangle A F E$. Without loss of generality, let the radius of this circle be 1.
+We then let $A=1, F=z=e^{i \theta}$ and $E=z e^{2 i \alpha}=e^{i(\theta+2 \alpha)}$.
+Then $\angle A=\alpha, 0<\alpha \leq 90^{\circ}$, and $0<\theta<360^{\circ}-2 \alpha$.
+Thus,
+
+$$
+B=2 z-1
+$$
+
+and
+
+$$
+\begin{aligned}
+G & =\frac{1}{3}(2 z-1)+\frac{2}{3}\left(z e^{2 i \alpha}\right) \\
+& =\frac{1}{3}\left(2 e^{i \theta}+2 e^{i(\theta+2 \alpha)}-1\right)
+\end{aligned}
+$$
+
+For quadrilateral $A F G E$ to be cyclic, it is now necessary that
+
+$$
+|G|=1 .
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-03.jpg?height=1123&width=1148&top_left_y=162&top_left_x=456)
+
+For $|G|=1$, we must have
+
+$$
+\begin{aligned}
+9= & (2 \cos (\theta)+2 \cos (\theta+2 \alpha)-1)^{2}+(2 \sin (\theta)+2 \sin (\theta+2 \alpha))^{2} \\
+= & 4\left(\cos ^{2}(\theta)+\sin ^{2}(\theta)\right)+4\left(\cos ^{2}(\theta+2 \alpha)+\sin ^{2}(\theta+2 \alpha)\right)+1 \\
+& +8(\cos (\theta) \cos (\theta+2 \alpha)+\sin (\theta) \sin (\theta+2 \alpha))-4 \cos (\theta)-4 \cos (\theta+2 \alpha) \\
+= & 9+8 \cos (2 \alpha)-8 \cos (\alpha) \cos (\theta+\alpha)
+\end{aligned}
+$$
+
+so that
+
+$$
+\cos (\theta+\alpha)=\frac{\cos (2 \alpha)}{\cos (\alpha)}
+$$
+
+] 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 $\alpha=60^{\circ}$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-03.jpg?height=350&width=641&top_left_y=1950&top_left_x=661)
+$\square$ 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 $360^{\circ}-\alpha-\theta$.
+
+## Continuation "B"
+
+Let $O$ 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.
+
+It 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.
+
+Let $\theta$ 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 $\alpha \in\left(0,90^{\circ}\right.$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-04.jpg?height=1091&width=1484&top_left_y=669&top_left_x=478)
+
+Now, $D$ lies on an arc of a circle, centre $P$, radius $P D$ exterior to the circle, centre $O$, radius $A O$.
+By similarity, $G$, lies on an arc of a circle, centre $Q$, radius $Q G$ where $A Q=\frac{2}{3} A P$ and $Q G=\frac{2}{3} P D$.
+For the quadrilateral $A F G E$ to be cyclic, we must have that the radius $Q G$ is greater than or equal to $Q P$.
+The easiest way to calulate these radii is to consider the case in which the diameter $A P$ bisects the angle $\angle B A C$.
+Thus we re-draw the diagram as below. Let $A H$ be a diameter of the larger circle.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-05.jpg?height=1126&width=1061&top_left_y=163&top_left_x=472)
+
+Thus we have $A H=4$ and by similar triangles,
+
+$$
+\frac{A D}{A B}=\frac{A B}{A H}=\cos \left(\frac{\alpha}{2}\right)
+$$
+
+so that
+
+$$
+\begin{aligned}
+A D & =4 \cos ^{2}\left(\frac{\alpha}{2}\right) \\
+& =2+2 \cos (\alpha) .
+\end{aligned}
+$$
+
+Thus $P D=2 \cos (\alpha)$ and $Q G=\frac{2}{3} 2 \cos (\alpha)=\frac{4}{3} \cos (\alpha)$.
+The necessary condition for a cyclic quadrilateral is then
+
+$$
+\frac{4}{3}(1+\cos (\alpha)) \geq 2
+$$
+
+[5
+
+$$
+\cos (\alpha) \geq \frac{1}{2}
+$$
+
+：７
+Thus it is clear that there is precisely one (up to similarity) solution for $0<\alpha \leq 60^{\circ}$ and no solutions otherwise.
+
+## Question 2
+
+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.
+Show that
+
+$$
+S_{k} S_{n-k} \geq\binom{ n}{k}^{2} a_{1} a_{2} \ldots a_{n}, \quad \text { for } \quad k=1,2, \ldots, n-1
+$$
+
+## FIRST SOLUTION
+
+$$
+\binom{n}{k} a_{1} a_{2} \ldots a_{n}
+$$
+
+$2=\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}} \cdot a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}$
+(and using the Cauchy-Schwarz inequality)
+
+$$
+\begin{aligned}
+& \leq\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \cdot\left(\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{1} a_{2} \ldots a_{n} / a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}\right)^{\frac{1}{2}} \\
+& =S_{k}^{\frac{1}{2}} \cdot S_{n-k}^{\frac{1}{2}}
+\end{aligned}
+$$
+
+Therefore
+
+$$
+\binom{n}{k}^{2} a_{1} a_{2} \ldots a_{n} \leq S_{k} S_{n-k}
+$$
+
+q.e.d.
+
+SECOND SOLUTION (provided by the Canadian Problems Committee).
+Write $S_{k}$ as $\sum_{i=1}^{\binom{n}{k}} t_{i}$. Then
+주
+
+$$
+S_{n-k}=\left(\prod_{m=1}^{n} a_{m}\right)\left(\sum_{i=1}^{\binom{n}{k}} \frac{1}{t_{i}}\right)
+$$
+
+$$
+\text { so that } \left.\begin{array}{rl}
+S_{k} S_{n-k} & =\left(\prod_{m=1}^{n} a_{m}\right) \cdot\left(\begin{array}{l}
+\binom{n}{k} \\
+i=1
+\end{array} t_{i}\right)\left(\begin{array}{l}
+\binom{n}{k} \\
+\sum_{j=1}^{2}
+\end{array} \frac{1}{t_{j}}\right) \\
+& =\left(\prod_{m=1}^{n} a_{m}\right)\left[\sum_{i=1}^{\binom{n}{k}} 1+\sum_{i=1}^{\binom{n}{k}} \sum_{j=1}^{n} \begin{array}{l}
+n \\
+k
+\end{array}\right) \\
+\frac{t_{i}}{} \\
+t_{j}
+\end{array}\right] .
+$$
+
+As there are
+
+$$
+\frac{\binom{n}{k}^{2}-\binom{n}{k}}{2}
+$$
+
+terms in the sum
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-07.jpg?height=98&width=91&top_left_y=355&top_left_x=63)
+
+$$
+\begin{aligned}
+S_{k} S_{n-k} & \geq\left(\prod_{m=1}^{n} a_{m}\right)\left[\binom{n}{k}+2 \cdot \frac{\binom{n}{k}^{2}-\binom{n}{k}}{2}\right] \\
+& =\binom{n}{k}^{2}\left(\prod_{m=1}^{n} a_{m}\right)
+\end{aligned}
+$$
+
+since $\frac{t_{i}}{t_{j}}+\frac{t_{j}}{t_{i}} \geq 2$ for $t_{i}, t_{j}>0$.
+
+## Question 3
+
+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
+
+$$
+\begin{aligned}
+A B \cdot h \cdot A C \cdot h_{b} \cdot B C \cdot h_{a} & =8 . \text { area of } \triangle A B C)^{3} \\
+& =(A B \cdot h)^{3},
+\end{aligned}
+$$
+
+园
+which 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.
+Since
+
+$$
+\begin{aligned}
+(\sin C) \cdot A C \cdot B C & =B C \cdot h_{a} \\
+& =2 \cdot \text { area of } \triangle A B C
+\end{aligned}
+$$
+
+(3)
+which is a constant, $A C \cdot B C$ attains its minimum when $\sin C$ reaches its maximum. There are two cases:
+(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 .
+(b) $h>A B / 2$. In this case the angle at $C$ is acute and assumes its maximum when the triangle is isosceles.
+
+Note that a solution using calculus obviously exists.
+
+## Question 4
+
+A set of 1990 persons is divided into non-intersecting subsets in such a way that
+(a) no one in a subset knows all the others in the subset;
+(b) among any three persons in a subset, there are always at least two who do not know each other; and
+(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.
+(i) Prove that within each subset, every person has the same number of acquaintances.
+(ii) Determine the maximum possible number of subsets.
+
+Note: 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 $S$.
+Assume 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).
+By (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 $x_{1}, y_{2}, \ldots, y_{n}$.
+Because $n$ 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 $S$.
+Letting $y_{1}$ 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 $S$.
+(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 $=398$.
+
+## Question 5
+
+Show that for every integer $n \geq 6$, there exists a convex hexagon which can be dissected into exactly $n$ congruent triangles.
+
+FIRST SOLUTION (provided by the Canadian Problems Committee).
+The basic building blocks will be right angled triangles with sides $p, q$ (which are positive integers) adjacent to the right angle.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=153&width=185&top_left_y=623&top_left_x=902)
+
+In the first instance, we take $p=q=1$ and construct five basic building blocks: $L_{1}, L_{2}, M, R_{1}$ and $R_{2}$ 。
+
+## [3)
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=186&width=115&top_left_y=956&top_left_x=411)
+$L_{1}$
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=186&width=112&top_left_y=956&top_left_x=649)
+$L_{2}$
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=183&width=107&top_left_y=955&top_left_x=933)
+M
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=179&width=107&top_left_y=954&top_left_x=1215)
+$R_{1}$
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=177&width=110&top_left_y=952&top_left_x=1452)
+$R_{2}$
+
+We 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 $n \geq 0$.
+
+Thus, with the obvious interpretation:
+$L_{1}+n M+R_{1}$ gives a convex hexagon containing $2+4 n+2=4 n+4 \quad(n \geq 1)$ congruent triangles;
+$L_{1}+n M+R_{2}$ gives a convex hexagon containing $2+4 n+3=4 n+5(n \geq 1)$ congruent triangles; and
+$L_{2}+n M+R_{2}$ 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.
+
+We 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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-10.jpg?height=438&width=1180&top_left_y=2101&top_left_x=407)
+
+So 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
+
+$$
+(n+1) p=(n+2) q
+$$
+
+so we take
+
+$$
+q=n+1 \quad \text { and } \quad p=n+2
+$$
+
+This completes the solution.
+SECOND SOLUTION (provided by the Canadian Problems Committee):
+The basic building blocks will be right angled triangles with sides $m, n$ (which are positive integers) adjacent to the right angle.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-11.jpg?height=251&width=180&top_left_y=758&top_left_x=910)
+
+We 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.
+
+We 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.
+
+Two 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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-12.jpg?height=1212&width=1356&top_left_y=260&top_left_x=341)
+
+Now 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 $m \geq 3$.
+
+We 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.
+
+Now count the total number of triangles in the full configuration.
+The upper configuration gives $2 m$ triangles. The lower configuration gives
+
+$$
+\sum_{k=1}^{n}(2 k-1)=n^{2} \quad \text { triangles. }
+$$
+
+Thus the total number of triangles in a hexagon in class 1 is
+
+$$
+2 m-2+n^{2}-1
+$$
+
+and the total number of triangle in a hexagon in class 2 is
+
+$$
+2 m-2+n^{2}-2
+$$
+
+These, together with the restrictions on $n$ and $m$, generate all positive integers greater than or equal to 11.
+
+For the integers $6,7,8,9$ and 10 , we give specific examples:
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=394&width=283&top_left_y=513&top_left_x=365)
+
+6
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=329&width=402&top_left_y=584&top_left_x=742)
+
+7
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=394&width=404&top_left_y=513&top_left_x=1240)
+
+8
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=388&width=410&top_left_y=1015&top_left_x=559)
+
+9
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0cc1366c8ef5af6fd96g-13.jpg?height=392&width=413&top_left_y=1013&top_left_x=1059)
+
+10
+
+This completes the solution.
+
+There 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
+
+## ④
+
+$$
+\frac{S_{k}}{\binom{n}{k}} \geq\left[\prod_{i=1}^{n} a_{i}^{\binom{n-1}{k-1}}\right]^{\frac{1}{n}\binom{n}{n}}
+$$
+
+But $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$, leading to
+6 S $\quad S_{k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}$.
+Hence
+田
+
+$$
+S_{k} S_{n-k} \geq\binom{ n}{k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{k}{n}}\binom{n}{n-k}\left(\prod_{i=1}^{n} a_{i}\right)^{\frac{n-k}{n}}=\binom{n}{k}^{2}\left(\prod_{1}^{n} a_{i}\right) .
+$$
+

APMO/md/en-apmo1991_sol.md

@@ -0,0 +1,181 @@
+# APMO 1991 - Problems and Solutions 
+
+## Problem 1
+
+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 1
+
+Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=663&top_left_x=673)
+
+It 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 $G X=\frac{1}{2} X Y=\frac{1}{3} B C$.
+Now look at the similarity between triangles $Q B C$ and $Q G X$ :
+
+$$
+\frac{Q G}{Q B}=\frac{G X}{B C}=\frac{1}{3} \Longrightarrow Q B=3 Q G \Longrightarrow Q B=\frac{3}{4} B G=\frac{3}{4} \cdot \frac{2}{3} B R=\frac{1}{2} B R .
+$$
+
+Finally, 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 $\frac{1}{4}$ ).
+
+## Solution 2
+
+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 $C S$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-1.jpg?height=558&width=647&top_left_y=1971&top_left_x=676)
+
+Due 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 $G X$.
+
+Now 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 $M S$.
+The 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 2
+
+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
+
+$$
+\frac{y_{1}+y_{2}}{2}<\frac{y_{1}+y_{3}}{2}<\frac{y_{2}+y_{3}}{2}<\frac{y_{2}+y_{4}}{2}<\cdots<\frac{y_{995}+y_{997}}{2}<\frac{y_{996}+y_{997}}{2}
+$$
+
+there 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 3
+
+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
+
+$$
+\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}} \geq \frac{a_{1}+a_{2}+\cdots+a_{n}}{2}
+$$
+
+## Solution
+
+By the Cauchy-Schwartz inequality,
+
+$$
+\left(\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}}\right)\left(\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\cdots+\left(a_{n}+b_{n}\right)\right) \geq\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} .
+$$
+
+Since $\left(\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\cdots+\left(a_{n}+b_{n}\right)\right)=2\left(a_{1}+a_{2}+\cdots+a_{n}\right)$,
+
+$$
+\frac{a_{1}^{2}}{a_{1}+b_{1}}+\frac{a_{2}^{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}^{2}}{a_{n}+b_{n}} \geq \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{2}
+$$
+
+## Problem 4
+
+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.
+
+Answer: All powers of 2 .
+
+## Solution 1
+
+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.
+If $n=2^{a} m, m>1$ 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,
+
+$$
+f(x) \equiv t \quad(\bmod n) \Longleftrightarrow \begin{cases}f(x) \equiv t & \left(\bmod 2^{a}\right) \\ f(x) \equiv t & (\bmod m)\end{cases}
+$$
+
+Therefore, $f$ is not a bijection modulo $n$.
+If $n=2^{a}$, then
+
+$$
+f(x)-f(y)=\frac{1}{2}(x(x+1)-y(y+1))=\frac{1}{2}\left(x^{2}-y^{2}+x-y\right)=\frac{(x-y)(x+y+1)}{2} .
+$$
+
+and
+
+$$
+f(x) \equiv f(y) \quad\left(\bmod 2^{a}\right) \Longleftrightarrow(x-y)(x+y+1) \equiv 0 \quad\left(\bmod 2^{a+1}\right)
+$$
+
+If $x$ 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,
+
+$$
+(*) \Longleftrightarrow x+y+1 \equiv 0 \quad\left(\bmod 2^{a+1}\right)
+$$
+
+However, $1 \leq x+y+1 \leq 2\left(2^{a}-1\right)+1=2^{a+1}-1$, so $x+y+1$ is not a multiple of $2^{a+1}$. Therefore $f$ is a bijection if $n$ is a power of 2 .
+
+## Solution 2
+
+We give a full description of $a_{n}$, the size of the range of $f$.
+Since 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 $n, a_{n}=\prod_{p^{\alpha} \| n} a_{p^{\alpha}}$.
+Refer to the first solution to check the case $p=2: a_{2^{\alpha}}=2^{\alpha}$.
+For an odd prime $p$,
+
+$$
+f(x)=\frac{x(x+1)}{2}=\frac{(2 x+1)^{2}-1}{8}
+$$
+
+and 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 $p^{\alpha}$.
+Let $g$ 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
+
+$$
+a_{p^{\alpha}}=\frac{p-1}{2} \cdot p^{\alpha-1}+a_{p^{\alpha}-2} .
+$$
+
+Since $a_{p}=\frac{p-1}{2}+1$ and $a_{p^{2}}=\frac{p-1}{2} \cdot p+1$, telescoping yields
+
+$$
+a_{p^{2 t}}=\frac{p-1}{2}\left(p^{2 t-1}+p^{2 t-3}+\cdots+p\right)+1=\frac{p\left(p^{2 t}-1\right)}{2(p+1)}+1
+$$
+
+and
+
+$$
+a_{p^{2 t-1}}=\frac{p-1}{2}\left(p^{2 t-2}+p^{2 t-4}+\cdots+1\right)+1=\frac{p^{2 t}-1}{2(p+1)}+1
+$$
+
+Now 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 5
+
+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.
+Let $\Gamma_{1}$ 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 $P$.
+Now 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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-7.jpg?height=918&width=729&top_left_y=997&top_left_x=635)
+
+We proceed with the construction with the aid of some macro constructions that will be detailed later.
+
+Step 1. Draw the common tangents to $\Gamma_{1}$ and $\Gamma_{2}$.
+Step 2. For each common tangent $t$, draw the projection $P_{t}$ of $P$ onto $t$.
+Step 3. Find the inverse $P_{1}$ of $P_{t}$ with respect to the circle with center $P$ and radius $P T$.
+Step 4. $\omega_{t}$ is the circle with diameter $P P_{1}$.
+Let's work out the details for steps 1 and 3 . Steps 2 and 4 are immediate.
+Step 1. In this particular case in which $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent, there is a small shortcut:
+
+- Draw the circle with diameter on the two centers $O_{1}$ of $\Gamma_{1}$ and $O_{2}$ of $\Gamma_{2}$, and find its center $O$.
+- 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 $R$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=572&width=729&top_left_y=282&top_left_x=635)
+
+Let'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
+
+$$
+\sin \angle T Q O=\frac{O T}{O Q}=\frac{R_{2}-R_{1}}{R_{1}+R_{2}}
+$$
+
+which is also the sine of the angle between $O_{1} O_{2}$ and the common tangent lines.
+Let $t$ 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
+
+$$
+d(O, t)=O Q=\frac{R_{1}+R_{2}}{2}=\frac{d\left(O_{1}, t\right)+d\left(O_{2}, t\right)}{2}
+$$
+
+and $O$ 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.
+Step 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.
+
+- If $X$ lies in $\Omega$, then its inverse is $X$.
+- 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 $O X \cdot O X^{\prime}=O Y^{2}$.
+- 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 $O X \cdot O X^{\prime}=O Y^{2}$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=347&width=481&top_left_y=2265&top_left_x=408)
+$X$ is inside $\Omega$
+![](https://cdn.mathpix.com/cropped/2024_11_22_5347031840c93d3f0aefg-8.jpg?height=398&width=469&top_left_y=2274&top_left_x=1096)
+

APMO/md/en-apmo1992_sol.md

@@ -0,0 +1,162 @@
+# APMO 1992 - Problems and Solutions 
+
+## Problem 1
+
+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.
+For which original triangles can this process be repeated indefinitely?
+Answer: 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.
+Suppose 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 2
+
+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 $A$.
+Prove 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:
+
+$$
+O_{1} ; O_{2} ; A, \quad O ; O_{1} ; A_{1}, \quad O ; O_{2} ; A_{2}
+$$
+
+Because of that we can ignore the circles and only draw their centers and tangency points.
+![](https://cdn.mathpix.com/cropped/2024_11_22_c7e8c80a7518426c71e5g-2.jpg?height=1018&width=1095&top_left_y=713&top_left_x=432)
+
+Now the problem is immediate from Ceva's theorem in triangle $O O_{1} O_{2}$, because
+
+$$
+\frac{O A_{1}}{A_{1} O_{1}} \cdot \frac{O_{1} A}{A O_{2}} \cdot \frac{O_{2} A_{2}}{A_{2} O}=\frac{r}{r_{1}} \cdot \frac{r_{1}}{r_{2}} \cdot \frac{r_{2}}{r}=1
+$$
+
+## Problem 3
+
+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.
+(a) Show that if we choose all three numbers greater than $n / 2$, then the values of these combinations are all distinct.
+(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
+
+$$
+x+y+z, \quad x+y z, \quad x y+z, \quad y+z x, \quad(x+y) z, \quad(z+x) y, \quad(x+y) z, \quad x y z
+$$
+
+Since, 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 $(t+m) n-(t+n) m=t(n-m)>0$,
+
+$$
+x+y+z<z+x y<y+z x<x+y z
+$$
+
+and
+
+$$
+(y+z) x<(x+z) y<(x+y) z<x y z .
+$$
+
+Also, $(y+z) x-(y+z x)=(x-1) y>0 \Longrightarrow(y+z) x>y+z x$ 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,
+
+$$
+x+y z=(y+z) x \Longleftrightarrow(y-x)(z-x)=x(x-1)
+$$
+
+Now we can solve the items.
+(a) if $n / 2<x<y<z$ then $z-x<n / 2$, and since $y-x<z-x, y-x<n / 2-1$; then
+
+$$
+(y-x)(z-x)<\frac{n}{2}\left(\frac{n}{2}-1\right)<x(x-1)
+$$
+
+and therefore $x+y z<(y+z) x$.
+(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,
+
+$$
+x=p, \quad, y=p+d, \quad z=p+\frac{p(p-1)}{d}
+$$
+
+which is a solution for every divisor $d$ of $p-1$ because
+
+$$
+x=p<y=p+d<2 p \leq p+p \cdot \frac{p-1}{d}=z .
+$$
+
+Comment: 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 4
+
+Determine all pairs $(h, s)$ of positive integers with the following property: If one draws $h$ horizontal lines and another $s$ lines which satisfy
+(i) they are not horizontal,
+(ii) no two of them are parallel,
+(iii) no three of the $h+s$ lines are concurrent,
+then the number of regions formed by these $h+s$ lines is 1992 .
+Answer: $(995,1),(176,10)$, 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 $n-1+2=n+1$.
+Then $a_{0,0}=1$ (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
+
+$$
+a_{0, s}=s+(s-1)+\cdots+1+a_{0,0}=\frac{s^{2}+s+2}{2} .
+$$
+
+Each horizontal line only intersects the $s$ non-horizontal lines, so $a_{h, s}=a_{h-1, s}+s+1$, which implies
+
+$$
+a_{h, s}=a_{0, s}+h(s+1)=\frac{s^{2}+s+2}{2}+h(s+1) .
+$$
+
+Our final task is solving
+
+$$
+a_{h, s}=1992 \Longleftrightarrow \frac{s^{2}+s+2}{2}+h(s+1)=1992 \Longleftrightarrow(s+1)(s+2 h)=2 \cdot 1991=2 \cdot 11 \cdot 181
+$$
+
+The 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 $(h, s)$ :
+
+$$
+(995,1), \quad(176,10), \quad \text { and }(80,21) .
+$$
+
+## Problem 5
+
+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.
+
+Answer: 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:
+
+| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ |
+| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
+| $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
+| $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ |
+| $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ |
+| $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ |
+| $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ |
+| $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |
+
+Let $S$ 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
+
+$$
+-a,-a, b,-a,-a,-a, b,-a,-a, b,-a,-a,-a, b,-a,-a .
+$$
+
+The 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,
+
+$$
+\frac{5 a}{2}<b<\frac{8 a}{3} \Longleftrightarrow 15 a<6 b<16 a
+$$
+
+Then we can choose, say, $a=7$ and $105<6 b<112 \Longleftrightarrow b=18$. A valid sequence is then
+
+$$
+-7,-7,18,-7,-7,-7,18,-7,-7,18,-7,-7,-7,18,-7,-7
+$$
+

APMO/md/en-apmo1993_sol.md

@@ -0,0 +1,173 @@
+# APMO 1993 - Problems and Solutions 
+
+## Problem 1
+
+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 $A F$.
+Prove 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)
+
+Triangles $A E D$ 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,
+
+$$
+\frac{A E}{C D}=\frac{A D}{C F} \Longleftrightarrow \frac{A E}{A C}=\frac{A C}{C F}
+$$
+
+The last equality combined with
+
+$$
+\angle E A C=180^{\circ}-\angle B A C=120^{\circ}=\angle A C F
+$$
+
+shows 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 2
+
+Find the total number of different integer values the function
+
+$$
+f(x)=[x]+[2 x]+\left[\frac{5 x}{3}\right]+[3 x]+[4 x]
+$$
+
+takes for real numbers $x$ with $0 \leq x \leq 100$.
+Note: $[t]$ is the largest integer that does not exceed $t$.
+Answer: 734.
+
+## Solution
+
+Note that, since $[x+n]=[x]+n$ for any integer $n$,
+
+$$
+f(x+3)=[x+3]+[2(x+3)]+\left[\frac{5(x+3)}{3}\right]+[3(x+3)]+[4(x+3)]=f(x)+35,
+$$
+
+one only needs to investigate the interval $[0,3)$.
+The 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
+
+- $0,1,2$ for $x$;
+- $\frac{n}{2}, 0 \leq n \leq 5$ for $2 x$;
+- $\frac{3 n}{5}, 0 \leq n \leq 4$ for $\frac{5 x}{3}$;
+- $\frac{n}{3}, 0 \leq n \leq 8$ for $3 x$;
+- $\frac{n}{4}, 0 \leq n \leq 11$ for $4 x$.
+
+Of these numbers there are
+
+- 3 integers $(0,1,2)$;
+- 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );
+- 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );
+- 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );
+- 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).
+
+Therefore $f(x)$ 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, $99 \frac{1}{2}, 99 \frac{1}{3}, 99 \frac{2}{3}, 99 \frac{1}{4}$, $99 \frac{3}{4}, 99 \frac{3}{5}$.
+The total is then $33 \cdot 22+8=734$.
+Comment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of
+
+$$
+0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30
+$$
+
+in 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 3
+
+Let
+
+$$
+f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0} \quad \text { and } \quad g(x)=c_{n+1} x^{n+1}+c_{n} x^{n}+\cdots+c_{0}
+$$
+
+be 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:
+
+- $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \ldots, n$, and $a=c \Longrightarrow \frac{a}{c}=1 \leq n+1$.
+- $|r| \geq 1$. Then
+
+$$
+\begin{gathered}
+\left|a_{0}\right|=\left|\frac{c_{0}}{r}\right| \leq c \\
+\left|a_{1}\right|=\left|\frac{c_{1}-a_{0}}{r}\right| \leq\left|c_{1}\right|+\left|a_{0}\right| \leq 2 c
+\end{gathered}
+$$
+
+and inductively if $\left|a_{k}\right| \leq(k+1) c$
+
+$$
+\left|a_{k+1}\right|=\left|\frac{c_{k+1}-a_{k}}{r}\right| \leq\left|c_{k+1}\right|+\left|a_{k}\right| \leq c+(k+1) c=(k+2) c
+$$
+
+Therefore, $\left|a_{k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c \Longleftrightarrow \frac{a}{c} \leq n+1$.
+
+- $0<|r|<1$. Now work backwards: $\left|a_{n}\right|=\left|c_{n+1}\right| \leq c$,
+
+$$
+\left|a_{n-1}\right|=\left|c_{n}-r a_{n}\right| \leq\left|c_{n}\right|+\left|r a_{n}\right|<c+c=2 c,
+$$
+
+and inductively if $\left|a_{n-k}\right| \leq(k+1) c$
+
+$$
+\left|a_{n-k-1}\right|=\left|c_{n-k}-r a_{n-k}\right| \leq\left|c_{n-k}\right|+\left|r a_{n-k}\right|<c+(k+1) c=(k+2) c .
+$$
+
+Therefore, $\left|a_{n-k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c$ again.
+
+## Problem 4
+
+Determine all positive integers $n$ for which the equation
+
+$$
+x^{n}+(2+x)^{n}+(2-x)^{n}=0
+$$
+
+has an integer as a solution.
+
+## Answer: $n=1$.
+
+## Solution
+
+If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd.
+For $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$.
+For $n>1$, 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
+
+$$
+y^{n}+(1+y)^{n}+(1-y)^{n}=0 .
+$$
+
+Looking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization
+
+$$
+a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+\cdots+b^{n-1}\right) \quad \text { for } n \text { odd, }
+$$
+
+which has a sum of $n$ terms as the second factor, the equation is now equivalent to
+
+$$
+y^{n}+(1+y+1-y)\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\cdots+(1-y)^{n-1}\right)=0
+$$
+
+or
+
+$$
+y^{n}=-2\left((1+y)^{n-1}-(1+y)^{n-2}(1-y)+\cdots+(1-y)^{n-1}\right) .
+$$
+
+Each 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 5
+
+Let $P_{1}, P_{2}, \ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties:
+(i) both coordinates of $P_{i}$ are integers, for $i=1,2, \ldots, 1993$;
+(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 $i=0,1, \ldots, 1992$.
+
+Prove 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 $Q$.
+In 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.
+

APMO/md/en-apmo1994_sol.md

@@ -0,0 +1,246 @@
+# APMO 1994 - Problems and Solutions 
+
+## Problem 1
+
+Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that
+(i) For all $x, y \in \mathbb{R}$,
+
+$$
+f(x)+f(y)+1 \geq f(x+y) \geq f(x)+f(y)
+$$
+
+(ii) For all $x \in[0,1), f(0) \geq f(x)$,
+(iii) $-f(-1)=f(1)=1$.
+
+Find all such functions $f$.
+Answer: $f(x)=\lfloor x\rfloor$, the largest integer that does not exceed $x$, is the only function.
+
+## Solution
+
+Plug $y \rightarrow 1$ in (i):
+
+$$
+f(x)+f(1)+1 \geq f(x+1) \geq f(x)+f(1) \Longleftrightarrow f(x)+1 \leq f(x+1) \leq f(x)+2
+$$
+
+Now plug $y \rightarrow-1$ and $x \rightarrow x+1$ in (i):
+
+$$
+f(x+1)+f(-1)+1 \geq f(x) \geq f(x+1)+f(-1) \Longleftrightarrow f(x) \leq f(x+1) \leq f(x)+1
+$$
+
+Hence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \Longrightarrow f(0)=0$.
+Condition (ii) states that $f(x) \leq 0$ in $[0,1)$.
+Now plug $y \rightarrow 1-x$ in (i):
+
+$$
+f(x)+f(1-x)+1 \leq f(x+(1-x)) \leq f(x)+f(1-x) \Longrightarrow f(x)+f(1-x) \geq 0
+$$
+
+If $x \in(0,1)$ 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
+$x+y=\lfloor x\rfloor+\lfloor y\rfloor+\{x\}+\{y\}$ and $0 \leq\{x\}+\{y\}<2 \Longrightarrow\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<\lfloor x\rfloor+\lfloor y\rfloor+2$ implies
+
+$$
+\lfloor x\rfloor+\lfloor y\rfloor+1 \geq\lfloor x+y\rfloor \geq\lfloor x\rfloor+\lfloor y\rfloor .
+$$
+
+## Problem 2
+
+Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.
+
+## Solution 1
+
+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
+
+$$
+O H=|a+b+c| \leq|a|+|b|+|c|=3 R .
+$$
+
+The equality cannot occur because $a, b$, and $c$ are not collinear, so $O H<3 R$.
+
+## Solution 2
+
+Suppose with loss of generality that $\angle A<90^{\circ}$. Let $B D$ be an altitude. Then
+
+$$
+A H=\frac{A D}{\cos \left(90^{\circ}-C\right)}=\frac{A B \cos A}{\sin C}=2 R \cos A
+$$
+
+By the triangle inequality,
+
+$$
+O H<A O+A H=R+2 R \cos A<3 R
+$$
+
+Comment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that
+
+$$
+O H^{2}=9 R^{2}-a^{2}-b^{2}-c^{2} .
+$$
+
+In fact, using vectors in a coordinate system with $O$ as origin, by the Euler line
+
+$$
+\overrightarrow{O H}=3 \overrightarrow{O G}=3 \cdot \frac{\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}}{3}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}
+$$
+
+so
+
+$$
+O H^{2}=\overrightarrow{O H} \cdot \overrightarrow{O H}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
+$$
+
+Expanding 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
+$\overrightarrow{O A} \cdot \overrightarrow{O B}=O A \cdot O B \cdot \cos \angle A O B=R^{2} \cos 2 C=R^{2}\left(1-2 \sin ^{2} C\right)=R^{2}\left(1-2\left(\frac{c}{2 R}\right)^{2}\right)=R^{2}-\frac{c^{2}}{2}$, we find that
+
+$$
+\begin{aligned}
+O H^{2} & =\overrightarrow{O A} \cdot \overrightarrow{O A}+\overrightarrow{O B} \cdot \overrightarrow{O B}+\overrightarrow{O C} \cdot \overrightarrow{O C}+2 \overrightarrow{O A} \cdot \overrightarrow{O B}+2 \overrightarrow{O A} \cdot \overrightarrow{O C}+2 \overrightarrow{O B} \cdot \overrightarrow{O C} \\
+& =3 R^{2}+\left(2 R^{2}-c^{2}\right)+\left(2 R^{2}-b^{2}\right)+\left(2 R^{2}-a^{2}\right) \\
+& =9 R^{2}-a^{2}-b^{2}-c^{2}
+\end{aligned}
+$$
+
+as required.
+This 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 3
+
+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 $n$.
+
+Answer: $n=2,5,13$.
+
+## 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.
+Suppose without loss of generality that $a \geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.
+
+- If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.
+- 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 $a b=b(b+1)$.
+One 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 $b>2$.
+Consider, 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.
+Since $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \geq 0$, 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.
+- 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.
+
+Hence the only solutions are $n=2,5,13$.
+
+## Problem 4
+
+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?
+Answer: Yes.
+
+## Solution 1
+
+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.
+Proof 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 $k \neq 0$.
+We prove a stronger result: let $\sin (k \theta)=\frac{a_{k}}{5^{k}}$. Then
+
+$$
+\begin{aligned}
+\sin ((k+1) \theta)+\sin ((k-1) \theta)=2 \sin (k \theta) \cos \theta & \Longleftrightarrow \frac{a_{k+1}}{5^{k+1}}+\frac{a_{k-1}}{5^{k-1}}=2 \cdot \frac{a_{k}}{5^{k}} \cdot \frac{3}{5} \\
+& \Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1}
+\end{aligned}
+$$
+
+Since $a_{0}=0$ 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.
+
+## Solution 2
+
+We present a different construction. Consider the (collinear) points
+
+$$
+P_{k}=\left(1, \frac{x_{k}}{y_{k}}\right),
+$$
+
+such that the distance $O P_{k}$ from the origin $O$,
+
+$$
+O P_{k}=\frac{\sqrt{x_{k}^{2}+y_{k}^{2}}}{y_{k}}
+$$
+
+is 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.
+Perform 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
+
+$$
+Q_{i} Q_{j}=\frac{1^{2} P_{i} P_{j}}{O P_{i} \cdot O P_{j}}
+$$
+
+is 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
+
+$$
+x_{k}=k^{2}-1, \quad y_{k}=2 k
+$$
+
+This implies $O P_{k}=\frac{k^{2}+1}{2 k}$, and then
+
+$$
+Q_{i} Q_{j}=\frac{\left|\frac{i^{2}-1}{i}-\frac{j^{2}-1}{j}\right|}{\frac{i^{2}+1}{2 i} \cdot \frac{j^{2}+1}{2 j}}=\frac{|4(i-j)(i j+1)|}{\left(i^{2}+1\right)\left(j^{2}+1\right)}
+$$
+
+## Problem 5
+
+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:
+
+| $A$ | $B$ | $C$ |
+| :--- | :--- | :--- |
+| 10 | 1010 | 20 |
+| 100 | 1100100 | 400 |
+| 1000 | 1111101000 | 13000 |
+| $\vdots$ | $\vdots$ | $\vdots$ |
+
+Prove 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
+
+$$
+2^{b_{k}-1} \leq 10^{k}<2^{b_{k}} \Longleftrightarrow \log _{2} 10^{k}<b_{k} \leq \log _{2} 10^{k}+1 \Longleftrightarrow b_{k}=\left\lfloor k \cdot \log _{2} 10\right\rfloor+1
+$$
+
+and, similarly
+
+$$
+c_{k}=\left\lfloor k \cdot \log _{5} 10\right\rfloor+1
+$$
+
+Beatty's theorem states that if $\alpha$ and $\beta$ are irrational positive numbers such that
+
+$$
+\frac{1}{\alpha}+\frac{1}{\beta}=1
+$$
+
+then the sequences $\lfloor k \alpha\rfloor$ and $\lfloor k \beta\rfloor, k=1,2, \ldots$, partition the positive integers.
+Then, since
+
+$$
+\frac{1}{\log _{2} 10}+\frac{1}{\log _{5} 10}=\log _{10} 2+\log _{10} 5=\log _{10}(2 \cdot 5)=1
+$$
+
+the 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.
+Comment: For the sake of completeness, a proof of Beatty's theorem follows.
+Let $x_{n}=\alpha n$ and $y_{n}=\beta n, n \geq 1$ integer. Note that, since $\alpha m=\beta n$ implies that $\frac{\alpha}{\beta}$ is rational but
+
+$$
+\frac{\alpha}{\beta}=\alpha \cdot \frac{1}{\beta}=\alpha\left(1-\frac{1}{\alpha}\right)=\alpha-1
+$$
+
+is irrational, the sequences have no common terms, and all terms in both sequences are irrational.
+The 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
+
+$$
+T(N)=\left\lfloor\frac{N}{\alpha}\right\rfloor+\left\lfloor\frac{N}{\beta}\right\rfloor
+$$
+
+However, $x-1<\lfloor x\rfloor<x$ for nonintegers $x$, so
+
+$$
+\begin{aligned}
+\frac{N}{\alpha}-1+\frac{N}{\beta}-1<T(N)<\frac{N}{\alpha}+\frac{N}{\beta} & \Longleftrightarrow N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)-2<T(N)<N\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\
+& \Longleftrightarrow N-2<T(N)<N,
+\end{aligned}
+$$
+
+that is, $T(N)=N-1$.
+Therefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.
+

APMO/md/en-apmo1995_sol.md

@@ -0,0 +1,158 @@
+# 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
+
+$$
+\sum_{n=1}^{1995} 2 \sqrt{a_{n}-(n-1)} \geq \sum_{n=1}^{1995} a_{n}-\sum_{n=1}^{1994}(n-1)+1=\sum_{n=1}^{1995} a_{n}-\sum_{n=1}^{1995}\{(n-1)+1\}
+$$
+
+i.e.
+
+$$
+0 \geq \sum_{n=1}^{1995}\left\{a_{n}-(n-1)+1-2 \sqrt{a_{n}-(n-1)}\right.
+$$
+
+[2 points]
+Next, note that
+
+$$
+\left[\sqrt{a_{n}-(n-1)}-1\right]^{2}=a_{n}-(n-1)-2 \sqrt{a_{n}-(n-1)}+1
+$$
+
+for $n=1,2, \ldots, 1995$.
+[1 point]
+Hence,
+$0 \geq \sum_{n=1}^{1995}\left[a_{n}-(n-1)-2 \sqrt{a_{n}-(n-1)}+1\right]=\sum_{n=1}^{1995}\left[\sqrt{a_{n}-(n-1)}-1\right]^{2} \geq 0$.
+Therefore, $\sqrt{a_{n}-(n-1)}=1$, for $\mathrm{n}=1,2, \ldots, 1995$. It follows that $a_{n}=\mathrm{n}$ for $\mathrm{n}=1,2, \ldots, 1995$.
+
+## [2 points]
+
+Conversely, if $\sqrt{a_{n}-(n-1)}=1$, for $\mathrm{n}=1,2, \ldots, 1995$, then
+
+$$
+2 \sqrt{n-(n-1)}=2=n+1-(n-1), \text { for } \mathrm{n}=1,2, \ldots, 1994
+$$
+
+and
+
+$$
+2 \sqrt{1995-1994}=2=1+1
+$$
+
+which shows that $a_{\mathrm{n}}=\mathrm{n}$, for $\mathrm{n}=1,2, \ldots, 1995$, is indeed a solution of the given system of inequalities.
+[2 points]
+Question 2. The answer is 14.
+[1 point]
+Denote the required number by M . We observe that the sequence $2 \cdot 101,3 \cdot 97$, $5 \cdot 89,7 \cdot 83,11 \cdot 79,13 \cdot 73,17 \cdot 71,19 \cdot 67,23 \cdot 61=1403,29 \cdot 59=1711,31 \cdot 53=1643$, $37 \cdot 47=1739,41 \cdot 43=1763$ satisfies conditions i) and ii) and contains no prime number. Hence, $M>13$.
+
+## [3 points]
+
+Now we show that a sequence with 14 elements that satisfies conditions i) and ii) will contain a prime number. We proceed by contradiction. Suppose the elements are $a_{1}, a_{2}, \ldots, a_{14}$. Since none of them is a prime number, each element will contain at least two prime factors. We take any two prime factors from each $a_{i}$, and list them in ascending order $\mathrm{p}_{1}<\mathrm{p}_{2}<\ldots<\mathrm{p}_{26}<\mathrm{p}_{27}<\mathrm{p}_{28}$. As the 14 th prime is 43 , this means $43 \leq p_{14}, 47 \leq p_{15}$ and so on. Now $43 \cdot 47=2021>1995$. This means that $p_{14}$ must pair up with one of the $p_{1}, p_{2} \ldots p_{13}$ to form a certain $a_{i}$. Likewise $p_{15}$ must pair up with one of the $p_{1}, p_{2} \ldots p_{13}$ to form another $a_{i}$, and so on (without repetition). Hence there exist $p_{i}, p_{j}, 13<i<j$, that must pair up together to form some $a_{i}$. But then $a_{i} \geq p_{i} p_{j} \geq 43 \cdot 47>1995$, a contradiction.
+[3 points]
+Question 3. Let T be the intersection of PQ and $\mathrm{RS}, \mathrm{T}$ lies outside C , the circle PQRS.
+i) Clearly any point on C belongs to the set A .
+ii) Let $\mathrm{r}=\sqrt{T P \cdot T Q}=\sqrt{T R \cdot T S}$, and consider the circle with center T and radius r. Let $V$ a point on this circle. Since $T V^{2}=T P \cdot T Q=T R \cdot T S$, TV is tangent to the circles PQV and RSV. Therefore, PQV is tangent to RSV. That means, V is in the set A :
+[4 points]
+Conversely, assume V is in A , i.e. PQV is tangent to RSV. If the circles PQV and RSV are the same, then $\mathrm{PQV}=\mathrm{RSV}=\mathrm{PQRS}$. Otherwise, let the line TV intersect $P Q V$ in $V_{1}$, and RSV in $V_{2}$. Then
+$\mathrm{TP} \cdot \mathrm{TQ}=\mathrm{TV} \cdot \mathrm{TV}_{1}$
+$\mathrm{TR} \cdot \mathrm{TS}=\mathrm{TV} \cdot \mathrm{TV}_{2}$.
+
+Due to the fact that $P Q R$ and $S$ are on a circle, we have $T P \cdot T Q=T R \cdot T S$, thus $\mathrm{TV} \cdot \mathrm{TV}_{1}=\mathrm{TV} \cdot \mathrm{TV}_{2}$. Moreover, since T does not lie on $\mathrm{C}, \mathrm{T} \neq \mathrm{V}$, which implies $T V_{1}=T V_{2}$, i.e., $V_{1}=V_{2}=V$.
+All this means that TV is tangent to the circles PQV and RSV, therefore V lies on the circle with center T and radius $\mathrm{r}=\sqrt{T P \cdot T Q}=\sqrt{T R \cdot T S}$.
+[3 points]
+Question 4. First, we will show that MS is perpendicular to A'B'. Since SAMB, SBN'A', SA'M'B' and SB'NA are rectangles, it follows that MNM'N' is a rectangle with its sides parallel to $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$.
+Moreover, the perpendicular bisectors of $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$ pass through O , and they coincide with those of $\mathrm{MN}^{\prime}$ and $\mathrm{NM}^{\prime}$. Therefore, O is the center of the rectangle. Let I and H be the intersections of MS with AB and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$. We then have $\angle \mathrm{HSA}^{\prime}=\angle \mathrm{ASI}$, $\angle \mathrm{ASI}=\angle \mathrm{SAI}$, $\angle \mathrm{SAI}=\angle \mathrm{A}^{\prime} \mathrm{AB}=\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{B}$.
+In the triangle $\mathrm{SA}^{\prime} \mathrm{B}^{\prime}, \angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{B}$ or $\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{S}$ is the complementary angle of $\angle \mathrm{SA}^{\prime} \mathrm{B}^{\prime}$. The angles $\mathrm{HSA}^{\prime}$ and $\mathrm{SA}^{\prime} \mathrm{B}$ are complementary angles and the triangle $\mathrm{SA}^{\prime} \mathrm{H}$ is a right-angled triangle with right angle at H . Therefore, $\mathrm{MS} \perp \mathrm{A}^{\prime} \mathrm{B}$ '.
+[1 point]
+Next, we will show that $A B^{2}+A^{\prime} B^{12}=4 R^{2}$ and that $M N^{\prime 2}+N^{\prime} M^{12}$ is constant.
+Let D be the second intersection of $\mathrm{MN}^{\prime}$ with the circle, then $\mathrm{AD}=\mathrm{AB}$, since they subtend equal angles. This implies
+
+$$
+\mathrm{AB}^{2}+\mathrm{AB}^{\prime} \mathrm{B}^{\prime 2}=\mathrm{A}^{\prime} \mathrm{D}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}
+$$
+
+But, we know $\mathrm{DA}^{\prime} \| \mathrm{MH}$, since $\angle \mathrm{BDA}^{\prime}=\angle \mathrm{BAA}^{\prime}=\angle \mathrm{BMH}$, that means $\angle \mathrm{DA}^{\prime} \mathrm{B}^{\prime}$ $=90^{\circ}$ and it is inscribed in the circle, therefore D and $\mathrm{B}^{\prime}$ are diametrically opposed, what finally implies
+
+$$
+\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\mathrm{A}^{\prime} \mathrm{D}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\mathrm{DB}^{12}=(2 \mathrm{R})^{2}=4 \mathrm{R}^{2}
+$$
+
+i.e.
+
+$$
+\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{\prime 2}=4 \mathrm{R}^{2}
+$$
+
+[2 points]
+To see that $\mathrm{MN}^{12}+\mathrm{N}^{\prime} \mathrm{M}^{12}$ is constant consider the following equalities
+
+$$
+\begin{aligned}
+\mathrm{MN}^{\prime 2}=\left(\mathrm{MB}+\mathrm{BN} N^{\prime}\right)^{2} & =\mathrm{MB}^{2}+\mathrm{BN}^{\prime 2}+2 \mathrm{MB} \cdot \mathrm{BN}^{\prime} \\
+& =\mathrm{SA}^{2}+\mathrm{SA}^{\prime 2}+2 \mathrm{SA} \cdot \mathrm{SA}^{\prime 2} \\
+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=\left(\mathrm{N}^{\prime} \mathrm{A}^{\prime}+\mathrm{A}^{\prime} \mathrm{M}^{\prime}\right)^{2} & =\mathrm{N}^{\prime} \mathrm{A}^{\prime 2}+\mathrm{A}^{\prime} \mathrm{M}^{\prime 2}+2 \mathrm{~N}^{\prime} \mathrm{A}^{\prime} \cdot \mathrm{A}^{\prime} \mathrm{M}^{\prime} \\
+& =\mathrm{SB}^{2}+\mathrm{SB}^{\prime 2}+2 \mathrm{SB} \cdot \mathrm{SB}^{\prime}
+\end{aligned}
+$$
+
+By Pythagoras, we have
+
+$$
+\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}=\left(\mathrm{SA}^{2}+\mathrm{SB}^{2}\right)+\left(\mathrm{SA}^{12}+\mathrm{SB}^{12}\right)
+$$
+
+This implies,
+
+$$
+\begin{aligned}
+\mathrm{MN}^{\prime 2}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}= & \mathrm{SA}^{2}+\mathrm{SB}^{2}+\mathrm{SA}^{12}+\mathrm{SB}^{12}+2 \mathrm{SA} \cdot \mathrm{SA}^{\prime}+2 \mathrm{SBSB}^{\prime} \\
+& =\mathrm{AB}^{2}+\mathrm{A}^{\prime} \mathrm{B}^{12}+4 \mathrm{SA} \cdot \mathrm{SA}^{\prime} \\
+& =8 \mathrm{R}^{2}-40 \mathrm{~S}^{2}
+\end{aligned}
+$$
+
+Additionally we know that
+
+$$
+\mathrm{MN}^{\prime 2}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=\mathrm{MM}^{12}=4 \mathrm{OM}^{2}
+$$
+
+[2 points]
+But, $40 M^{2}=8 R^{2}-40 S^{2}$.
+Therefore,
+
+$$
+\mathrm{MN}^{12}+\mathrm{M}^{\prime} \mathrm{N}^{\prime 2}=4 \mathrm{OM}^{2}
+$$
+
+This last quantity is clearly a constant.
+
+## [1 point]
+
+Finally, it is clear that the vertices of the rectangle $M N N^{\prime} N^{\prime}$ lie on the circle with center O and radius $\mathrm{OM}=\sqrt{2 R^{2}-O S^{2}}$. Therefore, the set of points consists of a circle.
+
+## [1 point]
+
+Question 5. The minimum value of $k$ is $\mathrm{k}^{*}=4$.
+
+## [1 point]
+
+First, we define a function $f$ from $Z$ to $\{1,2,3,4\}$ recursively as follows: $f(0)=1$. For any positive integer $i, f(i)$ is defined to be the minimum positive integer not in $A_{i}:=\{\mathrm{f}(\mathrm{j}): \mathrm{i}-\mathrm{j} \in\{5,7,12\}$ and $-\mathrm{i}<\mathrm{j}<\mathrm{i}\}$, and $\mathrm{f}(-\mathrm{i})$ the minimum positive integer not in $B_{i}:=\{f(j): j+i \in\{5,7,12\}$ and $-i<j<i\}$. Note that $\left|A_{i}\right| \leq 3$ and $\left|B_{i}\right| \leq 3$ for any $i$. So, f is a function from Z to $\{1,2,3,4\}$ such that $\mathrm{f}(\mathrm{x}) \neq f(\mathrm{y})$ whenever $\mid \mathrm{x}$ $-\mathrm{y} \mid \in\{5,7,12\}$. This gives that $\mathrm{k}^{*} \leq 4$.
+[3 points]
+Next, we claim that $k^{*} \geq 4$. Suppose it is not the case. Then there exists a function f from Z to $\{1,2,3\}$ with the property that $\mathrm{f}(\mathrm{x}) \neq \mathrm{f}(\mathrm{y})$ whenever $|\mathrm{x}-\mathrm{y}| \in$ $\{5,7,12\}$. For any integer $x$, consider the values $f(x), f(x-5)$, and $f(x+7)$. These three values are different. Now consider $f(x+2)$. Since $f(x+2) \notin\{f(x-5), f(x+$ 7)\}
+
+$$
+f(x)=f(x+2) \text { for any integer } x
+$$
+
+Hence,
+
+$$
+\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{x}+2)=\mathrm{f}(\mathrm{x}+4)=\ldots=\mathrm{f}(\mathrm{x}+12)
+$$
+
+which is impossible. Thus $\mathrm{k}^{*}>4$.
+[3 points]
+

APMO/md/en-apmo1996_sol.md

@@ -0,0 +1,207 @@
+# Solutions 
+
+Problem 1. Let M'N' and $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}$ be two segments perpendicular to BD at a distance d . We need to prove that the perimeters $A M N C Q P$ and $A M^{\prime} N^{\prime} C Q^{\prime} P^{\prime}$ are equal. Denote by $\mathrm{S}^{\prime}$ the projection of N into M'N' and by S the projection of $\mathrm{P}^{\prime}$ into PQ . The triangles $\mathrm{NS}^{\prime} \mathrm{N}^{\prime}$ and $\mathrm{P}^{\prime} \mathrm{SP}$ are equal. (right triangles such that NS' = P'S and equal angles $\angle \mathrm{P}^{\prime} \mathrm{PS}=\angle \mathrm{NN}^{\prime} \mathrm{S}^{\prime}$.) So $\mathrm{PP}^{\prime}=\mathrm{NN}^{\prime}$. Since it is clear that $\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}$, we have $\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}=\mathrm{PP}^{\prime}=\mathrm{QQ} \mathrm{Q}^{\prime}$. On the other hand, since $\mathrm{SP}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}$, using $\mathrm{T}^{\prime}$ and T the projections of M and $\mathrm{Q}^{\prime}$ into $\mathrm{M}^{\prime} \mathrm{N}^{\prime}$ and PQ , respectively, we have
+
+$$
+\mathrm{T}^{\prime} \mathrm{M}^{\prime}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}=\mathrm{PS}=\mathrm{TQ}
+$$
+
+So
+
+$$
+\begin{aligned}
+\mathrm{P}^{\prime} \mathrm{Q}^{\prime}+\mathrm{M}^{\prime} \mathrm{N}^{\prime}= & \mathrm{ST}+\mathrm{M}^{\prime} \mathrm{T}^{\prime}+\mathrm{T}^{\prime} \mathrm{S}^{\prime}+\mathrm{S}^{\prime} \mathrm{N}^{\prime} \\
+& =\mathrm{ST}+\mathrm{PS}+\mathrm{TQ}+\mathrm{MN} \\
+& =\mathrm{PQ}+\mathrm{MN} .
+\end{aligned}
+$$
+
+(Without loss of generality $\mathrm{M}^{\prime}$ lies between M and A )
+So the perimeter of $A M N C Q P$ is:
+
+$$
+\begin{aligned}
+& \mathrm{AM}+\mathrm{MN}+\mathrm{NC}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
+& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{M}+\mathrm{MN}+\mathrm{NN}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
+& \mathrm{AM}^{\prime}+\mathrm{PP}^{\prime}+\mathrm{MN}+\mathrm{QQ}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QP}+\mathrm{PA}= \\
+& \mathrm{AM}^{\prime}+\mathrm{MN}+\mathrm{QP}+\mathrm{PP}^{\prime}+\mathrm{PA}+\mathrm{QQ}^{\prime}+\mathrm{CQ}+\mathrm{N}^{\prime} \mathrm{C}= \\
+& \mathrm{AM}^{\prime}+\mathrm{MN}+\mathrm{QT}+\mathrm{TS}+\mathrm{SP}+\mathrm{PP}^{\prime}+\mathrm{PA}+\mathrm{QQ}^{\prime}+\mathrm{CQ}+\mathrm{N}^{\prime} \mathrm{C}= \\
+& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{T} \text { '+ T'S' + S'N' } \mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}+\mathrm{QQ}^{\prime}+\mathrm{Q}^{\prime} \mathrm{P}^{\prime}+\mathrm{P}^{\prime} \mathrm{P}+\mathrm{PA}= \\
+& \mathrm{AM}^{\prime}+\mathrm{M}^{\prime} \mathrm{N}^{\prime}+\mathrm{N}^{\prime} \mathrm{C}+\mathrm{CQ}^{\prime}+\mathrm{Q}^{\prime} \mathrm{P}^{\prime}+\mathrm{P}^{\prime} \mathrm{A},
+\end{aligned}
+$$
+
+which is the perimeter of $\mathrm{AM}^{\prime} \mathrm{N}^{\prime} \mathrm{CQ}^{\prime} \mathrm{P}$ '.
+
+Points to be given for showing that:
+$\mathrm{MM}^{\prime}=\mathrm{NN}^{\prime}=\mathrm{PP}^{\prime}=\mathrm{QQ}^{\prime}$
+1 point
+$\mathrm{T}^{\prime} \mathrm{M}^{\prime}=\mathrm{S}^{\prime} \mathrm{N}^{\prime}=\mathrm{PS}=\mathrm{TQ} . \quad 1$ point
+$P^{\prime} Q^{\prime}+M^{\prime} N^{\prime}=P Q+M N$.
+2 points
+
+The perimeter of $A M N C Q P$ isthe same as the perimeter of $A M^{\prime} N^{\prime} \mathrm{CQ}^{\prime} \mathrm{P}^{\prime}$
+
+3 points
+Problem 2. We first prove by induction on n that:
+
+$$
+\frac{(m+n)!}{(m-n)!}=\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)
+$$
+
+1. $\frac{(m+1)!}{(m-1)!}=m(m+1)=m^{2}+m$.
+2. $\frac{(m+n+1)!}{(m-n-1)!}=\left(\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)\right)(m+n+1)(m-n)$ (by induction)
+
+$$
+\begin{gathered}
+=\left(\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)\right)\left(m^{2}+m-n^{2}-n\right) \\
+=\prod_{i=1}^{n+1}\left(m^{2}+m-i^{2}+i\right)
+\end{gathered}
+$$
+
+But $m^{2}+m \geq m^{2}+m-i^{2}+i \geq i^{2}+i-i^{2}+i=2 i$, for $i \geq m$.
+Therefore
+
+$$
+2^{n} n!\leq \frac{(m+n)!}{(m-n)!} \leq\left(m^{2}+m\right)^{n}
+$$
+
+Points to be given for showing that:
+the innequlities hold for special values
+up to 1 point
+$\frac{(m+n)!}{(m-n)!}=\prod_{i=1}^{n}\left(m^{2}+m-i^{2}+i\right)$ for all $n$
+4 points
+$m^{2}+m \geq m^{2}+m-i^{2}+i \geq i^{2}+i-i^{2}+i=2 i$, for $i \geq m$, and therefore the innequalities hold
+
+2 points
+
+Problem 3. Let C be the given circle. Draw four circles $\mathrm{C}_{12}, \mathrm{C}_{23}, \mathrm{C}_{34}, \mathrm{C}_{41}$ with centers $\mathrm{O}_{12}, \mathrm{O}_{23}$, $\mathrm{O}_{34}, \mathrm{O}_{41}$, respectively, on the circle C such that $\mathrm{C}_{12}$ passes through $\mathrm{P}_{1}$ and $\mathrm{P}_{2}, \mathrm{C}_{23}$ passes through $\mathrm{P}_{2}$ and $P_{3}, C_{34}$ passes through $P_{3}$ and $P_{4}, C_{41}$ passes through $P_{4}$ and $P_{1}$. Let the other point of intersection of $\mathrm{C}_{12}$ and $\mathrm{C}_{23}$ be $\mathrm{Q}_{4}$, the other point of intersection of $\mathrm{C}_{23}$ and $\mathrm{C}_{34}$ be $\mathrm{Q}_{11}$, the other point of intersection of $C_{34}$ and $C_{41}$ be $Q_{2}$, and the other point of intersection of $C_{41}$ and $C_{12}$ be $Q_{3}$. Then
+
+$$
+\angle \mathrm{Q}_{4} \mathrm{P}_{1} \mathrm{P}_{2}=\frac{1}{2} \angle \mathrm{Q}_{4} \mathrm{O}_{12} \mathrm{P}_{2} \text { and } \angle \mathrm{O}_{23} \mathrm{P}_{1} \mathrm{P}_{2}=\frac{1}{2} \angle \mathrm{P}_{3} \mathrm{O}_{12} \mathrm{P}_{2}
+$$
+
+Clearly, $\mathrm{O}_{23}, \mathrm{Q}_{4}$ and $\mathrm{P}_{1}$ are collinear.
+It follows that $\angle \mathrm{Q}_{4} \mathrm{Q}_{3} \mathrm{P}_{2}=\angle \mathrm{Q}_{4} \mathrm{P}_{1} \mathrm{P}_{2}=\angle \mathrm{O}_{23} \mathrm{P}_{1} \mathrm{P}_{2} \angle \mathrm{O}_{23} \mathrm{O}_{41} \mathrm{P}_{2}$. Since also $\mathrm{O}_{41}, \mathrm{Q}_{3}$ and $\mathrm{P}_{2}$ are collinear, it follows that $\mathrm{Q}_{3} \mathrm{Q}_{4}$ and $\mathrm{O}_{41} \mathrm{O}_{23}$ are parallel.
+Since $\mathrm{O}_{\mathrm{ij}}$ bisects The arcs $\mathrm{P}_{\mathrm{i}} \mathrm{P}_{\mathrm{j}}$, for $(\mathrm{i}, \mathrm{j})=(1,2),(2,3),(3,4),(4,1)$ we conclude that $\mathrm{O}_{41} \mathrm{O}_{23}$ and $\mathrm{O}_{12} \mathrm{O}_{34}$ are perpendicular, and hence $\mathrm{Q}_{3} \mathrm{Q}_{4}$ and $\mathrm{O}_{12} \mathrm{O}_{34}$ are perpendicular.
+Since both $\left(\mathrm{O}_{12}, \mathrm{Q}_{3}, \mathrm{P}_{4}\right)$ and $\left(\mathrm{O}_{12}, \mathrm{Q}_{4}, \mathrm{P}_{3}\right)$ are collinear triples of points, we have $\angle \mathrm{P}_{4} \mathrm{O}_{12} \mathrm{P}_{3}=\angle$ $\mathrm{Q}_{3} \mathrm{O}_{12} \mathrm{Q}_{4}$, and this angle is bisected by $\mathrm{O}_{12} \mathrm{O}_{34}$.
+Thus $Q_{3}$ and $Q_{4}$ are reflections through the axis $\mathrm{O}_{12} \mathrm{O}_{34}$, and so are, by a similar argument $\mathrm{Q}_{1}$ and $\mathrm{Q}_{2}$.
+We have thus shown: $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ form a rectangle. But as $Q_{4}$ lies on both the angle bisector $\mathrm{O}_{12} \mathrm{P}_{3}$ and the angle bisector $\mathrm{O}_{23} \mathrm{P}_{1}$ of the triangle $\mathrm{P}_{1} \mathrm{P}_{2} \mathrm{P}_{3}$, the point $\mathrm{Q}_{4}$ must coincide with the incenter $\mathrm{I}_{4}$ of the triangle $\mathrm{P}_{1} \mathrm{P}_{2} \mathrm{P}_{3}$, and by a similar argument, $\mathrm{Q}_{1}=\mathrm{I}_{1}, \mathrm{Q}_{2}=\mathrm{I}_{2}$ and $\mathrm{Q}_{3}=\mathrm{I}_{3}$.
+
+Points to be given for showing that
+$\mathrm{Q}_{3} \mathrm{Q}_{4}$ and $\mathrm{O}_{41} \mathrm{O}_{23}$ are parallel
+$\mathrm{Q}_{3} \mathrm{Q}_{4}$ and $\mathrm{O}_{12} \mathrm{O}_{34}$ are perpendicular 1 points
+$\mathrm{Q}_{1}, \mathrm{Q}_{2}, \mathrm{Q}_{3}, \mathrm{Q}_{4}$ form a rectangle 2 points
+$\mathrm{Q}_{4}=\mathrm{I}_{4}, \mathrm{Q}_{1}=\mathrm{I}_{1}, \mathrm{Q}_{2}=\mathrm{I}_{2}$ and $\mathrm{Q}_{3}=\mathrm{I}_{3} \quad 1$ point
+
+Problem 4. . We may assume that the n couples will form x male groups and y female groups. Without loss of generality, let $x \geq y$, and
+
+$$
+x+y=17
+$$
+
+Then, by the pigeonhole theorem, there exists a male group of size $\leq\left[\frac{n}{x}\right]$ and a female group of size $\geq\left[\frac{n}{y}\right]$. By condition (2), we have
+
+$$
+\left[\frac{n}{y}\right]-\left[\frac{n}{x}\right] \leq 1
+$$
+
+From the conditions $\mathrm{x}+\mathrm{y}=17$ and $\mathrm{x} \geq \mathrm{y}$ follows that $\mathrm{x} \geq 9$, and $\mathrm{y} \leq 8$, which in turn implies
+
+$$
+\left[\frac{n}{y}\right]-\left[\frac{n}{x}\right]>\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]
+$$
+
+Therefore, we only need to exclude those n such that $\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1$. Let $\mathrm{n}=9 \mathrm{u}+\mathrm{s}, 0 \leq \mathrm{s}<9$. Then
+
+$$
+\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1 \Leftrightarrow\left[\frac{u+s}{8}\right]>1
+$$
+
+By analyzing this condition it is clear that the only values of n that are allowed are
+
+$$
+\begin{gathered}
+\mathrm{n}=9,10,11,12,13,14,15,16,18,19,20,21,22,23,24,27,28,29,30, \\
+\quad 31,32,36,37,38,39,40,45,46,46,47,48,54,55,56,63,63,72 .
+\end{gathered}
+$$
+
+Conversely, conditions $\left(^{*}\right)$ and $\left({ }^{* *}\right)$ give rise to a set of discussing groups according to the following description:
+
+Let $\left[\frac{n}{x}\right]=p$ and $\left[\frac{n}{y}\right]=q$, then $p \leq q \leq p+1$
+We have
+
+$$
+\mathrm{n}=\mathrm{px}+\alpha, 0 \leq \alpha<\mathrm{x} \quad \text { and } \mathrm{n}=\mathrm{qy}-\beta \quad 0 \leq \beta<\mathrm{y}
+$$
+
+So we can arrange the males in $\alpha$ discussing groups of size $p+1$ and $x-\alpha$ groups of $p$ elements. The females are distributed in $\beta$ discussing groups of size $q-1$, and $(y-\beta)$ discussing groups of size q.
+
+Points to be given for showing that
+$x \geq 9$, and $y \leq 8 \quad 1$ points
+one only needs to exclude those n such that $\left[\frac{n}{8}\right]-\left[\frac{n}{9}\right]>1 \quad 2$ points
+analyzing the condition ian giving the values of $n$ that are allowed 2 points
+given p and q as described in the solution one can arrange the males in $\alpha$ discussion groups of size $p+1$
+and $\mathrm{x}-\alpha$ groups of p elements and the females in $\beta$ discussion groups of size $q-1$, and $(y-\beta)$ groups of size $q$. 2 points
+
+Problem 5. Without loss of generality we can assume that $a \geq b \geq c$. Note that if $x \geq y>0$, then $\sqrt{y} \leq 1 / 2(\sqrt{x}+\sqrt{y})$, i.e.,
+
+$$
+\frac{1}{2 \sqrt{y}}(\sqrt{x}+\sqrt{y}) \geq 1
+$$
+
+Similarly, if $y \geq x>0$, then
+
+$$
+\frac{1}{2 \sqrt{y}}(\sqrt{x}+\sqrt{y}) \leq 1
+$$
+
+Multiplying both inequalities by $\sqrt{x}-\sqrt{y}$ we obtain
+
+$$
+\sqrt{x}-\sqrt{y} \leq \frac{1}{2 \sqrt{y}}(x-y)
+$$
+
+for every $\mathrm{x}, \mathrm{y}>0$. Moreover, it is easily seen that equality occurs if and only if $\mathrm{x}=\mathrm{y}$. By applying this last inequality we obtain
+
+$$
+\sqrt{a+b-c}-\sqrt{a} \leq \frac{1}{2 \sqrt{a}}(b-c)
+$$
+
+$$
+\begin{aligned}
+\sqrt{c+a-b}-\sqrt{b} & \leq \frac{1}{2 \sqrt{b}}(c+a-2 b) \\
+\sqrt{b+c-a}-\sqrt{c} & \leq \frac{1}{2 \sqrt{c}}(b-a)
+\end{aligned}
+$$
+
+and by adding up the left hand and the right hand sides of these inequalities we have
+
+$$
+\begin{aligned}
+& \sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}-(\sqrt{a}+\sqrt{b}+\sqrt{c}) \\
+& \frac{1}{2}\left(\frac{1}{\sqrt{a}}(b-c)+\frac{1}{\sqrt{b}}(c+a-2 b)+\frac{1}{\sqrt{c}}(b-a)\right) \\
+& \frac{1}{2}\left((b-c)\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)+(a-b)\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{c}}\right)\right) \leq 0,
+\end{aligned}
+$$
+
+i.e.,
+
+$$
+\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b} \leq \sqrt{a}+\sqrt{b}+\sqrt{c},
+$$
+
+and equality occurs if and only if the three relations in $\left({ }^{* *}\right)$ are equalities, i.e., if and only if $a=b=c$.
+
+Points to be given for showing that
+$\sqrt{x}-\sqrt{y} \leq \frac{1}{2 \sqrt{y}}(x-y)$
+2 points
+the relations in $\left({ }^{* *}\right)$ hold true
+1 point each
+$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}-(\sqrt{a}+\sqrt{b}+\sqrt{c}) \leq$
+1 point
+equality occurs if and only if the three relations in $\left({ }^{(*)}\right.$ are
+equalities, i.e., if and only $a=b=c$
+1 point
+

APMO/md/en-apmo1997_sol.md

@@ -0,0 +1,232 @@
+# Solutions 
+
+Note: The points to be awarded for each part of the solution are indicated on the right side.
+Problem 1.
+
+$$
+\begin{gathered}
+1=\frac{1 \times 2}{2} \\
+1+\frac{1}{3}=\frac{2 \times 2}{3} \\
+1+\frac{1}{3}+\frac{1}{6}+\ldots+\frac{1}{r_{n}}=\frac{n<2}{n+1}
+\end{gathered}
+$$
+
+which is easily shown by induction.
+(up to 3 points)
+Now 5 is the sum of the reciprocals of these numbers where the last, $1993006=$
+$1996 \times 1997$ $\frac{1996 \times 1997}{2}=8996$. Thus we have
+
+$$
+\begin{gathered}
+S=\frac{1}{2}\left(\frac{2}{1}+\frac{3}{2}+\ldots+\frac{1997}{1996}\right) \\
+=\frac{1}{2}\left(1996+\left(1+\frac{1}{2}+\ldots+\frac{1}{1996}\right)\right)
+\end{gathered}
+$$
+
+(ap to 3 points)
+
+$$
+\therefore \frac{1}{2}(1996+6)
+$$
+
+$$
+=1001
+$$
+
+P'roblem 2. Note that $2^{n+2}=2\left(2^{n-1}+1\right)$ so that $n$ is of the form 2 r with r odd. We will consider two cases.
+i) $n=2 p$ with $p$ prime. $2 p \mid 2^{2 p}+2$, implies that $p!2^{2 n-1}+1$ and hence, hence $p \mid 2^{40-2}-1$. On the
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-1.jpg?height=82&width=1678&top_left_y=1935&top_left_x=229) follows that $p \mid 2^{d}-1$. But $d \mid p-1$ and $d \mid 4 p-2=4(p-1)+2$. Fence $d \mid 2$ and since $p-1,4 p-2$ are even $d=2$. Then $p=3$ and $n=6<100$.
+(up to 2 points)
+ii) $n=2 p q$ where $p, q$ are odd primes, $p<q$ and $p q<\frac{1997}{2}$. Now $n!2 r+2$ impites that $p i 2^{n-1}$ +1 and therefore that $\mathrm{p}^{\prime} 2^{2 \mathrm{~m}-2} . .1=2^{4 \mathrm{~m}-2}-1$. Once again by Femat's theorern we have pl $2^{n-1}$ 1 which implies that $\mathrm{p}-1 / 4 \mathrm{pq}$-2. The same holds tole for q so that
+
+$$
+q-1 \mid 4 p q-2
+$$
+
+Both $p-1$ and $q-1$ are thus multiples of 2 but not of 4 su that $p \equiv q \equiv 3$ (mod 4).
+( points)
+laking $p=3$, we huve $4 p q-2=12 q-2$. Now from (1) we have
+
+$$
+12=\frac{12 q-12}{q-1}<\frac{12 q-2}{q-1}=\frac{12(q-1)+10}{q-1}=12+\frac{10}{q-1} \leq 1
+$$
+
+if $q \geq 11$, and clearly $\frac{12 q-?}{q-1}=13$ if $q=11$. But this gives $n=2(3)(11)=66<100$. Furthermore $(p, \varphi)=:(3,7)$ does not salisty (1).
+
+Taking $p=7$ we observe that $4 \mathrm{pq}-2=28 \mathrm{q}-2$, and from (1) we have
+
+$$
+28<\frac{28 q-2}{q-1}=\frac{28(q-1)+26}{q-1}=28+\frac{26}{q-1} \leq 2
+$$
+
+if $q \geq 27$ and clearly $\frac{28 q-2}{q-1}=29$ if $q=27$. But 27 is not prime and the cases $(p, q)=(7$. 11), (7,19) and (7,23) do not satisfy (1).
+
+Taking $p=11$, then $4 p q-2=44 q-2$, and
+
+$$
+44<\frac{44 q-2}{y-1} \text { and } \frac{44 q-2}{q-1} \leq 45 \text { if } q \geq 43 .
+$$
+
+Now clearly $\frac{44 q-2}{q-1}=45$ when $q=43$. In this case we have $n:=2 p q=2(11)(43)=946$.
+Furthermore, $\frac{2^{44 x}+2}{946}$ is incieed an integer. The cases $(p, q)=(11,19),(11,23)$ and 111,31$)$ do not satisfy (1).
+(2 points)
+[Aditionally for completeness, if $p=19$ then $4 p y-2=76 q-2$ and $76<\frac{76 q-2}{4-1}=77$ if $q$ $\geqq 75$. Now 75 is not prime and for the cases $(p, 4)=(19,23),(19,31),(19,43)$ and $(19,47), q$ -1 is not a divisor of $74=2 \times 37$.
+
+Similarly, if $p=23$ then $4 \mathrm{pq}-2=92 \mathrm{q}-2$ and $92<\frac{92 q-2}{q-\Gamma} \leq 93$ if $q \geq 91$ and $\frac{92 q-2}{q-1}=93$ if $q=91$. But 91 is not prime and of the cases $(p, q)=(23,31),(23,43)$, when q $=31$ all of the conditions are satisfied. But, $n=2 p q=1426$ is not a solution because $\frac{2^{1426}+2}{1+20}$
+is not an integer.
+
+No other pairs of F, q yield numbers within the required range.]
+(1) point)
+
+## Problem 3.
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=864&width=1050&top_left_y=690&top_left_x=565)
+$\angle A L D=\frac{1}{2}(\overparen{M C}+\widehat{A B})$
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=147&width=386&top_left_y=1114&top_left_x=1718)
+$\angle \triangle N M$
+
+It is known (see Geometry Revisited) or easily derivable that
+
+$$
+m_{a}^{2}=(A L)^{2}=b c\left(1-\left(\frac{a}{b+c}\right)^{2}\right)
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=275&width=570&top_left_y=1559&top_left_x=1520)
+(1 point)
+From $\triangle$ ADL a MAN we have
+
+$$
+\begin{gathered}
+\frac{A D}{A L}=\frac{4 M}{M N} \Rightarrow \\
+h_{a} \cdot 2 R=A L \cdot A M=m_{a} \cdot M a \\
+h_{a}=\frac{2(A B C)}{a}
+\end{gathered}
+$$
+
+$$
+\begin{gathered}
+\frac{2(A B C)}{a} \cdot 2 R=m_{a} M_{a} \\
+\frac{\frac{u b c}{d} \cdot 4 R}{a}=m_{a} M_{a} \\
+b c=m_{a} M_{u}
+\end{gathered}
+$$
+
+So that
+
+$$
+l_{a}=\frac{m_{a}^{2}}{m_{a} I_{a}}=1-(b+c)^{2}
+$$
+
+with similar expressions for $l_{b}$ and $l_{c}$.
+(2 points)
+$\therefore$ Given that $\sin A=\frac{t^{2}}{2 k}$, etc. the expression we are working with becomes
+
+$$
+\begin{aligned}
+& \frac{l_{a}}{\sin ^{2} \cdot A}+\frac{l_{b}}{\sin ^{2} A}+\frac{l_{b}}{\sin ^{2} C^{2}}=\frac{4 R^{2}}{a^{2}}\left(1-\left(\frac{b^{2}}{b+c}\right)^{2}\right)+\frac{4 R^{2}}{b^{2}}\left(1-\left(\frac{b}{a+c}\right)^{2}\right)+\frac{4 R^{2}}{c^{2}}\left(1-\left(\frac{i}{a-b}\right)^{2}\right) \\
+& =4 R^{2}\left[\left(\frac{1}{a^{2}}-\frac{1}{(b+c)^{2}}\right)+\left(\frac{1}{b^{2}}-\frac{1}{(a+c)^{2}}\right)+\left(\frac{1}{c^{2}}-\frac{1}{(a+b)^{2}}\right)\right] \\
+& \geq 4 R^{2}\left[\frac{1}{a^{2}}-\frac{1}{4 b c}+\frac{1}{b^{2}}-\frac{1}{4 a c}+\frac{1}{c^{2}}-\frac{1}{4 a b}\right] \\
+& =2 \mathrm{R}^{2}\left[\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+\left(\frac{1}{a^{2}}+\frac{1}{c^{2}}\right)+\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-\frac{1}{2 a b}-\frac{1}{2 a c}-\frac{1}{2 b c}\right] \\
+& 2\left[\frac{2}{a b}+\frac{2}{a c}+\frac{2}{b c}-\frac{1}{2 a b}-\frac{1}{2 a c}-\frac{1}{2 b c}\right] \\
+& =2 R^{2}\left[2 \frac{3}{2 b}+\frac{3}{2 a c}+\frac{3}{2 b c}\right] \\
+& =3 R^{2}\left[\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}\right]=3 R^{2}\left[\frac{a+b+c}{a b c}\right]
+\end{aligned}
+$$
+
+But $a b c=f R(.4 B C)$ so that this last expression becomes
+
+$$
+\frac{3 R(a+b+c)}{4(A B C)}=\frac{3 R \cdot 2 s}{4 s r}=3 \cdot \frac{\beta}{2 r}=3
+$$
+
+since $R \geq 2 r$. All of the inequalitics are equalities iff $a=b=c$.
+(1 point)
+
+## Problem 4.
+
+- (a) Consider the gequence of triangles on the plane 4 . $42,4, A_{2}, 4,4,4,4,4, \ldots$ It is easy to see that any pair of them are similar. Let's prove that triangles $A_{1} A_{3} A_{5}$ and $A_{3} A$; are similar.
+Triangles $A_{2} A_{3} A_{4}$ and $A_{4} A_{4} A_{6}$ are similar and their altitudes are $A_{4} A_{5}$ and $A_{6}, A_{0}$, then
+
+$$
+\frac{A_{2} A_{3}}{A_{4}}=\frac{A_{4} A_{5}}{A_{6} A_{7}} .
+$$
+
+Triangles $A A_{*} A_{s}$ and $A_{s} A_{*} A$ - are similar. then
+
+$$
+\frac{A+A_{5}}{A_{0} A_{7}}=\frac{A 3 A s}{A_{5} A_{7}}
+$$
+
+Now we can conclude that
+
+$$
+\frac{A_{1} A_{3}}{A_{3} A_{s}}=\frac{A_{3} A_{s}}{A_{4}}
+$$
+
+and triangles $A, A: A$ and $A ; A: A-$ are similar.
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-5.jpg?height=63&width=1644&top_left_y=1389&top_left_x=227) $\triangle \quad A \cdot A_{s} A_{1}=\triangle \quad A-A_{3} P=\triangle \quad A_{A} A_{1}$, so triangle $A-A, P$ has a righr angle at $P$ and lines $A_{1} A_{5}$ and $A A^{A}$ - are perpendicular. In the same way lines $A A_{2}$. and $A$ are perpendicular and lines $A, A y$ and $A_{-1} A_{11}$ are perpendicular, hence $A_{1}, A_{3}, A_{9}$ are collinear and $A_{-}, A_{1 i} . A_{j}$ are collincar. It follows that triangle $A_{l A} A_{2} A_{;}$and $A_{0} A_{10} A_{l \mid}$ are homothetic and the center of homothety is P . Moreover, all
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-5.jpg?height=56&width=1630&top_left_y=1639&top_left_x=231) an interior point to any of these triangles and there is no other point distinct from $P$ that is interior to any of these triangles. So this is the point we are looking for.
+
+$$
+\text { (up to }+ \text { points) }
+$$
+
+(b) Since $\triangle \quad A_{1} P A_{3} 90^{\prime \prime}$ then $P$ lies on the circle with diameter $A_{i}, A_{3}$. Let $A_{i} A_{3}=1, A_{2} A_{2}=s, A_{3} A_{3}$ $=r$ and let $A_{i} A_{2} A_{3}$ be clockwise. Triangles $A_{1} A_{2} A_{3}$ and $A_{3} A_{+} A_{5}$ are similar, thus $A_{2} A_{5}: r=s: 1$, and so $A_{2} A_{5}=r s$. Besides $A_{3} A_{y}=r \sqrt{1+s^{2}}$ (Pythagoras), and area of triangle $A_{1} A_{2} A_{3}=$ $\frac{1}{2} r \sqrt{1+s^{2}} \cdot \sqrt{1+s^{2}}=\frac{1}{2} s \cdot 1$. Thus $r=\frac{s}{1+s^{2}}$. By the arithmetic-geometric mean $\frac{s}{1+s^{2}} \leq \frac{1}{2}$, thus $r \leq \frac{1}{4}$ and the set of all possible values of $r$ consists of two real intervals $\left[-\frac{1}{2}, 0\right)$ and $\left(0, \frac{1}{2}\right]$. $\triangle A_{2} A_{1} P$ takes the maximum value when $r= \pm$ thus the locus of $P$
+consists of two continuous arcs from the circle with dianeter $4, A_{i}$ with two extreme positions corresponding to $r=-\frac{1}{2}$ and $r=\frac{1}{2}$.
+(up to 3 points)
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-6.jpg?height=1558&width=920&top_left_y=390&top_left_x=597)
+
+## Problem 5.
+
+A redistribution can be written as $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ where $x_{1}$ denotes the number of objects transferred from $A_{i}$ to $A_{i+1}$. Our objective is to minimize the function
+$F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n}\left|x_{1}\right|$
+After redistribution we should have at each $A_{i}, a_{i}-x_{1}+x_{i-1}=N$ for $i \in\{1,2, \ldots, n\}$ where $x_{0}$ means $x_{n}$.
+(1 point)
+Solving this system of linear equations we obtain:
+$x_{i}=x_{1}-\left[(i-1) N-a_{2}-a_{3}-\ldots-a_{i}\right]$
+for $i \in\{1,2, \ldots, n\}$.
+Hence
+
+$$
+\begin{aligned}
+F\left(x_{1}, x_{2}, \ldots, x_{n}\right)= & \left|x_{1}\right|+\left|x_{1}-\left(N-a_{2}\right)\right|+\left|x_{1}-2 N-a_{2}-a_{3}\right| \\
+& +\ldots+\left|x_{1}-\left[(n-1) N-a_{2}-a_{3}-\ldots-a_{n}\right]\right|
+\end{aligned}
+$$
+
+Basically the problem reduces to find the minimum of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|$ where $\alpha_{i}=(i-1) N-\sum_{j=2}^{i} a_{j}$.
+(up to. 3 points)
+
+First rearrange $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ in non decreasing order. Collecting terms which are equal to one another we write the ordered sequence $\beta_{1}<\beta_{2}<\cdots<\beta_{m}$, each $\beta_{i}$ occurs $k_{i}$ times in the family $\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right\}$. Thus $k_{1}+k_{2}+\cdots+k_{m}=n$.
+
+Consider the intervals $\left(-\infty, \beta_{1}\right],\left[\begin{array}{l}\beta_{1}, \beta_{2}\end{array}\right], \cdots,\left[\begin{array}{cc}\beta_{m-1}, \beta_{m}\end{array}\right],\left[\begin{array}{l}\left.\beta_{m}, \infty\right) \\ \end{array}\right.$ the graph of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|=\sum_{i=1}^{m} k_{i}\left|x-\beta_{i}\right|$ is a continuos piece wise linear graph define in the following way:
+
+$$
+F(x)=\left\{\begin{array}{c}
+k_{1}\left(\beta_{1}-x\right)+k_{2}\left(\beta_{2}-x\right)+\cdots+k_{m}\left(\beta_{m}-x\right) \text { if } \mathrm{x} \in\left(-\infty, \beta_{1}\right] \\
+k_{1}\left(x-\beta_{1}\right)+k_{2}\left(\beta_{2}-x\right)+\cdots+k_{m}\left(\beta_{m}-x\right) \text { if } \mathrm{x} \in\left[\beta_{1}, \beta_{2}\right] \\
+\vdots \\
+\left.k_{1}\left(x-\beta_{1}\right)+k_{2}\left(x-\beta_{2}\right)+\cdots+k_{m}\right)\left(x-\beta_{m}\right) \text { if } \mathrm{x} \in\left[\beta_{m}, \infty\right)
+\end{array}\right.
+$$
+
+The slopes of each line segment on each interval are respectively: $S_{0}=-k_{1}-k_{2}-k_{3}-\cdots-k_{m}$
+$S_{1}=k_{1}-k_{2}-k_{3}-\cdots-k_{m}$
+$S_{2}=k_{1}+k_{2}-k_{3}-\cdots-k_{m}$
+$S_{m}=k_{1}+k_{2}+k_{3}+\cdots+k_{m}$
+Note that this sequence of increasing numbers goes from a negative to a positive number, hence for some $t \geq 1$ there is an
+
+$$
+S_{t}=0 \text { or } S_{t-1}<0<S_{t}
+$$
+
+In the first case the minimum occurs at $x=\beta_{t}$ or $\beta_{t+1}$ and in the second case the minimum occurs at $x=\beta_{t}$
+(Up to 7 points)
+We can rephrase the computations above in terms of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ rather than $\beta_{1}, \beta_{2}, \cdots, \beta_{m}$. After rearranging the $\alpha^{\prime} s$ in non decreasing order, pick $x=\alpha \quad$ if $n$ is odd and take $x=\alpha$ or $\alpha \quad$ if $n$ is even.
+
+$$
+\frac{n+1}{2} \quad \frac{n}{2} \quad \frac{n}{2}+1
+$$
+
+If no justification is given for the choice of $x$, give up to 4 points.
+

APMO/md/en-apmo1998_sol.md

@@ -0,0 +1,202 @@
+# APMO SOLUTIONS 
+
+PROBLEM 1. Let $F$ be the set of all $n$-luples $\left(A_{1}, \Lambda_{2}, \ldots, A_{n}\right)$ where each $\Lambda_{i}$, $i=$ $1,2, \ldots, n$ is a subset of $\{1,2, \ldots, 1998\}$. Let $|\Lambda|$ denote the number of elements of the set $A$. Find the number
+
+$$
+\sum_{\left(A_{1}, A_{2}, \ldots, A_{n}\right)}\left|A_{1} \cup A_{2} \cup \ldots \cup A_{n}\right|
+$$
+
+## MARKING SCHEME:
+
+Let $M$ be a subset of the set $\{1,2 \ldots, 1998\}$ and let $|M|=k$. Then the set $M$ call be obtained as the union of $t$ sets $A_{1}, A_{2}, \ldots, A_{t}$ in $\left(2^{t}-1\right)^{k}$ different ways since each clement $x \in M$ can belong to $2^{t}-1$ nonempty families of subsets $A_{1}, I_{2}, \ldots, A_{t}$.
+
+$$
+3 \text { points for describing the correct counting method }
+$$
+
+Thus we have
+
+$$
+\begin{aligned}
+& \sum_{\left(A_{1}, A_{2}, \ldots, A_{t}\right) \in F}\left|A_{1} \cup A_{2} \cup \ldots \cup A_{t}\right|=\sum_{k=1}^{1998} k\binom{1998}{k}\left(2^{t}-1\right)^{k} \\
+& 2 \text { points for setting up the above formula } \\
+& =1998\left(2^{\prime}-1\right) \sum_{k=0}^{1997}\binom{1997}{k}\left(2^{t}-1\right)^{k}=1998\left(2^{t}-1\right) 2^{1997 t} . \\
+& 2 \text { points for corrcctly carrying out the computation }
+\end{aligned}
+$$
+
+PROBLEM 2. Show that for any positive integers $a$ and $b,(36 a+b)(a+36 b)$ cannot be a power of 2 .
+
+MARKING SCCHEME:
+Suppose that $(36 a+b)(a+36 b)$ is a power of 2 for some positive integers $a$ and $b$. Write $36 a+b=2^{m}=r$ and $a+36 b=2^{n}=s$. Then
+
+$$
+36 r-s=35 \times 3 \overline{7} a, a n d, 36 s-r=35 \times 37 b
+$$
+
+Hence
+
+$$
+\begin{aligned}
+1 / 36<r / s= & 2^{m-n}<36, \text { or }-6<m-n<6
+\end{aligned}
+$$
+
+Furthermore,
+
+$$
+4^{n}\left(4^{m-n}-1\right)=r^{2}-s^{2}=35 \times 37\left(a^{2}-b^{2}\right)
+$$
+
+Thus
+
+$$
+4^{m-n} \equiv 1(\bmod 37)
+$$
+
+## 2 points for this congruence
+
+Observe that the 9 th power of 4 is the smallest power of 4 that is congruent 10 $1($ mod $3 \overline{1})$. Thus $9 \mid(m-n)$. Aso note that $m \neq n$. Hence $|m-n| \geq 9$, which is unt possible because $|m-n|<6$. 3 points for the final conclusion
+
+Problem 3. Let $a, b, c$ be positive real numbers. Prove that
+
+$$
+\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right) \geq 2\left(1+\frac{a+b+c}{\sqrt[3]{a b c}}\right)
+$$
+
+## MARKING SCHEME:
+
+Lol
+
+$$
+x=\frac{a}{\sqrt[3]{a b c}} \cdot \quad y=\frac{b}{\sqrt[3]{a b c}} . \quad z=\frac{r}{\sqrt[3]{a b c}}
+$$
+
+We need to show that
+
+$$
+\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2(1+x+y+z)
+$$
+
+or, since $x y z=1$,
+
+$$
+(x+y)(y+z)(z+x) \geq 2+2(x+y+z)(1)
+$$
+
+2 points for (1) by making change of variables or by assuming abe $=1$ without loss of generality
+which is equivalent to
+
+$$
+(x+y+z)(x y+y z+z x-2)-x y z \geq 2(2)
+$$
+
+$\therefore$ points for reducing to (2)
+Since $x y z=1$, the AM-GMS inequality implies
+
+$$
+x+y+z \geq 3 \text { and } x y+y z+z x \geq 3
+$$
+
+So. (2) follows.
+3 points for surcessful application of AM-CMM inequality
+
+## ALTERNATE SOLUTION
+
+2 points for (1) by making change of variables or by assuming abc $=1$ without loss of generality
+i.From (1), one can proceed as follows:
+
+$$
+(x+y)(y+z)(z+x)=2 x y z+x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)=2+x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)
+$$
+
+Thus, (1) is equivalent to
+
+$$
+x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) \geq 2(x+y+z) \cdot 3
+$$
+
+$$
+2 \text { points for reducing to (3) }
+$$
+
+Assume that $x \geq y \geq=$ Then
+
+$$
+\frac{1}{y}+\frac{1}{z} \geq \frac{1}{z}+\frac{1}{x} \geq \frac{1}{x}+\frac{1}{y}
+$$
+
+hence we can apply (heloyshev's incquality
+1 poinls for checking the hypothesis of C'hevyshev's inequality
+t.0 get.
+
+$$
+x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) \geq \frac{1}{3}(x+y+z)\left(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\right)
+$$
+
+i.From the AM-GM inequality and the fact that $x y z=1$ we obtain
+
+$$
+\left(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\right) \geq 6
+$$
+
+and hence (3) follows. 2 points for successful applications of Chevyshev's and AM-GM inequalities
+
+Problem 4. Let. $A B C^{\prime}$ be a triangle and $D$ the foot of the altitude from $A$ Lel $E$ and $F$ be on a line passing through $D$ such that $A E$ is perpendicular to $B E$. $A F$ is perpendicular to $C F$. and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $B C$ and $E F$, respectively. Prove that.$L N$ is perpendicular to $N M$.
+
+## MARIING SCHEME:
+
+Let $P$ be such that $A D M P$ is a rectangle. Choose points $Q$ and $R$ on the line $A P$ such that $Q B D A$ and $A D C R$ are eectangles. Points $Q, B$ and $D$ lie on the rircle of diameter $A B$, hence $A D E Q$ is a cyclic quadrilateral. Similarly, $R, C$ and $D$ lic on the circle of diameter $A C$, hence $A D F R$ is a cyclic quadrilateral.
+
+1 point for proving $A D E Q$ and/or $A D F R$ are ryclic
+The two quadrilaterals share a side, and have the same supporting lines for other two sides. Since they are cyclic, the remaining two sides $E Q$ and $R F$ must be parallel. Thus $E, Q, R$ and $F$ are vertices of a trapezoid.
+
+1 point for proving $E Q / / R F$
+On the other hand. in rectangle $Q B C R M$ is the midpoint of $B C$. and $M P$ is parallel to $Q B$, so $P$ is the midpoint of $Q R$. Since $N$ is the midpoint of $E F$, we obtain that, in trapezoid $Q E F R, N P$ is parallel to $Q E$.
+
+2 points for proving NP//QE
+This implies that quadrilateral $A D N P$ is cyclic, having the sides parallel to the sides of $A D F R$. Moreover, $A$ lies on the circle circumscribed to this quadrilateral, because the other three vertices of the rectangle $A D M P$ lie on it. Hence the quadrilateral $A D M N$ is cyclic.
+
+2 proints for proving $A D N P$ and $A D M N$ are cyclic
+(consquently. $\angle . V M=180^{\circ}-\angle A D M=90^{\circ}$.
+1 point for proving $\angle A N M=180^{\circ}-\angle A D M=90^{\circ}$
+
+Problem 5. Determine the largest of all integers $n$ with the property that $n$ is divisible by all positive integers that are less than $\sqrt[3]{n}$.
+
+## MARKINC SCHEME:
+
+Observation from that $\operatorname{lcm}(2.3,4.5,6.7)=420$ is divisible by every integer less than or equal to $7=[\sqrt[3]{420}]$ and that $\operatorname{lrm}(2,3,4,5,6.7,8)=840$ is not. divisible by $9=[\sqrt[3]{840}]$, one may guess 420 is the required integer. 2 points for the correct guess
+
+Lert $N$ be the required integer and supposic $V>120$. Put $t=[\sqrt[3]{\sqrt{2}}]$. Then
+
+$$
+\therefore \leq 1\left(1^{3}+31+3\right)(1)
+$$
+
+Sinte $t \geq 7, \quad \operatorname{cm}(2.3 .4 .5 .6 .7)=420$ should divide .1 and hence $N \geq 840$. which implies $1 \geq 9$. But then $\operatorname{lcm}(2,3,4,5,6,7.8 .9)=2520$ should divide $N$, which implies $t \geq 13=[\sqrt[3]{2520}]$.
+
+1 points for $t \geq 13$
+Observe that any four consecutive integers are divisible by 8 and that any t.wo out of four consecutive integers have ged either 1.2 , or 3 . So, we have $1(1-1)(t-2)(1-3)$ divides 6 N and in particular.
+
+$$
+\begin{aligned}
+& t(t-1)(t-2)(t-3) \leq 6 \times(2) \\
+& 2 \text { points for } t(t-1)(t-2)(t-3) \leq 6 N
+\end{aligned}
+$$
+
+¿.From (1) and (2) follows
+
+$$
+t(t-1)(t-2)(t-3) \leq 6 t\left(t^{3}+3 t+3\right) \frac{12}{t}+\frac{7}{t^{2}}+\frac{24}{t^{3}} \geq 1
+$$
+
+Since $t \geq 13$.
+
+$$
+\frac{12}{t}+\frac{T}{t^{2}}+\frac{24}{t^{3}}<1
+$$
+
+which is a contradiction.
+$\therefore$ points for the contradiction
+

APMO/md/en-apmo1999_sol.md

@@ -0,0 +1,304 @@
+![](https://cdn.mathpix.com/cropped/2024_11_22_808e310d9512bffbfcbfg-01.jpg?height=378&width=1687&top_left_y=212&top_left_x=243)
+
+# XI APMO - SOLUTIONS AND MARKING SCHEMES 
+
+Problem 1. 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:
+We 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 $a_{i+2 j}=a_{i+j}+\left(a_{i+j}-a_{i}\right)$.
+
+1 POINT for realizing that the integers must be "equally spaced".
+Thus, 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
+
+$$
+1,1+\frac{1}{k}, 1+\frac{2}{k}, \cdots, 1+\frac{k-1}{k}, 2,2+\frac{1}{k}, \cdots, n-1, \cdots, n-1+\frac{k-1}{k}, n
+$$
+
+This 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.
+
+2 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.
+
+Thus 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 $(k+1) n-(k+1)+1 \geq 2000$.
+
+4 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).
+
+That is, putting $k=\lfloor 1999 /(n+1)\rfloor$, we want the smallest integer $n$ so that
+
+$$
+\left\lfloor\frac{1999}{n+1}\right\rfloor(n-1)+n \geq 2000
+$$
+
+This inequality does not hold if
+
+$$
+\frac{1999}{n+1} \cdot(n-1)+n<2000
+$$
+
+2 POINTS for setting up an inequality for $n$.
+This simplifies to $n^{2}<3999$, that is, $n \leq 63$. Now we check integers from $n=64$ on:
+
+$$
+\begin{aligned}
+& \text { for } n=64,\left\lfloor\frac{1999}{65}\right\rfloor \cdot 63+64=30 \cdot 63+64=1954<2000 ; \\
+& \text { for } n=65,\left\lfloor\frac{1999}{66}\right\rfloor \cdot 64+65=30 \cdot 64+65=1985<2000 ; \\
+& \text { for } n=66,\left\lfloor\frac{1999}{67}\right\rfloor \cdot 65+66=29 \cdot 65+66=1951<2000 ; \\
+& \text { for } n=67,\left\lfloor\left\lfloor\frac{1999}{68}\right\rfloor \cdot 66+67=29 \cdot 66+67=1981<2000 ;\right. \\
+& \text { for } n=68,\left\lfloor\frac{1999}{69}\right\rfloor \cdot 67+68=28 \cdot 67+68=1944<2000 ; \\
+& \text { for } n=69,\left\lfloor\frac{1999}{70}\right\rfloor \cdot 68+69=28 \cdot 68+69=1973<2000 ; \\
+& \text { for } n=70,\left\lfloor\frac{1999}{71}\right\rfloor \cdot 69+70=28 \cdot 69+70=2002 \geq 2000 .
+\end{aligned}
+$$
+
+Thus the answer is $n=70$.
+1.POINT for checking these rumbers and finding that $n=70$.
+
+Problem 2. 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
+
+$$
+a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geq a_{n}
+$$
+
+for each positive integer $n$.
+
+## Solution and Marking Scheme:
+
+Letting $b_{i}=a_{i} / i,(i=1,2, \cdots)$, we prove that
+
+$$
+b_{1}+\cdots+b_{n} \geq a_{n} \quad(n=1,2, \cdots)
+$$
+
+by induction on $n$. For $n=1, b_{1}=a_{1} \geq a_{1}$, and the induction starts. Assume that
+
+$$
+b_{1}+\cdots+b_{k} \geq a_{k}
+$$
+
+for all $k=1,2, \cdots, n-1$. It suffices to prove that $b_{1}+\cdots+b_{n} \geq a_{n}$ or equivalently that
+
+$$
+\begin{aligned}
+& n b_{1}+\cdots+n b_{n-1} \geq(n-1) a_{n} . \\
+& 3 \text { POINTS for separating } a_{n} \text { from } b_{1}, \cdots, b_{n-1} \text {. } \\
+& n b_{1}+\cdots+n b_{n-1}=(n-1) b_{1}+(n-2) b_{2}+\cdots+b_{n-1}+b_{1}+2 b_{2}+\cdots+(n-1) b_{n-1} \\
+& =b_{1}+\left(b_{1}+b_{2}\right)+\cdots+\left(b_{1}+b_{2}+\cdots+b_{n-1}\right)+\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\
+& \geq 2\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)=\sum_{i=1}^{n-1}\left(a_{i}+a_{n-i}\right) \geq(n-1) a_{n} .
+\end{aligned}
+$$
+
+3 POINTS for the first inequaliny and 1 POINT for the rest.
+
+Problem 3. 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:
+
+Let $\alpha=\angle P A B, \beta=\angle A B P$ 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.
+
+3 POINTS for proving that $A, B, R, Q$ are concyclic.
+Since $A B$ 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 $\angle P Q R=\angle P Q B+$ $\angle B Q R=\alpha+\beta$.
+
+$$
+2 \text { POINTS for proving that } \angle P Q R=\angle P R B=\alpha+\beta
+$$
+
+Since $\angle B P R$ is an exterior angle of triangle $A B P, \angle B P R=\alpha+\beta$. We have
+
+$$
+\angle P Q R=\angle B P R=\angle B R P
+$$
+
+1 POINT for proving $\angle B P R=\alpha+\beta$.
+So circumcircle of $P Q R$ is tangent to $B P$ and $B R$.
+1 POINT for concluding.
+Remark. 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 4. 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.
+
+## First Solution and Marking Scheme:
+
+Without 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.
+Now we consider the case $b \neq 0$. Because $a^{2}+4 b$ is a perfect square, the quadratic equation
+
+$$
+x^{2}+a x-b=0
+$$
+
+has two non-zero integral roots $x_{1}, x_{2}$.
+2 POINTS for noticing that this equation has integral roots.
+Then $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=-b$, and from this it follows that
+
+$$
+\frac{1}{\left|x_{1}\right|}+\frac{1}{\left|x_{2}\right|} \geq\left|\frac{1}{x_{1}}+\frac{1}{x_{2}}\right|=\frac{|a|}{|b|} \geq 1
+$$
+
+Hence there is at least one root, say $x_{1}$, such that $\left|x_{1}\right| \leq 2$.
+3 POINTS for finding that $\left|x_{1}\right| \leq 2$.
+
+There are the following possibilities.
+(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 $|a| \geq|b|$.
+(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 $a=3, b=-2$.
+(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 $a=-6, b=-5$.
+(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.
+
+Testing these solutions and by symmetry we obtain the following solutions
+
+$$
+(-4,-4),(-5,-6),(-6,-5),\left(0, k^{2}\right),\left(k^{2}, 0\right),(k, 1-k)
+$$
+
+where $k$ 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.)
+
+1 POINT for writing up the correct answer.
+
+## Second Solution and Marking Scheme:
+
+Without 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
+
+$$
+a^{2}+4 b \leq a^{2}
+$$
+
+2 POINTS for proving (1).
+Let us consider three cases.
+Case 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.
+Case 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.
+If $a>0$ 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.
+If $a=0$ then $b=1$, but from (1) $b$ must be non-positive.
+If $a<0$ then $b^{2}+4 a=m^{2}-6 m+1$ must be a perfect square, where $m=-a>0$. For $m \geq 8$
+
+$$
+(m-3)^{2}>m^{2}-6 m+1>(m-4)^{2}
+$$
+
+therefore $m<8$. 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.
+
+2 POINTS for case 1 and case 2.
+Case 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:
+
+## 1 POINT for finding that $|a| \leq 4$ in this case.
+
+(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 $b=-4$.
+(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 $a=3$.
+(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.
+(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.
+(c) $|a|=0$. Since $|b| \leq|a|$ then $b=0$.
+
+1 POINT for concluding case 3.
+
+Testing these solutions and by symmetry we obtain the following solutions:
+
+$$
+\left(k^{2}, 0\right),\left(0, k^{2}\right),(k, 1-k),(-6,-5),(-5,-6),(-4,-4)
+$$
+
+where $k$ 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.
+
+1 POINT for writing up the correct answer.
+Remark: 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 5. 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:
+
+Lemma 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.
+
+## Proof of Lemma 1.
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_808e310d9512bffbfcbfg-07.jpg?height=521&width=939&top_left_y=520&top_left_x=639)
+
+Let $N$ 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 $\beta_{1}>\beta_{2}>\ldots>\beta_{m}$.
+Note 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.
+Observe 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 $B_{j}$.
+
+1 POINT for characterizing the points that lie inside a given circle ins terms of these angles, or for similar considerations.
+Order 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
+
+$$
+\alpha_{1}>\alpha_{2}>\ldots>\alpha_{i-1}>\alpha_{i}^{\prime}>\alpha_{i+1}>\ldots>\alpha_{k}
+$$
+
+and
+
+$$
+\beta_{1}>\beta_{2}>\ldots>\beta_{j-1}>\beta_{j}^{\prime}>\beta_{j+1}>\ldots>\beta_{m}
+$$
+
+1 POINT for this or another useful transformation of the set $S$.
+Analyze 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:
+a) If the circle $P, Q, A_{i}$ was good in $S$, the circle $P, Q, A_{i}^{\prime}$ may not be good in $S^{\prime}$.
+b) If the circle $P, Q, B_{j}$ was good in $S$, the circle $P, Q, B_{j}^{\prime}$ may not be good in $S^{\prime}$.
+c) If the circle $P, Q, A_{i}$ was not good in $S$, the circle $P, Q, A_{i}^{\prime}$ may be good in $S^{\prime}$.
+d) If the circle $P, Q, B_{j}$ was not good in $S$, the circle $P, Q, B_{j}^{\prime}$ may be good in $S^{\prime}$.
+
+1 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 $j-i=\frac{1}{2}(m-k+1)$.
+Therefore, 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 $S^{\prime}$.
+
+1 POINT for realizing that the "goodness" of these circles is changed in pairs.
+Continuing 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
+
+$$
+\beta_{1}^{\prime}>\beta_{2}^{\prime}>\ldots>\beta_{n}^{\prime}>\alpha_{1}^{\prime}>\alpha_{2}^{\prime}>\ldots>\alpha_{k}^{\prime}
+$$
+
+and 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.
+
+1 POINT for showing that this configuration has exactly one good circle.
+
+Now 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
+
+$$
+a_{1}+a_{3}+a_{5}+\ldots=\binom{2 n+1}{2}
+$$
+
+But then the number of good circles in $S$ is
+
+$$
+\begin{aligned}
+\frac{1}{3}\left(a_{1}+3 a_{3}+5 a_{5}+7 a_{7}+\ldots\right) & \equiv a_{1}+3 a_{3}+5 a_{5}+7 a_{7}+\ldots \\
+& \equiv a_{1}+a_{3}+a_{5}+a_{7}+\ldots \\
+& \equiv\binom{2 n+1}{2} \\
+& \equiv n(2 n+1) \\
+& \equiv n(\bmod 2) .
+\end{aligned}
+$$
+
+Here 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.
+
+2 POINTS for this computation.
+
+## Alteraative Proof of Lemma 1.
+
+Let, $A_{1}, A_{2}, \ldots, A_{2 n-1}$ be the $2 n-1$ given points other than $P$ and $Q$.
+Invert 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.
+
+1 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 $O$.
+
+1 POINT for this or a similar simplification.
+Let $C_{1}, C_{2}, \ldots, C_{2 n-1}$ 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".
+Now 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.
+
+1 POINT for this or a similar transformation of the set.
+It is easy to verify the following two claims:
+
+1. Point $B_{i}$ is good in configuration 1 if and only if point $B_{j}$ is good in configuration 1.
+2. Let $k \neq i, j$. Then point $B_{k}$ is good in configuration 1 if and only if it is good in configuration 2.
+
+It 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.
+
+## Another Approach to Lemma 1.
+
+One 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.
+
+1 POINT
+Then we transform the set $S$ by removing the points $A_{i}$ and $B_{j}$ instead of replacing then by $A_{i}^{p}$ and $B_{j}^{\prime}$.
+
+1 POINT
+It 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 $P, Q, B_{j}$.
+
+2 POINTS
+Finally, 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.
+
+1 POINT
+
+Remark: 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.
+

APMO/md/en-apmo2000_sol.md

@@ -0,0 +1,299 @@
+# APMO 2000 - Problems and Solutions 
+
+## Problem 1
+
+Compute the sum $S=\sum_{i=0}^{101} \frac{x_{i}^{3}}{1-3 x_{i}+3 x_{i}^{2}}$ for $x_{i}=\frac{i}{101}$.
+Answer: $S=51$.
+
+## Solution
+
+Since $x_{101-i}=\frac{101-i}{101}=1-\frac{i}{101}=1-x_{i}$ and
+
+$$
+1-3 x_{i}+3 x_{i}^{2}=\left(1-3 x_{i}+3 x_{i}^{2}-x_{i}^{3}\right)+x_{i}^{3}=\left(1-x_{i}\right)^{3}+x_{i}^{3}=x_{101-i}^{3}+x_{i}^{3},
+$$
+
+we have, by replacing $i$ by $101-i$ in the second sum,
+
+$$
+2 S=S+S=\sum_{i=0}^{101} \frac{x_{i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}+\sum_{i=0}^{101} \frac{x_{101-i}^{3}}{x_{i}^{3}+x_{101-i}^{3}}=\sum_{i=0}^{101} \frac{x_{i}^{3}+x_{101-i}^{3}}{x_{101-i}^{3}+x_{i}^{3}}=102,
+$$
+
+so $S=51$.
+
+## Problem 2
+
+Given the following arrangement of circles:
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=358&width=401&top_left_y=264&top_left_x=793)
+
+Each 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
+(i) the sums of the four numbers on each side of the triangle are equal;
+(ii) the sums of squares of the four numbers on each side of the triangle are equal.
+
+Find all ways in which this can be done.
+Answer: The only solutions are
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=361&width=404&top_left_y=1076&top_left_x=792)
+and 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):
+
+$$
+\begin{gathered}
+3 s=a+b+c+(1+2+\cdots+9)=a+b+c+45 \\
+3 t=a^{2}+b^{2}+c^{2}+\left(1^{2}+2^{2}+\cdots+9^{2}\right)=a^{2}+b^{2}+c^{2}+285
+\end{gathered}
+$$
+
+At 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,
+
+$$
+\{a, b, c\}=\{3,6,9\}, \quad\{1,4,7\}, \quad \text { or } \quad\{2,5,8\}
+$$
+
+Case 1: $\{a, b, c\}=\{3,6,9\}$. Then $3 t=3^{2}+6^{2}+9^{2}+285 \Longleftrightarrow t=137$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-2.jpg?height=372&width=409&top_left_y=2395&top_left_x=789)
+
+In 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.
+Hence there are no solutions in this case.
+Case 2: $\{a, b, c\}=\{1,4,7\}$. Then $3 t=1^{2}+4^{2}+7^{2}+285 \Longleftrightarrow t=117$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=360&width=398&top_left_y=434&top_left_x=795)
+
+In 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.
+Case 3: $\{a, b, c\}=\{2,5,8\}$. Then $3 t=2^{2}+5^{2}+8^{2}+285 \Longleftrightarrow t=126$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=355&width=401&top_left_y=1025&top_left_x=793)
+
+Then
+
+$$
+\left\{\begin{array} { c } 
+{ x ^ { 2 } + y ^ { 2 } + 2 ^ { 2 } + 8 ^ { 2 } = 1 2 6 } \\
+{ t ^ { 2 } + u ^ { 2 } + 2 ^ { 2 } + 5 ^ { 2 } = 1 2 6 } \\
+{ m ^ { 2 } + n ^ { 2 } + 5 ^ { 2 } + 8 ^ { 2 } = 1 2 6 }
+\end{array} \Longleftrightarrow \left\{\begin{array}{c}
+x^{2}+y^{2}=58 \\
+t^{2}+u^{2}=97 \\
+m^{2}+n^{2}=37
+\end{array}\right.\right.
+$$
+
+The 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
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-3.jpg?height=353&width=392&top_left_y=1751&top_left_x=798)
+and 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 3
+
+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 1
+
+Let $A N$ meet the circumcircle of $A B C$ at point $K$, the midpoint of arc $B C$ that does not contain $A$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_a2c6ce53d2466fb5c44ag-4.jpg?height=775&width=869&top_left_y=572&top_left_x=562)
+
+The 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 $P Q$.
+Now 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 $K B=K C$.)
+
+## Solution 2
+
+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 $N=(n, 0)$.
+Point $M$ is the midpoint $\left(\frac{b+c}{2}, \frac{m b-m c}{2}\right)$ of $B C$, so $A M$ has slope $\frac{m(b-c)}{b+c}$.
+The line through $N$ that is perpendicular to the $x$-axis $A N$ is $x=n$. Therefore
+
+$$
+P=(n, m n) \quad \text { and } \quad Q=\left(n, \frac{m(b-c) n}{b+c}\right) .
+$$
+
+In the right triangle $A P O$, with altitude $A N, A N \cdot A O=A P^{2}$. Thus
+
+$$
+n \cdot A O=(0-n)^{2}+(0-m n)^{2} \Longleftrightarrow A O=n\left(m^{2}+1\right) \Longrightarrow O=\left(n\left(m^{2}+1\right), 0\right)
+$$
+
+Finally, the slope of $O Q$ is
+
+$$
+\frac{\frac{m(b-c) n}{b+c}-0}{n-n\left(m^{2}+1\right)}=-\frac{b-c}{(b+c) m}
+$$
+
+Since the product of the slopes of $O Q$ and $B C$ is
+
+$$
+-\frac{b-c}{(b+c) m} \cdot \frac{m(b+c)}{b-c}=-1
+$$
+
+$O Q$ and $B C$ are perpendicular, and we are done.
+Comment: 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 4
+
+Let $n, k$ be given positive integers with $n>k$. Prove that
+
+$$
+\frac{1}{n+1} \cdot \frac{n^{n}}{k^{k}(n-k)^{n-k}}<\frac{n!}{k!(n-k)!}<\frac{n^{n}}{k^{k}(n-k)^{n-k}} .
+$$
+
+## Solution
+
+The inequality is equivalent to
+
+$$
+\frac{n^{n}}{n+1}<\binom{n}{k} k^{k}(n-k)^{n-k}<n^{n}
+$$
+
+which suggests investigating the binomial expansion of
+
+$$
+n^{n}=((n-k)+k)^{n}=\sum_{i=0}^{n}\binom{n}{i}(n-k)^{n-i} k^{i}
+$$
+
+The $(k+1)$ 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.
+Now, for $1 \leq i \leq n$,
+
+$$
+\frac{T_{i+1}}{T_{i}}=\frac{\binom{n}{i}(n-k)^{n-i} k^{i}}{\binom{n}{i-1}(n-k)^{n-i+1} k^{i-1}}=\frac{(n-i+1) k}{i(n-k)}
+$$
+
+and
+
+$$
+\frac{T_{i+1}}{T_{i}}>1 \Longleftrightarrow(n-i+1) k>i(n-k) \Longleftrightarrow i<k+\frac{k}{n} \Longleftrightarrow i \leq k
+$$
+
+This means that
+
+$$
+T_{1}<T_{2}<\cdots<T_{k+1}>T_{k+2}>\cdots>T_{n+1}
+$$
+
+that 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
+
+$$
+\binom{n}{k} k^{k}(n-k)^{n-k}>\frac{n^{n}}{n+1}
+$$
+
+as required.
+Comment: If we divide further by $n^{n}$ one finds
+
+$$
+\frac{1}{n+1}<\binom{n}{k}\left(\frac{k}{n}\right)^{k}\left(1-\frac{k}{n}\right)^{n-k}<1
+$$
+
+The 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 5
+
+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?
+
+Answer: $n=2$ 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 $n$.
+If $n=1$ 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.
+Call 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
+
+$$
+\begin{aligned}
+& (1,15,14,13,12,11,10,9,8,7,6,5,4,3,2,0) \\
+& (1,15,14,13,12,11,10,9,8,7,6,5,4,0,2,3) \\
+& (1,15,14,13,12,11,10,9,8,7,6,0,4,5,2,3) \\
+& (1,15,14,13,12,11,10,9,8,0,6,7,4,5,2,3) \\
+& (1,15,14,13,12,11,10,0,8,9,6,7,4,5,2,3) \\
+& (1,15,14,13,12,0,10,11,8,9,6,7,4,5,2,3) \\
+& (1,15,14,0,12,13,10,11,8,9,6,7,4,5,2,3) \\
+& (1,0,14,15,12,13,10,11,8,9,6,7,4,5,2,3)
+\end{aligned}
+$$
+
+After exchanging 0 and 2, the second pass is
+
+$$
+\begin{aligned}
+& (1,2,14,15,12,13,10,11,8,9,6,7,4,5,0,3) \\
+& (1,2,14,15,12,13, \mathbf{1 0}, 11,8,9,0,7,4,5,6,3) \\
+& (1,2, \mathbf{1 4}, 15,12,13,0,11,8,9,10,7,4,5,6,3) \\
+& (1,2,0,15,12,13,14,11,8,9,10,7,4,5,6,3)
+\end{aligned}
+$$
+
+After exchanging 0 and 3 , the third pass is
+
+$$
+\begin{aligned}
+& (1,2,3,15,12,13,14,11,8,9,10,7,4,5,6,0) \\
+& (1,2,3,15,12,13,14,11,8,9,10,0,4,5,6,7) \\
+& (1,2,3,15,12,13,14,0,8,9,10,11,4,5,6,7) \\
+& (1,2,3,0,12,13,14,15,8,9,10,11,4,5,6,7)
+\end{aligned}
+$$
+
+After exchanging 0 and 4, the fourth pass is
+
+$$
+\begin{aligned}
+& (1,2,3,4,12,13,14,15,8,9,10,11,0,5,6,7) \\
+& (1,2,3,4,0,13,14,15,8,9,10,11,12,5,6,7)
+\end{aligned}
+$$
+
+And then one can successively perform the operations to eventually find
+
+$$
+(1,2,3,4,5,6,7,0,8,9,10,11,12,13,14,15)
+$$
+
+after which 0 will move one unit to the right with each transposition, and $n=15$ is a solution. The general case follows.
+
+Case 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.
+Case 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
+
+$$
+[1: R-1],(0),[N-R: N-1],[N-2 R: N-R-1], \ldots,[2 R: 3 R-1],[R: 2 R-1]
+$$
+
+$P_{r}$ 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.
+Then 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 $P_{1}$.
+Starting 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
+
+$$
+\begin{gathered}
+{[1: R],(0),[N-R+1: N-1],[N-2 R: N-R],[N-3 R+1: N-2 R-1], \ldots,} \\
+{[3 R+1: 4 R-1],[2 R: 3 R],[R+1: 2 R-1]}
+\end{gathered}
+$$
+
+Note 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.
+Now $0, N-R+1, N-3 R+1, \ldots, 3 R+1, R+1$ are shifted. Note that, for every $i$ th block, $i$ odd greater than 1 , the first number is cyclically shifted, and the blocks become
+
+$$
+\begin{gathered}
+{[1: R+1],(0),[N-R+2: N-1],[N-2 R: N-R+1],[N-3 R+2: N-2 R-1], \ldots,} \\
+{[3 R+1: 4 R-1],[2 R: 3 R+1],[R+2: 2 R-1]}
+\end{gathered}
+$$
+
+The 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
+
+$$
+[1: 2 R-1],(0),[N-2 R: N-1], \ldots,[2 R: 4 R-1]
+$$
+
+which is precisely $P_{r+1}$.
+Since $P_{k}=[1: N-1],(0), n=2^{k}-1$ is a solution.
+Case 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 $P_{a}$ :
+
+$$
+\left[1: 2^{a}-1\right],(0),\left[N-2^{a}: N-1\right],\left[N-2^{a+1}: N-2^{a}-1\right], \ldots,\left[2^{a+1}: 3 \cdot 2^{a}-1\right],\left[2^{a}: 2^{a+1}-1\right]
+$$
+
+But 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.
+
+Comment: The general problem of finding the number of regular permutations for any $n$ seems to be difficult. A computer finds the first few values
+
+$$
+1,2,5,14,47,189,891,4815,29547
+$$
+
+which is not catalogued at oeis.org.
+

APMO/md/en-apmo2001_sol.md

@@ -0,0 +1,357 @@
+# XI APMO - SOLUTIONS AND MARKING SCHEMES 
+
+## Problem 1.
+
+## First solution.
+
+Let us prove that $n=S(n)+9 T(n)$ proceding by induction over the number of digits of $n$.
+1 POINT.
+If $n$ has one digit then the result is trivial. Suppose that $n=S(n)+9 T(n)$ is true for any integer $n$ of $k$ digits. Now any number $m$ of $k+1$ digits can be writen as $m=10 n+a$ where $n$ is a number of $k$ digits. Obviously,
+
+$$
+\begin{aligned}
+T(m)=n+T(n) & \text { and }(m)=S(n)+a \\
+& 2 \text { POINTS (1 POINT for each of the last equalities). }
+\end{aligned}
+$$
+
+Therefore
+
+$$
+\begin{aligned}
+m-S(m) & =10 n+a-S(n)-a \\
+& =10 n-S(n) \\
+& =(n-S(n))+9 n \\
+& =9 T(n)+9 n \\
+& =9 T(m)
+\end{aligned}
+$$
+
+as required.
+\& POINTS.
+
+## Second solution.
+
+Let $n=\overline{a_{k} a_{k-1} \ldots a_{1} a_{0}}=10^{k} a_{k}+10^{k-1} a_{k-1}+\ldots+10 a_{1}+a_{0}$, where $a_{k}, a_{k-1}, \ldots, a_{1}, a_{0}$ are digits. The stumps of $n$ are
+
+$$
+\begin{aligned}
+\overline{a_{k} a_{k-1} \cdots a_{1}} & =10^{k-1} a_{k}+10^{k-2} a_{k-1}+\ldots+10 a_{2}+a_{1} \\
+\overline{a_{k} a_{k-1} \cdots a_{2}} & =10^{k-2} a_{k}+10^{k-3} a_{k-1}+\ldots+a_{2} \\
+& \vdots \\
+\overline{a_{k} a_{k-1}} & =10 a_{k}+a_{k-1} \\
+a_{k} & =a_{k} .
+\end{aligned}
+$$
+
+1 POINT.
+Since $10^{m-1}+10^{m-2}+\ldots+10+1=\frac{10^{m}-1}{9}$ then the sum of all stumps of $n$ is
+
+$$
+T(n)=\frac{10^{k}-1}{9} a_{k}+\frac{10^{k-1}-1}{9} a_{k-1}+\ldots+\frac{10-1}{9} a_{1}
+$$
+
+3 POINTS.
+and hence
+
+$$
+\begin{aligned}
+9 T(n) & =10^{k} a_{k}+10^{k-1} a_{k-1}+\ldots+10 a_{1}+a_{0}-\left(a_{k}+a_{k-1}+\ldots+a_{1}+a_{0}\right) \\
+& =n-S(n) .
+\end{aligned}
+$$
+
+Consequently $n=S(n)+9 T(n)$.
+3 POINTS.
+
+## Third Solution.
+
+Let $k(n)$ be the number of zeros at the end of the decimal representation of $n$ or, which is the same, the largest power of 10 which divides $n$. The the following two observations are straightforward:
+
+1. $S(n)=S(n-1)-(9 k(n)-1)$,
+2. $k(1)+k(2)+\ldots+k(n)=T(n)$.
+
+4 POINTS (2 POINTS for each of these equalities).
+Then summing the following equalities up
+
+$$
+\begin{aligned}
+S(1) & =S(0)-(9 k(1)-1), \\
+S(2) & =S(1)-(9 k(2)-1), \\
+& \vdots \\
+S(n-1) & =S(n-2)-(9 k(n-1)-1) \\
+S(n) & =S(n-1)-(9 k(n-1)-1)
+\end{aligned}
+$$
+
+we get $S(n)=n-(k(1)+k(2)+\ldots+k(n))=n-9 T(n)$.
+3 POINT for concluding.
+
+## Problem 2.
+
+## First Solution.
+
+This is equivalent to find the largest positive integer solution of the equation
+
+$$
+\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor
+$$
+
+or
+
+$$
+\left\lfloor\frac{N}{3}\right\rfloor+\left\lfloor\frac{N}{35}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\left\lfloor\frac{N}{7}\right\rfloor .\right.
+$$
+
+1 POINT for equality (1).
+For $N$ to be a solution of (1) it is necessary that
+
+$$
+\frac{N-2}{3}+\frac{N-34}{35} \leq \frac{N}{5}+\frac{N}{7}
+$$
+
+which simplifies to $N \leq 86$.
+1 POINTS for finding that $N \leq 86$.
+However, if $N \geq 70$ then because $N \leq 86$, (1) implies that
+
+$$
+\frac{N-2}{3}+\frac{N-16}{35} \leq \frac{N}{5}+\frac{N}{7}
+$$
+
+which simplifies to $N \leq 59$, contradicting $N \geq 70$. it follows that $N$ must be at most 69 .
+4 POINTS for finding that $N \leq 69$.
+Checking (1) for $N \leq 69$ we find that
+
+$$
+\begin{array}{llll}
+\text { when } \quad N=69, & \text { (1) is } & 23+1=13+9, & \text { false } \\
+\text { when } N=68,67,66, & \text { (1) is } 22+1=13+9, & \text { false } \\
+\text { when } N=65, & \text { (1) is } 21+1=13+9, & \text { true. }
+\end{array}
+$$
+
+Thus the answer is $N=65$.
+1 POINT for concluding.
+
+## Second solution.
+
+This is equivalent to find the largest positive integer solution of the equation
+
+$$
+\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor,
+$$
+
+Let $N=35 k+r(0 \leq r<35)$ be a solution of (1). Then (1) can be writen as
+
+$$
+\left\lfloor\frac{35 k+r}{3}\right\rfloor=11 k+\left\lfloor\frac{r}{5}\right\rfloor+\left\lfloor\frac{r}{7}\right\rfloor .
+$$
+
+Now $\quad \frac{35 k+r-2}{3} \leq\left\lfloor\frac{35 k+r}{3}\right\rfloor,\left\lfloor\frac{r}{5}\right\rfloor \leq \frac{r}{5}$ and $\left\lfloor\frac{r}{7}\right\rfloor \leq \frac{\tau}{7}$. Therefore r
+
+$$
+\frac{35 k+r-2}{3} \leq 11 k+\frac{r}{5}+\frac{r}{7}=\frac{12 r}{35}
+$$
+
+which implies that $70 k \leq r+70<35+70=105$. Then $k \leq 1$ or equivalently $N \leq 69$.
+4 POINTS for finding that $N \leq 69$.
+As in the first solution, checking (1) for $N \leq 69$ we find the answer $N=65$.
+
+## 1 POINT for concluding.
+
+## Remark.
+
+2 POINTS can be given for finding a good upper bound for example $N \leq 100$ (in the first solution we found $n \leq 86$ ). 6 POINTS for proving that $N \leq 69$.
+1 POINT can be given for the correct answer.
+
+## Problem 3.
+
+## First Solution.
+
+It is easy to see that the intersection $S \cap T$ has $2 n$ sides only if the sides of $S \cap T$ alternate: blue, red, blue, red, etc.
+
+## 1 POINT.
+
+Denote the vertices of $S \cap T$ clockwise as $C_{1} D_{1} C_{2} D_{2} \ldots C_{n} D_{n}$ so that the sides $C_{1} D_{1}, C_{2} D_{2}, \ldots C_{n} D_{n}$ are blue. Denote the vertices of $S$ by $A_{1}, A_{2}, \ldots, A_{n}$ and the vertices of $T$ by $B_{1}, B_{2}, \ldots, B_{n}$ so that $C_{1} D_{1} \subset B_{1} B_{2}, \ldots, C_{n} D_{n} \subset B_{n} B_{1}$ and $D_{n} C_{1} \subset A_{1} A_{2}, \ldots, D_{n-1} C_{n} \subset A_{n} A_{1}$.
+It is easy to check that all the triangles $D_{n} B_{1} C_{1}, D_{1} B_{2} C_{2}, \ldots, D_{n-1} B_{n} C_{n}$ and $C_{1} A_{2} D_{1}, C_{2} A_{3} D_{2}, \ldots, C_{n} A_{1} D_{n}$ are similar.
+
+## 1 POINT.
+
+Therefore,
+
+$$
+\begin{gathered}
+\frac{D_{n} C_{1}}{D_{n} B_{1}+B_{1} C_{1}}=\frac{D_{1} C_{2}}{D_{1} B_{2}+B_{2} C_{2}}=\ldots=\frac{D_{n-1} C_{n}}{D_{n-1} B_{n}+B_{n} C_{n}}= \\
+=\frac{C_{1} D_{1}}{C_{1} A_{2}+A_{2} D_{1}}=\frac{C_{1} D_{2}}{C_{2} A_{3}+A_{3} D_{2}}=\ldots=\frac{C_{n} D_{n}}{C_{n} A_{1}+A_{1} D_{n}}
+\end{gathered}
+$$
+
+1 POINT.
+Hence,
+
+$$
+\frac{D_{n} C_{1}+D_{1} C_{2}+\ldots+D_{n-1} C_{n}}{D_{n} B_{1}+B_{1} C_{1}+D_{1} B_{2}+B_{2} C_{2}+\ldots+D_{n-1} B_{n}+B_{n} C_{n}}=
+$$
+
+$$
+=\frac{C_{1} D_{1}+C_{1} D_{2}+\ldots+C_{n} D_{n}}{C_{1} A_{2}+A_{2} D_{1}+C_{2} A_{3}+A_{3} D_{2}+\ldots+C_{n} A_{1}+A_{1} D_{n}}
+$$
+
+Let $x=D_{n} C_{1}+D_{1} C_{2}+\ldots+D_{n-1} C_{n}$ and $y=C_{1} D_{1}+C_{1} \tilde{D}_{2}+\ldots+C_{n} D_{n}$ then $x$ is the sum of the blue sides of $S \cap T$ and $y$ is the sum of the red sides. If $a$ is the length of a side of $S$ (or $T$ ), then the equality (1) can be written in the following form
+
+$$
+\frac{x}{n a-y}=\frac{y}{n a-x} .
+$$
+
+It follows that
+
+$$
+\begin{aligned}
+n a x-x^{2} & =n a y-y^{2} \\
+n a(x-y) & =(x+y)(x-y) \\
+(n a-x-y)(x-y) & =0
+\end{aligned}
+$$
+
+Since the perimeter $x+y$ of $S \cap T$ is strictly less than the perimeter $n a$ of $S$ or $T, n a-x-y>0$. We obtain $x-y=0$ and $x=y$ q.e.d.
+
+4 POINTS for concluding.
+
+## Second Solution.
+
+As in the first solution, $S \cap T$ has $2 n$ sides only if the sides of $S \cap T$ alternate.
+1 POINT.
+Label the vertices of the red $n$-gon $R_{1}, R_{2}, \ldots R_{n}$ and the vertices of the blue $n$-gon $B_{1}, B_{2}, \ldots, B_{n}$. Place the $n$-gons so that the vertices are in the following clockwise order: $B_{1}, R_{1}, B_{2}, R_{2}, \ldots, B_{n}, R_{n}$. Each of these vertices together with the opposite side determines a triangle and all these triangles are similar.
+
+1 POINT.
+For each $i=1, \ldots, n$, we let the lengths of the sides of the triangle determined by $B_{i}$ be $b_{i}, c_{i}, d_{i}$ in the clockwise order where $b_{i}$ is the side opposite $B_{i}$ such that $b_{i} / b_{1}=c_{i} / c_{1}=d_{i} / d_{1}=p_{i}$. We also let the lengths of the sides of the triangle determined by $R_{i}$ be $r_{i}, s_{i}, t_{i}$ in the counter clockwise order such that $r_{i} / b_{1}=s_{i} / c_{1}=t_{i} / d_{1}=q_{i}$.
+
+1 POINT.
+Then we want to prove that $b_{1}+\cdots+b_{n}=r_{1}+\cdots+r_{n}$ or $b_{1}\left(p_{1}+\cdots+p_{n}\right)=b_{1}\left(q_{1}+\cdots+q_{n}\right)$, where $p_{1}=1$, or $p=q$ where $p=\left(p_{1}+\cdots+p_{n}\right), q=\left(q_{1}+\cdots+q_{n}\right)$. The perimeter of the blue $n$-gon is $c_{1}+d_{1}+r_{1}+\cdots+c_{n}+d_{n}+r_{n}=p c_{1}+p d_{1}+q b_{1}$. Likewise the perimeter of the red $n$ gon is $p b_{1}+q c_{1}+q d_{1}$. Equating the two we have $p\left(c_{1}+d_{1}-b_{1}\right)=q\left(c_{1}+d_{1}-b_{1}\right)$, which implies $p=q$ as
+required.
+![](https://cdn.mathpix.com/cropped/2024_11_22_d0bd4d1accc9b46debb3g-06.jpg?height=440&width=1134&top_left_y=555&top_left_x=753)
+
+## Problem 4.
+
+Answer. All the polynomials of degree 1 with rational coefficients.
+
+## First Solution.
+
+Note that a polynomial that satisfies the conditions of the problem takes rational values for rational numbers and irrational values for irrational numbers. Let $f(x)$ be a polynomial of degree $n$ such that $f(r) \in Q$ for every $r \in Q$. For distinct rational numbers $r_{0}, r_{1}, \ldots, r_{n}$, where $n=\operatorname{deg} f(x)$ let us define the polynomial
+
+$$
+\begin{aligned}
+g(x)= & c_{0}\left(x-r_{1}\right)\left(x-r_{2}\right) \ldots\left(x-r_{n}\right)+c_{1}\left(x-r_{0}\right)\left(x-r_{2}\right) \ldots\left(x-r_{n}\right)+\ldots \\
+& +c_{i}\left(x-r_{0}\right) \ldots\left(x-r_{i-1}\right)\left(x-r_{i+1}\right) \ldots\left(x-r_{n}\right)+\ldots \\
+& +c_{n}\left(x-r_{0}\right)\left(x-r_{1}\right) \ldots\left(x-r_{n-1}\right)
+\end{aligned}
+$$
+
+where $c_{0}, c_{1}, \ldots, c_{n}$ are real numbers.
+Suppose that $g\left(r_{i}\right)=f\left(r_{i}\right), i=0,1, \ldots, n$. Since $g\left(r_{i}\right)=c_{i}\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right) \ldots\left(r_{i}-r_{n}\right)$ then
+
+$$
+c_{i}=\frac{g\left(r_{i}\right)}{\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right)}=\frac{f\left(r_{i}\right)}{\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right)} .
+$$
+
+Clearly $c_{i}$ is a rational number for $i=0,1, \ldots, n$ and therefore the coefficients of $g(x)$ are rational. The polynomial $g(x)$ defined in (1) with coefficients $c_{0}, c_{1}, \ldots, c_{n}$ satisfying (2) coincides with $f(x)$ in $n+1$ points and both polynomials $f$ and $g$ have degree $n$. It follows that for every real $x, f(x)=g(x)$. Therefore the coefficients of $f(x)$ are rational.
+Thus if a polynomial satisfies the conditions, all its coefficients are rational.
+1 POINT for proving that the polinomial has rational coefficients.
+It is easy to see that all the polynomials of first degree with rational coefficients satisfy the conditions of the problem and polynomials of degree 0 do not satisfy it. Let us prove that no other polynomials exist.
+
+Suppose that $f(x)=a_{0}+a_{1} x+\ldots+a_{n} x^{n}$ is a polynomial with rational coefficients and degree $n \geq 2$ that satisfies the conditions of the problem. We may assume that the coefficients of $f(x)$ are integers, because the sets of solutions of equations $f(x)=r$ and $a f^{\prime}(x)=a r$, where $a$ is an integer, coincide. Moreover let us denote $g(x)=a_{n}^{n-1} f\left(\frac{x}{a_{n}}\right) \cdot g(x)$ is a polynomial with integer coefficients whose leading coefficient is 1 . The equation $f(x)=r$ has an irrational root if and only if $g(x)=a_{n}^{n-1} r$ has an irrational root. Therefore, we may assume WLOG that $f(x)$ has integer coefficients and $a_{n}=1$.
+
+1 POINT more for proving that it is suficient to consider polinomials with integer and leading coefficient equal to 1 .
+
+Let $r$ be a sufficiently large prime, such that
+
+$$
+r>\max \left\{f(1)-f(0), x_{1}, x_{2}, \ldots, x_{k}\right\}
+$$
+
+where $\left\{x_{1}, x_{2}, \ldots, x_{k}\right\}$ denote the set of all real roots of $f(x)-f(0)-x=0$. Putting $q=r+f(0) \in Z$, and considering the equality
+
+$$
+f(x)-q=f(x)-f(0)-r
+$$
+
+we then have
+
+$$
+f(1)-q=f(1)-f(0)-r<0 .
+$$
+
+On the other hand, by the choice of $r$, we have
+
+$$
+f(1)-q=f(r)-f(0)-r>0 .
+$$
+
+It follows from the intermediate value theorem that there is at least one real root $p$ of $f(x)-q=0$ between 1 and $r$. Notice that from the criterion theorem for rational roots, the possible positive rational roots of the equation $f(x)-q=f(x)-f(0)-r=0$ are 1 and $r$. Thus $p$ must be irrational.
+
+## 5 POINTS for concluding.
+
+## Second Solution.
+
+As in the first solution we may assume WLOG that $f(x)$ has integer coefficients and the leading coefficient is $a_{n}=1$.
+
+2 POINTS.
+Observing the graph of $f(x)$, it is easy to see that there exists a sufficiently great integer $r$ such that $f(x)=r$ has one possitive root $x_{0}$ and for $x \geq x_{0}$ the derivative $f^{\prime}(x)$ is greater than 1 . The equation $f(x)=r+1$ has also one positive root $x_{1}>x_{0}$. Since $a_{n}=1$, rational roots $x_{0}$ and $x_{1}$ must be integers. Then $x_{1}-x_{0} \geq 1$ and
+
+$$
+\frac{f\left(x_{1}\right)-f\left(x_{0}\right)}{x_{1}-x_{0}} \leq 1
+$$
+
+It follows that $f^{\prime}(z)<1$ for $z \in\left[x_{0}, x_{1}\right]$. This contradiction proves that $f(x)=r$ necessarily has an irrational root for at least one integer $r$.
+
+## Remark.
+
+No points can be given just for the answer.
+
+## Problem 5.
+
+## First Solution.
+
+One of the sides $A X_{i}$ or $B X_{i}$ is equal to $C D$, thus $X_{i}$ is on one of the circles of radius $C D$ and center $A$ or $B$. In the same way $X_{i}$ is on one of circles of radius $A B$ with center $C$ or $D$. The intersection of these four circles has no more than 8 points so that $n \leq 8$.
+
+## 1 POINT for finding that $n \leq 8$.
+
+Suppose that circle $S_{B}$ with center $B$ and radius $C D$ intersects circle $S_{C}$ with center $C$ and radius $A B$ in two points $X_{1}$ and $X_{2}$ which satisfy the conditions of the problem. Then in triangles $A B X_{1}$ and $C D X_{1}$ we have $B X_{1}=C D$ and $C X_{1}=A B$. Since these triangles are congruent then $A X_{1}=$ $D X_{1}$, therefore $X_{1}$ and $X_{2}$ are on the perpendicular bisector of $A D$. On the other hand $X_{1} X_{2}$ is perpendicular to segment $B C$. Then $B C \| A D$ and $A B$ and $C D$ are the diagonals or nonparallel sides of a trapezoid.
+![](https://cdn.mathpix.com/cropped/2024_11_22_d0bd4d1accc9b46debb3g-08.jpg?height=302&width=513&top_left_y=1193&top_left_x=795)
+
+Suppose that $A B<C D$. Then $B X_{1}=C D>A B=C X_{1}$. It follows that the distance from $A$ to the perpendicular bisector of $B C$ must be less than the distance from $D$ to this line otherwise we obtain a contradiction to the condition $A B<C D$. Then for any point $X$ in the perpendicular bisector of $B C$ we have $A X<D X$ and it is not possible to have $A X=C D, D X=A B$. Thus if the circle with center $A$ and radius $C D$ intersects the circle with center $D$ and radius $A B$, then the points of intersection do not satisfy the condition of congruence. Therefore if the points of intersection of $S_{B}$ with $S_{C}$ satisfy the condition of congruence, then the points of intersection of $S_{A}$ with $S_{D}$ do not. Thus no more than half of the 8 points of intersection of these circles can satisfy the condition of congruence, i.e. $n \leq 4$.
+
+4 POINTS for proving that $n \leq 4$.
+If $n=4$ we have the following example of a regular hexagon.
+![](https://cdn.mathpix.com/cropped/2024_11_22_d0bd4d1accc9b46debb3g-08.jpg?height=310&width=489&top_left_y=2070&top_left_x=815)
+
+2 POINTS for proving that $n \geq 4$.
+
+## Second Solution.
+
+The greatest possible value of $n$ is 4 . First we will prover that $n \geq 4$ and finally we will prove that $n \leq 4$.
+
+- To prove that $n \geq 4$ it is enough to show a configuration of 8 points that satisfies the conditions of the problem. Let us choose 6 points $A, B, Y, C, D, X$ on a circle such that triangle $A Y D$ is equilateral and $B, C, X$ are points on arcs $A Y, Y D, D A$ respectively, such that arcs $A B, Y C, D X$ are equal and less than $60^{\circ}$. Then it is easy to see that $A B, C D$ and $X Y$ are parallel and $\angle D X Y=\angle C Y X=60^{\circ}$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_d0bd4d1accc9b46debb3g-09.jpg?height=282&width=347&top_left_y=954&top_left_x=889)
+
+If $X Y$ intersects $A C$ and $B D$ at $X^{\prime}, Y^{\prime}$, respectively, then $X, Y, X^{\prime} Y^{\prime}$ satisfy the conditions of the problem.
+
+2 POINTS for proving that $n \geq 4$.
+
+- To show that $n \leq 4$ let us consider the following figure.
+![](https://cdn.mathpix.com/cropped/2024_11_22_d0bd4d1accc9b46debb3g-09.jpg?height=508&width=611&top_left_y=1546&top_left_x=849)
+
+Suppose that $A B<C D$. The trace of point $K$ such that $(A B K)=(C D K)$ is two lines $l$ and $m$ through the point of intersection $O$ of $A B$ with $C D$. (If $A B \| C D$ then the trace of $K$ is formed by two parallel lines.)
+If $X$ is a point such that $\triangle A B X \cong \triangle C D X$ then $X$ must be on one or more of the perpendicular bisectors of the segments $A C, A D, B C, B D$. Since $X$ lies on $l$ or $m$, on each of the perpendicular bisectors of $A C, A D, B C, B D$ there can be no more than 2 intersection points, i.e. $n \leq 8$.
+
+1 POINT for proving that $n \leq 8$.
+We will prove that at most one point on each perpendicular bisector satisfies the conditions of the problem.
+
+WLOG Suposse that $X_{1}$ and $X_{2}$ are points on the perpendicular bisector of $B C$ such that $\triangle A B X_{1} \cong \triangle C D X_{1}$ and $\triangle A B X_{2} \cong \triangle C D X_{2}$ with $X_{1} \in l$ and $X_{2} \in m$.
+Let us prove that $A X_{1}=C D$. Since $A B \neq C D$ one of the segments $A X_{1}$ or $B X_{1}$ is equal to $C D$. If $B X_{1}=C D$ then $C X_{1}=C D$ by construction, and $\triangle C D X_{1}$ has two sides equal to $C D$, thus $A X_{1}=C D$.
+In the same way we have $D X_{1}=A B$
+The same argument can be used if we consider the point $X_{2}$. In this case we conclude that $A X_{1}=$ $A X_{2}$ and $D X_{1}=D X_{2}$, then $A D$ and $X_{1} X_{2}$ are perpendicular. But $B C$ and $X_{1} X_{2}$ perpendicular and therefore $A D \| B C$. We are considering the case when $A D$ and $B C$ are diagonals of a cuadrilateral, therefore they are not parallel. This is a contradiction.
+If we consider the perpendicular bisector of $B D$, the same argument allows us to conclude that $A C$ and $B D$ are parallel. Therefore a point $X$ on $l$ or $m$ (i.e $(X A B)=(X C D))$ satisfies $(X O B)=(X O D)$ because triangles $O A C$ and $O B D$ are similar. Thus, the points of intersection of $A C$ with $l$ and $B D$ with $l$ are midpoints of the respective segments. Then $X_{1} X_{2} \subset l$ and, since $l$ is the perpendicular bisector and median, triangle $A B D$ is isosceles with $O B=O C$. So, $A B=C D$, which is a contradiction. Therefore $n \leq 4$.
+
+4 POINT for proving that $n \leq 4$.
+

APMO/md/en-apmo2002_sol.md

@@ -0,0 +1,399 @@
+# XIV APMO: Solutions and Marking Schemes 
+
+1. Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer.
+
+Let
+
+$$
+A_{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}
+$$
+
+Prove that
+
+$$
+a_{1}!a_{2}!\ldots a_{n}!\geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
+$$
+
+where $\left\lfloor A_{n}\right\rfloor$ is the greatest integer less than or equal to $A_{n}$, and $a!=1 \times 2 \times \cdots \times a$ for $a \geq 1$ (and $0!=1$ ). When does equality hold?
+
+## Solution 1.
+
+Assume without loss of generality that $a_{1} \geq a_{2} \geq \cdots \geq a_{n} \geq 0$, and let $s=\left\lfloor A_{n}\right\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \geq s \geq a_{k+1}$.
+
+Our inequality is equivalent to proving that
+
+$$
+\frac{a_{1}!}{s!} \cdot \frac{a_{2}!}{s!} \cdot \ldots \cdot \frac{a_{k}!}{s!} \geq \frac{s!}{a_{k+1}!} \cdot \frac{s!}{a_{k+2}!} \cdot \ldots \cdot \frac{s!}{a_{n}!}
+$$
+
+Now for $i=1,2, \ldots, k, a_{i}!/ s!$ is the product of $a_{i}-s$ factors. For example, 9!/5! $=9 \cdot 8 \cdot 7 \cdot 6$. The left side of inequality (1) therefore is the product of $A=a_{1}+a_{2}+\cdots+a_{k}-k s$ factors, all of which are greater than $s$. Similarly, the right side of (1) is the product of $B=(n-k) s-$ $\left(a_{k+1}+a_{k+2}+\cdots+a_{n}\right)$ factors, all of which are at most $s$. Since $\sum_{i=1}^{n} a_{i}=n A_{n} \geq n s, A \geq B$. This proves the inequality. [ 5 marks to here.]
+
+Equality in (1) holds if and only if either:
+(i) $A=B=0$, that is, both sides of (1) are the empty product, which occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}$; or
+(ii) $a_{1}=1$ and $s=0$, that is, the only factors on either side of (1) are 1 's, which occurs if and only if $a_{i} \in\{0,1\}$ for all $i$. [2 marks for both (i) and (ii), no marks for (i) only.]
+
+Solution 2.
+Assume without loss of generality that $0 \leq a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Let $d=a_{n}-a_{1}$ and $m=\left|\left\{i: a_{i}=a_{1}\right\}\right|$. Our proof is by induction on $d$.
+
+We first do the case $d=a_{n}-a_{1}=0$ or 1 separately. Then $a_{1}=a_{2}=\cdots=a_{m}=a$ and $a_{m+1}=\cdots=a_{n}=a+1$ for some $1 \leq m \leq n$ and $a \geq 0$. In this case we have $\left\lfloor A_{n}\right\rfloor=a$, so the inequality to be proven is just $a_{1}!a_{2}!\ldots a_{n}!\geq(a!)^{n}$, which is obvious. Equality holds if and only if either $m=n$, that is, $a_{1}=a_{2}=\cdots=a_{n}=a$; or if $a=0$, that is, $a_{1}=\cdots=a_{m}=0$ and $a_{m+1}=\cdots=a_{n}=1$. [ 2 marks to here.]
+
+So assume that $d=a_{n}-a_{1} \geq 2$ and that the inequality holds for all sequences with smaller values of $d$, or with the same value of $d$ and smaller values of $m$. Then the sequence
+
+$$
+a_{1}+1, a_{2}, a_{3}, \ldots, a_{n-1}, a_{n}-1
+$$
+
+though not necessarily in non-decreasing order any more, does have either a smaller value of $d$, or the same value of $d$ and a smaller value of $m$, but in any case has the same value of $A_{n}$. Thus, by induction and since $a_{n}>a_{1}+1$,
+
+$$
+\begin{aligned}
+a_{1}!a_{2}!\ldots a_{n}! & =\left(a_{1}+1\right)!a_{2}!\ldots a_{n-1}!\left(a_{n}-1\right)!\cdot \frac{a_{n}}{a_{1}+1} \\
+& \geq\left(\left\lfloor A_{n}\right\rfloor!\right)^{n} \cdot \frac{a_{n}}{a_{1}+1} \\
+& >\left(\left\lfloor A_{n}\right\rfloor!\right)^{n}
+\end{aligned}
+$$
+
+which completes the proof. Equality cannot hold in this case.
+2. Find all positive integers $a$ and $b$ such that
+
+$$
+\frac{a^{2}+b}{b^{2}-a} \text { and } \frac{b^{2}+a}{a^{2}-b}
+$$
+
+are both integers.
+Solution.
+By the symmetry of the problem, we may suppose that $a \leq b$. Notice that $b^{2}-a \geq 0$, so that if $\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \geq 0$. Since $a, b>0$, we must have $a \geq b-1$. [3 marks to here.] We therefore have two cases:
+
+Case 1: $a=b$. Substituting, we have
+
+$$
+\frac{a^{2}+a}{a^{2}-a}=\frac{a+1}{a-1}=1+\frac{2}{a-1},
+$$
+
+which is an integer if and only if $(a-1) \mid 2$. As $a>0$, the only possible values are $a-1=1$ or 2. Hence, $(a, b)=(2,2)$ or $(3,3)$. [1 mark.]
+
+Case 2: $a=b-1$. Substituting, we have
+
+$$
+\frac{b^{2}+a}{a^{2}-b}=\frac{(a+1)^{2}+a}{a^{2}-(a+1)}=\frac{a^{2}+3 a+1}{a^{2}-a-1}=1+\frac{4 a+2}{a^{2}-a-1} .
+$$
+
+Once again, notice that $4 a+2>0$, and hence, for $\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \leq \bar{a} \leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. [3 marks.]
+
+Thus, the ordered pairs that work are
+
+$$
+(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)
+$$
+
+where the last two pairs follow by symmetry. [2 marks if these solutions are found without proof that there are no others.]
+3. Let $A B C$ be an equilateral triangle. Let $P$ be a point on the side $A C$ and $Q$ be a point on the side $A B$ so that both triangles $A B P$ and $A C Q$ are acute. Let $R$ be the orthocentre of triangle $A B P$ and $S$ be the orthocentre of triangle $A C Q$. Let $T$ be the point common to the segments $B P$ and $C Q$. Find all possible values of $\angle C B P$ and $\angle B C Q$ such that triangle $T R S$ is equilateral.
+
+## Solution.
+
+We are going to show that this can only happen when
+
+$$
+\angle C B P=\angle B C Q=15^{\circ} .
+$$
+
+Lemma. If $\angle C B P>\angle B C Q$, then $R T>S T$.
+Proof. Let $A D, B E$ and $C F$ be the altitudes of triangle $A B C$ concurrent at its centre $G$. Then $P$ lies on $C E, Q$ lies on $B F$, and thus $T$ lies in triangle $B D G$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-3.jpg?height=590&width=679&top_left_y=968&top_left_x=636)
+
+Note that -
+
+$$
+\angle F A S=\angle F C Q=30^{\circ}-\angle B C Q>30^{\circ}-\angle C B P=\angle E B P=\angle E A R
+$$
+
+Since $A F=A E$, we have $F S>E R$ so that
+
+$$
+G S=G F-F S<G E-E R=G R .
+$$
+
+Let $T_{x}$ be the projection of $T$ onto $B C$ and $T_{y}$ be the projection of $T$ onto $A D$, and similarly for $R$ and $S$. We have
+
+$$
+R_{x} T_{x}=D R_{x}+D T_{x}>\left|D S_{x}-D T_{x}\right|=S_{x} T_{x}
+$$
+
+and
+
+$$
+R_{y} T_{y}=G R_{y}+G T_{y}>G S_{y}+G T_{y}=S_{y} T_{y}
+$$
+
+It follows that $R T>S T$.
+[1 mark for stating the Lemma, 3 marks for proving it.]
+Thus, if $\triangle T R S$ is equilateral, we must have $\angle C B P=\angle B C Q$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_f9f048d50025eb9b7941g-4.jpg?height=589&width=663&top_left_y=291&top_left_x=639)
+
+It is clear from the symmetry of the figure that $T R=T S$, so $\triangle T R S$ is equilateral if and only if $\angle R T A=30^{\circ}$. Now, as $B R$ is an altitude of the triangle $A B C, \angle R B A=30^{\circ}$. So $\triangle T R S$ is equilateral if and only if $R T B A$ is a cyclic quadrilateral. Therefore, $\triangle T R S$ is equilateral if and only if $\angle T B R=\angle T A R$. But
+
+$$
+\begin{aligned}
+90^{\circ} & =\angle T B A+\angle B A R \\
+& =(\angle T B R+\angle R B A)+(\angle B A T+\angle T A R) \\
+& =\left(\angle T B R+30^{\circ}\right)+\left(30^{\circ}+\angle T A R\right)
+\end{aligned}
+$$
+
+and so
+
+$$
+30^{\circ}=\angle T A R+\angle T B R
+$$
+
+But these angles must be equal, so $\angle T A R=\angle T B R=15^{\circ}$. Therefore $\angle C B P=\angle B C Q=15^{\circ}$. [3 marks for finishing the proof with the assumption that $\angle C B P=\angle B C Q$.]
+4. Let $x, y, z$ be positive numbers such that
+
+$$
+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
+$$
+
+Show that
+
+$$
+\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}
+$$
+
+Solution 1.
+
+$$
+\begin{aligned}
+\sum_{\text {cyclic }} \sqrt{x+y z} & =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}+\frac{1}{y z}} \\
+& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\frac{1}{x}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{y z}} \quad[1 \text { mark. }] \\
+& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)} \quad[1 \text { mark.] }
+\end{aligned}
+$$
+
+$$
+\begin{aligned}
+& =\sqrt{x y z} \sum_{\text {cyclic }} \sqrt{\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right)^{2}+\frac{(\sqrt{y}-\sqrt{z})^{2}}{x y z}} \quad[2 \text { marks. }] \\
+& \geq \sqrt{x y z} \sum_{\text {cyclic }}\left(\frac{1}{x}+\frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
+& =\sqrt{x y z}\left(1+\sum_{\text {cyclic }} \frac{1}{\sqrt{y z}}\right) \quad[1 \text { mark. }] \\
+& =\sqrt{x y z}+\sum_{\text {cyclic }} \sqrt{x} \quad[1 \text { mark. }]
+\end{aligned}
+$$
+
+Note. It is easy to check that equality holds if and only if $x=y=z=3$.
+Solution 2.
+Squaring both sides of the given inequality, we obtain
+
+$$
+\begin{aligned}
+& \sum_{\text {cyclic }} x+\sum_{\text {cyclic }} y z+2 \sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \\
+& \quad \geq x y z+2 \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} x+2 \sum_{\text {cyclic }} \sqrt{x y} \quad \text { [1 mark.] }
+\end{aligned}
+$$
+
+It follows from the given condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ that $x y z=\sum_{\text {cyclic }} x y$. Therefore, the given inequality is equivalent to
+
+$$
+\sum_{\text {cyclic }} \sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y z} \sum_{\text {cyclic }} \sqrt{x}+\sum_{\text {cyclic }} \sqrt{x y} . \quad[2 \text { marks.] }
+$$
+
+Using the Cauchy-Schwarz inequality [or just $x^{2}+y^{2} \geq 2 x y$ ], we see that
+
+$$
+(x+y z)(y+z x) \geq\left(\sqrt{x y}+\sqrt{x y z^{2}}\right)^{2}, \quad[1 \text { mark. }]
+$$
+
+or
+
+$$
+\sqrt{x+y z} \sqrt{y+z x} \geq \sqrt{x y}+\sqrt{z} \sqrt{x y z} . \quad[1 \text { mark. }]
+$$
+
+Taking the cyclic sum of this inequality over $x, y$ and $z$, we get the desired inequality. [2 marks.]
+
+Solution 3.
+This is another way of presenting the idea in the first solution.
+Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ and the AM-GM inequality, we have
+
+$$
+\begin{aligned}
+x+y z-\left(\sqrt{\frac{y z}{x}}+\sqrt{x}\right)^{2} & =y z\left(1-\frac{1}{x}\right)-2 \sqrt{y z} \\
+& =y z\left(\frac{1}{y}+\frac{1}{z}\right)-2 \sqrt{y z}=y+z-2 \sqrt{y z} \geq 0
+\end{aligned}
+$$
+
+which gives
+
+$$
+\sqrt{x+y z} \geq \sqrt{\frac{y z}{x}}+\sqrt{x} . \quad[3 \text { marks. }]
+$$
+
+Similarly, we have
+
+$$
+\sqrt{y+z x} \geq \sqrt{\frac{z x}{y}}+\sqrt{y} \text { and } \sqrt{z+x y} \geq \sqrt{\frac{x y}{z}}+\sqrt{z}
+$$
+
+Addition yields
+
+$$
+\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}+\sqrt{x}+\sqrt{y}+\sqrt{z}
+$$
+
+[2 marks.] Using the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ again, we have
+
+$$
+\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}=\sqrt{x y z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt{x y z}, \quad[1 \text { mark. }]
+$$
+
+and thus
+
+$$
+\sqrt{x+y z}+\sqrt{y+z x}+\sqrt{z+x y} \geq \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z} . \quad[1 \text { mark. }]
+$$
+
+## Solution 4.
+
+This is also another way of presenting the idea in the first solution.
+We make the substitution $a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z}$. Then it is enough to show that
+
+$$
+\sqrt{\frac{1}{a}+\frac{1}{b c}}+\sqrt{\frac{1}{b}+\frac{1}{c a}}+\sqrt{\frac{1}{c}+\frac{1}{a b}} \geq \sqrt{\frac{1}{a b c}}+\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}},
+$$
+
+where $a+b+c=1$. Multiplying this inequality by $\sqrt{a b c}$, we find that it can be written
+
+$$
+\sqrt{a+b c}+\sqrt{b+c a}+\sqrt{c+a b} \geq 1+\sqrt{b c}+\sqrt{c a}+\sqrt{a b} . \quad[1 \text { mark. }]
+$$
+
+This is equivalent to
+
+$$
+\begin{aligned}
+& \sqrt{a(a+b+c)+b c}+\sqrt{b(a+b+c)+c a}+\sqrt{c(a+b+c)+a b} \\
+& \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}, \quad[1 \text { mark. }]
+\end{aligned}
+$$
+
+which in turn is equivalent to
+
+$$
+\sqrt{(a+b)(a+c)}+\sqrt{(b+c)(b+a)}+\sqrt{(c+a)(c+b)} \geq a+b+c+\sqrt{b c}+\sqrt{c a}+\sqrt{a b}
+$$
+
+[1 mark.] (This is a homogeneous version of the original inequality.) By the Cauchy-Schwarz inequality (or since $b+c \geq 2 \sqrt{b c}$ ), we have
+
+$$
+\left[(\sqrt{a})^{2}+(\sqrt{b})^{2}\right]\left[(\sqrt{a})^{2}+(\sqrt{c})^{2}\right] \geq(\sqrt{a} \sqrt{a}+\sqrt{b} \sqrt{c})^{2}
+$$
+
+or
+
+$$
+\sqrt{(a+b)(a+c)} \geq a+\sqrt{b c} . \quad[2 \text { marks. }]
+$$
+
+Taking the cyclic sum of this inequality over $a, b, c$, we get the desired inequality. [2 marks.]
+5. Let R denote the set of all real numbers. Find all functions $f$ from R to R satisfying:
+(i) there are only finitely many $s$ in R such that $f(s)=0$, and
+(ii) $f\left(x^{4}+y\right)=x^{3} f(x)+f(f(y))$ for all $x, y$ in $\mathbf{R}$.
+
+## Solution 1.
+
+The only such function is the identity function on $R$.
+Setting $(x, y)=(1,0)$ in the given functional equation (ii), we have $f(f(0))=0$. Setting $x=0$ in (ii), we find
+
+$$
+f(y)=f(f(y))
+$$
+
+[1 mark.] and thus $f(0)=f(f(0))=0$ [1 mark.]. It follows from (ii) that $f\left(x^{4}+y\right)=$ $x^{3} f(x)+f(y)$ for all $x, y \in \mathbf{R}$. Set $y=0$ to obtain
+
+$$
+f\left(x^{4}\right)=x^{3} f(x)
+$$
+
+for all $x \in \mathrm{R}$, and so
+
+$$
+f\left(x^{4}+y\right)=f\left(x^{4}\right)+f(y)
+$$
+
+for all $x, y \in \mathbf{R}$. The functional equation (3) suggests that $f$ is additive, that is, $f(a+b)=$ $f(a)+f(b)$ for all $a, b \in \mathbf{R}$. [1 mark.] We now show this.
+
+First assume that $a \geq 0$ and $b \in R$. It follows from (3) that
+
+$$
+f(a+b)=f\left(\left(a^{1 / 4}\right)^{4}+b\right)=f\left(\left(a^{1 / 4}\right)^{4}\right)+f(b)=f(a)+f(b)
+$$
+
+We next note that $f$ is an odd function, since from (2)
+
+$$
+f(-x)=\frac{f\left(x^{4}\right)}{(-x)^{3}}=\frac{f\left(x^{4}\right)}{-x^{3}}=-f(x), \quad x \neq 0
+$$
+
+Since $f$ is odd, we have that, for $a<0$ and $b \in R$,
+
+$$
+\begin{aligned}
+f(a+b) & =-f((-a)+(-b))=-(f(-a)+f(-b)) \\
+& =-(-f(a)-f(b))=f(a)+f(b)
+\end{aligned}
+$$
+
+Therefore, we conclude that $f(a+b)=f(a)+f(b)$ for all $a, b \in R$. [2 mers.].]
+We now show that $\{s \in \mathrm{R} \mid f(s)=0\}=\{0\}$. Recall that $f(0)=0$. Assume that there is a nonzero $h \in \mathrm{R}$ such that $f(h)=0$. Then, using the fact that $f$ is additive, we inductively have $f(n h)=0$ or $n h \in\{s \in R \mid f(s)=0\}$ for all $n \in \mathbf{N}$. However, this is a contradiction to the given condition (i). [1 mark.]
+
+It's now easy to check that $f$ is one-to-one. Assume that $f(a)=f^{\prime}(b)$ for some $a, b \in \mathbb{P}$. Then, we have $f(b)=f(a)=f(a-b)+f(b)$ or $f(a-b)=0$. This implies that $a-b \in\{s \in$ $\mathbf{R} \mid f(s)=0\}=\{0\}$ or $a=b$, as desired. From (1) and the fact that $f$ is one-to-one, we deduce that $f(x)=x$ for all $x \in \mathbf{R}$. [1 mark.] This completes the proof.
+
+## Solution 2.
+
+Again, the only such function is the identity function on R .
+As in Solution 1, we first show that $f(f(y))=f(y), f(0)=0$, and $f\left(x^{4}\right)=x^{3} f(x)$. [2 marks.] From the latter follows
+
+$$
+f(x)=0 \Longrightarrow f\left(x^{4}\right)=0
+$$
+
+and from condition (i) we get that $f(x)=0$ only possibly for $x \in\{0,1,-1\}$. [1 mark.]
+Next we prove
+
+$$
+f(a)=b \Longrightarrow f(\sqrt[4]{|a-b|})=0
+$$
+
+This is clear if $a=b$. If $a>b$ then
+
+$$
+\begin{aligned}
+f(a) & =f((a-b)+b)=(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(b)) \\
+& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(b) \\
+& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(f(a)) \\
+& =(a-b)^{3 / 4} f(\sqrt[4]{a-b})+f(a),
+\end{aligned}
+$$
+
+so $(a-b)^{3 / 4} f(\sqrt[4]{a-b})=0$ which means $f(\sqrt[4]{|a-b|})=0$. If $a<b$ we get similarly
+
+$$
+\begin{aligned}
+f(b) & =f((b-a)+a)=(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(f(a)) \\
+& =(b-a)^{3 / 4} f(\sqrt[4]{b-a})+f(b)
+\end{aligned}
+$$
+
+and again $f(\sqrt[4]{|a-b|})=0$. [2 marks.]
+Thus $f(a)=b \Longrightarrow|a-b| \in\{0,1\}$. Suppose that $f(x)=x+b$ for some $x$, where $|b|=1$. Then from $f\left(x^{4}\right)=x^{3} f(x)$ and $f\left(x^{4}\right)=x^{4}+a$ for some $|a| \leq 1$ we get $x^{3}=a / b$, so $|x| \leq 1$. Thus $f(x)=x$ for all $x$ except possibly $x= \pm 1$. [ 1 mark.] But for example,
+
+$$
+f(1)=f\left(2^{4}-15\right)=2^{3} f(2)+f(f(-15))=2^{3} \cdot 2-15=1
+$$
+
+and
+
+$$
+f(-1)=f\left(2^{4}-17\right)=2^{3} f(2)+f(f(-17))=2^{3} \cdot 2-17=-1
+$$
+
+[1 mark.] This finishes the proof.
+

APMO/md/en-apmo2003_sol.md

@@ -0,0 +1,246 @@
+# XV APMO: Solutions and Marking Schemes 
+
+1. Let $a, b, c, d, e, f$ be real numbers such that the polynomial
+
+$$
+p(x)=x^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f
+$$
+
+factorises into eight linear factors $x-x_{i}$, with $x_{i}>0$ for $i=1,2, \ldots, 8$. Determine all possible values of $f$.
+
+## Solution.
+
+From
+
+$$
+x^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{8}\right)
+$$
+
+we have
+
+$$
+\sum_{i=1}^{8} x_{i}=4 \quad \text { and } \quad \sum x_{i} x_{j}=7
+$$
+
+where the second sum is over all pairs $(i, j)$ of integers where $1 \leq i<j \leq 8$. Since this sum can also be written
+
+$$
+\frac{1}{2}\left[\left(\sum_{i=1}^{8} x_{i}\right)^{2}-\sum_{i=1}^{8} x_{i}^{2}\right]
+$$
+
+we get
+
+$$
+14=\left(\sum_{i=1}^{8} x_{i}\right)^{2}-\sum_{i=1}^{8} x_{i}^{2}=16-\sum_{i=1}^{8} x_{i}^{2}
+$$
+
+so
+
+$$
+\sum_{i=1}^{8} x_{i}^{2}=2 \quad \text { while } \quad \sum_{i=1}^{8} x_{i}=4 . \quad[3 \text { marks }]
+$$
+
+Now
+
+$$
+\sum_{i=1}^{8}\left(2 x_{i}-1\right)^{2}=4 \sum_{i=1}^{8} x_{i}^{2}-4 \sum_{i=1}^{8} x_{i}+8=4(2)-4(4)+8=0
+$$
+
+which forces $x_{i}=1 / 2$ for all $i$. [3 marks] Therefore
+
+$$
+f=\prod_{i=1}^{8} x_{i}=\left(\frac{1}{2}\right)^{8}=\frac{1}{256} . \quad[1 m x]
+$$
+
+Alternate solution: After obtaining (1) [3 marks], use Cauchy's inequality to get
+
+$$
+16=\left(x_{1} \cdot 1+x_{2} \cdot 1+\cdots+x_{8} \cdot 1\right)^{2} \leq\left(x_{1}^{2-}+x_{2}^{2}+\cdots+x_{8}^{2}\right)\left(1^{2}+1^{2}+\cdots+1^{2}\right)=8 \cdot 2=16
+$$
+
+or the power mean inequality to get
+
+$$
+\frac{1}{2}=\frac{1}{8} \sum_{i=1}^{8} x_{i} \leq\left(\frac{1}{8} \sum_{i=1}^{8} x_{i}^{2}\right)^{1 / 2}=\frac{1}{2} . \quad[2 \text { marks }]
+$$
+
+Either way, equality must hold, which can only happen if all the terms $x_{i}$ are equal, that is, if $x_{i}=1 / 2$ for all $i$. [1 mark] Thus $f=1 / 256$ as above. [ 1 mark]
+2. Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_{1}$ and $\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\triangle A E F$ and $\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.
+
+Solution 1.
+Let $E H$ intersect $F G$ at $O$. The distance from $G$ to line $F D$ and line $E F$ are both $a$. So $F G$ bisects $\angle E F D$. Similarly, $E H$ bisects $\angle B E F$. So $O$ is an excentre of $\triangle A E F$. Similarly, $O$ is an excentre of $\triangle C G H$. [2 marks] Construct these excircles with centre $O$. Let $M, N, P, Q$ be on sides $A B, B C, C D, D A$ respectively, where these excircles touch the square. Then $O M \perp A B, O N \perp B C, O P \perp C D$, and $O Q \perp D A$. Since $A B \| C D$ and $A D \| B C, M, O, P$ are collinear and $N, O, Q$ are collinear. Now $M P=N Q=a$. [2 marks] Using the fact that the two tangents from a point to a circle have the same length, we get $E F=E M+F Q$ and $G H=G N+H P$. [1 mark] Then
+
+$$
+m_{1}=A E+A F+E F=A E+A F+(E M+F Q)=A M+A Q=O Q+O M
+$$
+
+and
+
+$$
+m_{2}=C G+C H+G H=C G+C H+(G N+H P)=C N+C P=O P+O N . \quad[1 \text { mark }]
+$$
+
+Therefore
+
+$$
+m_{1}+m_{2}=(O Q+O M)+(O P+O N)=M P+N Q=2 a . \quad[1 \mathrm{mark}]
+$$
+
+## Solution 2.
+
+Extend $A B$ to $I$ and $D C$ to $J$ so that $A E=B I=C J$. Let $\ell_{2}$ intersect $I J$ at $M$, and let $K$ lie on $I J$ so that $G K \perp I J$. Then, since $A E=G K, \triangle A E F$ and $\triangle K G M$ are congruent. [1 mark] Thus, since $G K=C J$ and $G C=K J$,
+
+$$
+m_{1}+m_{2}=\operatorname{perimeter}(K G M)+\operatorname{perimeter}(C G H)=\operatorname{perimeter}(H M J) . \quad[\mathbf{2} \text { marks }]
+$$
+
+Let $L$ lie on $C D$ so that $E L \perp C D$. Then a circle with centre $E$ and radius $a$ will touch $D C$ at $L, I J$ at $I$, and the interior of $H M$ at some point $N$, so
+
+$$
+\operatorname{perimeter}(H M J)=J H+(H N+N M)+J M=(J H+H L)+(M I+J M)=J L+I J=a+a=2 a
+$$
+
+[4 marks] Thus $m_{1}+m_{2}=2 a$.
+Solution 3.
+Without loss of generality, assume the square has side $a=1$. Let $\theta$ be the acute angle between $\ell_{1}$ (or $\ell_{2}$ ) and the sides $A B$ and $C D$ of the square. Then, letting $E F=x$ and $G H=y$, we have
+
+$$
+E A=x \cos \theta, \quad A F=x \sin \theta, \quad C H=y \cos \theta, \quad C G=y \sin \theta
+$$
+
+Thus
+
+$$
+m_{1}+m_{2}=(x+y)(\sin \theta+\cos \theta+1) . \quad[2 \text { marks }]
+$$
+
+Draw lines parallel to $\ell_{1}, \ell_{2}$ through $A$ and $C$ respectively. The distance between these lines is $\sin \theta+\cos \theta$ [1 mark], as can be seen by drawing a mutual perpendicular to these lines through $B$, say. Also, the altitudes from $A$ to $E F$ and from $C$ to $G H$ have lengths $x \sin \theta \cos \theta$ and $y \sin \theta \cos \theta$ respectively [ 1 mark]. Therefore the distance between $\ell_{1}$ and $\ell_{2}$ must be
+
+$$
+(\sin \theta+\cos \theta)-x \sin \theta \cos \theta-y \sin \theta \cos \theta
+$$
+
+But we are given that this distance is $a=1$, so
+
+$$
+(x+y) \sin \theta \cos \theta+1=\sin \theta+\cos \theta
+$$
+
+or
+
+$$
+x+y=\frac{\sin \theta+\cos \theta-1}{\sin \theta \cos \theta} \cdot \quad[1 \text { mark }]
+$$
+
+Therefore, by (1),
+
+$$
+\begin{aligned}
+m_{1}+m_{2} & =\frac{(\sin \theta+\cos \theta-1)(\sin \theta+\cos \theta+1)}{\sin \theta \cos \theta} \\
+& =\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)-1}{\sin \theta \cos \theta} \\
+& =\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}=2 . \quad[2 \text { marks }]
+\end{aligned}
+$$
+
+3. Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove:
+(a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !;
+(b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.
+
+Solution.
+(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \nmid(n-k)$ !, so $2 p_{k} \nless(n-k)$ !. [1 mark]
+(b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$.
+
+If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. [ 1 mark] Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.
+
+Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]
+
+Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\{1,2, \ldots, n-k\}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks]
+4. Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \geq 2$ be an integer. Show that
+
+$$
+\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+\frac{\sqrt[n]{2}}{2}
+$$
+
+## Solution.
+
+Without loss of generality, assume $a \leq b \leq c$. As $a+b>c$, we have
+
+$$
+\frac{\sqrt[n]{2}}{2}=\frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c)=\sqrt[n]{2 c^{n}} \geq \sqrt[n]{b^{n}+c^{n}} \quad \quad[2 \text { marks }]
+$$
+
+As $a \leq c$ and $n \geq 2$, we have
+
+$$
+\begin{aligned}
+\left(c^{n}+a^{n}\right)-\left(c+\frac{a}{2}\right)^{n} & =a^{n}-\sum_{k=1}^{n}\binom{n}{k} c^{n-k}\left(\frac{a}{2}\right)^{k} \\
+& \leq\left[1-\sum_{k=1}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n} \quad\left(\text { since } c^{n-k} \geq a^{n-k}\right) \\
+& =\left[\left(1-\frac{n}{2}\right)-\sum_{k=2}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n}<0
+\end{aligned}
+$$
+
+Thus
+
+$$
+\sqrt[n]{c^{n}+a^{n}}<c+\frac{a}{2} . \quad[3 \text { marks }]
+$$
+
+Likewise
+
+$$
+\sqrt[n]{b^{n}+a^{n}}<b+\frac{a}{2} \quad \quad[1 \text { mark }]
+$$
+
+Adding (1), (2) and (3), we get
+
+$$
+\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<\frac{\sqrt[n]{2}}{2}+c+\frac{a}{2}+b+\frac{a}{2}=1+\frac{\sqrt[n]{2}}{2} . \quad[1 \text { mark }]
+$$
+
+5. Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2 m$ of them who form $m$ pairs of mutually acquainted people or there are $2 n$ of them forming $n$ pairs of mutually unacquainted people.
+
+## Solution.
+
+Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that
+
+$$
+r(m, n)=2(m+n)-\min \{m, n\}-1
+$$
+
+Observe that, by symmetry, $r(m, n)=r(n, m)$. Therefore it suffices to consider the case where $m \geq n$, and to prove that
+
+$$
+r(m, n)=2 m+n-1 . \quad[1 \text { mark }]
+$$
+
+First we prove that
+
+$$
+r(m, n) \geq 2 m+n-1
+$$
+
+by an example. Call a group of $k$ people, every two of whom are mutually acquainted, a $k$-clique. Consider a set of $2 m+n-2$ people consisting of a $(2 m-1)$-clique together with an additional $n-1$ people none of whom know anyone else. (Call such people isolated.) Then there are not $2 m$ people forming $m$ mutually acquainted pairs, and there also are not $2 n$ people forming $n$ mutually unacquainted pairs. Thus $r(m, n) \geq$ $(2 m-1)+(n-1)+1=2 m+n-1$ by the definition of $r(m, n)$. [1 mark]
+
+To establish (1), we need to prove that $r(m, n) \leq 2 m+n-1$. To do this, we now show that
+
+$$
+r(m, n) \leq r(m-1, n-1)+3 \quad \text { for all } m \geq n \geq 2
+$$
+
+Let $G$ be a group of $t=r(m-1, n-1)+3$ people. Notice that
+
+$$
+t \geq 2(m-1)+(n-1)-1+3=2 m+n-1 \geq 2 m \geq 2 n
+$$
+
+If $G$ is a $t$-clique, then $G$ contains $2 m$ people forming $m$ mutually acquainted pairs, and if $G$ has only isolated people, then $G$ contains $2 n$ people forming $n$ mutually unacquainted pairs. Otherwise, there are three people in $G$, say $a, b$ and $c$, such that $a, b$ are acquainted but $a, c$ are not. Now consider the group $A$ obtained byremoving $a, b$ and $c$ from $G$. A has $t-3=r(m-1, n-1)$ people, so by the definition of $r(m-1, n-1)$, A either contains $2(m-1)$ people forming $m-1$ mutually acquainted pairs, or else contains $2(n-1)$ people forming $n-1$ mutually unacquainted pairs. In the former case, we add the acquainted pair $a, b$ to $A$ to form $m$ mutually acquainted pairs in $G$. In the latter case, we add the unacquainted pair $a, c$ to $A$ to form $n$ mutually unacquainted pairs in $G$. This proves (2). [3 marks]
+
+Trivially, $r(s, 1)=2 s$ for all $s[\mathbf{1}$ mark], so $r(m, n) \leq 2 m+n-1$ holds whenever $n=1$. Proceeding by induction on $n$, by (2) we obtain
+
+$$
+r(m, n) \leq r(m-1, n-1)+3 \leq 2(m-1)+(n-1)-1+3=2 m+n-1
+$$
+
+which completes the proof. [1 mark]
+Note. Give an additional 1 mark to any student who gets at most 5 marks by the above marking scheme, but in addition gives a valid argument that $r(2,2)=5$.
+

APMO/md/en-apmo2004_sol.md

@@ -0,0 +1,283 @@
+# APMO 2004 - Problems and Solutions 
+
+## Problem 1
+
+Determine all finite nonempty sets $S$ of positive integers satisfying
+
+$$
+\frac{i+j}{(i, j)} \text { is an element of } S \text { for all } i, j \text { in } S \text {, }
+$$
+
+where $(i, j)$ is the greatest common divisor of $i$ and $j$.
+Answer: $S=\{2\}$.
+
+## Solution
+
+Let $k \in S$. Then $\frac{k+k}{(k, k)}=2$ is in $S$ as well.
+Suppose 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.
+Now 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.
+Therefore $S$ can only contain 2 , and $S=\{2\}$ is the only solution.
+
+## Problem 2
+
+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 1
+
+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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_7992a4504bd2a9e623e4g-2.jpg?height=444&width=618&top_left_y=523&top_left_x=685)
+
+Let $M$ 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
+
+$$
+[B O H]+[C O H]=\frac{O H \cdot d(B, O H)}{2}+\frac{O H \cdot d(C, O H)}{2}=\frac{O H \cdot 2 d(M, O H)}{2} .
+$$
+
+Since $A G=2 G M, d(A, O H)=2 d(M, O H)$. Hence
+
+$$
+[B O H]+[C O H]=\frac{O H \cdot d(A, O H)}{2}=[A O H]
+$$
+
+and the result follows.
+
+## Solution 2
+
+One can use barycentric coordinates: it is well known that
+
+$$
+\begin{gathered}
+A=(1: 0: 0), \quad B=(0: 1: 0), \quad C=(0: 0: 1), \\
+O=(\sin 2 A: \sin 2 B: \sin 2 C) \quad \text { and } \quad H=(\tan A: \tan B: \tan C) .
+\end{gathered}
+$$
+
+Then the (signed) area of $A O H$ is proportional to
+
+$$
+\left|\begin{array}{ccc}
+1 & 0 & 0 \\
+\sin 2 A & \sin 2 B & \sin 2 C \\
+\tan A & \tan B & \tan C
+\end{array}\right|
+$$
+
+Adding all three expressions we find that the sum of the signed sums of the areas is a constant times
+
+$$
+\left|\begin{array}{ccc}
+1 & 0 & 0 \\
+\sin 2 A & \sin 2 B & \sin 2 C \\
+\tan A & \tan B & \tan C
+\end{array}\right|+\left|\begin{array}{ccc}
+0 & 1 & 0 \\
+\sin 2 A & \sin 2 B & \sin 2 C \\
+\tan A & \tan B & \tan C
+\end{array}\right|+\left|\begin{array}{ccc}
+0 & 0 & 1 \\
+\sin 2 A & \sin 2 B & \sin 2 C \\
+\tan A & \tan B & \tan C
+\end{array}\right|
+$$
+
+By multilinearity of the determinant, this sum equals
+
+$$
+\left|\begin{array}{ccc}
+1 & 1 & 1 \\
+\sin 2 A & \sin 2 B & \sin 2 C \\
+\tan A & \tan B & \tan C
+\end{array}\right|
+$$
+
+which 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.
+Comment: 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 3
+
+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.
+Note: 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.
+Now 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 $q r$ :
+![](https://cdn.mathpix.com/cropped/2024_11_22_7992a4504bd2a9e623e4g-3.jpg?height=515&width=809&top_left_y=950&top_left_x=586)
+
+Any 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 $p q r$.
+Let $n_{i}$ 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
+
+$$
+\begin{aligned}
+n(p, q)+n(q, r)+n(p, r) & \equiv\left(n_{2}+n_{5}+n_{7}\right)+\left(n_{1}+n_{4}+n_{7}\right)+\left(n_{3}+n_{6}+n_{7}\right) \\
+& \equiv n_{1}+n_{2}+n_{3}+n_{4}+n_{5}+n_{6}+n_{7}=2004-3 \equiv 1 \quad(\bmod 2)
+\end{aligned}
+$$
+
+and the result follows.
+Comment: 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 4
+
+For a real number $x$, let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$. Prove that
+
+$$
+\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor
+$$
+
+is even for every positive integer $n$.
+
+## Solution
+
+Consider four cases:
+
+- $n \leq 5$. Then $\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=0$ is an even number.
+- $n$ 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.
+- $n \geq 7$ 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.
+Also, $n$ 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
+
+$$
+\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n+1}{n(n+1)}-1
+$$
+
+is even.
+
+- $n+1 \geq 7$ 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
+
+$$
+\left\lfloor\frac{(n-1)!}{n(n+1)}\right\rfloor=\frac{(n-1)!+n}{n(n+1)}-1
+$$
+
+is even.
+
+## Problem 5
+
+Prove that
+
+$$
+\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(a b+b c+c a)
+$$
+
+for all real numbers $a, b, c>0$.
+
+## Solution 1
+
+Let $p=a+b+c, q=a b+b c+c a$, and $r=a b c$. The inequality simplifies to
+
+$$
+a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8-9(a b+b c+c a) \geq 0
+$$
+
+Since $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=q^{2}-2 p r$ and $a^{2}+b^{2}+c^{2}=p^{2}-2 q$,
+
+$$
+r^{2}+2 q^{2}-4 p r+4 p^{2}-8 q+8-9 q \geq 0
+$$
+
+which simplifies to
+
+$$
+r^{2}+2 q^{2}+4 p^{2}-17 q-4 p r+8 \geq 0
+$$
+
+Bearing in mind that equality occurs for $a=b=c=1$, which means that, for instance, $p=3 r$, one can rewrite $(I)$ as
+
+$$
+\left(r-\frac{p}{3}\right)^{2}-\frac{10}{3} p r+\frac{35}{9} p^{2}+2 q^{2}-17 q+8 \geq 0
+$$
+
+Since $(a b-b c)^{2}+(b c-c a)^{2}+(c a-a b)^{2} \geq 0$ is equivalent to $q^{2} \geq 3 p r$, rewrite $(I I)$ as
+
+$$
+\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9} p^{2}+\frac{8}{9} q^{2}-17 q+8 \geq 0
+$$
+
+Finally, $a=b=c=1$ implies $q=3$; then rewrite (III) as
+
+$$
+\left(r-\frac{p}{3}\right)^{2}+\frac{10}{9}\left(q^{2}-3 p r\right)+\frac{35}{9}\left(p^{2}-3 q\right)+\frac{8}{9}(q-3)^{2} \geq 0
+$$
+
+This final inequality is true because $q^{2} \geq 3 p r$ and $p^{2}-3 q=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \geq 0$.
+
+## Solution 2
+
+We prove the stronger inequality
+
+$$
+\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 3(a+b+c)^{2}
+$$
+
+which 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.
+The inequality $(*)$ is equivalent to
+
+$$
+\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right) a^{2}-6(b+c) a+2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2} \geq 0
+$$
+
+Seeing 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
+
+$$
+(3(b+c))^{2}-\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)\left(2\left(b^{2}+2\right)\left(c^{2}+2\right)-3(b+c)^{2}\right) \leq 0
+$$
+
+This simplifies to
+
+$$
+-2\left(b^{2}+2\right)\left(c^{2}+2\right)+3(b+c)^{2}+6 \leq 0
+$$
+
+Now we look $(* *)$ as a quadratic inequality in $b$ with negative leading coefficient $-2 c^{2}-1$ :
+
+$$
+\left(-2 c^{2}-1\right) b^{2}+6 c b-c^{2}-2 \leq 0
+$$
+
+If suffices to show that the discriminant of $(* *)$ is non-positive, which is equivalent to
+
+$$
+9 c^{2}-\left(2 c^{2}+1\right)\left(c^{2}+2\right) \leq 0
+$$
+
+It 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 $a=\frac{6(b+c)}{2\left(\left(b^{2}+2\right)\left(c^{2}+2\right)-3\right)}=1$.
+
+## Solution 3
+
+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
+
+$$
+4 \sec ^{2} A \sec ^{2} B \sec ^{2} C \geq 9(\tan A \tan B+\tan B \tan C+\tan C \tan A)
+$$
+
+Substituting $\sec x=\frac{1}{\cos x}$ for $x \in\{A, B, C\}$ and clearing denominators, the inequality is equivalent to
+
+$$
+\cos A \cos B \cos C(\sin A \sin B \cos C+\cos A \sin B \sin C+\sin A \cos B \sin C) \leq \frac{4}{9}
+$$
+
+Since
+
+$$
+\begin{aligned}
+& \cos (A+B+C)=\cos A \cos (B+C)-\sin A \sin (B+C) \\
+= & \cos A \cos B \cos C-\cos A \sin B \sin C-\sin A \cos B \sin C-\sin A \sin B \cos C,
+\end{aligned}
+$$
+
+we rewrite our inequality as
+
+$$
+\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \frac{4}{9}
+$$
+
+The 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,
+
+$$
+\cos A \cos B \cos C \leq\left(\frac{\cos A+\cos B+\cos C}{3}\right)^{3} \leq \cos ^{3} \frac{A+B+C}{3}=\cos ^{3} \theta
+$$
+
+Therefore, 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 $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$,
+$\cos A \cos B \cos C(\cos A \cos B \cos C-\cos (A+B+C)) \leq \cos ^{3} \theta\left(\cos ^{3} \theta-\cos 3 \theta\right)=3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)$. Finally, by AM-GM (notice that $1-\cos ^{2} \theta=\sin ^{2} \theta>0$ ),
+$3 \cos ^{4} \theta\left(1-\cos ^{2} \theta\right)=\frac{3}{2} \cos ^{2} \theta \cdot \cos ^{2} \theta\left(2-2 \cos ^{2} \theta\right) \leq \frac{3}{2}\left(\frac{\cos ^{2} \theta+\cos ^{2} \theta+\left(2-2 \cos ^{2} \theta\right)}{3}\right)^{3}=\frac{4}{9}$,
+and the result follows.
+

APMO/md/en-apmo2005_sol.md

@@ -0,0 +1,302 @@
+# XVII APMO - March, 2005 
+
+## Problems and Solutions
+
+Problem 1. Prove that for every irrational real number $a$, there are irrational real numbers $b$ and $b^{\prime}$ so that $a+b$ and $a b^{\prime}$ are both rational while $a b$ and $a+b^{\prime}$ are both irrational.
+(Solution) Let $a$ be an irrational number. If $a^{2}$ is irrational, we let $b=-a$. Then, $a+b=0$ is rational and $a b=-a^{2}$ is irrational. If $a^{2}$ is rational, we let $b=a^{2}-a$. Then, $a+b=a^{2}$ is rational and $a b=a^{2}(a-1)$. Since
+
+$$
+a=\frac{a b}{a^{2}}+1
+$$
+
+is irrational, so is $a b$.
+Now, we let $b^{\prime}=\frac{1}{a}$ or $b^{\prime}=\frac{2}{a}$. Then $a b^{\prime}=1$ or 2 , which is rational. Note that
+
+$$
+a+b^{\prime}=\frac{a^{2}+1}{a} \quad \text { or } \quad a+b^{\prime}=\frac{a^{2}+2}{a} .
+$$
+
+Since,
+
+$$
+\frac{a^{2}+2}{a}-\frac{a^{2}+1}{a}=\frac{1}{a}
+$$
+
+at least one of them is irrational.
+
+Problem 2. Let $a, b$ and $c$ be positive real numbers such that $a b c=8$. Prove that
+
+$$
+\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}}+\frac{b^{2}}{\sqrt{\left(1+b^{3}\right)\left(1+c^{3}\right)}}+\frac{c^{2}}{\sqrt{\left(1+c^{3}\right)\left(1+a^{3}\right)}} \geq \frac{4}{3} .
+$$
+
+(Solution) Observe that
+
+$$
+\frac{1}{\sqrt{1+x^{3}}} \geq \frac{2}{2+x^{2}}
+$$
+
+In fact, this is equivalent to $\left(2+x^{2}\right)^{2} \geq 4\left(1+x^{3}\right)$, or $x^{2}(x-2)^{2} \geq 0$. Notice that equality holds in (1) if and only if $x=2$.
+
+We substitute $x$ by $a, b, c$ in (1), respectively, to find
+
+$$
+\begin{gathered}
+\frac{a^{2}}{\sqrt{\left(1+a^{3}\right)\left(1+b^{3}\right)}}+\frac{b^{2}}{\sqrt{\left(1+b^{3}\right)\left(1+c^{3}\right)}}+\frac{c^{2}}{\sqrt{\left(1+c^{3}\right)\left(1+a^{3}\right)}} \\
+\geq \frac{4 a^{2}}{\left(2+a^{2}\right)\left(2+b^{2}\right)}+\frac{4 b^{2}}{\left(2+b^{2}\right)\left(2+c^{2}\right)}+\frac{4 c^{2}}{\left(2+c^{2}\right)\left(2+a^{2}\right)}
+\end{gathered}
+$$
+
+We combine the terms on the right hand side of (2) to obtain
+
+$$
+\text { Left hand side of }(2) \geq \frac{2 S(a, b, c)}{36+S(a, b, c)}=\frac{2}{1+36 / S(a, b, c)}
+$$
+
+where $S(a, b, c):=2\left(a^{2}+b^{2}+c^{2}\right)+(a b)^{2}+(b c)^{2}+(c a)^{2}$. By AM-GM inequality, we have
+
+$$
+\begin{aligned}
+a^{2}+b^{2}+c^{2} & \geq 3 \sqrt[3]{(a b c)^{2}}=12 \\
+(a b)^{2}+(b c)^{2}+(c a)^{2} & \geq 3 \sqrt[3]{(a b c)^{4}}=48
+\end{aligned}
+$$
+
+Note that the equalities holds if and only if $a=b=c=2$. The above inequalities yield
+
+$$
+S(a, b, c)=2\left(a^{2}+b^{2}+c^{2}\right)+(a b)^{2}+(b c)^{2}+(c a)^{2} \geq 72
+$$
+
+Therefore
+
+$$
+\frac{2}{1+36 / S(a, b, c)} \geq \frac{2}{1+36 / 72}=\frac{4}{3}
+$$
+
+which is the required inequality.
+
+Problem 3. Prove that there exists a triangle which can be cut into 2005 congruent triangles.
+(Solution) Suppose that one side of a triangle has length $n$. Then it can be cut into $n^{2}$ congruent triangles which are similar to the original one and whose corresponding sides to the side of length $n$ have lengths 1 .
+
+Since $2005=5 \times 401$ where 5 and 401 are primes and both primes are of the type $4 k+1$, it is representable as a sum of two integer squares. Indeed, it is easy to see that
+
+$$
+\begin{aligned}
+2005 & =5 \times 401=\left(2^{2}+1\right)\left(20^{2}+1\right) \\
+& =40^{2}+20^{2}+2^{2}+1 \\
+& =(40-1)^{2}+2 \times 40+20^{2}+2^{2} \\
+& =39^{2}+22^{2}
+\end{aligned}
+$$
+
+Let $A B C$ be a right-angled triangle with the legs $A B$ and $B C$ having lengths 39 and 22 , respectively. We draw the altitude $B K$, which divides $A B C$ into two similar triangles. Now we divide $A B K$ into $39^{2}$ congruent triangles as described above and $B C K$ into $22^{2}$ congruent triangles. Since $A B K$ is similar to $B K C$, all 2005 triangles will be congruent.
+
+Problem 4. In a small town, there are $n \times n$ houses indexed by $(i, j)$ for $1 \leq i, j \leq n$ with $(1,1)$ being the house at the top left corner, where $i$ and $j$ are the row and column indices, respectively. At time 0 , a fire breaks out at the house indexed by $(1, c)$, where $c \leq \frac{n}{2}$. During each subsequent time interval $[t, t+1]$, the fire fighters defend a house which is not yet on fire while the fire spreads to all undefended neighbors of each house which was on fire at time $t$. Once a house is defended, it remains so all the time. The process ends when the fire can no longer spread. At most how many houses can be saved by the fire fighters? A house indexed by $(i, j)$ is a neighbor of a house indexed by $(k, \ell)$ if $|i-k|+|j-\ell|=1$.
+(Solution) At most $n^{2}+c^{2}-n c-c$ houses can be saved. This can be achieved under the following order of defending:
+
+$$
+\begin{gathered}
+(2, c),(2, c+1) ;(3, c-1),(3, c+2) ;(4, c-2),(4, c+3) ; \ldots \\
+(c+1,1),(c+1,2 c) ;(c+1,2 c+1), \ldots,(c+1, n)
+\end{gathered}
+$$
+
+Under this strategy, there are
+2 columns (column numbers $c, c+1$ ) at which $n-1$ houses are saved
+2 columns (column numbers $c-1, c+2$ ) at which $n-2$ houses are saved
+...
+2 columns (column numbers $1,2 c$ ) at which $n-c$ houses are saved
+$n-2 c$ columns (column numbers $n-2 c+1, \ldots, n$ ) at which $n-c$ houses are saved
+Adding all these we obtain :
+
+$$
+2[(n-1)+(n-2)+\cdots+(n-c)]+(n-2 c)(n-c)=n^{2}+c^{2}-c n-c
+$$
+
+We say that a house indexed by $(i, j)$ is at level $t$ if $|i-1|+|j-c|=t$. Let $d(t)$ be the number of houses at level $t$ defended by time $t$, and $p(t)$ be the number of houses at levels greater than $t$ defended by time $t$. It is clear that
+
+$$
+p(t)+\sum_{i=1}^{t} d(i) \leq t \text { and } p(t+1)+d(t+1) \leq p(t)+1
+$$
+
+Let $s(t)$ be the number of houses at level $t$ which are not burning at time $t$. We prove that
+
+$$
+s(t) \leq t-p(t) \leq t
+$$
+
+for $1 \leq t \leq n-1$ by induction. It is obvious when $t=1$. Assume that it is true for $t=k$. The union of the neighbors of any $k-p(k)+1$ houses at level $k+1$ contains at least $k-p(k)+1$ vertices at level $k$. Since $s(k) \leq k-p(k)$, one of these houses at level $k$ is burning. Therefore, at most $k-p(k)$ houses at level $k+1$ have no neighbor burning. Hence we have
+
+$$
+\begin{aligned}
+s(k+1) & \leq k-p(k)+d(k+1) \\
+& =(k+1)-(p(k)+1-d(k+1)) \\
+& \leq(k+1)-p(k+1)
+\end{aligned}
+$$
+
+We now prove that the strategy given above is optimal. Since
+
+$$
+\sum_{t=1}^{n-1} s(t) \leq\binom{ n}{2}
+$$
+
+the maximum number of houses at levels less than or equal to $n-1$, that can be saved under any strategy is at most $\binom{n}{2}$, which is realized by the strategy above. Moreover, at levels bigger than $n-1$, every house is saved under the strategy above.
+
+The following is an example when $n=11$ and $c=4$. The houses with $\bigcirc$ mark are burned. The houses with $\otimes$ mark are blocked ones and hence those and the houses below them are saved.
+![](https://cdn.mathpix.com/cropped/2024_11_22_4451135f73c37463c161g-5.jpg?height=1023&width=1055&top_left_y=1055&top_left_x=519)
+
+Problem 5. In a triangle $A B C$, points $M$ and $N$ are on sides $A B$ and $A C$, respectively, such that $M B=B C=C N$. Let $R$ and $r$ denote the circumradius and the inradius of the triangle $A B C$, respectively. Express the ratio $M N / B C$ in terms of $R$ and $r$.
+(Solution) Let $\omega, O$ and $I$ be the circumcircle, the circumcenter and the incenter of $A B C$, respectively. Let $D$ be the point of intersection of the line $B I$ and the circle $\omega$ such that $D \neq B$. Then $D$ is the midpoint of the arc $A C$. Hence $O D \perp C N$ and $O D=R$.
+
+We first show that triangles $M N C$ and $I O D$ are similar. Because $B C=B M$, the line $B I$ (the bisector of $\angle M B C$ ) is perpendicular to the line $C M$. Because $O D \perp C N$ and $I D \perp M C$, it follows that
+
+$$
+\angle O D I=\angle N C M
+$$
+
+Let $\angle A B C=2 \beta$. In the triangle $B C M$, we have
+
+$$
+\frac{C M}{N C}=\frac{C M}{B C}=2 \sin \beta
+$$
+
+Since $\angle D I C=\angle D C I$, we have $I D=C D=A D$. Let $E$ be the point of intersection of the line $D O$ and the circle $\omega$ such that $E \neq D$. Then $D E$ is a diameter of $\omega$ and $\angle D E C=\angle D B C=\beta$. Thus we have
+
+$$
+\frac{D I}{O D}=\frac{C D}{O D}=\frac{2 R \sin \beta}{R}=2 \sin \beta .
+$$
+
+Combining equations (8), (9), and (10) shows that triangles $M N C$ and $I O D$ are similar. It follows that
+
+$$
+\frac{M N}{B C}=\frac{M N}{N C}=\frac{I O}{O D}=\frac{I O}{R} .
+$$
+
+The well-known Euler's formula states that
+
+$$
+O I^{2}=R^{2}-2 R r .
+$$
+
+Therefore,
+
+$$
+\frac{M N}{B C}=\sqrt{1-\frac{2 r}{R}}
+$$
+
+(Alternative Solution) Let $a$ (resp., $b, c$ ) be the length of $B C$ (resp., $A C, A B$ ). Let $\alpha$ (resp., $\beta, \gamma$ ) denote the angle $\angle B A C$ (resp., $\angle A B C, \angle A C B$ ). By introducing coordinates $B=(0,0), C=(a, 0)$, it is immediate that the coordinates of $M$ and $N$ are
+
+$$
+M=(a \cos \beta, a \sin \beta), \quad N=(a-a \cos \gamma, a \sin \gamma)
+$$
+
+respectively. Therefore,
+
+$$
+\begin{aligned}
+(M N / B C)^{2} & =\left[(a-a \cos \gamma-a \cos \beta)^{2}+(a \sin \gamma-a \sin \beta)^{2}\right] / a^{2} \\
+& =(1-\cos \gamma-\cos \beta)^{2}+(\sin \gamma-\sin \beta)^{2} \\
+& =3-2 \cos \gamma-2 \cos \beta+2(\cos \gamma \cos \beta-\sin \gamma \sin \beta) \\
+& =3-2 \cos \gamma-2 \cos \beta+2 \cos (\gamma+\beta) \\
+& =3-2 \cos \gamma-2 \cos \beta-2 \cos \alpha \\
+& =3-2(\cos \gamma+\cos \beta+\cos \alpha) .
+\end{aligned}
+$$
+
+Now we claim
+
+$$
+\cos \gamma+\cos \beta+\cos \alpha=\frac{r}{R}+1
+$$
+
+From
+
+$$
+\begin{aligned}
+& a=b \cos \gamma+c \cos \beta \\
+& b=c \cos \alpha+a \cos \gamma \\
+& c=a \cos \beta+b \cos \alpha
+\end{aligned}
+$$
+
+we get
+
+$$
+a(1+\cos \alpha)+b(1+\cos \beta)+c(1+\cos \gamma)=(a+b+c)(\cos \alpha+\cos \beta+\cos \gamma)
+$$
+
+Thus
+
+$$
+\begin{aligned}
+& \cos \alpha+\cos \beta+\cos \gamma \\
+& =\frac{1}{a+b+c}(a(1+\cos \alpha)+b(1+\cos \beta)+c(1+\cos \gamma)) \\
+& =\frac{1}{a+b+c}\left(a\left(1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)+b\left(1+\frac{a^{2}+c^{2}-b^{2}}{2 a c}\right)+c\left(1+\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\right) \\
+& =\frac{1}{a+b+c}\left(a+b+c+\frac{a^{2}\left(b^{2}+c^{2}-a^{2}\right)+b^{2}\left(a^{2}+c^{2}-b^{2}\right)+c^{2}\left(a^{2}+b^{2}-c^{2}\right)}{2 a b c}\right) \\
+& =1+\frac{2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}}{2 a b c(a+b+c)}
+\end{aligned}
+$$
+
+On the other hand, from $R=\frac{a}{2 \sin \alpha}$ it follows that
+
+$$
+\begin{aligned}
+R^{2} & =\frac{a^{2}}{4\left(1-\cos ^{2} \alpha\right)}=\frac{a^{2}}{4\left(1-\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}\right)} \\
+& =\frac{a^{2} b^{2} c^{2}}{2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}}
+\end{aligned}
+$$
+
+Also from $\frac{1}{2}(a+b+c) r=\frac{1}{2} b c \sin \alpha$, it follows that
+
+$$
+\begin{aligned}
+r^{2} & =\frac{b^{2} c^{2}\left(1-\cos ^{2} \alpha\right)}{(a+b+c)^{2}}=\frac{b^{2} c^{2}\left(1-\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}\right)}{(a+b+c)^{2}} \\
+& =\frac{2 a^{2} b^{2}+2 b^{2} c^{2}+2 c^{2} a^{2}-a^{4}-b^{4}-c^{4}}{4(a+b+c)^{2}}
+\end{aligned}
+$$
+
+Combining (19), (20) and (21), we get (16) as desired.
+Finally, by (15) and (16) we have
+
+$$
+\frac{M N}{B C}=\sqrt{1-\frac{2 r}{R}}
+$$
+
+Another proof of (16) from R.A. Johnson's "Advanced Euclidean Geometry" ${ }^{1}$ :
+Construct the perpendicular bisectors $O D, O E, O F$, where $D, E, F$ are the midpoints of $B C, C A, A B$, respectively. By Ptolemy's Theorem applied to the cyclic quadrilateral $O E A F$, we get
+
+$$
+\frac{a}{2} \cdot R=\frac{b}{2} \cdot O F+\frac{c}{2} \cdot O E
+$$
+
+Similarly
+
+$$
+\frac{b}{2} \cdot R=\frac{c}{2} \cdot O D+\frac{a}{2} \cdot O F, \quad \frac{c}{2} \cdot R=\frac{a}{2} \cdot O E+\frac{b}{2} \cdot O D .
+$$
+
+Adding, we get
+
+$$
+s R=O D \cdot \frac{b+c}{2}+O E \cdot \frac{c+a}{2}+O F \cdot \frac{a+b}{2}
+$$
+
+where $s$ is the semiperimeter. But also, the area of triangle $O B C$ is $O D \cdot \frac{a}{2}$, and adding similar formulas for the areas of triangles $O C A$ and $O A B$ gives
+
+$$
+r s=\triangle A B C=O D \cdot \frac{a}{2}+O E \cdot \frac{b}{2}+O F \cdot \frac{c}{2}
+$$
+
+Adding (23) and (24) gives $s(R+r)=s(O D+O E+O F)$, or
+
+$$
+O D+O E+O F=R+r .
+$$
+
+Since $O D=R \cos A$ etc., (16) follows.
+
+[^0]
+[^0]:    ${ }^{1}$ This proof was introduced to the coordinating country by Professor Bill Sands of Canada.
+

APMO/md/en-apmo2006_sol.md

@@ -0,0 +1,241 @@
+Problem 1. Let $n$ be a positive integer. Find the largest nonnegative real number $f(n)$ (depending on $n$ ) with the following property: whenever $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers such that $a_{1}+a_{2}+\cdots+a_{n}$ is an integer, there exists some $i$ such that $\left|a_{i}-\frac{1}{2}\right| \geq f(n)$.
+(Solution) The answer is
+
+$$
+f(n)=\left\{\begin{array}{cl}
+0 & \text { if } n \text { is even, } \\
+\frac{1}{2 n} & \text { if } n \text { is odd. }
+\end{array}\right.
+$$
+
+First, assume that $n$ is even. If $a_{i}=\frac{1}{2}$ for all $i$, then the sum $a_{1}+a_{2}+\cdots+a_{n}$ is an integer. Since $\left|a_{i}-\frac{1}{2}\right|=0$ for all $i$, we may conclude $f(n)=0$ for any even $n$.
+
+Now assume that $n$ is odd. Suppose that $\left|a_{i}-\frac{1}{2}\right|<\frac{1}{2 n}$ for all $1 \leq i \leq n$. Then, since $\sum_{i=1}^{n} a_{i}$ is an integer,
+
+$$
+\frac{1}{2} \leq\left|\sum_{i=1}^{n} a_{i}-\frac{n}{2}\right| \leq \sum_{i=1}^{n}\left|a_{i}-\frac{1}{2}\right|<\frac{1}{2 n} \cdot n=\frac{1}{2}
+$$
+
+a contradiction. Thus $\left|a_{i}-\frac{1}{2}\right| \geq \frac{1}{2 n}$ for some $i$, as required. On the other hand, putting $n=2 m+1$ and $a_{i}=\frac{m}{2 m+1}$ for all $i$ gives $\sum a_{i}=m$, while
+
+$$
+\left|a_{i}-\frac{1}{2}\right|=\frac{1}{2}-\frac{m}{2 m+1}=\frac{1}{2(2 m+1)}=\frac{1}{2 n}
+$$
+
+for all $i$. Therefore, $f(n)=\frac{1}{2 n}$ is the best possible for any odd $n$.
+
+Problem 2. Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden mean $\tau=\frac{1+\sqrt{5}}{2}$. Here, an integral power of $\tau$ is of the form $\tau^{i}$, where $i$ is an integer (not necessarily positive).
+(Solution) We will prove this statement by induction using the equality
+
+$$
+\tau^{2}=\tau+1
+$$
+
+If $n=1$, then $1=\tau^{0}$. Suppose that $n-1$ can be written as a finite sum of integral powers of $\tau$, say
+
+$$
+n-1=\sum_{i=-k}^{k} a_{i} \tau^{i}
+$$
+
+where $a_{i} \in\{0,1\}$ and $n \geq 2$. We will write (1) as
+
+$$
+n-1=a_{k} \cdots a_{1} a_{0} \cdot a_{-1} a_{-2} \cdots a_{-k}
+$$
+
+For example,
+
+$$
+1=1.0=0.11=0.1011=0.101011
+$$
+
+Firstly, we will prove that we may assume that in (2) we have $a_{i} a_{i+1}=0$ for all $i$ with $-k \leq i \leq k-1$. Indeed, if we have several occurrences of 11 , then we take the leftmost such occurrence. Since we may assume that it is preceded by a 0 , we can replace 011 with 100 using the identity $\tau^{i+1}+\tau^{i}=\tau^{i+2}$. By doing so repeatedly, if necessary, we will eliminate all occurrences of two 1's standing together. Now we have the representation
+
+$$
+n-1=\sum_{i=-K}^{K} b_{i} \tau^{i}
+$$
+
+where $b_{i} \in\{0,1\}$ and $b_{i} b_{i+1}=0$.
+If $b_{0}=0$ in (3), then we just add $1=\tau^{0}$ to both sides of (3) and we are done.
+Suppose now that there is 1 in the unit position of (3), that is $b_{0}=1$. If there are two 0 's to the right of it, i.e.
+
+$$
+n-1=\cdots 1.00 \cdots
+$$
+
+then we can replace 1.00 with 0.11 because $1=\tau^{-1}+\tau^{-2}$, and we are done because we obtain 0 in the unit position. Thus we may assume that
+
+$$
+n-1=\cdots 1.010 \cdots
+$$
+
+Again, if we have $n-1=\cdots 1.0100 \cdots$, we may rewrite it as
+
+$$
+n-1=\cdots 1.0100 \cdots=\cdots 1.0011 \cdots=\cdots 0.1111 \cdots
+$$
+
+and obtain 0 in the unit position. Therefore, we may assume that
+
+$$
+n-1=\cdots 1.01010 \cdots
+$$
+
+Since the number of 1's is finite, eventually we will obtain an occurrence of 100 at the end, i.e.
+
+$$
+n-1=\cdots 1.01010 \cdots 100
+$$
+
+Then we can shift all 1's to the right to obtain 0 in the unit position, i.e.
+
+$$
+n-1=\cdots 0.11 \cdots 11
+$$
+
+and we are done.
+
+Problem 3. Let $p \geq 5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p \times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^{5}$. Here, we assume that all the checkers are identical.
+(Solution) Note that $r=\binom{p^{2}}{p}-p$. Hence, it suffices to show that
+
+$$
+\left(p^{2}-1\right)\left(p^{2}-2\right) \cdots\left(p^{2}-(p-1)\right)-(p-1)!\equiv 0 \quad\left(\bmod p^{4}\right)
+$$
+
+Now, let
+
+$$
+f(x):=(x-1)(x-2) \cdots(x-(p-1))=x^{p-1}+s_{p-2} x^{p-2}+\cdots+s_{1} x+s_{0} .
+$$
+
+Then the congruence equation (1) is same as $f\left(p^{2}\right)-s_{0} \equiv 0\left(\bmod p^{4}\right)$. Therefore, it suffices to show that $s_{1} p^{2} \equiv 0\left(\bmod p^{4}\right)$ or $s_{1} \equiv 0\left(\bmod p^{2}\right)$.
+
+Since $a^{p-1} \equiv 1(\bmod p)$ for all $1 \leq a \leq p-1$, we can factor
+
+$$
+x^{p-1}-1 \equiv(x-1)(x-2) \cdots(x-(p-1)) \quad(\bmod p)
+$$
+
+Comparing the coefficients of the left hand side of (3) with those of the right hand side of (2), we obtain $p \mid s_{i}$ for all $1 \leq i \leq p-2$ and $s_{0} \equiv-1(\bmod p)$. On the other hand, plugging $p$ for $x$ in (2), we get
+
+$$
+f(p)=(p-1)!=s_{0}=p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{1} p+s_{0}
+$$
+
+which implies
+
+$$
+p^{p-1}+s_{p-2} p^{p-2}+\cdots+s_{2} p^{2}=-s_{1} p
+$$
+
+Since $p \geq 5, p \mid s_{2}$ and hence $s_{1} \equiv 0\left(\bmod p^{2}\right)$ as desired.
+
+Problem 4. Let $A, B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment $A B$. Let $O_{1}$ be the circle tangent to the line $A B$ at $P$ and tangent to the circle $O$. Let $\ell$ be the tangent line, different from the line $A B$, to $O_{1}$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $\ell$ and $O$. Let $Q$ be the midpoint of the line segment $B C$ and $O_{2}$ be the circle tangent to the line $B C$ at $Q$ and tangent to the line segment $A C$. Prove that the circle $O_{2}$ is tangent to the circle $O$.
+(Solution) Let $S$ be the tangent point of the circles $O$ and $O_{1}$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $S P$. Let $X$ be the tangent point of $\ell$ to $O_{1}$ and let $M$ be the midpoint of the line segment $X P$. Since $\angle T B P=\angle A S P$, the triangle $T B P$ is similar to the triangle $A S P$. Therefore,
+
+$$
+\frac{P T}{P B}=\frac{P A}{P S}
+$$
+
+Since the line $\ell$ is tangent to the circle $O_{1}$ at $X$, we have
+
+$$
+\angle S P X=90^{\circ}-\angle X S P=90^{\circ}-\angle A P M=\angle P A M
+$$
+
+which implies that the triangle $P A M$ is similar to the triangle $S P X$. Consequently,
+
+$$
+\frac{X S}{X P}=\frac{M P}{M A}=\frac{X P}{2 M A} \quad \text { and } \quad \frac{X P}{P S}=\frac{M A}{A P}
+$$
+
+From this and the above observation follows
+
+$$
+\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{X P}{2 M A} \cdot \frac{P A}{P S}=\frac{X P}{2 M A} \cdot \frac{M A}{X P}=\frac{1}{2} .
+$$
+
+Let $A^{\prime}$ be the intersection point of the circle $O$ and the perpendicular bisector of the chord $B C$ such that $A, A^{\prime}$ are on the same side of the line $B C$, and $N$ be the intersection point of the lines $A^{\prime} Q$ and $C T$. Since
+
+$$
+\angle N C Q=\angle T C B=\angle T C A=\angle T B A=\angle T B P
+$$
+
+and
+
+$$
+\angle C A^{\prime} Q=\frac{\angle C A B}{2}=\frac{\angle X A P}{2}=\angle P A M=\angle S P X,
+$$
+
+the triangle $N C Q$ is similar to the triangle $T B P$ and the triangle $C A^{\prime} Q$ is similar to the triangle $S P X$. Therefore
+
+$$
+\frac{Q N}{Q C}=\frac{P T}{P B} \quad \text { and } \quad \frac{Q C}{Q A^{\prime}}=\frac{X S}{X P} .
+$$
+
+and hence $Q A^{\prime}=2 Q N$ by (1). This implies that $N$ is the midpoint of the line segment $Q A^{\prime}$. Let the circle $O_{2}$ touch the line segment $A C$ at $Y$. Since
+
+$$
+\angle A C N=\angle A C T=\angle B C T=\angle Q C N
+$$
+
+and $|C Y|=|C Q|$, the triangles $Y C N$ and $Q C N$ are congruent and hence $N Y \perp A C$ and $N Y=N Q=N A^{\prime}$. Therefore, $N$ is the center of the circle $O_{2}$, which completes the proof.
+
+Remark: Analytic solutions are possible: For example, one can prove for a triangle $A B C$ inscribed in a circle $O$ that $A B=k(2+2 t), A C=k(1+2 t), B C=k(1+4 t)$ for some positive numbers $k, t$ if and only if there exists a circle $O_{1}$ such that $O_{1}$ is tangent to the side $A B$ at its midpoint, the side $A C$ and the circle $O$.
+
+One obtains $A B=k^{\prime}\left(1+4 t^{\prime}\right), A C=k^{\prime}\left(1+2 t^{\prime}\right), B C=k^{\prime}\left(2+2 t^{\prime}\right)$ by substituting $t=1 / 4 t^{\prime}$ and $k=2 k^{\prime} t^{\prime}$. So, there exists a circle $O_{2}$ such that $O_{2}$ is tangent to the side $B C$ at its midpoint, the side $A C$ and the circle $O$.
+
+In the above, $t=\tan ^{2} \alpha$ and $k=\frac{4 R \tan \alpha}{\left(1+\tan ^{2} \alpha\right)\left(1+4 \tan ^{2} \alpha\right)}$, where $R$ is the radius of $O$ and $\angle A=2 \alpha$. Furthermore, $t^{\prime}=\tan ^{2} \gamma$ and $k^{\prime}=\frac{4 R \tan \gamma}{\left(1+\tan ^{2} \gamma\right)\left(1+4 \tan ^{2} \gamma\right)}$, where $\angle C=2 \gamma$. Observe that $\sqrt{t t^{\prime}}=\tan \alpha \cdot \tan \gamma=\frac{X S}{X P} \cdot \frac{P T}{P B}=\frac{1}{2}$, which implies $t t^{\prime}=\frac{1}{4}$. It is now routine easy to check that $k=2 k^{\prime} t^{\prime}$.
+
+Problem 5. In a circus, there are $n$ clowns who dress and paint themselves up using a selection of 12 distinct colours. Each clown is required to use at least five different colours. One day, the ringmaster of the circus orders that no two clowns have exactly the same set
+of colours and no more than 20 clowns may use any one particular colour. Find the largest number $n$ of clowns so as to make the ringmaster's order possible.
+(Solution) Let $C$ be the set of $n$ clowns. Label the colours $1,2,3, \ldots, 12$. For each $i=1,2, \ldots, 12$, let $E_{i}$ denote the set of clowns who use colour $i$. For each subset $S$ of $\{1,2, \ldots, 12\}$, let $E_{S}$ be the set of clowns who use exactly those colours in $S$. Since $S \neq S^{\prime}$ implies $E_{S} \cap E_{S^{\prime}}=\emptyset$, we have
+
+$$
+\sum_{S}\left|E_{S}\right|=|C|=n
+$$
+
+where $S$ runs over all subsets of $\{1,2, \ldots, 12\}$. Now for each $i$,
+
+$$
+E_{S} \subseteq E_{i} \quad \text { if and only if } \quad i \in S
+$$
+
+and hence
+
+$$
+\left|E_{i}\right|=\sum_{i \in S}\left|E_{S}\right|
+$$
+
+By assumption, we know that $\left|E_{i}\right| \leq 20$ and that if $E_{S} \neq \emptyset$, then $|S| \geq 5$. From this we obtain
+
+$$
+20 \times 12 \geq \sum_{i=1}^{12}\left|E_{i}\right|=\sum_{i=1}^{12}\left(\sum_{i \in S}\left|E_{S}\right|\right) \geq 5 \sum_{S}\left|E_{S}\right|=5 n
+$$
+
+Therefore $n \leq 48$.
+Now, define a sequence $\left\{c_{i}\right\}_{i=1}^{52}$ of colours in the following way:
+$1234|5678| 9101112 \mid$
+$4123|8567| 1291011 \mid$
+$3412|7856| 1112910 \mid$
+$2341|6785| 1011129 \mid 1234$
+
+The first row lists $c_{1}, \ldots, c_{12}$ in order, the second row lists $c_{13}, \ldots, c_{24}$ in order, the third row lists $c_{25}, \ldots, c_{36}$ in order, and finally the last row lists $c_{37}, \ldots, c_{52}$ in order. For each $j, 1 \leq j \leq 48$, assign colours $c_{j}, c_{j+1}, c_{j+2}, c_{j+3}, c_{j+4}$ to the $j$-th clown. It is easy to check that this assignment satisfies all conditions given above. So, 48 is the largest for $n$.
+
+Remark: The fact that $n \leq 48$ can be obtained in a much simpler observation that
+
+$$
+5 n \leq 12 \times 20=240
+$$
+
+There are many other ways of constructing 48 distinct sets consisting of 5 colours. For example, consider the sets
+
+$$
+\begin{array}{cccc}
+\{1,2,3,4,5,6\}, & \{3,4,5,6,7,8\}, & \{5,6,7,8,9,10\}, & \{7,8,9,10,11,12\}, \\
+\{9,10,11,12,1,2\}, & \{11,12,1,2,3,4\}, & \{1,2,5,6,9,10\}, & \{3,4,7,8,11,12\} .
+\end{array}
+$$
+
+Each of the above 8 sets has 6 distinct subsets consisting of exactly 5 colours. It is easy to check that the 48 subsets obtained in this manner are all distinct.
+

APMO/md/en-apmo2007_sol.md

@@ -0,0 +1,325 @@
+# XIX Asian Pacific Mathematics Olympiad 
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_16a4255df619f73cfb01g-01.jpg?height=181&width=420&top_left_y=543&top_left_x=835)
+
+Problem 1. 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
+
+$$
+S=\left\{2^{a_{i}} 3^{b_{i}} \mid a_{i}, b_{i} \in \mathbb{Z}, a_{i}, b_{i} \geq 0,1 \leq i \leq 9\right\}
+$$
+
+It suffices to show that there are $1 \leq i_{1}, i_{2}, i_{3} \leq 9$ such that
+
+$$
+a_{i_{1}}+a_{i_{2}}+a_{i_{3}} \equiv b_{i_{1}}+b_{i_{2}}+b_{i_{3}} \equiv 0 \quad(\bmod 3) .
+$$
+
+For $n=2^{a} 3^{b} \in S$, let's call $(a(\bmod 3), b(\bmod 3))$ the type of $n$. Then there are 9 possible types:
+
+$$
+(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
+$$
+
+Let $N(i, j)$ be the number of integers in $S$ of type $(i, j)$. We obtain 3 distinct integers whose product is a perfect cube when
+(1) $N(i, j) \geq 3$ for some $i, j$, or
+(2) $N(i, 0) N(i, 1) N(i, 2) \neq 0$ for some $i=0,1,2$, or
+(3) $N(0, j) N(1, j) N(2, j) \neq 0$ for some $j=0,1,2$, or
+(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 $\left\{i_{1}, i_{2}, i_{3}\right\}=\left\{j_{1}, j_{2}, j_{3}\right\}=\{0,1,2\}$.
+
+Assume 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).)
+
+Second solution. Up to $(\dagger)$, we do the same as above and get 9 possible types:
+
+$$
+(a(\bmod 3), b(\bmod 3))=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)
+$$
+
+for $n=2^{a} 3^{b} \in S$.
+Note 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 $\{i, j, k\}=\{0,1,2\}$.
+
+Let's define
+$T$ : the set of types of the integers in $S$;
+$N(i)$ : the number of integers in $S$ of the type $(i, \cdot)$;
+$M(i)$ : the number of integers $j \in\{0,1,2\}$ such that $(i, j) \in T$.
+If $N(i) \geq 5$ for some $i$, the result follows from (i). Otherwise, for some permutation $(i, j, k)$ of $(0,1,2)$,
+
+$$
+N(i) \geq 3, \quad N(j) \geq 3, \quad N(k) \geq 1
+$$
+
+If $M(i)$ or $M(j)$ is 1 or 3 , the result follows from (ii). Otherwise $M(i)=M(j)=2$. Then either
+
+$$
+(i, x),(i, y),(j, x),(j, y) \in T \quad \text { or } \quad(i, x),(i, y),(j, x),(j, z) \in T
+$$
+
+for 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 2. 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
+
+$$
+2 \angle A H I=3 \angle A B C .
+$$
+
+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
+$\angle B I C=180^{\circ}-(\angle I B C+\angle I C B)=180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=90^{\circ}+\frac{1}{2} \angle B A C=120^{\circ}$
+and 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 $I E=E N$.
+
+Now, 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 $I H=N K$.
+
+Hence
+
+$$
+\angle A H I=180^{\circ}-\angle I H K=180^{\circ}-\angle A K N=\angle A B N .
+$$
+
+Since $I E=E N$ and $B E \perp I N$, the triangles $I B E$ and $N B E$ are congruent. Therefore
+
+$$
+\angle N B E=\angle I B E=\angle I B C=\angle I B A=\frac{1}{2} \angle A B C
+$$
+
+and thus
+
+$$
+\angle A H I=\angle A B N=\frac{3}{2} \angle A B C .
+$$
+
+Second 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,
+
+$$
+\angle I B H=\angle A B P-\angle A B I=30^{\circ}-\frac{1}{2} \angle A B C
+$$
+
+and
+
+$$
+\angle I C H=\angle A C I-\angle A C H=\frac{1}{2} \angle A C B-30^{\circ}=30^{\circ}-\frac{1}{2} \angle A B C,
+$$
+
+because $\angle A B H=\angle A C H=30^{\circ}$ 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
+
+$$
+\angle B H I=\angle B C I=\frac{1}{2} \angle A C B .
+$$
+
+On the other hand,
+
+$$
+\angle B H R=90^{\circ}-\angle H B R=90^{\circ}-(\angle A B C-\angle A B H)=120^{\circ}-\angle A B C
+$$
+
+Therefore,
+
+$$
+\begin{aligned}
+\angle A H I & =180^{\circ}-\angle B H I-\angle B H R=60^{\circ}-\frac{1}{2} \angle A C B+\angle A B C \\
+& =60^{\circ}-\frac{1}{2}\left(120^{\circ}-\angle A B C\right)+\angle A B C=\frac{3}{2} \angle A B C .
+\end{aligned}
+$$
+
+Problem 3. 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 $(n-1)(n-2) / 2$.
+Let'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 $\left|S_{\mathcal{C}}\right|$.
+
+We'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 $\mathcal{C}$.
+
+Let $C_{1}$ 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).
+
+For any $n$-configuration $\mathcal{C}, S_{\mathcal{C}}$ must satisfy the following properties:
+(1) $(i, i) \notin S_{\mathcal{C}}$,
+(2) $(i+1, i) \notin S_{\mathcal{C}},(1, n) \notin S_{\mathcal{C}}$,
+(3) if $(i, j),(j, k) \in S_{\mathcal{C}}$, then $(i, k) \in S_{\mathcal{C}}$,
+(4) if $(i, j) \in S_{\mathcal{C}}$, then $(j, i) \notin S_{\mathcal{C}}$.
+
+Now 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
+
+$$
+\binom{n}{2}-n=\frac{n(n-3)}{2}<\frac{(n-1)(n-2)}{2}
+$$
+
+elements. 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 $n-1$.
+
+We now claim that $\left|G-G^{\prime}\right| \leq n-2$ :
+Suppose 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.
+
+Since $\left|G-G^{\prime}\right| \leq n-2$, we obtain
+
+$$
+\left|G^{\prime}\right| \geq \frac{(n-1)(n-2)}{2}-(n-2)=\frac{(n-2)(n-3)}{2}
+$$
+
+This, however, contradicts the minimality of $n$, and hence proves (ii).
+
+Problem 4. Let $x, y$ and $z$ be positive real numbers such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove that
+
+$$
+\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq 1
+$$
+
+Solution. We first note that
+
+$$
+\begin{aligned}
+\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}} & =\frac{x^{2}-x(y+z)+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{x(y+z)}{\sqrt{2 x^{2}(y+z)}} \\
+& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\sqrt{\frac{y+z}{2}} \\
+& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{\sqrt{y}+\sqrt{z}}{2} .
+\end{aligned}
+$$
+
+Similarly, we have
+
+$$
+\begin{aligned}
+& \frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}} \geq \frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{\sqrt{z}+\sqrt{x}}{2}, \\
+& \frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geq \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{\sqrt{x}+\sqrt{y}}{2} .
+\end{aligned}
+$$
+
+We now add (1) (3) to get
+
+$$
+\begin{aligned}
+& \frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \\
+& \geq \frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\sqrt{x}+\sqrt{y}+\sqrt{z} \\
+& =\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+1 .
+\end{aligned}
+$$
+
+Thus, it suffices to show that
+
+$$
+\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}+\frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}} \geq 0 .
+$$
+
+Now, assume without loss of generality, that $x \geq y \geq z$. Then we have
+
+$$
+\frac{(x-y)(x-z)}{\sqrt{2 x^{2}(y+z)}} \geq 0
+$$
+
+and
+
+$$
+\begin{aligned}
+& \frac{(z-x)(z-y)}{\sqrt{2 z^{2}(x+y)}}+\frac{(y-z)(y-x)}{\sqrt{2 y^{2}(z+x)}}=\frac{(y-z)(x-z)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}} \\
+& \geq \frac{(y-z)(x-y)}{\sqrt{2 z^{2}(x+y)}}-\frac{(y-z)(x-y)}{\sqrt{2 y^{2}(z+x)}}=(y-z)(x-y)\left(\frac{1}{\sqrt{2 z^{2}(x+y)}}-\frac{1}{\sqrt{2 y^{2}(z+x)}}\right)
+\end{aligned}
+$$
+
+The last quantity is non-negative due to the fact that
+
+$$
+y^{2}(z+x)=y^{2} z+y^{2} x \geq y z^{2}+z^{2} x=z^{2}(x+y)
+$$
+
+This completes the proof.
+
+Second solution. By Cauchy-Schwarz inequality,
+
+$$
+\begin{aligned}
+& \left(\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right) \\
+& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1
+\end{aligned}
+$$
+
+and
+
+$$
+\begin{aligned}
+& \left(\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{z x}{\sqrt{2 y^{2}(z+x)}}+\frac{x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
+& \quad \times(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geq\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2}
+\end{aligned}
+$$
+
+We now combine (5) and (6) to find
+
+$$
+\begin{aligned}
+& \left(\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}}\right) \\
+& \quad \times(\sqrt{2(x+y)}+\sqrt{2(y+z)}+\sqrt{2(z+x)}) \\
+& \geq 1+\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right)^{2} \geq 2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) .
+\end{aligned}
+$$
+
+Thus, it suffices to show that
+
+$$
+2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}
+$$
+
+Consider the following inequality using AM-GM inequality
+
+$$
+\left[\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)\right]^{2} \geq 4 \sqrt{\frac{y z}{x}}\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right)=2(y+z)
+$$
+
+or equivalently
+
+$$
+\sqrt{\frac{y z}{x}}+\left(\frac{1}{2} \sqrt{\frac{z x}{y}}+\frac{1}{2} \sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)} .
+$$
+
+Similarly, we have
+
+$$
+\begin{aligned}
+& \sqrt{\frac{z x}{y}}+\left(\frac{1}{2} \sqrt{\frac{x y}{z}}+\frac{1}{2} \sqrt{\frac{y z}{x}}\right) \geq \sqrt{2(z+x)} \\
+& \sqrt{\frac{x y}{z}}+\left(\frac{1}{2} \sqrt{\frac{y z}{x}}+\frac{1}{2} \sqrt{\frac{z x}{y}}\right) \geq \sqrt{2(x+y)}
+\end{aligned}
+$$
+
+Adding the last three inequalities, we get
+
+$$
+2\left(\sqrt{\frac{y z}{x}}+\sqrt{\frac{z x}{y}}+\sqrt{\frac{x y}{z}}\right) \geq \sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)} .
+$$
+
+This completes the proof.
+
+Problem 5. 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:
+
+| 1 | 1 | 0 | 1 | 1 |
+| :--- | :--- | :--- | :--- | :--- |
+| 0 | 0 | 0 | 0 | 0 |
+| 1 | 1 | 0 | 1 | 1 |
+| 0 | 0 | 0 | 0 | 0 |
+| 1 | 1 | 0 | 1 | 1 |
+
+For 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.
+
+The $90^{\circ}$ 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.
+
+| 1 | 0 | 1 | 0 | 1 |
+| :--- | :--- | :--- | :--- | :--- |
+| 1 | 0 | 1 | 0 | 1 |
+| 0 | 0 | 0 | 0 | 0 |
+| 1 | 0 | 1 | 0 | 1 |
+| 1 | 0 | 1 | 0 | 1 |
+
+Since 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:
+
+|  |  |  |  |  |
+| :--- | :--- | :--- | :--- | :--- |
+|  | $*_{2}$ |  | $*_{1}$ |  |
+|  |  | $*_{0}$ |  |  |
+|  | $*_{3}$ |  | $*_{4}$ |  |
+|  |  |  |  |  |
+
+Now we demonstrate that all five positions are possible:
+Toggling 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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_16a4255df619f73cfb01g-11.jpg?height=287&width=326&top_left_y=705&top_left_x=708)
+
+|  | t |  | t |  |
+| :---: | :---: | :---: | :---: | :---: |
+| t | t |  | t | t |
+|  | t |  |  |  |
+|  |  | t | t | t |
+|  |  |  | t |  |
+

APMO/md/en-apmo2008_sol.md

@@ -0,0 +1,228 @@
+# XX Asian Pacific Mathematics Olympiad APMO: 
+
+March, 2008
+
+Problem 1. Let $A B C$ be a triangle with $\angle A<60^{\circ}$. Let $X$ and $Y$ be the points on the sides $A B$ and $A C$, respectively, such that $C A+A X=C B+B X$ and $B A+A Y=B C+C Y$. Let $P$ be the point in the plane such that the lines $P X$ and $P Y$ are perpendicular to $A B$ and $A C$, respectively. Prove that $\angle B P C<120^{\circ}$.
+(Solution) Let $I$ be the incenter of $\triangle A B C$, and let the feet of the perpendiculars from $I$ to $A B$ and to $A C$ be $D$ and $E$, respectively. (Without loss of generality, we may assume that $A C$ is the longest side. Then $X$ lies on the line segment $A D$. Although $P$ may or may not lie inside $\triangle A B C$, the proof below works for both cases. Note that $P$ is on the line perpendicular to $A B$ passing through $X$.) Let $O$ be the midpoint of $I P$, and let the feet of the perpendiculars from $O$ to $A B$ and to $A C$ be $M$ and $N$, respectively. Then $M$ and $N$ are the midpoints of $D X$ and $E Y$, respectively.
+![](https://cdn.mathpix.com/cropped/2024_11_22_654364ab29d2349bbfb4g-1.jpg?height=943&width=1263&top_left_y=1493&top_left_x=408)
+
+The conditions on the points $X$ and $Y$ yield the equations
+
+$$
+A X=\frac{A B+B C-C A}{2} \quad \text { and } \quad A Y=\frac{B C+C A-A B}{2} .
+$$
+
+From $A D=A E=\frac{C A+A B-B C}{2}$, we obtain
+
+$$
+B D=A B-A D=A B-\frac{C A+A B-B C}{2}=\frac{A B+B C-C A}{2}=A X .
+$$
+
+Since $M$ is the midpoint of $D X$, it follows that $M$ is the midpoint of $A B$. Similarly, $N$ is the midpoint of $A C$. Therefore, the perpendicular bisectors of $A B$ and $A C$ meet at $O$, that is, $O$ is the circumcenter of $\triangle A B C$. Since $\angle B A C<60^{\circ}, O$ lies on the same side of $B C$ as the point $A$ and
+
+$$
+\angle B O C=2 \angle B A C
+$$
+
+We can compute $\angle B I C$ as follows:
+
+$$
+\begin{aligned}
+\angle B I C & =180^{\circ}-\angle I B C-\angle I C B=180^{\circ}-\frac{1}{2} \angle A B C-\frac{1}{2} \angle A C B \\
+& =180^{\circ}-\frac{1}{2}(\angle A B C+\angle A C B)=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}+\frac{1}{2} \angle B A C
+\end{aligned}
+$$
+
+It follows from $\angle B A C<60^{\circ}$ that
+
+$$
+2 \angle B A C<90^{\circ}+\frac{1}{2} \angle B A C, \quad \text { i.e., } \quad \angle B O C<\angle B I C \text {. }
+$$
+
+From this it follows that $I$ lies inside the circumcircle of the isosceles triangle $B O C$ because $O$ and $I$ lie on the same side of $B C$. However, as $O$ is the midpoint of $I P, P$ must lie outside the circumcircle of triangle $B O C$ and on the same side of $B C$ as $O$. Therefore
+
+$$
+\angle B P C<\angle B O C=2 \angle B A C<120^{\circ} .
+$$
+
+Remark. If one assumes that $\angle A$ is smaller than the other two, then it is clear that the line $P X$ (or the line perpendicular to $A B$ at $X$ if $P=X$ ) runs through the excenter $I_{C}$ of the excircle tangent to the side $A B$. Since $2 \angle A C I_{C}=\angle A C B$ and $B C<A C$, we have $2 \angle P C B>\angle C$. Similarly, $2 \angle P B C>\angle B$. Therefore,
+
+$$
+\angle B P C=180^{\circ}-(\angle P B C+\angle P C B)<180^{\circ}-\left(\frac{\angle B+\angle C}{2}\right)=90+\frac{\angle A}{2}<120^{\circ}
+$$
+
+In this way, a special case of the problem can be easily proved.
+
+Problem 2. Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is 46 , there is a set of 10 students in which no group is properly contained.
+(Solution) We let $C$ be the set of all 46 students in the class and let
+
+$$
+s:=\max \{|S|: S \subseteq C \text { such that } S \text { contains no group properly }\}
+$$
+
+Then it suffices to prove that $s \geq 10$. (If $|S|=s>10$, we may choose a subset of $S$ consisting of 10 students.)
+
+Suppose that $s \leq 9$ and let $S$ be a set of size $s$ in which no group is properly contained. Take any student, say $v$, from outside $S$. Because of the maximality of $s$, there should be a group containing the student $v$ and two other students in $S$. The number of ways to choose two students from $S$ is
+
+$$
+\binom{s}{2} \leq\binom{ 9}{2}=36
+$$
+
+On the other hand, there are at least $37=46-9$ students outside of $S$. Thus, among those 37 students outside, there is at least one student, say $u$, who does not belong to any group containing two students in $S$ and one outside. This is because no two distinct groups have two members in common. But then, $S$ can be enlarged by including $u$, which is a contradiction.
+
+Remark. One may choose a subset $S$ of $C$ that contains no group properly. Then, assuming $|S|<10$, prove that there is a student outside $S$, say $u$, who does not belong to any group containing two students in $S$. After enlarging $S$ by including $u$, prove that the enlarged $S$ still contains no group properly.
+
+Problem 3. Let $\Gamma$ be the circumcircle of a triangle $A B C$. A circle passing through points $A$ and $C$ meets the sides $B C$ and $B A$ at $D$ and $E$, respectively. The lines $A D$ and $C E$ meet $\Gamma$ again at $G$ and $H$, respectively. The tangent lines of $\Gamma$ at $A$ and $C$ meet the line $D E$ at $L$ and $M$, respectively. Prove that the lines $L H$ and $M G$ meet at $\Gamma$.
+(Solution) Let $M G$ meet $\Gamma$ at $P$. Since $\angle M C D=\angle C A E$ and $\angle M D C=\angle C A E$, we have $M C=M D$. Thus
+
+$$
+M D^{2}=M C^{2}=M G \cdot M P
+$$
+
+and hence $M D$ is tangent to the circumcircle of $\triangle D G P$. Therefore $\angle D G P=\angle E D P$.
+Let $\Gamma^{\prime}$ be the circumcircle of $\triangle B D E$. If $B=P$, then, since $\angle B G D=\angle B D E$, the tangent lines of $\Gamma^{\prime}$ and $\Gamma$ at $B$ should coincide, that is $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside. Let $B \neq P$. If $P$ lies in the same side of the line $B C$ as $G$, then we have
+
+$$
+\angle E D P+\angle A B P=180^{\circ}
+$$
+
+because $\angle D G P+\angle A B P=180^{\circ}$. That is, the quadrilateral $B P D E$ is cyclic, and hence $P$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_654364ab29d2349bbfb4g-4.jpg?height=1212&width=1014&top_left_y=1287&top_left_x=538)
+
+Otherwise,
+
+$$
+\angle E D P=\angle D G P=\angle A G P=\angle A B P=\angle E B P .
+$$
+
+Therefore the quadrilateral $P B D E$ is cyclic, and hence $P$ again is on the intersection of $\Gamma^{\prime}$ with $\Gamma$.
+
+Similarly, if $L H$ meets $\Gamma$ at $Q$, we either have $Q=B$, in which case $\Gamma^{\prime}$ is tangent to $\Gamma$ from inside, or $Q \neq B$. In the latter case, $Q$ is on the intersection of $\Gamma^{\prime}$ with $\Gamma$. In either case, we have $P=Q$.
+
+Problem 4. Consider the function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$, where $\mathbb{N}_{0}$ is the set of all non-negative integers, defined by the following conditions:
+(i) $f(0)=0$,
+(ii) $f(2 n)=2 f(n)$ and
+(iii) $f(2 n+1)=n+2 f(n)$ for all $n \geq 0$.
+(a) Determine the three sets $L:=\{n \mid f(n)<f(n+1)\}, E:=\{n \mid f(n)=f(n+1)\}$, and $G:=\{n \mid f(n)>f(n+1)\}$.
+(b) For each $k \geq 0$, find a formula for $a_{k}:=\max \left\{f(n): 0 \leq n \leq 2^{k}\right\}$ in terms of $k$.
+(Solution) (a) Let
+
+$$
+L_{1}:=\{2 k: k>0\}, \quad E_{1}:=\{0\} \cup\{4 k+1: k \geq 0\}, \quad \text { and } G_{1}:=\{4 k+3: k \geq 0\} .
+$$
+
+We will show that $L_{1}=L, E_{1}=E$, and $G_{1}=G$. It suffices to verify that $L_{1} \subseteq E, E_{1} \subseteq E$, and $G_{1} \subseteq G$ because $L_{1}, E_{1}$, and $G_{1}$ are mutually disjoint and $L_{1} \cup E_{1} \cup G_{1}=\mathbb{N}_{0}$.
+
+Firstly, if $k>0$, then $f(2 k)-f(2 k+1)=-k<0$ and therefore $L_{1} \subseteq L$.
+Secondly, $f(0)=0$ and
+
+$$
+\begin{aligned}
+& f(4 k+1)=2 k+2 f(2 k)=2 k+4 f(k) \\
+& f(4 k+2)=2 f(2 k+1)=2(k+2 f(k))=2 k+4 f(k)
+\end{aligned}
+$$
+
+for all $k \geq 0$. Thus, $E_{1} \subseteq E$.
+Lastly, in order to prove $G_{1} \subset G$, we claim that $f(n+1)-f(n) \leq n$ for all $n$. (In fact, one can prove a stronger inequality : $f(n+1)-f(n) \leq n / 2$.) This is clearly true for even $n$ from the definition since for $n=2 t$,
+
+$$
+f(2 t+1)-f(2 t)=t \leq n
+$$
+
+If $n=2 t+1$ is odd, then (assuming inductively that the result holds for all nonnegative $m<n$ ), we have
+
+$$
+\begin{aligned}
+f(n+1)-f(n) & =f(2 t+2)-f(2 t+1)=2 f(t+1)-t-2 f(t) \\
+& =2(f(t+1)-f(t))-t \leq 2 t-t=t<n .
+\end{aligned}
+$$
+
+For all $k \geq 0$,
+
+$$
+\begin{aligned}
+& f(4 k+4)-f(4 k+3)=f(2(2 k+2))-f(2(2 k+1)+1) \\
+& =4 f(k+1)-(2 k+1+2 f(2 k+1))=4 f(k+1)-(2 k+1+2 k+4 f(k)) \\
+& =4(f(k+1)-f(k))-(4 k+1) \leq 4 k-(4 k+1)<0 .
+\end{aligned}
+$$
+
+This proves $G_{1} \subseteq G$.
+(b) Note that $a_{0}=a_{1}=f(1)=0$. Let $k \geq 2$ and let $N_{k}=\left\{0,1,2, \ldots, 2^{k}\right\}$. First we claim that the maximum $a_{k}$ occurs at the largest number in $G \cap N_{k}$, that is, $a_{k}=f\left(2^{k}-1\right)$. We use mathematical induction on $k$ to prove the claim. Note that $a_{2}=f(3)=f\left(2^{2}-1\right)$.
+
+Now let $k \geq 3$. For every even number $2 t$ with $2^{k-1}+1<2 t \leq 2^{k}$,
+
+$$
+f(2 t)=2 f(t) \leq 2 a_{k-1}=2 f\left(2^{k-1}-1\right)
+$$
+
+by induction hypothesis. For every odd number $2 t+1$ with $2^{k-1}+1 \leq 2 t+1<2^{k}$,
+
+$$
+\begin{aligned}
+f(2 t+1) & =t+2 f(t) \leq 2^{k-1}-1+2 f(t) \\
+& \leq 2^{k-1}-1+2 a_{k-1}=2^{k-1}-1+2 f\left(2^{k-1}-1\right)
+\end{aligned}
+$$
+
+again by induction hypothesis. Combining ( $\dagger$ ), ( $\ddagger$ ) and
+
+$$
+f\left(2^{k}-1\right)=f\left(2\left(2^{k-1}-1\right)+1\right)=2^{k-1}-1+2 f\left(2^{k-1}-1\right)
+$$
+
+we may conclude that $a_{k}=f\left(2^{k}-1\right)$ as desired.
+Furthermore, we obtain
+
+$$
+a_{k}=2 a_{k-1}+2^{k-1}-1
+$$
+
+for all $k \geq 3$. Note that this recursive formula for $a_{k}$ also holds for $k \geq 0,1$ and 2 . Unwinding this recursive formula, we finally get
+
+$$
+\begin{aligned}
+a_{k} & =2 a_{k-1}+2^{k-1}-1=2\left(2 a_{k-2}+2^{k-2}-1\right)+2^{k-1}-1 \\
+& =2^{2} a_{k-2}+2 \cdot 2^{k-1}-2-1=2^{2}\left(2 a_{k-3}+2^{k-3}-1\right)+2 \cdot 2^{k-1}-2-1 \\
+& =2^{3} a_{k-3}+3 \cdot 2^{k-1}-2^{2}-2-1 \\
+& \vdots \\
+& =2^{k} a_{0}+k 2^{k-1}-2^{k-1}-2^{k-2}-\ldots-2-1 \\
+& =k 2^{k-1}-2^{k}+1 \quad \text { for all } k \geq 0 .
+\end{aligned}
+$$
+
+Problem 5. Let $a, b, c$ be integers satisfying $0<a<c-1$ and $1<b<c$. For each $k$, $0 \leq k \leq a$, let $r_{k}, 0 \leq r_{k}<c$, be the remainder of $k b$ when divided by $c$. Prove that the two sets $\left\{r_{0}, r_{1}, r_{2}, \ldots, r_{a}\right\}$ and $\{0,1,2, \ldots, a\}$ are different.
+(Solution) Suppose that two sets are equal. Then $\operatorname{gcd}(b, c)=1$ and the polynomial
+
+$$
+f(x):=\left(1+x^{b}+x^{2 b}+\cdots+x^{a b}\right)-\left(1+x+x^{2}+\cdots+x^{a-1}+x^{a}\right)
+$$
+
+is divisible by $x^{c}-1$. (This is because: $m=n+c q \Longrightarrow x^{m}-x^{n}=x^{n+c q}-x^{n}=x^{n}\left(x^{c q}-1\right)$ and $\left(x^{c q}-1\right)=\left(x^{c}-1\right)\left(\left(x^{c}\right)^{q-1}+\left(x^{c}\right)^{q-2}+\cdots+1\right)$.) From
+
+$$
+f(x)=\frac{x^{(a+1) b}-1}{x^{b}-1}-\frac{x^{a+1}-1}{x-1}=\frac{F(x)}{(x-1)\left(x^{b}-1\right)}
+$$
+
+where $F(x)=x^{a b+b+1}+x^{b}+x^{a+1}-x^{a b+b}-x^{a+b+1}-x$, we have
+
+$$
+F(x) \equiv 0 \quad\left(\bmod x^{c}-1\right)
+$$
+
+Since $x^{c} \equiv 1\left(\bmod x^{c}-1\right)$, we may conclude that
+
+$$
+\{a b+b+1, b, a+1\} \equiv\{a b+b, a+b+1,1\} \quad(\bmod c)
+$$
+
+Thus,
+
+$$
+b \equiv a b+b, a+b+1 \text { or } 1(\bmod c)
+$$
+
+But neither $b \equiv 1(\bmod c)$ nor $b \equiv a+b+1(\bmod c)$ are possible by the given conditions. Therefore, $b \equiv a b+b(\bmod c)$. But this is also impossible because $\operatorname{gcd}(b, c)=1$.
+

APMO/md/en-apmo2009_sol.md

@@ -0,0 +1,238 @@
+# XXI Asian Pacific Mathematics Olympiad 
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_74363b62d2fdb79556feg-1.jpg?height=183&width=441&top_left_y=471&top_left_x=770)
+
+March, 2009
+
+Problem 1. Consider the following operation on positive real numbers written on a blackboard:
+
+Choose a number $r$ written on the blackboard, erase that number, and then write a pair of positive real numbers $a$ and $b$ satisfying the condition $2 r^{2}=a b$ on the board.
+
+Assume that you start out with just one positive real number $r$ on the blackboard, and apply this operation $k^{2}-1$ times to end up with $k^{2}$ positive real numbers, not necessarily distinct. Show that there exists a number on the board which does not exceed $k r$.
+(Solution) Using AM-GM inequality, we obtain
+
+$$
+\frac{1}{r^{2}}=\frac{2}{a b}=\frac{2 a b}{a^{2} b^{2}} \leq \frac{a^{2}+b^{2}}{a^{2} b^{2}} \leq \frac{1}{a^{2}}+\frac{1}{b^{2}}
+$$
+
+Consequently, if we let $S_{\ell}$ be the sum of the squares of the reciprocals of the numbers written on the board after $\ell$ operations, then $S_{\ell}$ increases as $\ell$ increases, that is,
+
+$$
+S_{0} \leq S_{1} \leq \cdots \leq S_{k^{2}-1}
+$$
+
+Therefore if we let $s$ be the smallest real number written on the board after $k^{2}-1$ operations, then $\frac{1}{s^{2}} \geq \frac{1}{t^{2}}$ for any number $t$ among $k^{2}$ numbers on the board and hence
+
+$$
+k^{2} \times \frac{1}{s^{2}} \geq S_{k^{2}-1} \geq S_{0}=\frac{1}{r^{2}}
+$$
+
+which implies that $s \leq k r$ as desired.
+
+Remark. The nature of the problem does not change at all if the numbers on the board are restricted to be positive integers. But that may mislead some contestants to think the problem is a number theoretic problem rather than a combinatorial problem.
+
+Problem 2. Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers satisfying the following equations:
+
+$$
+\frac{a_{1}}{k^{2}+1}+\frac{a_{2}}{k^{2}+2}+\frac{a_{3}}{k^{2}+3}+\frac{a_{4}}{k^{2}+4}+\frac{a_{5}}{k^{2}+5}=\frac{1}{k^{2}} \text { for } k=1,2,3,4,5
+$$
+
+Find the value of $\frac{a_{1}}{37}+\frac{a_{2}}{38}+\frac{a_{3}}{39}+\frac{a_{4}}{40}+\frac{a_{5}}{41}$. (Express the value in a single fraction.)
+(Solution) Let $R(x):=\frac{a_{1}}{x^{2}+1}+\frac{a_{2}}{x^{2}+2}+\frac{a_{3}}{x^{2}+3}+\frac{a_{4}}{x^{2}+4}+\frac{a_{5}}{x^{2}+5}$. Then $R( \pm 1)=1$, $R( \pm 2)=\frac{1}{4}, R( \pm 3)=\frac{1}{9}, R( \pm 4)=\frac{1}{16}, R( \pm 5)=\frac{1}{25}$ and $R(6)$ is the value to be found. Let's put $P(x):=\left(x^{2}+1\right)\left(x^{2}+2\right)\left(x^{2}+3\right)\left(x^{2}+4\right)\left(x^{2}+5\right)$ and $Q(x):=R(x) P(x)$. Then for $k= \pm 1, \pm 2, \pm 3, \pm 4, \pm 5$, we get $Q(k)=R(k) P(k)=\frac{P(k)}{k^{2}}$, that is, $P(k)-k^{2} Q(k)=0$. Since $P(x)-x^{2} Q(x)$ is a polynomial of degree 10 with roots $\pm 1, \pm 2, \pm 3, \pm 4, \pm 5$, we get
+
+$$
+P(x)-x^{2} Q(x)=A\left(x^{2}-1\right)\left(x^{2}-4\right)\left(x^{2}-9\right)\left(x^{2}-16\right)\left(x^{2}-25\right)
+$$
+
+Putting $x=0$, we get $A=\frac{P(0)}{(-1)(-4)(-9)(-16)(-25)}=-\frac{1}{120}$. Finally, dividing both sides of $(*)$ by $P(x)$ yields
+
+$$
+1-x^{2} R(x)=1-x^{2} \frac{Q(x)}{P(x)}=-\frac{1}{120} \cdot \frac{\left(x^{2}-1\right)\left(x^{2}-4\right)\left(x^{2}-9\right)\left(x^{2}-16\right)\left(x^{2}-25\right)}{\left(x^{2}+1\right)\left(x^{2}+2\right)\left(x^{2}+3\right)\left(x^{2}+4\right)\left(x^{2}+5\right)}
+$$
+
+and hence that
+
+$$
+1-36 R(6)=-\frac{35 \times 32 \times 27 \times 20 \times 11}{120 \times 37 \times 38 \times 39 \times 40 \times 41}=-\frac{3 \times 7 \times 11}{13 \times 19 \times 37 \times 41}=-\frac{231}{374699}
+$$
+
+which implies $R(6)=\frac{187465}{6744582}$.
+Remark. We can get $a_{1}=\frac{1105}{72}, a_{2}=-\frac{2673}{40}, a_{3}=\frac{1862}{15}, a_{4}=-\frac{1885}{18}, a_{5}=\frac{1323}{40}$ by solving the given system of linear equations, which is extremely messy and takes a lot of time.
+
+Problem 3. Let three circles $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, which are non-overlapping and mutually external, be given in the plane. For each point $P$ in the plane, outside the three circles, construct six points $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$ as follows: For each $i=1,2,3, A_{i}, B_{i}$ are distinct points on the circle $\Gamma_{i}$ such that the lines $P A_{i}$ and $P B_{i}$ are both tangents to $\Gamma_{i}$. Call the point $P$ exceptional if, from the construction, three lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.
+(Solution) Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$. We now claim that all exceptional points lie on $\Gamma$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_74363b62d2fdb79556feg-3.jpg?height=1120&width=909&top_left_y=1239&top_left_x=531)
+
+Let $P O_{1}$ intersect $A_{1} B_{1}$ in $X_{1}$. As $P O_{1} \perp A_{1} B_{1}$, we see that $X_{1}$ lies on $\Gamma$. As $P A_{1}$ is a tangent to $\Gamma_{1}$, triangle $P A_{1} O_{1}$ is right-angled and similar to triangle $A_{1} X_{1} O_{1}$. It follows that
+
+$$
+\frac{O_{1} X_{1}}{O_{1} A_{1}}=\frac{O_{1} A_{1}}{O_{1} P}, \quad \text { i.e., } \quad O_{1} X_{1} \cdot O_{1} P=O_{1} A_{1}^{2}=r_{1}^{2}
+$$
+
+On the other hand, $O_{1} X_{1} \cdot O_{1} P$ is also the power of $O_{1}$ with respect to $\Gamma$, so that
+
+$$
+r_{1}^{2}=O_{1} X_{1} \cdot O_{1} P=\left(O_{1} O-r\right)\left(O_{1} O+r\right)=O_{1} O^{2}-r^{2}
+$$
+
+and hence
+
+$$
+r^{2}=O O_{1}^{2}-r_{1}^{2}=\left(O O_{1}-r_{1}\right)\left(O O_{1}+r_{1}\right)
+$$
+
+Thus, $r^{2}$ is the power of $O$ with respect to $\Gamma_{1}$. By the same token, $r^{2}$ is also the power of $O$ with respect to $\Gamma_{2}$ and $\Gamma_{3}$. Hence $O$ must be the radical center of the three given circles. Since $r$, as the square root of the power of $O$ with respect to the three given circles, does not depend on $P$, it follows that all exceptional points lie on $\Gamma$.
+
+Remark. In the event of the radical point being at infinity (and hence the three radical axes being parallel), there are no exceptional points in the plane, which is consistent with the statement of the problem.
+
+Problem 4. Prove that for any positive integer $k$, there exists an arithmetic sequence
+
+$$
+\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
+$$
+
+of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{k}, b_{k}$ are all distinct.
+(Solution) For $k=1$, there is nothing to prove. Henceforth assume $k \geq 2$.
+Let $p_{1}, p_{2}, \ldots, p_{k}$ be $k$ distinct primes such that
+
+$$
+k<p_{k}<\cdots<p_{2}<p_{1}
+$$
+
+and let $N=p_{1} p_{2} \cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying
+
+$$
+x \equiv-i \quad\left(\bmod p_{i}\right)
+$$
+
+for all $i=1,2, \ldots, k$ and $x>N^{2}$. Consider the following sequence:
+
+$$
+\frac{x+1}{N}, \frac{x+2}{N}, \quad, \ldots, \frac{x+k}{N}
+$$
+
+This sequence is obviously an arithmetic sequence of positive rational numbers of length $k$. For each $i=1,2, \ldots, k$, the numerator $x+i$ is divisible by $p_{i}$ but not by $p_{j}$ for $j \neq i$, for otherwise $p_{j}$ divides $|i-j|$, which is not possible because $p_{j}>k>|i-j|$. Let
+
+$$
+a_{i}:=\frac{x+i}{p_{i}}, \quad b_{i}:=\frac{N}{p_{i}} \quad \text { for all } i=1,2, \ldots, k
+$$
+
+Then
+
+$$
+\frac{x+i}{N}=\frac{a_{i}}{b_{i}}, \quad \operatorname{gcd}\left(a_{i}, b_{i}\right)=1 \quad \text { for all } i=1,2, \ldots, k
+$$
+
+and all $b_{i}$ 's are distinct from each other. Moreover, $x>N^{2}$ implies
+
+$$
+a_{i}=\frac{x+i}{p_{i}}>\frac{N^{2}}{p_{i}}>N>\frac{N}{p_{j}}=b_{j} \quad \text { for all } i, j=1,2, \ldots, k
+$$
+
+and hence all $a_{i}$ 's are distinct from $b_{i}$ 's. It only remains to show that all $a_{i}$ 's are distinct from each other. This follows from
+
+$$
+a_{j}=\frac{x+j}{p_{j}}>\frac{x+i}{p_{j}}>\frac{x+i}{p_{i}}=a_{i} \quad \text { for all } i<j
+$$
+
+by our choice of $p_{1}, p_{2}, \ldots, p_{k}$. Thus, the arithmetic sequence
+
+$$
+\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \ldots, \quad \frac{a_{k}}{b_{k}}
+$$
+
+of positive rational numbers satisfies the conditions of the problem.
+
+Remark. Here is a much easier solution :
+
+For any positive integer $k \geq 2$, consider the sequence
+
+$$
+\frac{(k!)^{2}+1}{k!}, \frac{(k!)^{2}+2}{k!}, \ldots, \frac{(k!)^{2}+k}{k!}
+$$
+
+Note that $\operatorname{gcd}\left(k!,(k!)^{2}+i\right)=i$ for all $i=1,2, \ldots, k$. So, taking
+
+$$
+a_{i}:=\frac{(k!)^{2}+i}{i}, \quad b_{i}:=\frac{k!}{i} \quad \text { for all } i=1,2, \ldots, k
+$$
+
+we have $\operatorname{gcd}\left(a_{i}, b_{i}\right)=1$ and
+
+$$
+a_{i}=\frac{(k!)^{2}+i}{i}>a_{j}=\frac{(k!)^{2}+j}{j}>b_{i}=\frac{k!}{i}>b_{j}=\frac{k!}{j}
+$$
+
+for any $1 \leq i<j \leq k$. Therefore this sequence satisfies every condition given in the problem.
+
+Problem 5. Larry and Rob are two robots travelling in one car from Argovia to Zillis. Both robots have control over the steering and steer according to the following algorithm: Larry makes a $90^{\circ}$ left turn after every $\ell$ kilometer driving from start; Rob makes a $90^{\circ}$ right turn after every $r$ kilometer driving from start, where $\ell$ and $r$ are relatively prime positive integers. In the event of both turns occurring simultaneously, the car will keep going without changing direction. Assume that the ground is flat and the car can move in any direction.
+
+Let the car start from Argovia facing towards Zillis. For which choices of the pair $(\ell, r)$ is the car guaranteed to reach Zillis, regardless of how far it is from Argovia?
+(Solution) Let Zillis be $d$ kilometers away from Argovia, where $d$ is a positive real number. For simplicity, we will position Argovia at $(0,0)$ and Zillis at $(d, 0)$, so that the car starts out facing east. We will investigate how the car moves around in the period of travelling the first $\ell r$ kilometers, the second $\ell r$ kilometers, ..., and so on. We call each period of travelling $\ell r$ kilometers a section. It is clear that the car will have identical behavior in every section except the direction of the car at the beginning.
+
+Case 1: $\underline{\ell-r \equiv 2(\bmod 4)}$. After the first section, the car has made $\ell-1$ right turns and $r-1$ left turns, which is a net of $2(\equiv \ell-r(\bmod 4))$ right turns. Let the displacement vector for the first section be $(x, y)$. Since the car has rotated $180^{\circ}$, the displacement vector for the second section will be $(-x,-y)$, which will take the car back to $(0,0)$ facing east again. We now have our original situation, and the car has certainly never travelled further than lr kilometers from Argovia. So, the car cannot reach Zillis if it is further apart from Argovia.
+
+Case 2: $\ell-r \equiv 1(\bmod 4)$. After the first section, the car has made a net of 1 right turn. Let the displacement vector for the first section again be $(x, y)$. This time the car has rotated $90^{\circ}$ clockwise. We can see that the displacements for the second, third and fourth section will be $(y,-x),(-x,-y)$ and $(-y, x)$, respectively, so after four sections the car is back at $(0,0)$ facing east. Since the car has certainly never travelled further than $2 \ell r$ kilometers from Argovia, the car cannot reach Zillis if it is further apart from Argovia.
+
+Case 3: $\quad \ell-r \equiv 3(\bmod 4)$. An argument similar to that in Case 2 (switching the roles of left and right) shows that the car cannot reach Zillis if it is further apart from Argovia.
+
+Case 4: $\quad \ell \equiv r(\bmod 4)$. The car makes a net turn of $0^{\circ}$ after each section, so it must be facing east. We are going to show that, after traversing the first section, the car will be at $(1,0)$. It will be useful to interpret the Cartesian plane as the complex plane, i.e. writing $x+i y$ for $(x, y)$, where $i=\sqrt{-1}$. We will denote the $k$-th kilometer of movement by $m_{k-1}$,
+which takes values from the set $\{1, i,-1,-i\}$, depending on the direction. We then just have to show that
+
+$$
+\sum_{k=0}^{\ell r-1} m_{k}=1
+$$
+
+which implies that the car will get to Zillis no matter how far it is apart from Argovia.
+Case $4 \mathrm{a}: \underline{\ell \equiv r \equiv 1(\bmod 4)}$. First note that for $k=0,1, \ldots, \ell r-1$,
+
+$$
+m_{k}=i^{\lfloor k / \ell\rfloor}(-i)^{\lfloor k / r\rfloor}
+$$
+
+since $\lfloor k / \ell\rfloor$ and $\lfloor k / r\rfloor$ are the exact numbers of left and right turns before the $(k+1)$ st kilometer, respectively. Let $a_{k}(\equiv k(\bmod \ell))$ and $b_{k}(\equiv k(\bmod r))$ be the remainders of $k$ when divided by $\ell$ and $r$, respectively. Then, since
+
+$$
+a_{k}=k-\left\lfloor\frac{k}{\ell}\right\rfloor \ell \equiv k-\left\lfloor\frac{k}{\ell}\right\rfloor \quad(\bmod 4) \quad \text { and } \quad b_{k}=k-\left\lfloor\frac{k}{r}\right\rfloor r \equiv k-\left\lfloor\frac{k}{r}\right\rfloor \quad(\bmod 4),
+$$
+
+we have $\lfloor k / \ell\rfloor \equiv k-a_{k}(\bmod 4)$ and $\lfloor k / r\rfloor \equiv k-b_{k}(\bmod 4)$. We therefore have
+
+$$
+m_{k}=i^{k-a_{k}}(-i)^{k-b_{k}}=\left(-i^{2}\right)^{k} i^{-a_{k}}(-i)^{-b_{k}}=(-i)^{a_{k}} i^{b_{k}} .
+$$
+
+As $\ell$ and $r$ are relatively prime, by Chinese Remainder Theorem, there is a bijection between pairs $\left(a_{k}, b_{k}\right)=(k(\bmod \ell), k(\bmod r))$ and the numbers $k=0,1,2, \ldots, \ell r-1$. Hence
+
+$$
+\sum_{k=0}^{\ell r-1} m_{k}=\sum_{k=0}^{\ell r-1}(-i)^{a_{k}} i^{b_{k}}=\left(\sum_{k=0}^{\ell-1}(-i)^{a_{k}}\right)\left(\sum_{k=0}^{r-1} i^{b_{k}}\right)=1 \times 1=1
+$$
+
+as required because $\ell \equiv r \equiv 1(\bmod 4)$.
+Case $4 \mathrm{~b}: \underline{\ell \equiv r \equiv 3(\bmod 4)}$. In this case, we get
+
+$$
+m_{k}=i^{a_{k}}(-i)^{b_{k}}
+$$
+
+where $a_{k}(\equiv k(\bmod \ell))$ and $b_{k}(\equiv k(\bmod r))$ for $k=0,1, \ldots, \ell r-1$. Then we can proceed analogously to Case 4a to obtain
+
+$$
+\sum_{k=0}^{\ell r-1} m_{k}=\sum_{k=0}^{\ell r-1}(-i)^{a_{k}} i^{b_{k}}=\left(\sum_{k=0}^{\ell-1}(-i)^{a_{k}}\right)\left(\sum_{k=0}^{r-1} i^{b_{k}}\right)=i \times(-i)=1
+$$
+
+as required because $\ell \equiv r \equiv 3(\bmod 4)$.
+Now clearly the car traverses through all points between $(0,0)$ and $(1,0)$ during the first section and, in fact, covers all points between $(n-1,0)$ and $(n, 0)$ during the $n$-th section. Hence it will eventually reach $(d, 0)$ for any positive $d$.
+
+To summarize: $(\ell, r)$ satisfies the required conditions if and only if
+
+$$
+\ell \equiv r \equiv 1 \quad \text { or } \quad \ell \equiv r \equiv 3 \quad(\bmod 4)
+$$
+
+Remark. In case $\operatorname{gcd}(\ell, r)=d \neq 1$, the answer is :
+
+$$
+\frac{\ell}{d} \equiv \frac{r}{d} \equiv 1 \quad \text { or } \quad \frac{\ell}{d} \equiv \frac{r}{d} \equiv 3 \quad(\bmod 4)
+$$
+

APMO/md/en-apmo2010_sol.md

@@ -0,0 +1,218 @@
+# SOLUTIONS FOR 2010 APMO PROBLEMS 
+
+Problem 1. Let $A B C$ be a triangle with $\angle B A C \neq 90^{\circ}$. Let $O$ be the circumcenter of the triangle $A B C$ and let $\Gamma$ be the circumcircle of the triangle $B O C$. Suppose that $\Gamma$ intersects the line segment $A B$ at $P$ different from $B$, and the line segment $A C$ at $Q$ different from $C$. Let $O N$ be a diameter of the circle $\Gamma$. Prove that the quadrilateral $A P N Q$ is a parallelogram.
+Solution: From the assumption that the circle $\Gamma$ intersects both of the line segments $A B$ and $A C$, it follows that the 4 points $N, C, Q, O$ are located on $\Gamma$ in the order of $N, C, Q, O$ or in the order of $N, C, O, Q$. The following argument for the proof of the assertion of the problem is valid in either case. Since $\angle N Q C$ and $\angle N O C$ are subtended by the same arc $\widehat{N C}$ of $\Gamma$ at the points $Q$ and $O$, respectively, on $\Gamma$, we have $\angle N Q C=\angle N O C$. We also have $\angle B O C=2 \angle B A C$, since $\angle B O C$ and $\angle B A C$ are subtended by the same arc $\widehat{B C}$ of the circum-circle of the triangle $A B C$ at the center $O$ of the circle and at the point $A$ on the circle, respectively. From $O B=O C$ and the fact that $O N$ is a diameter of $\Gamma$, it follows that the triangles $O B N$ and $O C N$ are congruent, and therefore we obtain $2 \angle N O C=\angle B O C$. Consequently, we have $\angle N Q C=\frac{1}{2} \angle B O C=\angle B A C$, which shows that the 2 lines $A P, Q N$ are parallel.
+
+In the same manner, we can show that the 2 lines $A Q, P N$ are also parallel. Thus, the quadrilateral $A P N Q$ is a parallelogram.
+
+Problem 2. For a positive integer $k$, call an integer a pure $k$-th power if it can be represented as $m^{k}$ for some integer $m$. Show that for every positive integer $n$ there exist $n$ distinct positive integers such that their sum is a pure 2009-th power, and their product is a pure 2010-th power.
+Solution: For the sake of simplicity, let us set $k=2009$.
+First of all, choose $n$ distinct positive integers $b_{1}, \cdots, b_{n}$ suitably so that their product is a pure $k+1$-th power (for example, let $b_{i}=i^{k+1}$ for $i=1, \cdots, n$ ). Then we have $b_{1} \cdots b_{n}=t^{k+1}$ for some positive integer $t$. Set $b_{1}+\cdots+b_{n}=s$.
+
+Now we set $a_{i}=b_{i} s^{k^{2}-1}$ for $i=1, \cdots, n$, and show that $a_{1}, \cdots, a_{n}$ satisfy the required conditions. Since $b_{1}, \cdots, b_{n}$ are distinct positive integers, it is clear that so are $a_{1}, \cdots, a_{n}$. From
+
+$$
+\begin{aligned}
+a_{1}+\cdots+a_{n} & =s^{k^{2}-1}\left(b_{1}+\cdots+b_{n}\right)=s^{k^{2}}=\left(s^{k}\right)^{2009} \\
+a_{1} \cdots a_{n} & =\left(s^{k^{2}-1}\right)^{n} b_{1} \cdots b_{n}=\left(s^{k^{2}-1}\right)^{n} t^{k+1}=\left(s^{(k-1) n} t\right)^{2010}
+\end{aligned}
+$$
+
+we can see that $a_{1}, \cdots, a_{n}$ satisfy the conditions on the sum and the product as well. This ends the proof of the assertion.
+Remark: We can find the appropriate exponent $k^{2}-1$ needed for the construction of the $a_{i}$ 's by solving the simultaneous congruence relations: $x \equiv 0(\bmod k+1), x \equiv-1(\bmod k)$.
+
+Problem 3. Let $n$ be a positive integer. $n$ people take part in a certain party. For any pair of the participants, either the two are acquainted with each other or they are not. What is the maximum possible number of the pairs for which the two are not acquainted but have a common acquaintance among the participants?
+
+Solution: When 1 participant, say the person $A$, is mutually acquainted with each of the remaining $n-1$ participants, and if there are no other acquaintance relationships among the participants, then for any pair of participants not involving $A$, the two are not mutual acquaintances, but they have a common acquaintance, namely $A$, so any such pair satisfies the requirement. Thus, the number desired in this case is $\frac{(n-1)(n-2)}{2}=\frac{n^{2}-3 n+2}{2}$.
+
+Let us show that $\frac{n^{2}-3 n+2}{2}$ is the maximum possible number of the pairs satisfying the requirement of the problem. First, let us observe that in the process of trying to find the maximum possible number of such pairs, if we split the participants into two non-empty subsets $T$ and $S$ which are disjoint, we may assume that there is a pair consisting of one person chosen from $T$ and the other chosen from $S$ who are mutual acquaintances. This is so, since if there are no such pair for some splitting $T$ and $S$, then among the pairs consisting of one person chosen from $T$ and the other chosen from $S$, there is no pair for which the two have a common acquaintance among participants, and therefore, if we arbitrarily choose a person $A \in T$ and $B \in S$ and declare that $A$ and $B$ are mutual acquaintances, the number of the pairs satisfying the requirement of the problem does not decrease.
+
+Let us now call a set of participants a group if it satisfies the following 2 conditions:
+
+- One can connect any person in the set with any other person in the set by tracing a chain of mutually acquainted pairs. More precisely, for any pair of people $A, B$ in the set there exists a sequence of people $A_{0}, A_{1}, \cdots, A_{n}$ for which $A_{0}=A, A_{n}=B$ and, for each $i: 0 \leq i \leq n-1, A_{i}$ and $A_{i+1}$ are mutual acquaintances.
+- No person in this set can be connected with a person not belonging to this set by tracing a chain of mutually acquainted pairs.
+
+In view of the discussions made above, we may assume that the set of all the participants to the party forms a group of $n$ people. Let us next consider the following lemma.
+Lemma. In a group of $n$ people, there are at least $n-1$ pairs of mutual acquaintances.
+Proof: If you choose a mutually acquainted pair in a group and declare the two in the pair are not mutually acquainted, then either the group stays the same or splits into 2 groups. This means that by changing the status of a mutually acquainted pair in a group to that of a non-acquainted pair, one can increase the number of groups at most by 1 . Now if in a group of $n$ people you change the status of all of the mutually acquainted pairs to that of non-acquainted pairs, then obviously, the number of groups increases from 1 to $n$. Therefore, there must be at least $n-1$ pairs of mutually acquainted pairs in a group consisting of $n$ people.
+
+The lemma implies that there are at most $\frac{n(n-1)}{2}-(n-1)=\frac{n^{2}-3 n+2}{2}$ pairs satisfying the condition of the problem. Thus the desired maximum number of pairs satisfying the requirement of the problem is $\frac{n^{2}-3 n+2}{2}$.
+Remark: One can give a somewhat different proof by separating into 2 cases depending on whether there are at least $n-1$ mutually acquainted pairs, or at most $n-2$ such pairs. In the former case, one can argue in the same way as the proof above, while in the latter case, the Lemma above implies that there would be 2 or more groups to start with, but then, in view of the comment made before the definition of a group above, these groups can be combined to form one group, thereby one can reduce the argument to the former case.
+
+Alternate Solution 1: The construction of an example for the case for which the number $\frac{n^{2}-3 n+2}{2}$ appears, and the argument for the case where there is only 1 group would be the same as in the preceding proof.
+
+Suppose, then, $n$ participants are separated into $k(k \geq 2)$ groups, and the number of people in each group is given by $a_{i}, i=1, \cdots, k$. In such a case, the number of pairs for which paired people are not mutually acquainted but have a common acquaintance is at most $\sum_{i=1}^{k} a_{i} C_{2}$, where we set ${ }_{1} C_{2}=0$ for convenience. Since ${ }_{a} C_{2}+{ }_{b} C_{2} \leq{ }_{a+b} C_{2}$ holds for any pair of positive integers $a, b$, we have $\sum_{i=1}^{k} a_{i} C_{2} \leq{ }_{a_{1}} C_{2}+{ }_{n-a_{1}} C_{2}$. From
+
+$$
+{ }_{a_{1}} C_{2}+{ }_{n-a_{1}} C_{2}=a_{1}^{2}-n a_{1}+\frac{n^{2}-n}{2}=\left(a_{1}-\frac{n}{2}\right)^{2}+\frac{n^{2}-2 n}{4}
+$$
+
+it follows that ${ }_{a} C_{2}+{ }_{n-a_{1}} C_{2}$ takes its maximum value when $a_{1}=1, n-1$. Therefore, we have $\sum_{i=1}^{k}{ }_{a} C_{2} \leq{ }_{n-1} C_{2}$, which shows that in the case where the number of groups are 2 or more, the number of the pairs for which paired people are not mutually acquainted but have a common acquaintance is at most ${ }_{n-1} C_{2}=\frac{n^{2}-3 n+2}{2}$, and hence the desired maximum number of the pairs satisfying the requirement is $\frac{n^{2}-3 n+2}{2}$.
+Alternate Solution 2: Construction of an example would be the same as the preceding proof.
+
+For a participant, say $A$, call another participant, say $B$, a familiar face if $A$ and $B$ are not mutually acquainted but they have a common acquaintance among the participants, and in this case call the pair $A, B$ a familiar pair.
+
+Suppose there is a participant $P$ who is mutually acquainted with $d$ participants. Denote by $S$ the set of these $d$ participants, and by $T$ the set of participants different from $P$ and not belonging to the set $S$. Suppose there are $e$ pairs formed by a person in $S$ and a person in $T$ who are mutually acquainted.
+
+Then the number of participants who are familiar faces to $P$ is at most $e$. The number of pairs formed by two people belonging to the set $S$ and are mutually acquainted is at most ${ }_{d} C_{2}$. The number of familiar pairs formed by two people belonging to the set $T$ is at most ${ }_{n-d-1} C_{2}$. Since there are $e$ pairs formed by a person in the set $S$ and a person in the set $T$ who are mutually acquainted (and so the pairs are not familiar pairs), we have at most $d(n-1-d)-e$ familiar pairs formed by a person chosen from $S$ and a person chosen from $T$. Putting these together we conclude that there are at most $e+{ }_{d} C_{2}+{ }_{n-1-d} C_{2}+d(n-1-d)-e$ familiar pairs. Since
+
+$$
+e+{ }_{d} C_{2}+{ }_{n-1-d} C_{2}+d(n-1-d)-e=\frac{n^{2}-3 n+2}{2}
+$$
+
+the number we seek is at most $\frac{n^{2}-3 n+2}{2}$, and hence this is the desired solution to the problem.
+Problem 4. Let $A B C$ be an acute triangle satisfying the condition $A B>B C$ and $A C>B C$. Denote by $O$ and $H$ the circumcenter and the orthocenter, respectively, of the triangle $A B C$. Suppose that the circumcircle of the triangle $A H C$ intersects the line $A B$ at $M$ different from $A$, and that the circumcircle of the triangle $A H B$ intersects the line $A C$ at $N$ different from $A$. Prove that the circumcenter of the triangle $M N H$ lies on the line $O H$.
+Solution: In the sequel, we denote $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Let $O^{\prime}$ be the circumcenter of the triangle $M N H$. The lengths of line segments starting from the point $H$ will be treated as signed quantities.
+
+Let us denote by $M^{\prime}, N^{\prime}$ the point of intersection of $C H, B H$, respectively, with the circumcircle of the triangle $A B C$ (distinct from $C, B$, respectively.) From the fact that 4 points $A, M, H, C$ lie on the same circle, we see that $\angle M H M^{\prime}=\alpha$ holds. Furthermore, $\angle B M^{\prime} C, \angle B N^{\prime} C$ and $\alpha$ are all subtended by the same arc $\widehat{B C}$ of the circumcircle of the triangle $A B C$ at points on the circle, and therefore, we have $\angle B M^{\prime} C=\alpha$, and $\angle B N^{\prime} C=\alpha$ as well. We also have $\angle A B H=\angle A C N^{\prime}$ as they are subtended by the same $\operatorname{arc} A N^{\prime}$ of the circumcircle of the triangle $A B C$ at points on the circle. Since $H M^{\prime} \perp B M, H N^{\prime} \perp A C$, we conclude that
+
+$$
+\angle M^{\prime} H B=90^{\circ}-\angle A B H=90^{\circ}-\angle A C N^{\prime}=\alpha
+$$
+
+is valid as well. Putting these facts together, we obtain the fact that the quadrilateral $H B M^{\prime} M$ is a rhombus. In a similar manner, we can conclude that the quadrilateral $H C N^{\prime} N$ is also a rhombus. Since both of these rhombuses are made up of 4 right triangles with an angle of magnitude $\alpha$, we also see that these rhombuses are similar.
+
+Let us denote by $P, Q$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O^{\prime}$. Since $O^{\prime}$ is the circumcenter of the triangle $M N H, P, Q$ are respectively, the midpoints of the line segments $H M, H N$. Furthermore, if we denote by $R, S$ the feet of the perpendicular lines on $H M$ and $H N$, respectively, drawn from the point $O$, then since $O$ is the circumcenter of both the triangle $M^{\prime} B C$ and the triangle $N^{\prime} B C$, we see that $R$ is the intersection point of $H M$ and the perpendicular bisector of $B M^{\prime}$, and $S$ is the intersection point of $H N$ and the perpendicular bisector of $C N^{\prime}$.
+
+We note that the similarity map $\phi$ between the rhombuses $H B M^{\prime} M$ and $H C N^{\prime} N$ carries the perpendicular bisector of $B M^{\prime}$ onto the perpendicular bisector of $C N^{\prime}$, and straight line $H M$ onto the straight line $H N$, and hence $\phi$ maps $R$ onto $S$, and $P$ onto $Q$. Therefore, we get $H P: H R=H Q: H S$. If we now denote by $X, Y$ the intersection points of the line $H O^{\prime}$ with the line through $R$ and perpendicular to $H P$, and with the line through $S$ and perpendicular to $H Q$, respectively, then we get
+
+$$
+H O^{\prime}: H X=H P: H R=H Q: H S=H O^{\prime}: H Y
+$$
+
+so that we must have $H X=H Y$, and therefore, $X=Y$. But it is obvious that the point of intersection of the line through $R$ and perpendicular to $H P$ with the line through $S$ and perpendicular to $H Q$ must be $O$, and therefore, we conclude that $X=Y=O$ and that the points $H, O^{\prime}, O$ are collinear.
+
+Alternate Solution: Deduction of the fact that both of the quadrilaterals $H B M^{\prime} M$ and $H C N^{\prime} N$ are rhombuses is carried out in the same way as in the preceding proof.
+
+We then see that the point $M$ is located in a symmetric position with the point $B$ with respect to the line $C H$, we conclude that we have $\angle C M B=\beta$. Similarly, we have $\angle C N B=\gamma$. If we now put $x=\angle A H O^{\prime}$, then we get
+
+$$
+\angle O^{\prime}=\beta-\alpha-x, \angle M N H=90^{\circ}-\beta-\alpha+x
+$$
+
+from which it follows that
+
+$$
+\angle A N M=180^{\circ}-\angle M N H-\left(90^{\circ}-\alpha\right)=\beta-x
+$$
+
+Similarly, we get
+
+$$
+\angle N M A=\gamma+x
+$$
+
+Using the laws of sines, we then get
+
+$$
+\begin{aligned}
+\frac{\sin (\gamma+x)}{\sin (\beta-x)} & =\frac{A N}{A M}=\frac{A C}{A M} \cdot \frac{A B}{A C} \cdot \frac{A N}{A B} \\
+& =\frac{\sin \beta}{\sin (\beta-\alpha)} \cdot \frac{\sin \gamma}{\sin \beta} \cdot \frac{\sin (\gamma-\alpha)}{\sin \gamma}=\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}
+\end{aligned}
+$$
+
+On the other hand, if we let $y=\angle A H O$, we then get
+
+$$
+\angle O H B=180^{\circ}-\gamma-y, \angle C H O=180^{\circ}-\beta+y,
+$$
+
+and since
+
+$$
+\angle H B O=\gamma-\alpha, \angle O C H=\beta-\alpha,
+$$
+
+using the laws of sines and observing that $O B=O C$, we get
+
+$$
+\begin{aligned}
+\frac{\sin (\gamma-\alpha)}{\sin (\beta-\alpha)}=\frac{\sin \angle H B O}{\sin \angle O C H} & =\frac{\sin \left(180^{\circ}-\gamma-y\right) \cdot \frac{O H}{O B}}{\sin \left(180^{\circ}-\beta+y\right) \cdot \frac{O H}{O C}} \\
+& =\frac{\sin \left(180^{\circ}-\gamma-y\right)}{\sin \left(180^{\circ}-\beta+y\right)}=\frac{\sin (\gamma+y)}{\sin (\beta-y)}
+\end{aligned}
+$$
+
+We then get $\sin (\gamma+x) \sin (\beta-y)=\sin (\beta-x) \sin (\gamma+y)$. Expanding both sides of the last identity by using the addition formula for the sine function and after factoring and using again the addition formula we obtain that $\sin (x-y) \sin (\beta+\gamma)=0$. This implies that $x-y$ must be an integral multiple of $180^{\circ}$, and hence we conclude that $H, O, O^{\prime}$ are collinear.
+
+Problem 5. Find all functions $f$ from the set $\mathbf{R}$ of real numbers into $\mathbf{R}$ which satisfy for all $x, y, z \in \mathbf{R}$ the identity
+
+$$
+f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2 x y+f(z))+2 f(x z-y z) .
+$$
+
+Solution: It is clear that if $f$ is a constant function which satisfies the given equation, then the constant must be 0 . Conversely, $f(x)=0$ clearly satisfies the given equation, so, the identically 0 function is a solution. In the sequel, we consider the case where $f$ is not a constant function.
+
+Let $t \in \mathbf{R}$ and substitute $(x, y, z)=(t, 0,0)$ and $(x, y, z)=(0, t, 0)$ into the given functional equation. Then, we obtain, respectively,
+
+$$
+\begin{aligned}
+& f(f(t)+2 f(0))=f(f(t)-f(0))+f(f(0))+2 f(0), \\
+& f(f(t)+2 f(0))=f(f(0)-f(t))+f(f(0))+2 f(0)
+\end{aligned}
+$$
+
+from which we conclude that $f(f(t)-f(0))=f(f(0)-f(t))$ holds for all $t \in \mathbf{R}$. Now, suppose for some pair $u_{1}, u_{2}, f\left(u_{1}\right)=f\left(u_{2}\right)$ is satisfied. Then by substituting $(x, y, z)=\left(s, 0, u_{1}\right)$ and $(x, y, z)=\left(s, 0, u_{2}\right)$ into the functional equation and comparing the resulting identities, we can easily conclude that
+
+$$
+f\left(s u_{1}\right)=f\left(s u_{2}\right)
+$$
+
+holds for all $s \in \mathbf{R}$. Since $f$ is not a constant function there exists an $s_{0}$ such that $f\left(s_{0}\right)-f(0) \neq$ 0 . If we put $u_{1}=f\left(s_{0}\right)-f(0), u_{2}=-u_{1}$, then $f\left(u_{1}\right)=f\left(u_{2}\right)$, so we have by $(*)$
+
+$$
+f\left(s u_{1}\right)=f\left(s u_{2}\right)=f\left(-s u_{1}\right)
+$$
+
+for all $s \in \mathbf{R}$. Since $u_{1} \neq 0$, we conclude that
+
+$$
+f(x)=f(-x)
+$$
+
+holds for all $x \in \mathbf{R}$.
+Next, if $f(u)=f(0)$ for some $u \neq 0$, then by $(*)$, we have $f(s u)=f(s 0)=f(0)$ for all $s$, which implies that $f$ is a constant function, contradicting our assumption. Therefore, we must have $f(s) \neq f(0)$ whenever $s \neq 0$.
+
+We will now show that if $f(x)=f(y)$ holds, then either $x=y$ or $x=-y$ must hold. Suppose on the contrary that $f\left(x_{0}\right)=f\left(y_{0}\right)$ holds for some pair of non-zero numbers $x_{0}, y_{0}$ for which $x_{0} \neq y_{0}, x_{0} \neq-y_{0}$. Since $f\left(-y_{0}\right)=f\left(y_{0}\right)$, we may assume, by replacing $y_{0}$ by $-y_{0}$ if necessary, that $x_{0}$ and $y_{0}$ have the same sign. In view of $(*)$, we see that $f\left(s x_{0}\right)=f\left(s y_{0}\right)$ holds for all $s$, and therefore, there exists some $r>0, r \neq 1$ such that
+
+$$
+f(x)=f(r x)
+$$
+
+holds for all $x$. Replacing $x$ by $r x$ and $y$ by $r y$ in the given functional equation, we obtain
+
+$$
+f(f(r x)+f(r y)+f(z))=f(f(r x)-f(r y))+f\left(2 r^{2} x y+f(z)\right)+2 f(r(x-y) z)
+$$
+
+and replacing $x$ by $r^{2} x$ in the functional equation, we get
+
+$$
+f\left(f\left(r^{2} x\right)+f(y)+f(z)\right)=f\left(f\left(r^{2} x\right)-f(y)\right)+f\left(2 r^{2} x y+f(z)\right)+2 f\left(\left(r^{2} x-y\right) z\right)
+$$
+
+Since $f(r x)=f(x)$ holds for all $x \in \mathbf{R}$, we see that except for the last term on the right-hand side, all the corresponding terms appearing in the identities (i) and (ii) above are equal, and hence we conclude that
+
+$$
+\left.f(r(x-y) z)=f\left(\left(r^{2} x-y\right) z\right)\right)
+$$
+
+must hold for arbitrary choice of $x, y, z \in \mathbf{R}$. For arbitrarily fixed pair $u, v \in \mathbf{R}$, substitute $(x, y, z)=\left(\frac{v-u}{r^{2}-1}, \frac{v-r^{2} u}{r^{2}-1}, 1\right)$ into the identity (iii). Then we obtain $f(v)=f(r u)=f(u)$, since $x-y=u, r^{2} x-y=v, z=1$. But this implies that the function $f$ is a constant, contradicting our assumption. Thus we conclude that if $f(x)=f(y)$ then either $x=y$ or $x=-y$ must hold.
+
+By substituting $z=0$ in the functional equation, we get
+
+$$
+f(f(x)+f(y)+f(0))=f(f(x)-f(y)+f(0))=f((f(x)-f(y))+f(2 x y+f(0))+2 f(0)
+$$
+
+Changing $y$ to $-y$ in the identity above and using the fact that $f(y)=f(-y)$, we see that all the terms except the second term on the right-hand side in the identity above remain the same. Thus we conclude that $f(2 x y+f(0))=f(-2 x y+f(0))$, from which we get either $2 x y+f(0)=-2 x y+f(0)$ or $2 x y+f(0)=2 x y-f(0)$ for all $x, y \in \mathbf{R}$. The first of these alternatives says that $4 x y=0$, which is impossible if $x y \neq 0$. Therefore the second alternative must be valid and we get that $f(0)=0$.
+
+Finally, let us show that if $f$ satisfies the given functional equation and is not a constant function, then $f(x)=x^{2}$. Let $x=y$ in the functional equation, then since $f(0)=0$, we get
+
+$$
+f(2 f(x)+f(z))=f\left(2 x^{2}+f(z)\right)
+$$
+
+from which we conclude that either $2 f(x)+f(z)=2 x^{2}+f(z)$ or $2 f(x)+f(z)=-2 x^{2}-f(z)$ must hold. Suppose there exists $x_{0}$ for which $f\left(x_{0}\right) \neq x_{0}^{2}$, then from the second alternative, we see that $f(z)=-f\left(x_{0}\right)-x_{0}^{2}$ must hold for all $z$, which means that $f$ must be a constant function, contrary to our assumption. Therefore, the first alternative above must hold, and we have $f(x)=x^{2}$ for all $x$, establishing our claim.
+
+It is easy to check that $f(x)=x^{2}$ does satisfy the given functional equation, so we conclude that $f(x)=0$ and $f(x)=x^{2}$ are the only functions that satisfy the requirement.
+

APMO/md/en-apmo2011_sol.md

@@ -0,0 +1,134 @@
+# SOLUTIONS FOR 2011 APMO PROBLEMS 
+
+## Problem 1.
+
+Solution: Suppose all of the 3 numbers $a^{2}+b+c, b^{2}+c+a$ and $c^{2}+a+b$ are perfect squares. Then from the fact that $a^{2}+b+c$ is a perfect square bigger than $a^{2}$ it follows that $a^{2}+b+c \geq(a+1)^{2}$, and therefore, $b+c \geq 2 a+1$. Similarly we obtain $c+a \geq 2 b+1$ and $a+b \geq 2 c+1$.
+
+Adding the corresponding sides of the preceding 3 inequalities, we obtain $2(a+b+c) \geq 2(a+b+c)+3$, a contradiction. This proves that it is impossible to have all the 3 given numbers to be perfect squares.
+Alternate Solution: Since the given conditions of the problem are symmetric in $a, b, c$, we may assume that $a \geq b \geq c$ holds. From the assumption that $a^{2}+b+c$ is a perfect square, we can deduce as in the solution above the inequality $b+c \geq 2 a+1$. But then we have
+
+$$
+2 a \geq b+c \geq 2 a+1
+$$
+
+a contradiction, which proves the assertion of the problem.
+
+## Problem 2.
+
+Solution: We will show that $36^{\circ}$ is the desired answer for the problem.
+First, we observe that if the given 5 points form a regular pentagon, then the minimum of the angles formed by any triple among the five vertices is $36^{\circ}$, and therefore, the answer we seek must be bigger than or equal to $36^{\circ}$.
+
+Next, we show that for any configuration of 5 points satisfying the condition of the problem, there must exist an angle smaller than or equal to $36^{\circ}$ formed by a triple chosen from the given 5 points. For this purpose, let us start with any 5 points, say $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, on the plane satisfying the condition of the problem, and consider the smallest convex subset, call it $\Gamma$, in the plane containing all of the 5 points. Since this convex subset $\Gamma$ must be either a triangle or a quadrilateral or a pentagon, it must have an interior angle with $108^{\circ}$ or less. We may assume without loss of generality that this angle is $\angle A_{1} A_{2} A_{3}$. By the definition of $\Gamma$ it is clear that the remaining 2 points $A_{4}$ and $A_{5}$ lie in the interior of the angular region determined by $\angle A_{1} A_{2} A_{3}$, and therefore, there must be an angle smaller than or equal to $\frac{1}{3} \cdot 108^{\circ}=36^{\circ}$, which is formed by a triple chosen from the given 5 points, and this proves that $36^{\circ}$ is the desired maximum.
+
+## Problem 3.
+
+Solution: Since $\angle B_{1} B B_{2}=90^{\circ}$, the circle having $B_{1} B_{2}$ as its diameter goes through the points $B, B_{1}, B_{2}$. From $B_{1} A: B_{1} C=B_{2} A: B_{2} C=B A: B C$, it follows that this circle is the Apolonius circle with the ratio of the distances from the points $A$ and $C$ being $B A: B C$. Since the point $P$ lies on this circle, we have
+
+$$
+P A: P C=B A: B C=\sin C: \sin A,
+$$
+
+from which it follows that $P A \sin A=P C \sin C$. Similarly, we have $P A \sin A=$ $P B \sin B$, and therefore, $P A \sin A=P B \sin B=P C \sin C$.
+
+Let us denote by $D, E, F$ the foot of the perpendicular line drawn from $P$ to the line segment $B C, C A$ and $A B$, respectively. Since the points $E, F$ lie on a circle having $P A$ as its diameter, we have by the law of sines $E F=P A \sin A$. Similarly, we have $F D=P B \sin B$ and $D E=P C \sin C$. Consequently, we conclude that $D E F$ is an equilateral triangle. Furthermore, we have $\angle C P E=\angle C D E$, since the quadrilateral $C D P E$ is cyclic. Similarly, we have $\angle F P B=\angle F D B$. Putting these together, we get
+
+$$
+\begin{aligned}
+\angle B P C & =360^{\circ}-(\angle C P E+\angle F P B+\angle E P F) \\
+& =360^{\circ}-\left\{(\angle C D E+\angle F D B)+\left(180^{\circ}-\angle F A E\right)\right\} \\
+& =360^{\circ}-\left(120^{\circ}+150^{\circ}\right)=90^{\circ},
+\end{aligned}
+$$
+
+which proves the assertion of the problem.
+Alternate Solution: Let $O$ be the midpoint of the line segment $B_{1} B_{2}$. Then the points $B$ and $P$ lie on the circle with center at $O$ and going through the point $B_{1}$. From
+
+$$
+\angle O B C=\angle O B B_{1}-\angle C B B_{1}=\angle O B_{1} B-\angle B_{1} B A=\angle B A C
+$$
+
+it follows that the triangles $O C D$ and $O B A$ are similar, and therefore we have that $O C \cdot O A=O B^{2}=O P^{2}$. Thus we conclude that the triangles $O C P$ and $O P A$ are similar, and therefore, we have $\angle O P C=\angle P A C$. Using this fact, we obtain
+
+$$
+\begin{gathered}
+\angle P B C-\angle P B A=\left(\angle B_{1} B C+\angle P B B_{1}\right)-\left(\angle A B B_{1}-\angle P B B_{1}\right) \\
+=2 \angle P B B_{1}=\angle P O B_{1}=\angle P C A-\angle O P C \\
+=\angle P C A-\angle P A C,
+\end{gathered}
+$$
+
+from which we conclude that $\angle P A C+\angle P B C=\angle P B A+\angle P C A$. Similarly, we get $\angle P A B+\angle P C B=\angle P B A+\angle P C A$. Putting these facts together and taking into account the fact that
+
+$$
+(\angle P A C+\angle P B C)+(\angle P A B+\angle P C B)+(\angle P B A+\angle P C A)=180^{\circ}
+$$
+
+we conclude that $\angle P B A+\angle P C A=60^{\circ}$, and finally that
+$\angle B P C=(\angle P B A+\angle P A B)+(\angle P C A+\angle P A C)=\angle B A C+(\angle P B A+\angle P C A)=90^{\circ}$, proving the assertion of the problem.
+
+## Problem 4.
+
+Solution: We will show that the desired maximum value for $m$ is $n(n-1)$.
+First, let us show that $m \leq n(n-1)$ always holds for any sequence $P_{0}, P_{1}, \cdots, P_{m+1}$ satisfying the conditions of the problem.
+
+Call a point a turning point if it coincides with $P_{i}$ for some $i$ with $1 \leq i \leq m$. Let us say also that 2 points $\{P, Q\}$ are adjacent if $\{P, Q\}=\left\{P_{i-1}, P_{i}\right\}$ for some $i$ with $1 \leq i \leq m$, and vertically adjacent if, in addition, $P Q$ is parallel to the $y$-axis.
+
+Any turning point is vertically adjacent to exactly one other turning point. Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of "vertical adjacency". Thus we can conclude that if we fix $k \in\{1,2, \cdots, n\}$, the number of turning points having the $x$-coordinate $k$ must be even, and hence it is less than or equal to $n-1$. Therefore, altogether there are less than or equal to $n(n-1)$ turning points, and this shows that $m \leq n(n-1)$ must be satisfied.
+
+It remains now to show that for any positive odd number $n$ one can choose a sequence for which $m=n(n-1)$. We will show this by using the mathematical induction on $n$. For $n=1$, this is clear. For $n=3$, choose
+
+$$
+\begin{array}{llll}
+P_{0}=(0,1), & P_{1}=(1,1), & P_{2}=(1,2), & P_{3}=(2,2), \\
+P_{4}=(2,1), & P_{5}=(3,1), & P_{6}=(3,3), & P_{7}=(4,3) .
+\end{array}
+$$
+
+It is easy to see that these points satisfy the requirements (See fig. 1 below).
+
+## figure 1
+
+Let $n$ be an odd integer $\geq 5$, and suppose there exists a sequence satisfying the desired conditions for $n-4$. Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram (fig. 2), where the configuration inside of the dotted square is given by the induction hypothesis:
+figure 2
+By the induction hypothesis, there are exactly $(n-4)(n-5)$ turning points for the configuration inside of the dotted square in the figure 2 above, and all of the lattice points in the figure 2 lying outside of the dotted square except for the 4 points $(n, 2),(n-1, n-2),(2,3),(1, n-1)$ are turning points. Therefore, the total
+number of turning points in this configuration is
+
+$$
+(n-4)(n-5)+\left(n^{2}-(n-4)^{2}-4\right)=n(n-1)
+$$
+
+showing that for this $n$ there exists a sequence satisfying the desired properties, and thus completing the induction process.
+
+## Problem 5.
+
+Solution: By substituting $x=1$ and $y=1$ into the given identity we obtain $f(f(1))=f(1)$. Next, by substituting $x=1$ and $y=f(1)$ into the given identity and using $f(f(1))=f(1)$, we get $f(1)^{2}=f(1)$, from which we conclude that either $f(1)=0$ or $f(1)=1$. But if $f(1)=1$, then substituting $y=1$ into the given identity, we get $f(x)=x$ for all $x$, which contradicts the condition (1). Therefore, we must have $f(1)=0$.
+
+By substituting $x=1$ into the given identity and using the fact $f(1)=0$, we then obtain $f(f(y))=2 f(y)$ for all $y$. This means that if a number $t$ belongs to the range of the function $f$, then so does $2 t$, and by induction we can conclude that for any non-negative integer $n, 2^{n} t$ belongs to the range of $f$ if $t$ does. Now suppose that there exists a real number $a$ for which $f(a)>0$, then for any non-negative integer $n 2^{n} f(a)$ must belong to the range of $f$, which leads to a contradiction to the condition (1). Thus we conclude that $f(x) \leq 0$ for any real number $x$.
+
+By substituting $\frac{x}{2}$ for $x$ and $f(y)$ for $y$ in the given identity and using the fact that $f(f(y))=2 f(y)$, we obtain
+
+$$
+f(x f(y))+f(y) f\left(\frac{x}{2}\right)=x f(y)+f\left(\frac{x}{2} f(y)\right)
+$$
+
+from which it follows that $x f(y)-f(x f(y))=f(y) f\left(\frac{x}{2}\right)-f\left(\frac{x}{2} f(y)\right) \geq 0$, since the values of $f$ are non-positive. Combining this with the given identity, we conclude that $y f(x) \geq f(x y)$. When $x>0$, by letting $y$ to be $\frac{1}{x}$ and using the fact that $f(1)=0$, we get $f(x) \geq 0$. Since $f(x) \leq 0$ for any real number $x$, we conclude that $f(x)=0$ for any positive real number $x$. We also have $f(0)=f(f(1))=2 f(1)=0$.
+
+If $f$ is identically 0 , i.e., $f(x)=0$ for all $x$, then clearly, this $f$ satisfies the given identity. If $f$ satisfies the given identity but not identically 0 , then there exists a $b<0$ for which $f(b)<0$. If we set $c=f(b)$, then we have $f(c)=f(f(b))=2 f(b)=$ $2 c$. For any negative real number $x$, we have $c x>0$ so that $f(c x)=f(2 c x)=0$, and by substituting $y=c$ into the given identity, we get
+
+$$
+f(2 c x)+c f(x)=2 c x+f(c x)
+$$
+
+from which it follows that $f(x)=2 x$ for any negative real $x$.
+We therefore conclude that if $f$ satisfies the given identity and is not identically 0 , then $f$ is of the form $f(x)=\left\{\begin{array}{ll}0 & \text { if } x \geq 0 \\ 2 x & \text { if } x<0 .\end{array}\right.$ Finally, let us show that the function $f$ of the form shown above does satisfy the conditions of the problem. Clearly, it satisfies the condition (1). We can check that $f$ satisfies the condition (2) as well by separating into the following 4 cases depending on whether $x, y$ are non-negative or negative.
+
+- when both $x$ and $y$ are non-negative, both sides of the given identity are 0 .
+- when $x$ is non-negative and $y$ is negative, we have $x y \leq 0$ and both sides of the given identity are $4 x y$.
+- when $x$ is negative and $y$ is non-negative, we have $x y \leq 0$ and both sides of the given identity are $2 x y$.
+- when both $x$ and $y$ are negative, we have $x y>0$ and both sides of the given identity are $2 x y$.
+Summarizing the arguments above, we conclude that the functions $f$ satisfying the conditions of the problem are
+
+$$
+f(x)=0 \quad \text { and } \quad f(x)= \begin{cases}0 & \text { if } x \geq 0 \\ 2 x & \text { if } x<0\end{cases}
+$$
+

APMO/md/en-apmo2012_sol.md

@@ -0,0 +1,181 @@
+# SOLUTIONS FOR 2012 APMO PROBLEMS 
+
+## Problem 1.
+
+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 $z=\triangle P C A$.
+
+From
+
+$$
+y: z=\triangle B C P: \triangle A C P=B F: A F=\triangle B P F: \triangle A P F=(x-1): 1
+$$
+
+follows 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 $(x+1) y=(y+1) z=$ $(z+1) x$.
+
+We 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 2.
+
+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 .
+
+| 0 | 1 | 0 |  |
+| :--- | :--- | :--- | :--- |
+| 1 | 1 | 1 |  |
+| 0 | 1 | 0 |  |
+|  |  |  |  |
+
+We 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 $x>y$ ).
+
+First 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 $a=c$.
+
+Similarly, 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
+
+$$
+C(1, b-1) \leq 1, \quad C(b+1, n) \leq 1, \quad 1 \leq b \leq n
+$$
+
+If 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
+
+$$
+\leq R(1, a-1)+R(a+1, n)+C(1, b-1)+C(b+1, n)+r \leq 5
+$$
+
+## Problem 3.
+
+## 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.
+
+Next, we consider the case where $p \geq 3$.
+Suppose $s$ is an integer satisfying $s \geq p$. If $s^{p} \leq p^{s}$ for such an $s$, then we have
+
+$$
+\begin{aligned}
+(s+1)^{p} & =s^{p}\left(1+\frac{1}{s}\right)^{p} \leq p^{s}\left(1+\frac{1}{p}\right)^{p} \\
+& =p^{s} \sum_{r=0}^{p}{ }_{p} C_{r} \frac{1}{p^{r}}<p^{s} \sum_{r=0}^{p} \frac{1}{r!} \\
+& \leq p^{s}\left(1+\sum_{r=1}^{p} \frac{1}{2^{r-1}}\right) \\
+& <p^{s}(1+2) \leq p^{s+1}
+\end{aligned}
+$$
+
+Thus 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 $p^{n} \leq n^{p}$.
+
+We 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, $n^{2 p} \equiv 1(\bmod (p+1))$.
+
+Now, 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
+
+$$
+1 \equiv n^{2 p}=\left(n^{e}\right)^{x} \cdot n^{y} \equiv n^{y} \quad(\bmod (p+1))
+$$
+
+which 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 $1,2, p, 2 p$.
+
+Now, 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.
+
+From $n^{2} \equiv 1(\bmod (p+1))$, we get
+
+$$
+-1 \equiv n^{p}=n^{\left(2 \cdot \frac{p-1}{2}+1\right)} \equiv n \quad(\bmod (p+1))
+$$
+
+which 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.
+
+It 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 $p$.
+
+Alternate 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
+
+$$
+n^{p}=n^{\left(2 \cdot \frac{p-1}{2}+1\right)} \equiv n \quad(\bmod q),
+$$
+
+and since $n^{p} \equiv-1(\bmod q)$, we have $n \equiv-1(\bmod q)$ as well.
+Now, 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.
+
+Suppose, 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
+
+$$
+\begin{gathered}
+n^{p}+1=(n+1)\left(n^{p-1}-n^{p-2}+\cdots-n+1\right) \quad \text { and } \\
+n^{p-1}-n^{p-2}+\cdots-n+1 \equiv(-1)^{p-1}-(-1)^{p-2}+\cdots-(-1)+1 \not \equiv 0 \quad(\bmod q)
+\end{gathered}
+$$
+
+we 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 4.
+
+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.
+
+Due 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
+
+$$
+\angle B C K=\angle A C K-\angle A C B=90^{\circ}-\angle A C B=\angle C B H
+$$
+
+and
+
+$$
+\angle C B K=\angle A B K-\angle A B C=90^{\circ}-\angle A B C=\angle B C H,
+$$
+
+from 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 $\angle A E M=\angle A E K=90^{\circ}$.
+
+From $\angle A E D=90^{\circ}=\angle A D M$, 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
+therefore, 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 5.
+
+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
+
+$$
+n-a_{i} a_{j} \geq n-\frac{a_{i}^{2}+a_{j}^{2}}{2} \geq n-\frac{n}{2}=\frac{n}{2}>0 .
+$$
+
+If 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.
+
+By multiplying by $n$ the both sides of the desired inequality we get the inequality:
+
+$$
+\sum_{1 \leq i<j \leq n} \frac{n}{n-a_{i} a_{j}} \leq \frac{n^{2}}{2}
+$$
+
+and 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:
+
+$$
+\sum_{1 \leq i<j \leq n} \frac{a_{i} a_{j}}{n-a_{i} a_{j}} \leq \frac{n}{2}
+$$
+
+We will show that this inequality (i) holds.
+If 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 $i$.
+
+Let 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
+
+$$
+\frac{a_{i} a_{j}}{n-a_{i} a_{j}} \leq \frac{a_{i} a_{j}}{n-\frac{a_{i}^{2}+a_{j}^{2}}{2}} \leq \frac{\left(\frac{a_{i}+a_{j}}{2}\right)^{2}}{n-\frac{a_{i}^{2}+a_{j}^{2}}{2}}=\frac{1}{2} \cdot \frac{\left(a_{i}+a_{j}\right)^{2}}{\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)}
+$$
+
+Since $n-a_{i}^{2}>0, n-a_{j}^{2}>0$, we also get from the Cauchy-Schwarz inequality that
+
+$$
+\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right)\left(\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)\right) \geq\left(a_{i}+a_{j}\right)^{2},
+$$
+
+from which it follows that
+
+$$
+\frac{\left(a_{i}+a_{j}\right)^{2}}{\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)} \leq\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right)
+$$
+
+holds. Combining the inequalities (ii) and (iii), we get
+
+$$
+\begin{aligned}
+\sum_{1 \leq i<j \leq n} \frac{a_{i} a_{j}}{n-a_{i} a_{j}} & \leq \frac{1}{2} \sum_{1 \leq i<j \leq n}\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right) \\
+& =\frac{1}{2} \sum_{i \neq j} \frac{a_{j}^{2}}{n-a_{i}^{2}} \\
+& =\frac{1}{2} \sum_{i=1}^{n} \frac{n-a_{i}^{2}}{n-a_{i}^{2}} \\
+& =\frac{n}{2}
+\end{aligned}
+$$
+
+which establishes the desired inequality (i).
+

APMO/md/en-apmo2013_sol.md

@@ -0,0 +1,142 @@
+# Solutions of APMO 2013 
+
+Problem 1. Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas.
+
+Solution. Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O N}{B D}=\frac{O A}{B A}$. Then $\frac{O M}{A E}=\frac{O N}{B D}$ or $B D \cdot O M=A E \cdot O N$.
+
+Denote by $S(\Phi)$ the area of the figure $\Phi$. So, we see that $S(O B D)=\frac{1}{2} B D \cdot O M=$ $\frac{1}{2} A E \cdot O N=S(O A E)$. Analogously, $S(O C D)=S(O A F)$ and $S(O C E)=S(O B F)$.
+
+Alternative solution. Let $R$ be the circumradius of triangle $A B C$, and as usual write $A, B, C$ for angles $\angle C A B, \angle A B C, \angle B C A$ respectively, and $a, b, c$ for sides $B C, C A, A B$ respectively. Then the area of triangle $O C D$ is
+
+$$
+S(O C D)=\frac{1}{2} \cdot O C \cdot C D \cdot \sin (\angle O C D)=\frac{1}{2} R \cdot C D \cdot \sin (\angle O C D)
+$$
+
+Now $C D=b \cos C$, and
+
+$$
+\angle O C D=\frac{180^{\circ}-2 A}{2}=90^{\circ}-A
+$$
+
+(since triangle $O B C$ is isosceles, and $\angle B O C=2 A$ ). So
+
+$$
+S(O C D)=\frac{1}{2} R b \cos C \sin \left(90^{\circ}-A\right)=\frac{1}{2} R b \cos C \cos A
+$$
+
+A similar calculation gives
+
+$$
+\begin{aligned}
+S(O A F) & =\frac{1}{2} O A \cdot A F \cdot \sin (\angle O A F) \\
+& =\frac{1}{2} R \cdot(b \cos A) \sin \left(90^{\circ}-C\right) \\
+& =\frac{1}{2} R b \cos A \cos C
+\end{aligned}
+$$
+
+so $O C D$ and $O A F$ have the same area. In the same way we find that $O B D$ and $O A E$ have the same area, as do $O C E$ and $O B F$.
+
+Problem 2. Determine all positive integers $n$ for which $\frac{n^{2}+1}{[\sqrt{n}]^{2}+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
+
+Solution. We will show that there are no positive integers $n$ satisfying the condition of the problem.
+
+Let $m=[\sqrt{n}]$ and $a=n-m^{2}$. We have $m \geq 1$ since $n \geq 1$. From $n^{2}+1=\left(m^{2}+a\right)^{2}+1 \equiv$ $(a-2)^{2}+1\left(\bmod \left(m^{2}+2\right)\right)$, it follows that the condition of the problem is equivalent to the fact that $(a-2)^{2}+1$ is divisible by $m^{2}+2$. Since we have
+
+$$
+0<(a-2)^{2}+1 \leq \max \left\{2^{2},(2 m-2)^{2}\right\}+1 \leq 4 m^{2}+1<4\left(m^{2}+2\right)
+$$
+
+we see that $(a-2)^{2}+1=k\left(m^{2}+2\right)$ must hold with $k=1,2$ or 3 . We will show that none of these can occur.
+
+Case 1. When $k=1$. We get $(a-2)^{2}-m^{2}=1$, and this implies that $a-2= \pm 1, m=0$ must hold, but this contradicts with fact $m \geq 1$.
+
+Case 2. When $k=2$. We have $(a-2)^{2}+1=2\left(m^{2}+2\right)$ in this case, but any perfect square is congruent to $0,1,4 \bmod 8$, and therefore, we have $(a-2)^{2}+1 \equiv 1,2,5(\bmod 8)$, while $2\left(m^{2}+2\right) \equiv 4,6(\bmod 8)$. Thus, this case cannot occur either.
+
+Case 3. When $k=3$. We have $(a-2)^{2}+1=3\left(m^{2}+2\right)$ in this case. Since any perfect square is congruent to 0 or $1 \bmod 3$, we have $(a-2)^{2}+1 \equiv 1,2(\bmod 3)$, while $3\left(m^{2}+2\right) \equiv 0$ $(\bmod 3)$, which shows that this case cannot occur either.
+
+Problem 3. For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by
+
+$$
+X_{n}=\sum_{i=1}^{k}\left[a_{i} n+b_{i}\right] \quad(n=1,2, \ldots)
+$$
+
+If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.
+
+Solution. Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,
+
+$$
+a_{i} n+b_{i}-1<\left[a_{i} n+b_{i}\right] \leq a_{i} n+b_{i}
+$$
+
+we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left\{X_{n}\right\}$ is an arithmetic progression with the common difference $d$, then we have $n d=X_{n+1}-X_{1}$ and $A+B-k<X_{1} \leq A+B$ Combining with the inequalities obtained above, we get
+
+$$
+A(n+1)+B-k<n d+X_{1}<A(n+1)+B,
+$$
+
+or
+
+$$
+A n-k \leq A n+\left(A+B-X_{1}\right)-k<n d<A n+\left(A+B-X_{1}\right)<A n+k,
+$$
+
+from which we conclude that $|A-d|<\frac{k}{n}$ must hold. Since this inequality holds for any positive integer $n$, we must have $A=d$. Since $\left\{X_{n}\right\}$ is a sequence of integers, $d$ must be an integer also, and thus we conclude that $A$ is also an integer.
+
+Problem 4. Let $a$ and $b$ be positive integers, and let $A$ and $B$ be finite sets of integers satisfying:
+(i) $A$ and $B$ are disjoint;
+(ii) if an integer $i$ belongs either to $A$ or to $B$, then $i+a$ belongs to $A$ or $i-b$ belongs to $B$.
+
+Prove that $a|A|=b|B|$. (Here $|X|$ denotes the number of elements in the set $X$.)
+Solution. Let $A^{*}=\{n-a: n \in A\}$ and $B^{*}=\{n+b: n \in B\}$. Then, by (ii), $A \cup B \subseteq A^{*} \cup B^{*}$ and by (i),
+
+$$
+|A \cup B| \leq\left|A^{*} \cup B^{*}\right| \leq\left|A^{*}\right|+\left|B^{*}\right|=|A|+|B|=|A \cup B|
+$$
+
+Thus, $A \cup B=A^{*} \cup B^{*}$ and $A^{*}$ and $B^{*}$ have no element in common. For each finite set $X$ of integers, let $\sum(X)=\sum_{x \in X} x$. Then
+
+$$
+\begin{aligned}
+\sum(A)+\sum(B) & =\sum(A \cup B) \\
+& =\sum\left(A^{*} \cup B^{*}\right)=\sum\left(A^{*}\right)+\sum\left(B^{*}\right) \\
+& =\sum(A)-a|A|+\sum(B)+b|B|
+\end{aligned}
+$$
+
+which implies $a|A|=b|B|$.
+Alternative solution. Let us construct a directed graph whose vertices are labelled by the members of $A \cup B$ and such that there is an edge from $i$ to $j$ iff $j \in A$ and $j=i+a$ or $j \in B$ and $j=i-b$. From (ii), each vertex has out-degree $\geq 1$ and, from (i), each vertex has in-degree $\leq 1$. Since the sum of the out-degrees equals the sum of the in-degrees, each vertex has in-degree and out-degree equal to 1. This is only possible if the graph is the union of disjoint cycles, say $G_{1}, G_{2}, \ldots, G_{n}$. Let $\left|A_{k}\right|$ be the number of elements of $A$ in $G_{k}$ and $\left|B_{k}\right|$ be the number of elements of $B$ in $G_{k}$. The cycle $G_{k}$ will involve increasing vertex labels by $a$ a total of $\left|A_{k}\right|$ times and decreasing them by $b$ a total of $\left|B_{k}\right|$ times. Since it is a cycle, we have $a\left|A_{k}\right|=b\left|B_{k}\right|$. Summing over all cycles gives the result.
+
+Problem 5. Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B, E, R$ are collinear.
+
+Solution. To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.
+
+Since $\triangle P A D$ is similar to $\triangle P D C$ and $\triangle P A B$ is similar to $\triangle P B C$, we have $A D / D C=$ $P A / P D=P A / P B=A B / B C$. Hence, $A B \cdot D C=B C \cdot A D$. By Ptolemy's theorem, $A B \cdot D C=B C \cdot A D=\frac{1}{2} C A \cdot D B$. Similarly $C A \cdot E D=C E \cdot A D=\frac{1}{2} A E \cdot D C$.
+
+Thus
+
+$$
+\frac{D B}{A B}=\frac{2 D C}{C A}
+$$
+
+and
+
+$$
+\frac{D C}{C A}=\frac{2 E D}{A E}
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_ec23cb88754609477308g-4.jpg?height=869&width=1161&top_left_y=235&top_left_x=512)
+
+Since the triangles $R D C$ and $R C A$ are similar, we have $\frac{R D}{R C}=\frac{D C}{C A}=\frac{R C}{R A}$. Thus using (4)
+
+$$
+\frac{R D}{R A}=\frac{R D \cdot R A}{R A^{2}}=\left(\frac{R C}{R A}\right)^{2}=\left(\frac{D C}{C A}\right)^{2}=\left(\frac{2 E D}{A E}\right)^{2}
+$$
+
+Using the similar triangles $A B R^{\prime}$ and $E D R^{\prime}$, we have $R^{\prime} D / R^{\prime} B=E D / A B$. Using the similar triangles $D B R^{\prime}$ and $E A R^{\prime}$ we have $R^{\prime} A / R^{\prime} B=E A / D B$. Thus using (3) and (4),
+
+$$
+\frac{R^{\prime} D}{R^{\prime} A}=\frac{E D \cdot D B}{E A \cdot A B}=\left(\frac{2 E D}{A E}\right)^{2}
+$$
+
+It follows from (5) and (6) that $R=R^{\prime}$.
+

APMO/md/en-apmo2014_sol.md

@@ -0,0 +1,109 @@
+# Solutions of APMO 2014 
+
+Problem 1. For a positive integer $m$ denote by $S(m)$ and $P(m)$ the sum and product, respectively, of the digits of $m$. Show that for each positive integer $n$, there exist positive integers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying the following conditions:
+
+$$
+S\left(a_{1}\right)<S\left(a_{2}\right)<\cdots<S\left(a_{n}\right) \text { and } S\left(a_{i}\right)=P\left(a_{i+1}\right) \quad(i=1,2, \ldots, n)
+$$
+
+(We let $\left.a_{n+1}=a_{1}.\right)$ (Problem Committee of the Japan Mathematical Olympiad Foundation)
+Solution. Let $k$ be a sufficiently large positive integer. Choose for each $i=2,3, \ldots, n$, $a_{i}$ to be a positive integer among whose digits the number 2 appears exactly $k+i-2$ times and the number 1 appears exactly $2^{k+i-1}-2(k+i-2)$ times, and nothing else. Then, we have $S\left(a_{i}\right)=2^{k+i-1}$ and $P\left(a_{i}\right)=2^{k+i-2}$ for each $i, 2 \leq i \leq n$. Then, we let $a_{1}$ be a positive integer among whose digits the number 2 appears exactly $k+n-1$ times and the number 1 appears exactly $2^{k}-2(k+n-1)$ times, and nothing else. Then, we see that $a_{1}$ satisfies $S\left(a_{1}\right)=2^{k}$ and $P\left(a_{1}\right)=2^{k+n-1}$. Such a choice of $a_{1}$ is possible if we take $k$ to be large enough to satisfy $2^{k}>2(k+n-1)$ and we see that the numbers $a_{1}, \ldots, a_{n}$ chosen this way satisfy the given requirements.
+
+Problem 2. Let $S=\{1,2, \ldots, 2014\}$. For each non-empty subset $T \subseteq S$, one of its members is chosen as its representative. Find the number of ways to assign representatives to all non-empty subsets of $S$ so that if a subset $D \subseteq S$ is a disjoint union of non-empty subsets $A, B, C \subseteq S$, then the representative of $D$ is also the representative of at least one of $A, B, C$. (Warut Suksompong, Thailand)
+
+Solution. Answer: 108 - 2014!.
+For any subset $X$ let $r(X)$ denotes the representative of $X$. Suppose that $x_{1}=r(S)$. First, we prove the following fact:
+
+$$
+\text { If } x_{1} \in X \text { and } X \subseteq S \text {, then } x_{1}=r(X) \text {. }
+$$
+
+If $|X| \leq 2012$, then we can write $S$ as a disjoint union of $X$ and two other subsets of $S$, which gives that $x_{1}=r(X)$. If $|X|=2013$, then let $y \in X$ and $y \neq x_{1}$. We can write $X$ as a disjoint union of $\left\{x_{1}, y\right\}$ and two other subsets. We already proved that $r\left(\left\{x_{1}, y\right\}\right)=x_{1}$ (since $\left|\left\{x_{1}, y\right\}\right|=2<2012$ ) and it follows that $y \neq r(X)$ for every $y \in X$ except $x_{1}$. We have proved the fact.
+
+Note that this fact is true and can be proved similarly, if the ground set $S$ would contain at least 5 elements.
+
+There are 2014 ways to choose $x_{1}=r(S)$ and for $x_{1} \in X \subseteq S$ we have $r(X)=x_{1}$. Let $S_{1}=S \backslash\left\{x_{1}\right\}$. Analogously, we can state that there are 2013 ways to choose $x_{2}=r\left(S_{1}\right)$ and for $x_{2} \in X \subseteq S_{1}$ we have $r(X)=x_{2}$. Proceeding similarly (or by induction), there are $2014 \cdot 2013 \cdots 5$ ways to choose $x_{1}, x_{2}, \ldots, x_{2010} \in S$ so that for all $i=1,2 \ldots, 2010$, $x_{i}=r(X)$ for each $X \subseteq S \backslash\left\{x_{1}, \ldots, x_{i-1}\right\}$ and $x_{i} \in X$.
+
+We are now left with four elements $Y=\left\{y_{1}, y_{2}, y_{3}, y_{4}\right\}$. There are 4 ways to choose $r(Y)$. Suppose that $y_{1}=r(Y)$. Then we clearly have $y_{1}=r\left(\left\{y_{1}, y_{2}\right\}\right)=r\left(\left\{y_{1}, y_{3}\right\}\right)=r\left(\left\{y_{1}, y_{4}\right\}\right)$. The only subsets whose representative has not been assigned yet are $\left\{y_{1}, y_{2}, y_{3}\right\},\left\{y_{1}, y_{2}, y_{4}\right\}$, $\left\{y_{1}, y_{3}, y_{4}\right\},\left\{y_{2}, y_{3}, y_{4}\right\},\left\{y_{2}, y_{3}\right\},\left\{y_{2}, y_{4}\right\},\left\{y_{3}, y_{4}\right\}$. These subsets can be assigned in any way, hence giving $3^{4} \cdot 2^{3}$ more choices.
+
+In conclusion, the total number of assignments is $2014 \cdot 2013 \cdots 4 \cdot 3^{4} \cdot 2^{3}=108 \cdot 2014$ !.
+Problem 3. Find all positive integers $n$ such that for any integer $k$ there exists an integer $a$ for which $a^{3}+a-k$ is divisible by $n$. (Warut Suksompong, Thailand)
+
+Solution. Answer: All integers $n=3^{b}$, where $b$ is a nonnegative integer.
+We are looking for integers $n$ such that the set $A=\left\{a^{3}+a \mid a \in \mathbf{Z}\right\}$ is a complete residue system by modulo $n$. Let us call this property by $\left(^{*}\right)$. It is not hard to see that $n=1$ satisfies $\left({ }^{*}\right)$ and $n=2$ does not.
+
+If $a \equiv b(\bmod n)$, then $a^{3}+a \equiv b^{3}+b(\bmod n)$. So $n$ satisfies $\left(^{*}\right)$ iff there are no $a, b \in\{0, \ldots, n-1\}$ with $a \neq b$ and $a^{3}+a \equiv b^{3}+b(\bmod n)$.
+
+First, let us prove that $3^{j}$ satisfies $\left(^{*}\right)$ for all $j \geq 1$. Suppose that $a^{3}+a \equiv b^{3}+b\left(\bmod 3^{j}\right)$ for $a \neq b$. Then $(a-b)\left(a^{2}+a b+b^{2}+1\right) \equiv 0\left(\bmod 3^{j}\right)$. We can easily check mod 3 that $a^{2}+a b+b^{2}+1$ is not divisible by 3 .
+
+Next note that if $A$ is not a complete residue system modulo integer $r$, then it is also not a complete residue system modulo any multiple of $r$. Hence it remains to prove that any prime $p>3$ does not satisfy (*).
+
+If $p \equiv 1(\bmod 4)$, there exists $b$ such that $b^{2} \equiv-1(\bmod p)$. We then take $a=0$ to obtain the congruence $a^{3}+a \equiv b^{3}+b(\bmod p)$.
+
+Suppose now that $p \equiv 3(\bmod 4)$. We will prove that there are integers $a, b \not \equiv 0(\bmod p)$ such that $a^{2}+a b+b^{2} \equiv-1(\bmod p)$. Note that we may suppose that $a \not \equiv b(\bmod p)$, since otherwise if $a \equiv b(\bmod p)$ satisfies $a^{2}+a b+b^{2}+1 \equiv 0(\bmod p)$, then $(2 a)^{2}+(2 a)(-a)+$ $a^{2}+1 \equiv 0(\bmod p)$ and $2 a \not \equiv-a(\bmod p)$. Letting $c$ be the inverse of $b$ modulo $p$ (i.e. $b c \equiv 1(\bmod p)$ ), the relation is equivalent to $(a c)^{2}+a c+1 \equiv-c^{2}(\bmod p)$. Note that $-c^{2}$ can take on the values of all non-quadratic residues modulo $p$. If we can find an integer $x$ such that $x^{2}+x+1$ is a non-quadratic residue modulo $p$, the values of $a$ and $c$ will follow immediately. Hence we focus on this latter task.
+
+Note that if $x, y \in\{0, \ldots, p-1\}=B$, then $x^{2}+x+1 \equiv y^{2}+y+1(\bmod p)$ iff $p$ divides $x+y+1$. We can deduce that $x^{2}+x+1$ takes on $(p+1) / 2$ values as $x$ varies in $B$. Since there are $(p-1) / 2$ non-quadratic residues modulo $p$, the $(p+1) / 2$ values that $x^{2}+x+1$ take on must be 0 and all the quadratic residues.
+
+Let $C$ be the set of quadratic residues modulo $p$ and 0 , and let $y \in C$. Suppose that $y \equiv z^{2}(\bmod p)$ and let $z \equiv 2 w+1(\bmod p)$ (we can always choose such $\left.w\right)$. Then $y+3 \equiv$ $4\left(w^{2}+w+1\right)(\bmod p)$. From the previous paragraph, we know that $4\left(w^{2}+w+1\right) \in C$. This means that $y \in C \Longrightarrow y+3 \in C$. Unless $p=3$, the relation implies that all elements of $B$ are in $C$, a contradiction. This concludes the proof.
+
+Problem 4. Let $n$ and $b$ be positive integers. We say $n$ is $b$-discerning if there exists a set consisting of $n$ different positive integers less than $b$ that has no two different subsets $U$ and $V$ such that the sum of all elements in $U$ equals the sum of all elements in $V$.
+(a) Prove that 8 is a 100 -discerning.
+(b) Prove that 9 is not 100-discerning.
+(Senior Problems Committee of the Australian Mathematical Olympiad Committee)
+
+## Solution.
+
+(a) Take $S=\{3,6,12,24,48,95,96,97\}$, i.e.
+
+$$
+S=\left\{3 \cdot 2^{k}: 0 \leq k \leq 5\right\} \cup\left\{3 \cdot 2^{5}-1,3 \cdot 2^{5}+1\right\}
+$$
+
+As $k$ ranges between 0 to 5 , the sums obtained from the numbers $3 \cdot 2^{k}$ are $3 t$, where $1 \leq t \leq 63$. These are 63 numbers that are divisible by 3 and are at most $3 \cdot 63=189$.
+
+Sums of elements of $S$ are also the numbers $95+97=192$ and all the numbers that are sums of 192 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all divisible by 3 and at least equal to 192. In addition, sums of elements of $S$ are the numbers 95 and all the numbers that are sums of 95 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $-1 \bmod$ 3.
+
+Finally, sums of elements of $S$ are the numbers 97 and all the numbers that are sums of 97 and sums obtained from the numbers $3 \cdot 2^{k}$ with $0 \leq k \leq 5$. These are 64 numbers that are all congruent to $1 \bmod 3$.
+
+Hence there are at least $63+64+64+64=255$ different sums from elements of $S$. On the other hand, $S$ has $2^{8}-1=255$ non-empty subsets. Therefore $S$ has no two different subsets with equal sums of elements. Therefore, 8 is 100 -discerning.
+(b) Suppose that 9 is 100 -discerning. Then there is a set $S=\left\{s_{1}, \ldots, s_{9}\right\}, s_{i}<100$ that has no two different subsets with equal sums of elements. Assume that $0<s_{1}<\cdots<s_{9}<$ 100.
+
+Let $X$ be the set of all subsets of $S$ having at least 3 and at most 6 elements and let $Y$ be the set of all subsets of $S$ having exactly 2 or 3 or 4 elements greater than $s_{3}$.
+
+The set $X$ consists of
+
+$$
+\binom{9}{3}+\binom{9}{4}+\binom{9}{5}+\binom{9}{6}=84+126+126+84=420
+$$
+
+subsets of $S$. The set in $X$ with the largest sums of elements is $\left\{s_{4}, \ldots, s_{9}\right\}$ and the smallest sums is in $\left\{s_{1}, s_{2}, s_{3}\right\}$. Thus the sum of the elements of each of the 420 sets in $X$ is at least $s_{1}+s_{2}+s_{3}$ and at most $s_{4}+\cdots+s_{9}$, which is one of $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1$ integers. From the pigeonhole principle it follows that $\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right)+1 \geq 420$, i.e.,
+
+$$
+\left(s_{4}+\cdots+s_{9}\right)-\left(s_{1}+s_{2}+s_{3}\right) \geq 419
+$$
+
+Now let us calculate the number of subsets in $Y$. Observe that $\left\{s_{4}, \ldots, s_{9}\right\}$ has $\binom{6}{2}$ 2-element subsets, $\binom{6}{3}$ 3-element subsets and $\binom{6}{4}$ 4-element subsets, while $\left\{s_{1}, s_{2}, s_{3}\right\}$ has exactly 8 subsets. Hence the number of subsets of $S$ in $Y$ equals
+
+$$
+8\left(\binom{6}{2}+\binom{6}{3}+\binom{6}{4}\right)=8(15+20+15)=400
+$$
+
+The set in $Y$ with the largest sum of elements is $\left\{s_{1}, s_{2}, s_{3}, s_{6}, s_{7}, s_{8}, s_{9}\right\}$ and the smallest sum is in $\left\{s_{4}, s_{5}\right\}$. Again, by the pigeonhole principle it follows that $\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+\right.$ $\left.s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right)+1 \geq 400$, i.e.,
+
+$$
+\left(s_{1}+s_{2}+s_{3}+s_{6}+s_{7}+s_{8}+s_{9}\right)-\left(s_{4}+s_{5}\right) \geq 399
+$$
+
+Adding (1) and (2) yields $2\left(s_{6}+s_{7}+s_{8}+s_{9}\right) \geq 818$, so that $s_{9}+98+97+96 \geq$ $s_{9}+s_{8}+s_{7}+s_{6} \geq 409$, i.e. $s_{9} \geq 118$, a contradiction with $s_{9}<100$. Therefore, 9 is not 100-discerning.
+
+Problem 5. Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the $\operatorname{arc} A B$ of circle $\omega$ ( $M$ lies inside $\Omega$ ). A chord $M P$ of circle $\omega$ intersects $\Omega$ at $Q(Q$ lies inside $\omega)$. Let $\ell_{P}$ be the tangent line to $\omega$ at $P$, and let $\ell_{Q}$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_{P}, \ell_{Q}$, and $A B$ is tangent to $\Omega$. (Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan)
+
+Solution. Denote $X=A B \cap \ell_{P}, Y=A B \cap \ell_{Q}$, and $Z=\ell_{P} \cap \ell_{Q}$. Without loss of generality we have $A X<B X$. Let $F=M P \cap A B$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_fa403cf18214215ea344g-4.jpg?height=1169&width=1161&top_left_y=630&top_left_x=487)
+
+Denote by $R$ the second point of intersection of $P Q$ and $\Omega$; by $S$ the point of $\Omega$ such that $S R \| A B$; and by $T$ the point of $\Omega$ such that $R T \| \ell_{P}$. Since $M$ is the midpoint of arc $A B$, the tangent $\ell_{M}$ at $M$ to $\omega$ is parallel to $A B$, so $\angle(A B, P M)=\angle\left(P M, \ell_{P}\right)$. Therefore we have $\angle P R T=\angle M P X=\angle P F X=\angle P R S$. Thus the point $Q$ is the midpoint of the $\operatorname{arc} T Q S$ of $\Omega$, hence $S T \| \ell_{Q}$. So the corresponding sides of the triangles $R S T$ and $X Y Z$ are parallel, and there exist a homothety $h$ mapping $R S T$ to $X Y Z$.
+
+Let $D$ be the second point of intersection of $X R$ and $\Omega$. We claim that $D$ is the center of the homothety $h$; since $D \in \Omega$, this implies that the circumcircles of triangles $R S T$ and $X Y Z$ are tangent, as required. So, it remains to prove this claim. In order to do this, it suffices to show that $D \in S Y$.
+
+By $\angle P F X=\angle X P F$ we have $X F^{2}=X P^{2}=X A \cdot X B=X D \cdot X R$. Therefore, $\frac{X F}{X D}=\frac{X R}{X F}$, so the triangles $X D F$ and $X F R$ are similar, hence $\angle D F X=\angle X R F=\angle D R Q=$ $\angle D Q Y$; thus the points $D, Y, Q$, and $F$ are concyclic. It follows that $\angle Y D Q=\angle Y F Q=$ $\angle S R Q=180^{\circ}-\angle S D Q$ which means exactly that the points $Y, D$, and $S$ are collinear, with $D$ between $S$ and $Y$.
+

APMO/md/en-apmo2015_sol.md

@@ -0,0 +1,247 @@
+# Solutions of APMO 2015 
+
+Problem 1. Let $A B C$ be a triangle, and let $D$ be a point on side $B C$. A line through $D$ intersects side $A B$ at $X$ and ray $A C$ at $Y$. The circumcircle of triangle $B X D$ intersects the circumcircle $\omega$ of triangle $A B C$ again at point $Z \neq B$. The lines $Z D$ and $Z Y$ intersect $\omega$ again at $V$ and $W$, respectively. Prove that $A B=V W$.
+
+Solution. Suppose $X Y$ intersects $\omega$ at points $P$ and $Q$, where $Q$ lies between $X$ and $Y$. We will show that $V$ and $W$ are the reflections of $A$ and $B$ with respect to the perpendicular bisector of $P Q$. From this, it follows that $A V W B$ is an isosceles trapezoid and hence $A B=V W$.
+
+First, note that
+
+$$
+\angle B Z D=\angle A X Y=\angle A P Q+\angle B A P=\angle A P Q+\angle B Z P,
+$$
+
+so $\angle A P Q=\angle P Z V=\angle P Q V$, and hence $V$ is the reflection of $A$ with respect to the perpendicular bisector of $P Q$.
+
+Now, suppose $W^{\prime}$ is the reflection of $B$ with respect to the perpendicular bisector of $P Q$, and let $Z^{\prime}$ be the intersection of $Y W^{\prime}$ and $\omega$. It suffices to show that $B, X, D, Z^{\prime}$ are concyclic. Note that
+
+$$
+\angle Y D C=\angle P D B=\angle P C B+\angle Q P C=\angle W^{\prime} P Q+\angle Q P C=\angle W^{\prime} P C=\angle Y Z^{\prime} C .
+$$
+
+So $D, C, Y, Z^{\prime}$ are concyclic. Next, $\angle B Z^{\prime} D=\angle C Z^{\prime} B-\angle C Z^{\prime} D=180^{\circ}-\angle B X D$ and due to the previous concyclicity we are done.
+
+Alternative solution 1. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
+
+Using cyclic quadrilaterals $A B Z C$ and $Z D C Y$ in turn, we have $\angle A Z B=\angle A C B=\angle W Z V$ (or $180^{\circ}-\angle W Z V$ if $Z$ lies between $W$ and $C$ ).
+
+So $A B=V W$ because they subtend equal (or supplementary) angles in $\omega$.
+Alternative solution 2. Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle Z D Y=\angle Z B A=\angle Z C Y$. So $Z D C Y$ is cyclic.
+
+Using cyclic quadrilaterals $B X D Z$ and $A B Z V$ in turn, we have $\angle D X A=\angle V Z B=$ $180^{\circ}-B A V$. So $X D \| A V$.
+
+Using cyclic quadrilaterals $Z D C Y$ and $B C W Z$ in turn, we have $\angle Y D C=\angle Y Z C=$ $\angle W B C$. So $X D \| B W$.
+
+Hence $B W \| A V$ which implies that $A V W B$ is an isosceles trapezium with $A B=V W$.
+Problem 2. Let $S=\{2,3,4, \ldots\}$ denote the set of integers that are greater than or equal to 2 . Does there exist a function $f: S \rightarrow S$ such that
+
+$$
+f(a) f(b)=f\left(a^{2} b^{2}\right) \text { for all } a, b \in S \text { with } a \neq b ?
+$$
+
+Solution. We prove that there is no such function. For arbitrary elements $a$ and $b$ of $S$, choose an integer $c$ that is greater than both of them. Since $b c>a$ and $c>b$, we have
+
+$$
+f\left(a^{4} b^{4} c^{4}\right)=f\left(a^{2}\right) f\left(b^{2} c^{2}\right)=f\left(a^{2}\right) f(b) f(c)
+$$
+
+Furthermore, since $a c>b$ and $c>a$, we have
+
+$$
+f\left(a^{4} b^{4} c^{4}\right)=f\left(b^{2}\right) f\left(a^{2} c^{2}\right)=f\left(b^{2}\right) f(a) f(c)
+$$
+
+Comparing these two equations, we find that for all elements $a$ and $b$ of $S$,
+
+$$
+f\left(a^{2}\right) f(b)=f\left(b^{2}\right) f(a) \quad \Longrightarrow \quad \frac{f\left(a^{2}\right)}{f(a)}=\frac{f\left(b^{2}\right)}{f(b)} .
+$$
+
+It follows that there exists a positive rational number $k$ such that
+
+$$
+f\left(a^{2}\right)=k f(a), \quad \text { for all } a \in S
+$$
+
+Substituting this into the functional equation yields
+
+$$
+f(a b)=\frac{f(a) f(b)}{k}, \quad \text { for all } a, b \in S \text { with } a \neq b .
+$$
+
+Now combine the functional equation with equations (1) and (2) to obtain
+
+$$
+f(a) f\left(a^{2}\right)=f\left(a^{6}\right)=\frac{f(a) f\left(a^{5}\right)}{k}=\frac{f(a) f(a) f\left(a^{4}\right)}{k^{2}}=\frac{f(a) f(a) f\left(a^{2}\right)}{k}, \quad \text { for all } a \in S .
+$$
+
+It follows that $f(a)=k$ for all $a \in S$. Substituting $a=2$ and $b=3$ into the functional equation yields $k=1$, however $1 \notin S$ and hence we have no solutions.
+
+Problem 3. A sequence of real numbers $a_{0}, a_{1}, \ldots$ is said to be good if the following three conditions hold.
+(i) The value of $a_{0}$ is a positive integer.
+(ii) For each non-negative integer $i$ we have $a_{i+1}=2 a_{i}+1$ or $a_{i+1}=\frac{a_{i}}{a_{i}+2}$.
+(iii) There exists a positive integer $k$ such that $a_{k}=2014$.
+
+Find the smallest positive integer $n$ such that there exists a good sequence $a_{0}, a_{1}, \ldots$ of real numbers with the property that $a_{n}=2014$.
+
+Answer: 60.
+Solution. Note that
+
+$$
+a_{i+1}+1=2\left(a_{i}+1\right) \text { or } a_{i+1}+1=\frac{a_{i}+a_{i}+2}{a_{i}+2}=\frac{2\left(a_{i}+1\right)}{a_{i}+2} .
+$$
+
+Hence
+
+$$
+\frac{1}{a_{i+1}+1}=\frac{1}{2} \cdot \frac{1}{a_{i}+1} \text { or } \frac{1}{a_{i+1}+1}=\frac{a_{i}+2}{2\left(a_{i}+1\right)}=\frac{1}{2} \cdot \frac{1}{a_{i}+1}+\frac{1}{2} .
+$$
+
+Therefore,
+
+$$
+\frac{1}{a_{k}+1}=\frac{1}{2^{k}} \cdot \frac{1}{a_{0}+1}+\sum_{i=1}^{k} \frac{\varepsilon_{i}}{2^{k-i+1}}
+$$
+
+where $\varepsilon_{i}=0$ or 1 . Multiplying both sides by $2^{k}\left(a_{k}+1\right)$ and putting $a_{k}=2014$, we get
+
+$$
+2^{k}=\frac{2015}{a_{0}+1}+2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)
+$$
+
+where $\varepsilon_{i}=0$ or 1 . Since $\operatorname{gcd}(2,2015)=1$, we have $a_{0}+1=2015$ and $a_{0}=2014$. Therefore,
+
+$$
+2^{k}-1=2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)
+$$
+
+where $\varepsilon_{i}=0$ or 1 . We now need to find the smallest $k$ such that $2015 \mid 2^{k}-1$. Since $2015=$ $5 \cdot 13 \cdot 31$, from the Fermat little theorem we obtain $5\left|2^{4}-1,13\right| 2^{12}-1$ and $31 \mid 2^{30}-1$. We also have $\operatorname{lcm}[4,12,30]=60$, hence $5\left|2^{60}-1,13\right| 2^{60}-1$ and $31 \mid 2^{60}-1$, which gives $2015 \mid 2^{60}-1$.
+
+But $5 \nmid 2^{30}-1$ and so $k=60$ is the smallest positive integer such that $2015 \mid 2^{k}-1$. To conclude, the smallest positive integer $k$ such that $a_{k}=2014$ is when $k=60$.
+
+Alternative solution 1. Clearly all members of the sequence are positive rational numbers. For each positive integer $i$, we have $a_{i}=\frac{a_{i+1}-1}{2}$ or $a_{i}=\frac{2 a_{i+1}}{1-a_{i+1}}$. Since $a_{i}>0$ we deduce that
+
+$$
+a_{i}=\left\{\begin{array}{cl}
+\frac{a_{i+1}-1}{2} & \text { if } a_{i+1}>1 \\
+\frac{2 a_{i+1}}{1-a_{i+1}} & \text { if } a_{i+1}<1
+\end{array}\right.
+$$
+
+Thus $a_{i}$ is uniquely determined from $a_{i+1}$. Hence starting from $a_{k}=2014$, we simply run the sequence backwards until we reach a positive integer. We compute as follows.
+
+$$
+\begin{aligned}
+& \frac{2014}{1}, \frac{2013}{2}, \frac{2011}{4}, \frac{2007}{8}, \frac{1999}{16}, \frac{1983}{32}, \frac{1951}{64}, \frac{1887}{128}, \frac{1759}{256}, \frac{1503}{512}, \frac{991}{1024}, \frac{1982}{33}, \frac{1949}{66}, \frac{1883}{132}, \frac{1751}{264}, \frac{1487}{528}, \frac{959}{1056}, \frac{1918}{97}, \frac{1821}{194}, \frac{1627}{388}, \\
+& \frac{1239}{776}, \frac{463}{1552}, \frac{926}{1089}, \frac{1852}{163}, \frac{1689}{326}, \frac{1363}{652}, \frac{711}{1304}, \frac{1422}{593}, \frac{829}{1186}, \frac{1658}{357}, \frac{1301}{714}, \frac{587}{1428}, \frac{1174}{841}, \frac{333}{1682}, \frac{666}{1349}, \frac{1332}{683}, \frac{649}{1366}, \frac{1298}{717}, \frac{581}{1434}, \frac{1162}{853}, \\
+& \frac{309}{1706}, \frac{618}{1397}, \frac{1236}{779}, \frac{457}{1558}, \frac{914}{1101}, \frac{1828}{187}, \frac{1641}{374}, \frac{1267}{748}, \frac{519}{1496}, \frac{1038}{977}, \frac{61}{1954}, \frac{122}{1893}, \frac{244}{1771}, \frac{488}{1527}, \frac{976}{1039}, \frac{1952}{63}, \frac{1889}{126}, \frac{1763}{252}, \frac{1511}{504}, \frac{1007}{1008}, \frac{2014}{1}
+\end{aligned}
+$$
+
+There are 61 terms in the above list. Thus $k=60$.
+Alternative solution 1 is quite computationally intensive. Calculating the first few terms indicates some patterns that are easy to prove. This is shown in the next solution.
+
+Alternative solution 2. Start with $a_{k}=\frac{m_{0}}{n_{0}}$ where $m_{0}=2014$ and $n_{0}=1$ as in alternative solution 1. By inverting the sequence as in alternative solution 1, we have $a_{k-i}=\frac{m_{i}}{n_{i}}$ for $i \geq 0$ where
+
+$$
+\left(m_{i+1}, n_{i+1}\right)= \begin{cases}\left(m_{i}-n_{i}, 2 n_{i}\right) & \text { if } m_{i}>n_{i} \\ \left(2 m_{i}, n_{i}-m_{i}\right) & \text { if } m_{i}<n_{i}\end{cases}
+$$
+
+Easy inductions show that $m_{i}+n_{i}=2015,1 \leq m_{i}, n_{i} \leq 2014$ and $\operatorname{gcd}\left(m_{i}, n_{i}\right)=1$ for $i \geq 0$. Since $a_{0} \in \mathbb{N}^{+}$and $\operatorname{gcd}\left(m_{k}, n_{k}\right)=1$, we require $n_{k}=1$. An easy induction shows that $\left(m_{i}, n_{i}\right) \equiv\left(-2^{i}, 2^{i}\right)(\bmod 2015)$ for $i=0,1, \ldots, k$.
+
+Thus $2^{k} \equiv 1(\bmod 2015)$. As in the official solution, the smallest such $k$ is $k=60$. This yields $n_{k} \equiv 1(\bmod 2015)$. But since $1 \leq n_{k}, m_{k} \leq 2014$, it follows that $a_{0}$ is an integer.
+
+Problem 4. Let $n$ be a positive integer. Consider $2 n$ distinct lines on the plane, no two of which are parallel. Of the $2 n$ lines, $n$ are colored blue, the other $n$ are colored red. Let $\mathcal{B}$ be the set of all points on the plane that lie on at least one blue line, and $\mathcal{R}$ the set of all points on the plane that lie on at least one red line. Prove that there exists a circle that intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points.
+
+Solution. Consider a line $\ell$ on the plane and a point $P$ on it such that $\ell$ is not parallel to any of the $2 n$ lines. Rotate $\ell$ about $P$ counterclockwise until it is parallel to one of the $2 n$ lines. Take note of that line and keep rotating until all the $2 n$ lines are met. The $2 n$ lines are now ordered according to which line is met before or after. Say the lines are in order $\ell_{1}, \ldots, \ell_{2 n}$. Clearly there must be $k \in\{1, \ldots, 2 n-1\}$ such that $\ell_{k}$ and $\ell_{k+1}$ are of different colors.
+
+Now we set up a system of $X-$ and $Y$ - axes on the plane. Consider the two angular bisectors of $\ell_{k}$ and $\ell_{k+1}$. If we rotate $\ell_{k+1}$ counterclockwise, the line will be parallel to one of the bisectors before the other. Let the bisector that is parallel to the rotation of $\ell_{k+1}$ first be the $X$-axis, and the other the $Y$-axis. From now on, we will be using the directed angle notation: for lines $s$ and $s^{\prime}$, we define $\angle\left(s, s^{\prime}\right)$ to be a real number in $[0, \pi)$ denoting the angle in radians such that when $s$ is rotated counterclockwise by $\angle\left(s, s^{\prime}\right)$ radian, it becomes parallel to $s^{\prime}$. Using this
+notation, we notice that there is no $i=1, \ldots, 2 n$ such that $\angle\left(X, l_{i}\right)$ is between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$.
+
+Because the $2 n$ lines are distinct, the set $S$ of all the intersections between $\ell_{i}$ and $\ell_{j}(i \neq j)$ is a finite set of points. Consider a rectangle with two opposite vertices lying on $\ell_{k}$ and the other two lying on $\ell_{k+1}$. With respect to the origin (the intersection of $\ell_{k}$ and $\ell_{k+1}$ ), we can enlarge the rectangle as much as we want, while all the vertices remain on the lines. Thus, there is one of these rectangles $R$ which contains all the points in $S$ in its interior. Since each side of $R$ is parallel to either $X$ - or $Y$ - axis, $R$ is a part of the four lines $x= \pm a, y= \pm b$. where $a, b>0$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0249b629a4c75802d95g-4.jpg?height=429&width=969&top_left_y=685&top_left_x=543)
+
+Consider the circle $\mathcal{C}$ tangent to the right of the $x=a$ side of the rectangle, and to both $\ell_{k}$ and $\ell_{k+1}$. We claim that this circle intersects $\mathcal{B}$ in exactly $2 n-1$ points, and also intersects $\mathcal{R}$ in exactly $2 n-1$ points. Since $\mathcal{C}$ is tangent to both $\ell_{k}$ and $\ell_{k+1}$ and the two lines have different colors, it is enough to show that $\mathcal{C}$ intersects with each of the other $2 n-2$ lines in exactly 2 points. Note that no two lines intersect on the circle because all the intersections between lines are in $S$ which is in the interior of $R$.
+
+Consider any line $L$ among these $2 n-2$ lines. Let $L$ intersect with $\ell_{k}$ and $\ell_{k+1}$ at the points $M$ and $N$, respectively ( $M$ and $N$ are not necessarily distinct). Notice that both $M$ and $N$ must be inside $R$. There are two cases:
+(i) $L$ intersects $R$ on the $x=-a$ side once and another time on $x=a$ side;
+(ii) $L$ intersects $y=-b$ and $y=b$ sides.
+
+However, if (ii) happens, $\angle\left(\ell_{k}, L\right)$ and $\angle\left(L, \ell_{k+1}\right)$ would be both positive, and then $\angle(X, L)$ would be between $\angle\left(X, \ell_{k}\right)$ and $\angle\left(X, \ell_{k+1}\right)$, a contradiction. Thus, only (i) can happen. Then $L$ intersects $\mathcal{C}$ in exactly two points, and we are done.
+
+Alternative solution. By rotating the diagram we can ensure that no line is vertical. Let $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ be the lines listed in order of increasing gradient. Then there is a $k$ such that lines $\ell_{k}$ and $\ell_{k+1}$ are oppositely coloured. By rotating our coordinate system and cyclicly relabelling our lines we can ensure that $\ell_{1}, \ell_{2}, \ldots, \ell_{2 n}$ are listed in order of increasing gradient, $\ell_{1}$ and $\ell_{2 n}$ are oppositely coloured, and no line is vertical.
+
+Let $\mathcal{D}$ be a circle centred at the origin and of sufficiently large radius so that
+
+- All intersection points of all pairs of lines lie strictly inside $\mathcal{D}$; and
+- Each line $\ell_{i}$ intersects $\mathcal{D}$ in two points $A_{i}$ and $B_{i}$ say, such that $A_{i}$ is on the right semicircle (the part of the circle in the positive $x$ half plane) and $B_{i}$ is on the left semicircle.
+
+Note that the anticlockwise order of the points $A_{i}, B_{i}$ around $\mathcal{D}$ is $A_{1}, A_{2}, \ldots, A_{n}, B_{1}, B_{2}, \ldots, B_{n}$.
+(If $A_{i+1}$ occurred before $A_{i}$ then rays $r_{i}$ and $r_{i+1}$ (as defined below) would intersect outside $\mathcal{D}$.)
+![](https://cdn.mathpix.com/cropped/2024_11_22_b0249b629a4c75802d95g-5.jpg?height=701&width=1266&top_left_y=272&top_left_x=406)
+
+For each $i$, let $r_{i}$ be the ray that is the part of the line $\ell_{i}$ starting from point $A_{i}$ and that extends to the right. Let $\mathcal{C}$ be any circle tangent to $r_{1}$ and $r_{2 n}$, that lies entirely to the right of $\mathcal{D}$. Then $\mathcal{C}$ intersects each of $r_{2}, r_{3}, \ldots, r_{2 n-1}$ twice and is tangent to $r_{1}$ and $r_{2 n}$. Thus $\mathcal{C}$ has the required properties.
+
+Problem 5. Determine all sequences $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers with $a_{0} \geq 2015$ such that for all integers $n \geq 1$ :
+(i) $a_{n+2}$ is divisible by $a_{n}$;
+(ii) $\left|s_{n+1}-(n+1) a_{n}\right|=1$, where $s_{n+1}=a_{n+1}-a_{n}+a_{n-1}-\cdots+(-1)^{n+1} a_{0}$.
+
+Answer: There are two families of answers:
+(a) $a_{n}=c(n+2) n$ ! for all $n \geq 1$ and $a_{0}=c+1$ for some integer $c \geq 2014$, and
+(b) $a_{n}=c(n+2) n$ ! for all $n \geq 1$ and $a_{0}=c-1$ for some integer $c \geq 2016$.
+
+Solution. Let $\left\{a_{n}\right\}_{n=0}^{\infty}$ be a sequence of positive integers satisfying the given conditions. We can rewrite (ii) as $s_{n+1}=(n+1) a_{n}+h_{n}$, where $h_{n} \in\{-1,1\}$. Substituting $n$ with $n-1$ yields $s_{n}=n a_{n-1}+h_{n-1}$, where $h_{n-1} \in\{-1,1\}$. Note that $a_{n+1}=s_{n+1}+s_{n}$, therefore there exists $\delta_{n} \in\{-2,0,2\}$ such that
+
+$$
+a_{n+1}=(n+1) a_{n}+n a_{n-1}+\delta_{n}
+$$
+
+We also have $\left|s_{2}-2 a_{1}\right|=1$, which yields $a_{0}=3 a_{1}-a_{2} \pm 1 \leq 3 a_{1}$, and therefore $a_{1} \geq \frac{a_{0}}{3} \geq 671$. Substituting $n=2$ in (1), we find that $a_{3}=3 a_{2}+2 a_{1}+\delta_{2}$. Since $a_{1} \mid a_{3}$, we have $a_{1} \mid 3 a_{2}+\delta_{2}$, and therefore $a_{2} \geq 223$. Using (1), we obtain that $a_{n} \geq 223$ for all $n \geq 0$.
+
+Lemma 1: For $n \geq 4$, we have $a_{n+2}=(n+1)(n+4) a_{n}$.
+Proof. For $n \geq 3$ we have
+
+$$
+a_{n}=n a_{n-1}+(n-1) a_{n-2}+\delta_{n-1}>n a_{n-1}+3 .
+$$
+
+By applying (2) with $n$ substituted by $n-1$ we have for $n \geq 4$,
+
+$$
+a_{n}=n a_{n-1}+(n-1) a_{n-2}+\delta_{n-1}<n a_{n-1}+\left(a_{n-1}-3\right)+\delta_{n-1}<(n+1) a_{n-1}
+$$
+
+Using (1) to write $a_{n+2}$ in terms of $a_{n}$ and $a_{n-1}$ along with (2), we obtain that for $n \geq 3$,
+
+$$
+\begin{aligned}
+a_{n+2} & =(n+3)(n+1) a_{n}+(n+2) n a_{n-1}+(n+2) \delta_{n}+\delta_{n+1} \\
+& <(n+3)(n+1) a_{n}+(n+2) n a_{n-1}+3(n+2) \\
+& <\left(n^{2}+5 n+5\right) a_{n} .
+\end{aligned}
+$$
+
+Also for $n \geq 4$,
+
+$$
+\begin{aligned}
+a_{n+2} & =(n+3)(n+1) a_{n}+(n+2) n a_{n-1}+(n+2) \delta_{n}+\delta_{n+1} \\
+& >(n+3)(n+1) a_{n}+n a_{n} \\
+& =\left(n^{2}+5 n+3\right) a_{n} .
+\end{aligned}
+$$
+
+Since $a_{n} \mid a_{n+2}$, we obtain that $a_{n+2}=\left(n^{2}+5 n+4\right) a_{n}=(n+1)(n+4) a_{n}$, as desired.
+Lemma 2: For $n \geq 4$, we have $a_{n+1}=\frac{(n+1)(n+3)}{n+2} a_{n}$.
+Proof. Using the recurrence $a_{n+3}=(n+3) a_{n+2}+(n+2) a_{n+1}+\delta_{n+2}$ and writing $a_{n+3}$, $a_{n+2}$ in terms of $a_{n+1}, a_{n}$ according to Lemma 1 we obtain
+
+$$
+(n+2)(n+4) a_{n+1}=(n+3)(n+1)(n+4) a_{n}+\delta_{n+2} .
+$$
+
+Hence $n+4 \mid \delta_{n+2}$, which yields $\delta_{n+2}=0$ and $a_{n+1}=\frac{(n+1)(n+3)}{n+2} a_{n}$, as desired.
+Suppose there exists $n \geq 1$ such that $a_{n+1} \neq \frac{(n+1)(n+3)}{n+2} a_{n}$. By Lemma 2, there exist a greatest integer $1 \leq m \leq 3$ with this property. Then $a_{m+2}=\frac{(m+2)(m+4)}{m+3} a_{m+1}$. If $\delta_{m+1}=0$, we have $a_{m+1}=\frac{(m+1)(m+3)}{m+2} a_{m}$, which contradicts our choice of $m$. Thus $\delta_{m+1} \neq 0$.
+
+Clearly $m+3 \mid a_{m+1}$. Write $a_{m+1}=(m+3) k$ and $a_{m+2}=(m+2)(m+4) k$. Then $(m+$ 1) $a_{m}+\delta_{m+1}=a_{m+2}-(m+2) a_{m+1}=(m+2) k$. So, $a_{m} \mid(m+2) k-\delta_{m+1}$. But $a_{m}$ also divides $a_{m+2}=(m+2)(m+4) k$. Combining the two divisibility conditions, we obtain $a_{m} \mid(m+4) \delta_{m+1}$. Since $\delta_{m+1} \neq 0$, we have $a_{m} \mid 2 m+8 \leq 14$, which contradicts the previous result that $a_{n} \geq 223$ for all nonnegative integers $n$.
+
+So, $a_{n+1}=\frac{(n+1)(n+3)}{n+2} a_{n}$ for $n \geq 1$. Substituting $n=1$ yields $3 \mid a_{1}$. Letting $a_{1}=3 c$, we have by induction that $a_{n}=n!(n+2) c$ for $n \geq 1$. Since $\left|s_{2}-2 a_{1}\right|=1$, we then get $a_{0}=c \pm 1$, yielding the two families of solutions. By noting that $(n+2) n!=n!+(n+1)!$, we have $s_{n+1}=c(n+2)!+(-1)^{n}\left(c-a_{0}\right)$. Hence both families of solutions satisfy the given conditions.
+

APMO/md/en-apmo2016_sol.md

@@ -0,0 +1,259 @@
+# Solutions of APMO 2016 
+
+Problem 1. We say that a triangle $A B C$ is great if the following holds: for any point $D$ on the side $B C$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $A B$ and $A C$, respectively, then the reflection of $D$ in the line $P Q$ lies on the circumcircle of the triangle $A B C$.
+
+Prove that triangle $A B C$ is great if and only if $\angle A=90^{\circ}$ and $A B=A C$.
+Solution. For every point $D$ on the side $B C$, let $D^{\prime}$ be the reflection of $D$ in the line $P Q$. We will first prove that if the triangle satisfies the condition then it is isosceles and right-angled at $A$.
+
+Choose $D$ to be the point where the angle bisector from $A$ meets $B C$. Note that $P$ and $Q$ lie on the rays $A B$ and $A C$ respectively. Furthermore, $P$ and $Q$ are reflections of each other in the line $A D$, from which it follows that $P Q \perp A D$. Therefore, $D^{\prime}$ lies on the line $A D$ and we may deduce that either $D^{\prime}=A$ or $D^{\prime}$ is the second point of the angle bisector at $A$ and the circumcircle of $A B C$. However, since $A P D Q$ is a cyclic quadrilateral, the segment $P Q$ intersects the segment $A D$. Therefore, $D^{\prime}$ lies on the ray $D A$ and therefore $D^{\prime}=A$. By angle chasing we obtain
+
+$$
+\angle P D^{\prime} Q=\angle P D Q=180^{\circ}-\angle B A C
+$$
+
+and since $D^{\prime}=A$ we also know $\angle P D^{\prime} Q=\angle B A C$. This implies that $\angle B A C=90^{\circ}$.
+Now we choose $D$ to be the midpoint of $B C$. Since $\angle B A C=90^{\circ}$, we can deduce that $D Q P$ is the medial triangle of triangle $A B C$. Therefore, $P Q \| B C$ from which it follows that $D D^{\prime} \perp B C$. But the distance from $D^{\prime}$ to $B C$ is equal to both the circumradius of triangle $A B C$ and to the distance from $A$ to $B C$. This can only happen if $A=D^{\prime}$. This implies that $A B C$ is isosceles and right-angled at $A$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_ed00b2f2e7e7ff36d38eg-1.jpg?height=447&width=1633&top_left_y=1561&top_left_x=239)
+
+We will now prove that if $A B C$ is isosceles and right-angled at $A$ then the required property in the problem holds. Let $D$ be any point on side $B C$. Then $D^{\prime} P=D P$ and we also have $D P=B P$. Hence, $D^{\prime} P=B P$ and similarly $D^{\prime} Q=C Q$. Note that $A P D Q D^{\prime}$ is cyclic with diameter $P Q$. Therefore, $\angle A P D^{\prime}=\angle A Q D^{\prime}$, from which we obtain $\angle B P D^{\prime}=\angle C Q D^{\prime}$. So triangles $D^{\prime} P B$ and $D^{\prime} Q C$ are similar. It follows that $\angle P D^{\prime} Q=\angle P D^{\prime} C+\angle C D^{\prime} Q=$ $\angle P D^{\prime} C+\angle B D^{\prime} P=\angle B D^{\prime} C$ and $\frac{D^{\prime} P}{D^{\prime} Q}=\frac{D^{\prime} B}{D^{\prime} C}$. So we also obtain that triangles $D^{\prime} P Q$ and $D^{\prime} B C$ are similar. But since $D P Q$ and $D^{\prime} P Q$ are congruent, we may deduce that $\angle B D^{\prime} C=$ $\angle P D^{\prime} Q=\angle P D Q=90^{\circ}$. Therefore, $D^{\prime}$ lies on the circle with diameter $B C$, which is the circumcircle of triangle $A B C$.
+
+Problem 2. A positive integer is called fancy if it can be expressed in the form
+
+$$
+2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{100}}
+$$
+
+where $a_{1}, a_{2}, \ldots, a_{100}$ are non-negative integers that are not necessarily distinct.
+Find the smallest positive integer $n$ such that no multiple of $n$ is a fancy number.
+Answer: The answer is $n=2^{101}-1$.
+Solution. Let $k$ be any positive integer less than $2^{101}-1$. Then $k$ can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer $r$ and non-negative integers $a_{1}, a_{2}, \ldots, a_{r}$ such that $r \leq 100$ and $k=2^{a_{1}}+\cdots+2^{a_{r}}$. Notice that for a positive integer $s$ we have:
+
+$$
+\begin{aligned}
+2^{s} k & =2^{a_{1}+s}+2^{a_{2}+s}+\cdots+2^{a_{r-1}+s}+\left(1+1+2+\cdots+2^{s-1}\right) 2^{a_{r}} \\
+& =2^{a_{1}+s}+2^{a_{2}+s}+\cdots+2^{a_{r-1}+s}+2^{a_{r}}+2^{a_{r}}+\cdots+2^{a_{r}+s-1} .
+\end{aligned}
+$$
+
+This shows that $k$ has a multiple that is a sum of $r+s$ powers of two. In particular, we may take $s=100-r \geq 0$, which shows that $k$ has a multiple that is a fancy number.
+
+We will now prove that no multiple of $n=2^{101}-1$ is a fancy number. In fact we will prove a stronger statement, namely, that no multiple of $n$ can be expressed as the sum of at most 100 powers of 2 .
+
+For the sake of contradiction, suppose that there exists a positive integer $c$ such that $c n$ is the sum of at most 100 powers of 2 . We may assume that $c$ is the smallest such integer. By repeatedly merging equal powers of two in the representation of $c n$ we may assume that
+
+$$
+c n=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{r}}
+$$
+
+where $r \leq 100$ and $a_{1}<a_{2}<\ldots<a_{r}$ are distinct non-negative integers. Consider the following two cases:
+
+- If $a_{r} \geq 101$, then $2^{a_{r}}-2^{a_{r}-101}=2^{a_{r}-101} n$. It follows that $2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{r}-1}+2^{a_{r}-101}$ would be a multiple of $n$ that is smaller than $c n$. This contradicts the minimality of $c$.
+- If $a_{r} \leq 100$, then $\left\{a_{1}, \ldots, a_{r}\right\}$ is a proper subset of $\{0,1, \ldots, 100\}$. Then
+
+$$
+n \leq c n<2^{0}+2^{1}+\cdots+2^{100}=n
+$$
+
+This is also a contradiction.
+From these contradictions we conclude that it is impossible for cn to be the sum of at most 100 powers of 2 . In particular, no multiple of $n$ is a fancy number.
+
+Problem 3. Let $A B$ and $A C$ be two distinct rays not lying on the same line, and let $\omega$ be a circle with center $O$ that is tangent to ray $A C$ at $E$ and ray $A B$ at $F$. Let $R$ be a point on segment $E F$. The line through $O$ parallel to $E F$ intersects line $A B$ at $P$. Let $N$ be the intersection of lines $P R$ and $A C$, and let $M$ be the intersection of line $A B$ and the line through $R$ parallel to $A C$. Prove that line $M N$ is tangent to $\omega$.
+
+Solution. We present two approaches. The first one introduces an auxiliary point and studies similarities in the figure. The second one reduces the problem to computations involving a particular exradius of a triangle. The second approach has two variants.
+
+## Solution 1.
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_ed00b2f2e7e7ff36d38eg-3.jpg?height=981&width=1192&top_left_y=246&top_left_x=426)
+
+Let the line through $N$ tangent to $\omega$ at point $X \neq E$ intersect $A B$ at point $M^{\prime}$. It suffices to show that $M^{\prime} R \| A C$, since this would yield $M^{\prime}=M$.
+
+Suppose that the line $P O$ intersects $A C$ at $Q$ and the circumcircle of $A M^{\prime} O$ at $Y$, respectively. Then
+
+$$
+\angle A Y M^{\prime}=\angle A O M^{\prime}=90^{\circ}-\angle M^{\prime} O P
+$$
+
+By angle chasing we have $\angle E O Q=\angle F O P=90^{\circ}-\angle A O F=\angle M^{\prime} A O=\angle M^{\prime} Y P$ and by symmetry $\angle E Q O=\angle M^{\prime} P Y$. Therefore $\triangle M^{\prime} Y P \sim \triangle E O Q$.
+
+On the other hand, we have
+
+$$
+\begin{aligned}
+\angle M^{\prime} O P & =\angle M^{\prime} O F+\angle F O P=\frac{1}{2}(\angle F O X+\angle F O P+\angle E O Q)= \\
+& =\frac{1}{2}\left(\frac{180^{\circ}-\angle X O E}{2}\right)=90^{\circ}-\frac{\angle X O E}{2} .
+\end{aligned}
+$$
+
+Since we know that $\angle A Y M^{\prime}$ and $\angle M^{\prime} O P$ are complementary this implies
+
+$$
+\angle A Y M^{\prime}=\frac{\angle X O E}{2}=\angle N O E
+$$
+
+Therefore, $\angle A Y M^{\prime}$ and $\angle N O E$ are congruent angles, and this means that $A$ and $N$ are corresponding points in the similarity of triangles $\triangle M^{\prime} Y P$ and $\triangle E O Q$. It follows that
+
+$$
+\frac{A M^{\prime}}{M^{\prime} P}=\frac{N E}{E Q}=\frac{N R}{R P}
+$$
+
+We conclude that $M^{\prime} R \| A C$, as desired.
+
+## Solution 2a.
+
+As in Solution 1, we introduce point $M^{\prime}$ and reduce the problem to proving $\frac{P R}{R N}=\frac{P M^{\prime}}{M^{\prime} A}$. Menelaus theorem in triangle $A N P$ with transversal line $F R E$ yields
+
+$$
+\frac{P R}{R N} \cdot \frac{N E}{E A} \cdot \frac{A F}{F P}=1
+$$
+
+Since $A F=E A$, we have $\frac{F P}{N E}=\frac{P R}{R N}$, so that it suffices to prove
+
+$$
+\frac{F P}{N E}=\frac{P M^{\prime}}{M^{\prime} A}
+$$
+
+This is a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$. Indeed, setting $a=M^{\prime} N, b=N A, c=M^{\prime} A, s=\frac{a+b+c}{2}, x=s-a, y=s-b$ and $z=s-c$, then $A E=A F=s, M^{\prime} F=z$ and $N E=y$. From $\triangle O F P \sim \triangle A F O$ we have $F P=\frac{r_{a}^{2}}{s}$, where $r_{a}=O F$ is the exradius opposite $A$. Combining the following two standard formulas for the area of a triangle
+
+$$
+\left|A M^{\prime} N\right|^{2}=x y z s \quad \text { (Heron's formula) and } \quad\left|A M^{\prime} N\right|=r_{a}(s-a),
+$$
+
+we have $r_{a}^{2}=\frac{y z s}{x}$. Therefore, $F P=\frac{y z}{x}$. We can now write everything in (1) in terms of $x, y, z$. We conclude that we have to verify
+
+$$
+\frac{\frac{y z}{x}}{y}=\frac{z+\frac{y z}{x}}{x+y}
+$$
+
+which is easily seen to be true.
+Note: Antoher approach using Menalaus theorem is to construct the tangent from $M$ to create a point $N^{\prime}$ in $A C$ and then prove, using the theorem, that $P, R$ and $N^{\prime}$ are collinear. This also reduces to an algebraic identity.
+
+## Solution 2b.
+
+As in Solution 1, we introduce point $M^{\prime}$. Let the line through $M^{\prime}$ and parallel to $A N$ intersect $E F$ at $R^{\prime}$. Let $P^{\prime}$ be the intersection of lines $N R^{\prime}$ and $A M$. It suffices to show that $P^{\prime} O \| F E$, since this would yield $P=P^{\prime}$, and then $R=R^{\prime}$ and $M=M^{\prime}$. Hence it is enough to prove that
+
+$$
+\frac{A F}{F P^{\prime}}=\frac{A D}{D O}
+$$
+
+where $D$ is the intersection of $A O$ and $E F$. Once again, this reduces to a computation regarding the triangle $A M^{\prime} N$ and its excircle opposite $A$.
+
+Let $u=P^{\prime} F$ and $x, y, z, s$ as in Solution 2a. Note that since $A E=A F$ and $M^{\prime} R^{\prime} \| A E$, we have $M^{\prime} R^{\prime}=M^{\prime} F=z$. Since $M^{\prime} R^{\prime} \| A N$, we have $\frac{P^{\prime} M^{\prime}}{P^{\prime} A}=\frac{M^{\prime} R^{\prime}}{N A}$, that is,
+
+$$
+\frac{u+z}{u+x+y+z}=\frac{z}{x+z}
+$$
+
+From this last equation we obtain $u=\frac{y z}{x}$. Hence $\frac{A F}{F P^{\prime}}=\frac{x s}{y z}$. Also, as in Solution 2a, we have $r_{a}^{2}=\frac{y z s}{x}$.
+
+Finally, using similar triangles $O D F, F D A$ and $O F A$, and the above equalities, we have
+
+$$
+\frac{A D}{D O}=\frac{A D}{D F} \cdot \frac{D F}{D O}=\frac{A F}{O F} \cdot \frac{A F}{O F}=\frac{s^{2}}{r_{a}^{2}}=\frac{s^{2}}{\frac{y z s}{x}}=\frac{x s}{y z}=\frac{A F}{F P^{\prime}}
+$$
+
+as required.
+
+Problem 4. The country Dreamland consists of 2016 cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer $k$ such that no matter how Starways establishes its flights, the cities can always be partitioned into $k$ groups so that from any city it is not possible to reach another city in the same group by using at most 28 flights.
+
+## Answer: 57
+
+Solution. The flights established by Starways yield a directed graph $G$ on 2016 vertices in which each vertex has out-degree equal to 1.
+
+We first show that we need at least 57 groups. For this, suppose that $G$ has a directed cycle of length 57. Then, for any two cities in the cycle, one is reachable from the other using at most 28 flights. So no two cities in the cycle can belong to the same group. Hence, we need at least 57 groups.
+
+We will now show that 57 groups are enough. Consider another auxiliary directed graph $H$ in which the vertices are the cities of Dreamland and there is an arrow from city $u$ to city $v$ if $u$ can be reached from $v$ using at most 28 flights. Each city has out-degree at most 28 . We will be done if we can split the cities of $H$ in at most 57 groups such that there are no arrows between vertices of the same group. We prove the following stronger statement.
+
+Lemma: Suppose we have a directed graph on $n \geq 1$ vertices such that each vertex has out-degree at most 28. Then the vertices can be partitioned into 57 groups in such a way that no vertices in the same group are connected by an arrow.
+
+Proof: We apply induction. The result is clear for 1 vertex. Now suppose we have more than one vertex. Since the out-degree of each vertex is at most 28 , there is a vertex, say $v$, with in-degree at most 28. If we remove the vertex $v$ we obtain a graph with fewer vertices which still satifies the conditions, so by inductive hypothesis we may split it into at most 57 groups with no adjacent vertices in the same group. Since $v$ has in-degree and out-degree at most 28 , it has at most 56 neighboors in the original directed graph. Therefore, we may add $v$ back and place it in a group in which it has no neighbors. This completes the inductive step.
+
+Problem 5. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
+
+$$
+(z+1) f(x+y)=f(x f(z)+y)+f(y f(z)+x)
+$$
+
+for all positive real numbers $x, y, z$.
+Answer: The only solution is $f(x)=x$ for all positive real numbers $x$.
+Solution. The identity function $f(x)=x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x=y=1$ into (3) we get $2 f(f(z)+1)=(z+1)(f(2))$ for all $z \in \mathbb{R}^{+}$. Hence, $f$ is not bounded above.
+
+Lemma. Let $a, b, c$ be positive real numbers. If $c$ is greater than $1, a / b$ and $b / a$, then the system of linear equations
+
+$$
+c u+v=a \quad u+c v=b
+$$
+
+has a positive real solution $u, v$.
+Proof. The solution is
+
+$$
+u=\frac{c a-b}{c^{2}-1} \quad v=\frac{c b-a}{c^{2}-1} .
+$$
+
+The numbers $u$ and $v$ are positive if the conditions on $c$ above are satisfied.
+
+We will now prove that
+
+$$
+f(a)+f(b)=f(c)+f(d) \quad \text { for all } a, b, c, d \in \mathbb{R}^{+} \text {with } a+b=c+d
+$$
+
+Consider $a, b, c, d \in \mathbb{R}^{+}$such that $a+b=c+d$. Since $f$ is not bounded above, we can choose a positive number $e$ such that $f(e)$ is greater than $1, a / b, b / a, c / d$ and $d / c$. Using the above lemma, we can find $u, v, w, t \in \mathbb{R}^{+}$satisfying
+
+$$
+\begin{array}{ll}
+f(e) u+v=a, & u+f(e) v=b \\
+f(e) w+t=c, & w+f(e) t=d .
+\end{array}
+$$
+
+Note that $u+v=w+t$ since $(u+v)(f(e)+1)=a+b$ and $(w+t)(f(e)+1)=c+d$. Plugging $x=u, y=v$ and $z=e$ into (3) yields $f(a)+f(b)=(e+1) f(u+v)$. Similarly, we have $f(c)+f(d)=(e+1) f(w+t)$. The claim follows immediately.
+
+We then have
+
+$$
+y f(x)=f(x f(y)) \quad \text { for all } x, y \in \mathbb{R}^{+}
+$$
+
+since by (3) and (4),
+
+$$
+(y+1) f(x)=f\left(\frac{x}{2} f(y)+\frac{x}{2}\right)+f\left(\frac{x}{2} f(y)+\frac{x}{2}\right)=f(x f(y))+f(x) .
+$$
+
+Now, let $a=f(1 / f(1))$. Plugging $x=1$ and $y=1 / f(1)$ into (5) yields $f(a)=1$. Hence $a=a f(a)$ and $f(a f(a))=f(a)=1$. Since $a f(a)=f(a f(a))$ by (5), we have $f(1)=a=1$. It follows from (5) that
+
+$$
+f(f(y))=y \quad \text { for all } y \in \mathbb{R}^{+} .
+$$
+
+Using (4) we have for all $x, y \in \mathbb{R}^{+}$that
+
+$$
+\begin{aligned}
+& f(x+y)+f(1)=f(x)+f(y+1), \quad \text { and } \\
+& f(y+1)+f(1)=f(y)+f(2)
+\end{aligned}
+$$
+
+Therefore
+
+$$
+f(x+y)=f(x)+f(y)+b \quad \text { for all } x, y \in \mathbb{R}^{+},
+$$
+
+where $b=f(2)-2 f(1)=f(2)-2$. Using (5), (7) and (6), we get
+
+$$
+4+2 b=2 f(2)=f(2 f(2))=f(f(2)+f(2))=f(f(2))+f(f(2))+b=4+b
+$$
+
+This shows that $b=0$ and thus
+
+$$
+f(x+y)=f(x)+f(y) \quad \text { for all } x, y \in \mathbb{R}^{+} .
+$$
+
+In particular, $f$ is strictly increasing.
+We conclude as follows. Take any positive real number $x$. If $f(x)>x$, then $f(f(x))>$ $f(x)>x=f(f(x))$, a contradiction. Similarly, it is not possible that $f(x)<x$. This shows that $f(x)=x$ for all positive real numbers $x$.
+

APMO/md/en-apmo2017_sol.md

@@ -0,0 +1,223 @@
+# Solutions of APMO 2017 
+
+Problem 1. We call a 5-tuple of integers arrangeable if its elements can be labeled $a$, $b, c, d, e$ in some order so that $a-b+c-d+e=29$. Determine all 2017-tuples of integers $n_{1}, n_{2}, \ldots, n_{2017}$ such that if we place them in a circle in clockwise order, then any 5 -tuple of numbers in consecutive positions on the circle is arrangeable.
+
+Answer: $n_{1}=\cdots=n_{2017}=29$.
+Solution. A valid 2017-tuple is $n_{1}=\cdots=n_{2017}=29$. We will show that it is the only solution.
+
+We first replace each number $n_{i}$ in the circle by $m_{i}:=n_{i}-29$. Since the condition $a-b+$ $c-d+e=29$ can be rewritten as $(a-29)-(b-29)+(c-29)-(d-29)+(e-29)=0$, we have that any five consecutive replaced integers in the circle can be labeled $a, b, c, d, e$ in such a way that $a-b+c-d+e=0$. We claim that this is possible only when all of the $m_{i}$ 's are 0 (and thus all of the original $n_{i}$ 's are 29).
+
+We work with indexes modulo 2017. Notice that for every $i, m_{i}$ and $m_{i+5}$ have the same parity. Indeed, this follows from $m_{i} \equiv m_{i+1}+m_{i+2}+m_{i+3}+m_{i+4} \equiv m_{i+5}(\bmod 2)$. Since $\operatorname{gcd}(5,2017)=1$, this implies that all $m_{i}$ 's are of the same parity. Since $m_{1}+m_{2}+m_{3}+m_{4}+m_{5}$ is even, all $m_{i}$ 's must be even as well.
+
+Suppose for the sake of contradiction that not all $m_{i}$ 's are zero. Then our condition still holds when we divide each number in the circle by 2 . However, by performing repeated divisions, we eventually reach a point where some $m_{i}$ is odd. This is a contradiction.
+
+Problem 2. Let $A B C$ be a triangle with $A B<A C$. Let $D$ be the intersection point of the internal bisector of angle $B A C$ and the circumcircle of $A B C$. Let $Z$ be the intersection point of the perpendicular bisector of $A C$ with the external bisector of angle $\angle B A C$. Prove that the midpoint of the segment $A B$ lies on the circumcircle of triangle $A D Z$.
+
+Solution 1. Let $N$ be the midpoint of $A C$. Let $M$ be the intersection point of the circumcircle of triangle $A D Z$ and the segment $A B$. We will show that $M$ is the midpoint of $A B$. To do this, let $D^{\prime}$ the reflection of $D$ with respect to $M$. It suffices to show that $A D B D^{\prime}$ is a parallelogram.
+
+The internal and external bisectors of an angle in a triangle are perpendicular. This implies that $Z D$ is a diameter of the circumcircle of $A Z D$ and thus $\angle Z M D=90^{\circ}$. This means that $Z M$ is the perpendicular bisector of $D^{\prime} D$ and thus $Z D^{\prime}=Z D$. By construction, $Z$ is in the perpendicular bisector of $A C$ and thus $Z A=Z C$.
+
+Now, let $\alpha$ be the angle $\angle B A D=\angle D A C$. In the cyclic quadrilateral $A Z D M$ we get $\angle M Z D=\angle M A D=\alpha$, and thus $\angle D^{\prime} Z D=2 \alpha$. By angle chasing we get
+
+$$
+\angle A Z N=90^{\circ}-\angle Z A N=\angle D A C=\alpha,
+$$
+
+which implies that $\angle A Z C=2 \alpha$. Therefore,
+
+$$
+\angle D^{\prime} Z A=\angle D^{\prime} Z D-\angle A Z D=2 \alpha-\angle A Z D=\angle A Z C-\angle A Z D=\angle D Z C .
+$$
+
+Combining $\angle D^{\prime} Z A=\angle D Z C, Z D^{\prime}=Z D$ and $Z A=Z C$, we obtain by the $S A S$ criterion that the triangles $D^{\prime} Z A$ and $D Z C$ are congruent. In particular, $D^{\prime} A=D C$ and $\angle D^{\prime} A Z=$ $\angle D C Z$. From here $D B=D C=D^{\prime} A$.
+
+Finally, let $\beta=\angle A B C=\angle A D C$. We get the first of the following equalities by the sum of angles around point $A$ and the second one by the sum of internal angles of quadrilateral $A Z C D$
+
+$$
+\begin{aligned}
+& 360^{\circ}=\angle D^{\prime} A Z+\angle Z A D+\angle D A B+\angle B A D^{\prime}=\angle D^{\prime} A Z+90^{\circ}+\alpha+\angle B A D^{\prime} \\
+& 360^{\circ}=\angle D C Z+\angle Z A D+\angle C Z A+\angle A D C=\angle D C Z+90^{\circ}+2 \alpha+\beta
+\end{aligned}
+$$
+
+By canceling equal terms we conclude that $\angle B A D^{\prime}=\alpha+\beta$. Also, $\angle A B D=\alpha+\beta$. Therefore, the segments $D^{\prime} A$ and $D B$ are parallel and have the same length. We conclude that $A D B D^{\prime}$ is a parallelogram. As the diagonals of a parallelogram intersect at their midpoints, we obtain that $M$ is the midpoint of $A B$ as desired.
+![](https://cdn.mathpix.com/cropped/2024_11_22_ccb78644ce22f9d3e772g-2.jpg?height=989&width=1257&top_left_y=736&top_left_x=388)
+
+Variant of solution. The solution above is indirect in the sense that it assumes that $M$ is in the circumcircle of $A Z D$ and then shows that $M$ is the midpoint of $A B$. We point out that the same ideas in the solution can be used to give a direct solution. Here we present a sketch on how to proceed in this manner.
+
+Now we know that $M$ is the midpoint of the side $A B$. We construct the point $D^{\prime}$ in the same way. Now we have directly that $A D B D^{\prime}$ is a parallelogram and thus $D^{\prime} A=D B=D C$. By construction $Z A=Z C$. Also, the two sums of angles equal to $360^{\circ}$ in the previous solution let us conclude that $\angle D^{\prime} A Z=\angle D C Z$. Once again, we use (differently) the $S A S$ criterion and obtain that the triangles $D^{\prime} A Z$ and $D C Z$ are congruent. Thus, $D^{\prime} Z=D Z$.
+
+We finish the problem by noting that $Z M$ is a median of the isosceles triangle $D^{\prime} Z D$, so it is also a perpendicular bisector. This shows that $\angle D M Z=90^{\circ}=\angle D A Z$, and therefore $M$ lies in the circumcircle of $D A Z$.
+
+Solution 2. We proceed directly. As above, we name
+
+$$
+\alpha=\angle D A C=\angle A Z N=\angle C Z N=\angle D C B .
+$$
+
+Let $L$ be the midpoint of the segment $B C$. Since $M$ and $N$ are midpoints of $A B$ and $A C$, we have that $M N=C L$ and that the segment $M N$ is parallel to $B C$. Thus, $\angle A N M=\angle A C B$.
+
+Therefore, $\angle Z N M=\angle A C B+90^{\circ}$. Also, $\angle D C Z=\angle D C B+\angle A C B+\angle Z C A=\angle A C B+90^{\circ}$. We conclude that $\angle Z N M=\angle D C Z$.
+
+Now, the triangles $Z N C$ and $C L D$ are similar since $\angle D C L=90^{\circ}=\angle C N Z$ and $\angle D C L=$ $\alpha=\angle C Z N$. We use this fact to obtain the following chain of equalities
+
+$$
+\frac{C D}{M N}=\frac{C D}{C L}=\frac{C Z}{Z N}
+$$
+
+We combine the equality above with $\angle Z N M=\angle D C Z$ to conclude that the triangles $Z N M$ and $Z C D$ are similar. In particular, $\angle M Z N=\angle D Z C$ and $\frac{Z M}{Z N}=\frac{Z D}{Z C}$. Since we also have $\angle M Z D=\angle M Z N+\angle N Z D=\angle D Z C+\angle N Z D=\angle N Z C$, we conclude that the triangles $M Z D$ and $N Z C$ are similar. This yields that $\angle Z M D=90^{\circ}$ and therefore $M$ is in the circumcircle of triangle $D A Z$.
+
+Solution 3. Let $m$ be the perpendicular bisector of $A D$; thus $m$ passes through the center $O$ of the circumcircle of triangle $A B C$. Since $A D$ is the internal angle bisector of $A$ and $O M$ and $O N$ are perpendicular to $A B$ and $A C$ respectively, we obtain that $O M$ and $O N$ form equal angles with $A D$. This implies that they are symmetric with respect to $M$.
+
+Therefore, the line $M O$ passes through the point $Z^{\prime}$ symmetrical to $Z$ with respect to $m$. Since $\angle D A Z=90^{\circ}$, then also $\angle A D Z^{\prime}=90^{\circ}$. Moreover, since $A Z=D Z^{\prime}$, we have that $\angle A Z Z^{\prime}=\angle D Z^{\prime} Z=90^{\circ}$, so $A Z Z^{\prime} D$ is a rectangle. Since $\angle A M Z^{\prime}=\angle A M O=90^{\circ}$, we conclude that $M$ is in the circle with diameter $A Z^{\prime}$, which is the circumcircle of the rectangle. Thus $M$ lies on the circumcircle of the triangle $A D Z$.
+
+Problem 3. Let $A(n)$ denote the number of sequences $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers for which $a_{1}+\cdots+a_{k}=n$ and each $a_{i}+1$ is a power of two $(i=1,2, \ldots, k)$. Let $B(n)$ denote the number of sequences $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integers for which $b_{1}+\cdots+b_{m}=n$ and each inequality $b_{j} \geq 2 b_{j+1}$ holds $(j=1,2, \ldots, m-1)$.
+
+Prove that $A(n)=B(n)$ for every positive integer $n$.
+Solution. We say that a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of positive integers has type $A$ if $a_{i}+1$ is a power of two for $i=1,2, \ldots, k$. We say that a sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$ of positive integer has type $B$ if $b_{j} \geq 2 b_{j+1}$ for $j=1,2, \ldots, m-1$.
+
+Recall that the binary representation of a positive integer expresses it as a sum of distinct powers of two in a unique way. Furthermore, we have the following formula for every positive integer $N$
+
+$$
+2^{N}-1=2^{N-1}+2^{N-2}+\cdots+2^{1}+2^{0}
+$$
+
+Given a sequence $a_{1} \geq a_{2} \geq \ldots \geq a_{k}$ of type $A$, use the preceding formula to express each term as a sum of powers of two. Write these powers of two in left-aligned rows, in decreasing order of size. By construction, $a_{i}$ is the sum of the numbers in the $i$ th row. For example, we obtain the following array when we start with the type A sequence $15,15,7,3,3,3,1$.
+
+| 8 | 4 | 2 | 1 |
+| :--- | :--- | :--- | :--- |
+| 8 | 4 | 2 | 1 |
+| 4 | 2 | 1 |  |
+| 2 | 1 |  |  |
+| 2 | 1 |  |  |
+| 2 | 1 |  |  |
+| 1 |  |  |  |
+| 27 | 13 | 5 | 2 |
+
+Define the sequence $b_{1}, b_{2}, \ldots, b_{n}$ by setting $b_{j}$ to be the sum of the numbers in the $j$ th column of the array. For example, we obtain the sequence $27,13,5,2$ from the array above. We now show that this new sequence has type B. This is clear from the fact that each column in the array contains at least as many entries as the column to the right of it and that each number larger than 1 in the array is twice the number to the right of it. Furthermore, it is clear that $a_{1}+a_{2}+\cdots+a_{k}=b_{1}+b_{2}+\cdots+b_{m}$, since both are equal to the sum of all the entries in the array.
+
+We now show that we can do this operation backwards. Suppose that we are given a type B sequence $b_{1} \geq b_{2} \geq \ldots \geq b_{m}$. We construct an array inductively as follows:
+
+- We fill $b_{m}$ left-aligned rows with the numbers $2^{m-1}, 2^{m-2}, \ldots, 2^{1}, 2^{0}$.
+- Then we fill $b_{m-1}-2 b_{m}$ left aligned rows with the numbers $2^{m-2}, 2^{m-3}, \ldots, 2^{1}, 2^{0}$.
+- Then we fill $b_{m-2}-2 b_{m-1}$ left aligned rows with the numbers $2^{m-3}, 2^{m-4}, \ldots, 2^{1}, 2^{0}$, and so on.
+- In the last step we fill $b_{1}-2 b_{2}$ left-aligned rows with the number 1 .
+
+For example, if we start with the type B sequence $27,13,5,2$, we obtain once again the array above. We define the sequence $a_{1}, a_{2}, \ldots, a_{k}$ by setting $a_{i}$ to be the sum of the numbers in the $i$ th row of the array. By construction, this sequence has type A. Furthermore, it is clear that $a_{1}+\cdots+a_{k}=b_{1}+\cdots+b_{m}$, since once again both sums are equal to the sum of all the entries in the array.
+
+We have defined an operation that starts with a sequence of type A , produces an array whose row sums are given by the sequence, and outputs a sequence of type B corresponding to the column sums. We have also defined an operation that starts with a sequence of type B, produces an array whose column sums are given by the sequence, and outputs a sequence of type A corresponding to the row sums. The arrays produced in both cases comprise leftaligned rows of the form $2^{N-1}, 2^{N-2}, \ldots, 2^{1}, 2^{0}$, with non-increasing lengths. Let us refer to arrays obeying these properties as marvelous.
+
+To show that these two operations are inverses of each other, it then suffices to prove that marvelous arrays are uniquely defined by either their row sums or their column sums. The former is obviously true and the latter arises from the observation that each step in the above inductive algorithm was forced in order to create a marvelous array with the prescribed column sums.
+
+Thus, we have produced a bijection between the sequences of type A with sum $n$ and the sequences of type B with sum $n$. So we can conclude that $A(n)=B(n)$ for every positive integer $n$.
+
+Remark The solution above provides a bijection between type A and type B sequences via an algorithm. There are alternative ways to provide such a bijection. For example, given the numbers $a_{1} \geq \ldots \geq a_{k}$ we may define the $b_{i}$ 's as
+
+$$
+b_{j}=\sum_{i}\left\lfloor\frac{a_{i}+1}{2^{j}}\right\rfloor .
+$$
+
+Conversely, given the numbers $b_{1} \geq \ldots \geq b_{m}$, one may define the $a_{i}$ 's by taking, as in the solution, $b_{m}$ numbers equal to $2^{m}-1, b_{m-1}-2 b_{m}$ numbers equal to $2^{m-1}-1, \ldots$, and $b_{1}-2 b_{2}$ numbers equal to $2^{1}-1$. One now needs to verify that these maps are mutually inverse.
+
+Problem 4. Call a rational number $r$ powerful if $r$ can be expressed in the form $\frac{p^{k}}{q}$ for some relatively prime positive integers $p, q$ and some integer $k>1$. Let $a, b, c$ be positive
+rational numbers such that $a b c=1$. Suppose there exist positive integers $x, y, z$ such that $a^{x}+b^{y}+c^{z}$ is an integer. Prove that $a, b, c$ are all powerful.
+
+Solution. Let $a=\frac{a_{1}}{b_{1}}, b=\frac{a_{2}}{b_{2}}$, where $\operatorname{gcd}\left(a_{1}, b_{1}\right)=\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$. Then $c=\frac{b_{1} b_{2}}{a_{1} a_{2}}$. The condition that $a^{x}+b^{y}+c^{z}$ is an integer becomes
+
+$$
+\frac{a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}}{a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y}} \in \mathbb{Z}
+$$
+
+which can be restated as
+
+$$
+a_{1}^{z} a_{2}^{z} b_{1}^{x} b_{2}^{y} \mid a_{1}^{x+z} a_{2}^{z} b_{2}^{y}+a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
+$$
+
+In particular, $a_{1}^{z}$ divides the right-hand side. Since it divides the first and second terms in the sum, we conclude that $a_{1}^{z} \mid b_{1}^{x+z} b_{2}^{y+z}$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=1$, we have $a_{1}^{z} \mid b_{2}^{y+z}$.
+
+Let $p$ be a prime that divides $a_{1}$. Let $m, n \geq 1$ be integers such that $p^{n} \| a_{1}$ (i.e. $p^{n} \mid a_{1}$ but $\left.p^{n+1} \nmid a_{1}\right)$ and $p^{m} \| b_{2}$. The fact that $a_{1}^{z} \mid b_{2}^{y+z}$ implies $n z \leq m(y+z)$. Since $\operatorname{gcd}\left(a_{1}, b_{1}\right)=$ $\operatorname{gcd}\left(a_{2}, b_{2}\right)=1$, we have $p$ does not divide $b_{1}$ and does not divide $a_{2}$. Thus
+
+$$
+p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x} \text { and } p^{m(y+z)} \| b_{1}^{x+z} b_{2}^{y+z}
+$$
+
+On the other hand, (1) implies that
+
+$$
+p^{n z+m y} \mid a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z} .
+$$
+
+If $n z<m(y+z)$, then (2) gives $p^{n z} \| a_{1}^{z} a_{2}^{y+z} b_{1}^{x}+b_{1}^{x+z} b_{2}^{y+z}$, which contradicts (3). Thus $n z=m(y+z)$ so $n$ is divisible by $k:=\frac{y+z}{\operatorname{gcd}(z, y+z)}>1$. Thus each exponent in the prime decomposition of $a_{1}$ must be divisible by $k$. Hence $a_{1}$ is a perfect $k$-power which means $a$ is powerful. Similarly, $b$ and $c$ are also powerful.
+
+Problem 5. Let $n$ be a positive integer. A pair of $n$-tuples $\left(a_{1}, \ldots, a_{n}\right)$ and $\left(b_{1}, \ldots, b_{n}\right)$ with integer entries is called an exquisite pair if
+
+$$
+\left|a_{1} b_{1}+\cdots+a_{n} b_{n}\right| \leq 1
+$$
+
+Determine the maximum number of distinct $n$-tuples with integer entries such that any two of them form an exquisite pair.
+
+Answer: The maximum is $n^{2}+n+1$.
+Solution. First, we construct an example with $n^{2}+n+1 n$-tuples, each two of them forming an exquisite pair. In the following list, $*$ represents any number of zeros as long as the total number of entries is $n$.
+
+- (*)
+- $(*, 1, *)$
+- $(*,-1, *)$
+- $(*, 1, *, 1, *)$
+- $(*, 1, *,-1, *)$
+
+For example, for $n=2$ we have the tuples $(0,0),(0,1),(1,0),(0,-1),(-1,0),(1,1),(1,-1)$. The total number of such tuples is $1+n+n+\binom{n}{2}+\binom{n}{2}=n^{2}+n+1$. For any two of them, at most two of the products $a_{i} b_{i}$ are non-zero. The only case in which two of them are non-zero is when we take a sequence $(*, 1, *, 1, *)$ and a sequence $(*, 1, *,-1, *)$ with zero entries in the same places. But in this case one $a_{i} b_{i}$ is 1 and the other -1 . This shows that any two of these sequences form an exquisite pair.
+
+Next, we claim that among any $n^{2}+n+2$ tuples, some two of them do not form an exquisite pair. We begin with lemma.
+
+Lemma. Given $2 n+1$ distinct non-zero $n$-tuples of real numbers, some two of them $\left(a_{1}, \ldots, a_{n}\right)$ and $\left(b_{1}, \ldots, b_{n}\right)$ satisfy $a_{1} b_{1}+\cdots+a_{n} b_{n}>0$.
+
+Proof of Lemma. We proceed by induction. The statement is easy for $n=1$ since for every three non-zero numbers there are two of them with the same sign. Assume that the statement is true for $n-1$ and consider $2 n+1$ tuples with $n$ entries. Since we are working with tuples of real numbers, we claim that we may assume that one of the tuples is $a=(0,0, \ldots, 0,-1)$. Let us postpone the proof of this claim for the moment.
+
+If one of the remaining tuples $b$ has a negative last entry, then $a$ and $b$ satisfy the desired condition. So we may assume all the remaining tuples has a non-negative last entry. Now, from each tuple remove the last number. If two $n$-tuples $b$ and $c$ yield the same $(n-1)$-tuple, then
+
+$$
+b_{1} c_{1}+\cdots+b_{n-1} c_{n-1}+b_{n} c_{n}=b_{1}^{2}+\cdots+b_{n-1}^{2}+b_{n} c_{n}>0
+$$
+
+and we are done. The remaining case is that all the $n$-tuples yield distinct ( $n-1$ )-tuples. Then at most one of them is the zero $(n-1)$-tuple, and thus we can use the inductive hypothesis on $2 n-1$ of them. So we find $b$ and $c$ for which
+
+$$
+\left(b_{1} c_{1}+\cdots+b_{n-1} c_{n-1}\right)+b_{n} c_{n}>0+b_{n} c_{n}>0
+$$
+
+The only thing that we are left to prove is that in the inductive step we may assume that one of the tuples is $a=(0,0, \ldots, 0,-1)$. Fix one of the tuples $x=\left(x_{1}, \ldots, x_{n}\right)$. Set a real number $\varphi$ for which $\tan \varphi=\frac{x_{1}}{x_{2}}$. Change each tuple $a=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ (including $x$ ), to the tuple
+
+$$
+\left(a_{1} \cos \varphi-a_{2} \sin \varphi, a_{1} \sin \varphi+a_{2} \cos \varphi, a_{3}, a_{4}, \ldots, a_{n}\right)
+$$
+
+A straightforward calculation shows that the first coordinate of the tuple $x$ becomes 0 , and that all the expressions of the form $a_{1} b_{1}+\cdots+a_{n} b_{n}$ are preserved. We may iterate this process until all the entries of $x$ except for the last one are equal to 0 . We finish by multiplying all the entries in all the tuples by a suitable constant that makes the last entry of $x$ equal to -1 . This preserves the sign of all the expressions of the form $a_{1} b_{1}+\cdots+a_{n} b_{n}$.
+
+We proceed to the proof of our claim. Let $A$ be a set of non-zero tuples among which any two form an exquisite pair. It suffices to prove that $|A| \leq n^{2}+n$. We can write $A$ as a disjoint union of subsets $A_{1} \cup A_{2} \cup \ldots \cup A_{n}$, where $A_{i}$ is the set of tuples in $A$ whose last non-zero entry appears in the $i$ th position. We will show that $\left|A_{i}\right| \leq 2 i$, which will finish our proof since $2+4+\cdots+2 n=n^{2}+n$.
+
+Proceeding by contradiction, suppose that $\left|A_{i}\right| \geq 2 i+1$. If $A_{i}$ has three or more tuples whose only non-zero entry is in the $i$ th position, then for two of them this entry has the same sign. Since the tuples are different and their entries are integers, this yields two tuples for which $\left|\sum a_{i} b_{i}\right| \geq 2$, a contradiction. So there are at most two such tuples. We remove them from $A_{i}$.
+
+Now, for each of the remaining tuples $a$, if it has a positive $i$ th coordinate, we keep $a$ as it is. If it has a negative $i$ th coordinate, we replace it with the opposite tuple $-a$ with entries with opposite signs. This does not changes the exquisite pairs condition.
+
+After making the necessary changes, we have two cases. The first case is that there are two tuples $a$ and $b$ that have the same first $i-1$ coordinates and thus
+
+$$
+a_{1} b_{1}+\cdots+a_{i-1} b_{i-1}=a_{1}^{2}+\cdots+a_{i-1}^{2}>0
+$$
+
+and thus is at least 1 (the entries are integers). The second case is that no two tuples have the same first $i-1$ coordinates, but then by the Lemma we find two tuples $a$ and $b$ for which
+
+$$
+a_{1} b_{1}+\cdots+a_{i-1} b_{i-1} \geq 1
+$$
+
+In any case, we obtain
+
+$$
+a_{1} b_{1}+\cdots+a_{i-1} b_{i-1}+a_{i} b_{i} \geq 2
+$$
+
+This yields a final contradiction to the exquisite pair hypothesis.
+

APMO/md/en-apmo2018_sol.md

@@ -0,0 +1,256 @@
+# Solutions of APMO 2018 
+
+Problem 1. Let $H$ be the orthocenter of the triangle $A B C$. Let $M$ and $N$ be the midpoints of the sides $A B$ and $A C$, respectively. Assume that $H$ lies inside the quadrilateral $B M N C$ and that the circumcircles of triangles $B M H$ and $C N H$ are tangent to each other. The line through $H$ parallel to $B C$ intersects the circumcircles of the triangles $B M H$ and $C N H$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $M K$ and $N L$ and let $J$ be the incenter of triangle $M H N$. Prove that $F J=F A$.
+
+## Solution.
+
+Lemma 1. In a triangle $A B C$, let $D$ be the intersection of the interior angle bisector at $A$ with the circumcircle of $A B C$, and let $I$ be the incenter of $\triangle A B C$. Then
+
+$$
+D I=D B=D C
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-1.jpg?height=637&width=618&top_left_y=938&top_left_x=725)
+
+Proof.
+
+$$
+\angle D B I=\frac{\angle B A C}{2}+\frac{\widehat{B}}{2}=\angle D I B \quad \Rightarrow \quad D I=D B
+$$
+
+Analogously $D I=D C$.
+We start solving the problem. First we state some position considerations. Since there is an arc of the circumcircle of $B H M$ outside the triangle $A B C$, it must happen that $K$ and $N$ lie on opposite sides of $A M$. Similarly, $L$ and $M$ lie on opposite sides of $A N$. Also, $K$ and $L$ lie on the same side of $M N$, and opposite to $A$. Therefore, $F$ lies inside the triangle $A M N$.
+
+Now, since $H$ is the orthocenter of $\triangle A B C$ and the circumcircles of $B M H$ and $C N H$ are tangent we have
+
+$$
+\angle A B H=90^{\circ}-\angle B A C=\angle A C H \quad \Rightarrow \quad \angle M H N=\angle M B H+\angle N C H=180^{\circ}-2 \angle B A C .
+$$
+
+So $\angle M B H=\angle M K H=\angle N C H=\angle N L H=90^{\circ}-\angle B A C$ and, since $M N \| K L$, we have
+
+$$
+\angle F M N=\angle F N M=90^{\circ}-\angle B A C \Rightarrow \angle M F N=2 \angle B A C .
+$$
+
+The relations (1) and (2) yield that the quadrilateral $M F N H$ is cyclic, with the vertices in this order around the circumference. Since $F M=F N, \angle M F N=2 \angle B A C$ and $F$ is the correct side of $M N$ we have that the point $F$ is the circumcenter of triangle $A M N$, and thus $F A=F M=F N$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-2.jpg?height=801&width=1029&top_left_y=179&top_left_x=519)
+
+Since the quadrilateral $M F N H$ is cyclic, $F M=F N$ and $H$ lies on the correct side of $M N$, we have that $H, J$ and $F$ are collinear. According to Lemma $1, F J=F M=F N$. So $F J=F A$.
+
+Solution 2: According to Solution 1, we have $\angle M H N=180^{\circ}-2 \angle B A C$ and since the point $J$ is the incenter of $\triangle M H N$, we have $\angle M J N=90^{\circ}+\frac{1}{2} \angle M H N=180^{\circ}-\angle B A C$. So the quadrilateral $A M J N$ is cyclic.
+
+According to Solution 1, the point $F$ is the circumcenter of $\triangle A M N$. So $F J=F A$.
+
+Problem 2. Let $f(x)$ and $g(x)$ be given by
+
+$$
+f(x)=\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x-4}+\cdots+\frac{1}{x-2018}
+$$
+
+and
+
+$$
+g(x)=\frac{1}{x-1}+\frac{1}{x-3}+\frac{1}{x-5}+\cdots+\frac{1}{x-2017} .
+$$
+
+Prove that
+
+$$
+|f(x)-g(x)|>2
+$$
+
+for any non-integer real number $x$ satisfying $0<x<2018$.
+Solution 1 There are two cases: $2 n-1<x<2 n$ and $2 n<x<2 n+1$. Note that $f(2018-x)=-f(x)$ and $g(2018-x)=-g(x)$, that is, a half turn about the point $(1009,0)$ preserves the graphs of $f$ and $g$. So it suffices to consider only the case $2 n-1<x<2 n$.
+
+Let $d(x)=g(x)-f(x)$. We will show that $d(x)>2$ whenever $2 n-1<x<2 n$ and $n \in\{1,2, \ldots, 1009\}$.
+
+For any non-integer $x$ with $0<x<2018$, we have
+
+$$
+d(x+2)-d(x)=\left(\frac{1}{x+1}-\frac{1}{x+2}\right)+\left(\frac{1}{x-2018}-\frac{1}{x-2017}\right)>0+0=0
+$$
+
+Hence it suffices to prove $d(x)>2$ for $1<x<2$. Since $x<2$, it follows that $\frac{1}{x-2 i-1}>$ $\frac{1}{x-2 i}$ for $i=2,3, \ldots, 1008$. We also have $\frac{1}{x-2018}<0$. Hence it suffices to prove the following
+for $1<x<2$.
+
+$$
+\begin{aligned}
+& \frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x}-\frac{1}{x-2}>2 \\
+\Leftrightarrow & \left(\frac{1}{x-1}+\frac{1}{2-x}\right)+\left(\frac{1}{x-3}-\frac{1}{x}\right)>2 \\
+\Leftrightarrow & \frac{1}{(x-1)(2-x)}+\frac{3}{x(x-3)}>2 .
+\end{aligned}
+$$
+
+By the $G M-H M$ inequality (alternatively, by considering the maximum of the quadratic $(x-1)(2-x))$ we have
+
+$$
+\frac{1}{x-1} \cdot \frac{1}{2-x}>\left(\frac{2}{(x-1)+(2-x)}\right)^{2}=4
+$$
+
+To find a lower bound for $\frac{3}{x(x-3)}$, note that $x(x-3)<0$ for $1<x<2$. So we seek an upper bound for $x(x-3)$. From the shape of the quadratic, this occurs at $x=1$ or $x=2$, both of which yield $\frac{3}{x(x-3)}>-\frac{3}{2}$.
+
+It follows that $d(x)>4-\frac{3}{2}>2$, as desired.
+
+## Solution 2
+
+As in Solution 1, we may assume $2 n-1<x<2 n$ for some $1 \leq n \leq 1009$. Let $d(x)=$ $f(x)-g(x)$, and note that
+
+$$
+d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}
+$$
+
+We split the sum into three parts: the terms before $m=n$, after $m=n$, and the term $m=n$. The first two are
+
+$$
+\begin{aligned}
+0 & \leq \sum_{m=1}^{n-1} \frac{1}{(x-2 m)(x-2 m+1)} \\
+& \leq \sum_{m=1}^{n-1} \frac{1}{(2 n-1-2 m)(2 n-2 m)}=\sum_{i=1}^{n-1} \frac{1}{(2 i)(2 i-1)} \leq \sum_{i=1}^{1008} \frac{1}{2 i-1}-\frac{1}{2 i} \\
+0 & \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-x)(2 m-1-x)} \\
+& \leq \sum_{m=n+1}^{1009} \frac{1}{(2 m-2 n+1)(2 m-2 n)}=\sum_{i=1}^{1009-n} \frac{1}{(2 i+1)(2 i)} \leq \sum_{i=1}^{1008} \frac{1}{2 i}-\frac{1}{2 i+1} .
+\end{aligned}
+$$
+
+When we add the two sums the terms telescope and we are left with
+
+$$
+0 \leq \sum_{1 \leq m \leq 1009, m \neq n} \frac{1}{(x-2 m)(x-2 m+1)} \leq 1-\frac{1}{2017}<1
+$$
+
+For the term $m=n$, we write
+
+$$
+0<-(x-2 n)(x-2 n+1)=0.25-(x-2 n+0.5)^{2} \leq 0.25
+$$
+
+whence
+
+$$
+-4 \geq \frac{1}{(x-2 n)(x-2 n+1)}
+$$
+
+Finally, $\frac{1}{x}<1$ since $x>2 n-1 \geq 1$. Combining these we get
+
+$$
+d(x)=\frac{1}{x}+\sum_{m=1}^{1009} \frac{1}{(x-2 m)(x-2 m+1)}<1+1-4<-2 .
+$$
+
+## Solution 3
+
+First notice that
+
+$$
+f(x)-g(x)=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-2}-\cdots-\frac{1}{x-2017}+\frac{1}{x-2018} .
+$$
+
+As in Solution 1, we may deal only with the case $2 n<x<2 n+1$. Then $x-2 k+1$ and $x-2 k$ never differ in sign for any integer $k$. Then
+
+$$
+\begin{aligned}
+-\frac{1}{x-2 k+1}+\frac{1}{x-2 k} & =\frac{1}{(x-2 k+1)(x-2 k)}>0 \quad \text { for } k=1,2, \ldots, n-1, n+2, \ldots, 1009 \\
+\frac{1}{x-2 n}-\frac{1}{x-2 n-1} & =\frac{1}{(x-2 n)(2 n+1-x)} \geq\left(\frac{2}{x-2 n+2 n+1-x}\right)^{2}=4
+\end{aligned}
+$$
+
+Therefore, summing all inequalities and collecting the remaining terms we find $f(x)-g(x)>$ $4+\frac{1}{x-2}>4-1=3$ for $0<x<1$ and, for $n>0$,
+
+$$
+\begin{aligned}
+f(x)-g(x) & >\frac{1}{x}-\frac{1}{x-2 n+1}+4+\frac{1}{x-2 n-2} \\
+& =\frac{1}{x}-\frac{1}{x-2 n+1}+4-\frac{1}{2 n+2-x} \\
+& >\frac{1}{x}-\frac{1}{2 n-2 n+1}+4-\frac{1}{2 n+2-2 n-1} \\
+& =2+\frac{1}{x}>2 .
+\end{aligned}
+$$
+
+Problem 3. A collection of $n$ squares on the plane is called tri-connected if the following criteria are satisfied:
+(i) All the squares are congruent.
+(ii) If two squares have a point $P$ in common, then $P$ is a vertex of each of the squares.
+(iii) Each square touches exactly three other squares.
+
+How many positive integers $n$ are there with $2018 \leq n \leq 3018$, such that there exists a collection of $n$ squares that is tri-connected?
+
+Answer: 501
+Solution. We will prove that there is no tri-connected collection if $n$ is odd, and that tri-connected collections exist for all even $n \geq 38$. Since there are 501 even numbers in the range from 2018 to 3018, this yields 501 as the answer.
+
+For any two different squares $A$ and $B$, let us write $A \sim B$ to mean that square $A$ touches square $B$. Since each square touches exactly three other squares, and there are $n$ squares in total, the total number of instances of $A \sim B$ is $3 n$. But $A \sim B$ if and only if $B \sim A$. Hence the total number of instances of $A \sim B$ is even. Thus $3 n$ and hence also $n$ is even.
+
+We now construct tri-connected collections for each even $n$ in the range. We show two
+Construction 1 The idea is to use the following two configurations. Observe that in each configuration every square is related to three squares except for the leftmost and rightmost squares which are related to two squares. Note that the configuration on the left is of variable length. Also observe that multiple copies of the configuration on the right can be chained together to end around corners.
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-5.jpg?height=164&width=835&top_left_y=752&top_left_x=625)
+
+Putting the above two types of configurations together as in the following figure yields a tri-connected collection for every even $n \geq 38$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-5.jpg?height=309&width=789&top_left_y=1099&top_left_x=631)
+
+Construction 2 Consider a regular $4 n$-gon $A_{1} A_{2} \cdots A_{4 n}$, and make $4 n$ squares on the outside of the $4 n$-gon with one side being on the $4 n$-gon. Reflect squares sharing sides $A_{4 m+2} A_{4 m+3}, A_{4 m+3} A_{4 m+4}$ across line $A_{4 m+2} A_{4 m+4}$, for $0 \leq m \leq n-1$. This will produce a tri-connected set of $6 n$ squares, as long as the squares inside the $4 n$-gon do not intersect. When $n \geq 4$, this will be true. The picture for $n=24$ is as follows:
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-5.jpg?height=830&width=824&top_left_y=1758&top_left_x=593)
+
+To treat the other cases, consider the following gadget
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-6.jpg?height=184&width=272&top_left_y=296&top_left_x=869)
+
+Two squares touch 3 other squares, and the squares containing $X, Y$ touch 2 other squares. Take the $4 n$-gon from above, and break it into two along the line $A_{1} A_{2 n}$, moving the two parts away from that line. Do so until the gaps can be exactly filled by inserting two copies of the above figure, so that the vertices $X, Y$ touch the two vertices which used to be $A_{1}$ in one instance, and the two vertices which used to be $A_{2 n}$ in the other.
+This gives us a valid configuration for $6 n+8$ squares, $n \geq 4$. Finally, if we had instead spread the two parts out more and inserted two copies of the above figure into each gap, we would get $6 n+16$ for $n \geq 4$, which finishes the proof for all even numbers at least 36 .
+
+Problem 4. Let $A B C$ be an equilateral triangle. From the vertex $A$ we draw a ray towards the interior of the triangle such that the ray reaches one of the sides of the triangle. When the ray reaches a side, it then bounces off following the law of reflection, that is, if it arrives with a directed angle $\alpha$, it leaves with a directed angle $180^{\circ}-\alpha$. After $n$ bounces, the ray returns to $A$ without ever landing on any of the other two vertices. Find all possible values of $n$.
+
+Answer: All $n \equiv 1,5 \bmod 6$ with the exception of 5 and 17
+Solution. Consider an equilateral triangle $A A_{1} A_{2}$ of side length $m$ and triangulate it with unitary triangles. See the figure. To each of the vertices that remain after the triangulation we can assign a pair of coordinates $(a, b)$ where $a, b$ are non-negative integers, $a$ is the number of edges we travel in the $A A_{1}$ direction and $b$ is the number of edges we travel in the $A A_{2}$ direction to arrive to the vertex, (we have $A=(0,0), A_{1}=(m, 0)$ and $A_{2}=(0, m)$ ). The unitary triangle with vertex $A$ will be our triangle $A B C,(B=(1,0), C=(0,1))$. We can obtain every unitary triangle by starting with $A B C$ and performing reflections with respect to a side (the vertex $(1,1)$ is the reflection of $A$ with respect to $B C$, the vertex $(0,2)$ is the reflection of $B=(1,0)$ with respect to the side formed by $C=(1,0)$ and $(1,1)$, and so on).
+
+When we reflect a vertex $(a, b)$ with respect to a side of one of the triangles, the congruence of $a-b$ is preserved modulo 3. Furthermore, an induction argument shows that any two vertices $(a, b)$ and $\left(a^{\prime}, b^{\prime}\right)$ with $a-b \equiv a^{\prime}-b^{\prime}$ mod 3 can be obtained from each other by a series of such reflections. Therefore, the set of vertices $V$ that result from the reflections of $A$ will be those of the form $(a, b)$ satisfying $a \equiv b \bmod 3$. See the green vertices in the figure.
+
+Now, let $U$ be the set of vertices $u$ that satisfy that the line segment between $u$ and $A$ does not pass through any other vertex. A pair $(a, b)$ is in $U$ if and only if $\operatorname{gcd}(a, b)=1$, since otherwise for $d=\operatorname{gcd}(a, b)$ we have that the vertex $(a / d, b / d)$ also lies on the line segment between $u$ and $A$.
+
+Observe that the rays that come out from $A$ and eventually return to $A$ are those that come out towards a vertex in $V \cap U$ (they would be in $V$ to be able to come back to $A$ and in $U$ so that they do not reach a vertex beforehand). In the diagram, a ray toward one such vertex $(a, b)$ will intersect exactly $(a-1)+(b-1)+(a+b-1)=2(a+b)-3$ lines: $a-1$ of them parallel to $A B, b-1$ parallel to $A C$ and $a+b-1$ parallel to $B C$. Therefore, in the triangle $A B C$ the ray will bounce $2(a+b)-3$ times before returning to $A$. So we want to find all
+![](https://cdn.mathpix.com/cropped/2024_11_22_c87ece96a341aa1506ccg-7.jpg?height=906&width=1044&top_left_y=175&top_left_x=506)
+$n=2(a+b)-3$ where $a \equiv b \bmod 3$ and $\operatorname{gcd}(a, b)=1$.
+If $a+b$ is a multiple of 3 then we cannot satisfy both conditions simultaneously, therefore $n$ is not a multiple of 3 . We also know that $n$ is odd. Therefore $n \equiv 1,5,7,11 \bmod 12$. Note that the pair $(1,3 k+1)$ satisfies the conditions and we can create $n=2(3 k+2)-3=6 k+1$ for all $k \geq 0$ (this settles the question for $n \equiv 1,7 \bmod 12$ ). For $n \equiv 5 \bmod 12$ consider the pair $(3 k-1,3 k+5)$ when $k$ is even or $(3 k-4,3 k+8)$ when $k$ is odd. This gives us all the integers of the form $12 k+5$ for $k \geq 2$. For $11 \bmod 12$, take the pairs $(3 k-1,3 k+2)($ with $k \geq 1)$, which yield all positive integers of the form $12 k-1$.
+
+Finally, to discard 5 and 17 note that the only pairs $(a, b)$ that are solutions to $2(a+b)-3=5$ or $2(a+b)-3=17$ with the same residue $\bmod 3$ in this range are the non-relatively prime pairs $(2,2),(2,8)$ and $(5,5)$.
+
+Problem 5. Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(s t)$ is also an integer.
+
+Answer: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer.
+Solution 1: $P(x)=x^{n}+k,-x^{n}+k$ for $n$ a non-negative integer and $k$ an integer.
+Notice that if $P(x)$ is a solution, then so is $P(x)+k$ and $-P(x)+k$ for any integer $k$, so we may assume that the leading coefficient of $P(x)$ is positive and that $P(0)=0$, i.e., we can assume that $P(x)=\sum_{i=1}^{n} a_{i} x^{i}$ with $a_{n}>0$. We are going to prove that $P(x)=x^{n}$ in this case.
+
+Let $p$ be a large prime such that $p>\sum_{i=1}^{n}\left|a_{i}\right|$. Because $P$ has a positive leading coefficient and $p$ is large enough, we can find $t \in \mathbb{R}$ such that $P(t)=p$. Denote the greatest common divisor of the polynomial $P(x)-p$ and $P(2 x)-P(2 t)$ as $f(x)$, and $t$ is a root of it, so $f$ is a non-constant polynomial. Notice that $P(2 t)$ is an integer by using the hypothesis for $s=2$ and $t$. Since $P(x)-p$ and $P(2 x)-P(2 t)$ are polynomials with integer coefficients, $f$ can be chosen as a polynomial with rational coefficients.
+
+In the following, we will prove that $f$ is the same as $P(x)-p$ up to a constant multiplier. Say $P(x)-p=f(x) g(x)$, where $f$ and $g$ are non-constant polynomials. By Gauss's lemma, we can get $f_{1}, g_{1}$ with $P(x)-p=f_{1}(x) g_{1}(x)$ where $f_{1}$ is a scalar multiple of $f$ and $g_{1}$ is a scalar multiple of $g$ and one of $f_{1}, g_{1}$ has constant term $\pm 1$ (this is because $-p=P(0)-p=f(0) g(0)$ with $p$ prime). So $P(x)-p$ has at least one root $r$ with absolute value not greater than 1 (using
+
+Vieta, the product of the roots of the polynomial with constant term $\pm 1$ is $\pm 1$ ), but
+
+$$
+|P(r)-p|=\left|\sum_{i=1}^{n} a_{i} r^{i}-p\right|>p-\sum_{i=1}^{n}\left|a_{i}\right|>0
+$$
+
+hence we get a contradiction!
+Therefore $f$ is a constant multiple of $P(x)-p$, so $P(2 x)-P(2 t)$ is a constant multiple of $P(x)-p$ because they both have the same degree. By comparing leading coefficients we get that $P(2 x)-P(2 t)=2^{n}(P(x)-p)$. Comparing the rest of the coefficients we get that $P(x)=a_{n} x^{n}$. If we let $a=b=\left(1 / a_{n}\right)^{1 / n}$, then $P(a)=P(b)=1$, so $P(a b)$ must also be an integer. But $P(a b)=\frac{1}{a_{n}}$. Therefore $a_{n}=1$ and the proof is complete.
+
+Solution 2: Assume $P(x)=\sum_{i=0}^{n} a_{i} x^{i}$. Consider the following system of equations
+
+$$
+\begin{aligned}
+& a_{0}=P(0) \\
+& a_{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{0}=P(t) \\
+& 2^{n} a_{n} t^{n}+2^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(2 t) \\
+& \vdots \\
+& n^{n} a_{n} t^{n}+n^{n-1} a_{n-1} t^{n-1}+\cdots+a_{0}=P(n t)
+\end{aligned}
+$$
+
+viewing $a_{k} t^{k}$ as variables. Note that if $P(t)$ is an integer, then by the hypothesis all the terms on the right hand side of the equations are integers as well. By using Cramer's rule, we can get that $a_{k} t^{k}=D / M$, where $D$ is an integer and $M$ is the following determinant
+
+$$
+\left|\begin{array}{ccccc}
+1 & 0 & 0 & \cdots & 0 \\
+1 & 1 & 1 & \cdots & 1 \\
+1 & 2 & 4 & \cdots & 2^{n} \\
+\vdots & \vdots & \vdots & & \vdots \\
+1 & n & n^{2} & \cdots & n^{n}
+\end{array}\right| \neq 0
+$$
+
+Thus, if we let $r$ be the smallest positive index such that $a_{r} \neq 0$, we can express each $t \in \mathbb{R}$ with $P(t) \in \mathbb{Z}$ in the form $\left(\frac{m}{M^{\prime}}\right)^{1 / r}$ for some integer $m$, and where $M^{\prime}=M \times a_{r}$ is a constant.
+
+We can choose $L$ large enough such that $\left.P\right|_{\mathbb{R}_{\geq L}}$ is injective, and for any larger $N$, the growth order of the number of values in the form $\left(\frac{m}{M^{\prime}}\right)^{1 / r}$ is $N^{r}$, while the growth order of the number of integers in $[P(L), P(N)]$ is $N^{n}$, so $r=n$. Therefore $P(x)$ is of the form $a_{n} x^{n}+k$. The problem can be finished as in Solution 1.
+

APMO/md/en-apmo2019_sol.md

@@ -0,0 +1,288 @@
+# Solutions of APMO 2019 
+
+Problem 1. Let $\mathbb{Z}^{+}$be the set of positive integers. Determine all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ such that $a^{2}+f(a) f(b)$ is divisible by $f(a)+b$ for all positive integers $a$ and $b$.
+
+Answer: The answer is $f(n)=n$ for all positive integers $n$.
+Clearly, $f(n)=n$ for all $n \in \mathbb{Z}^{+}$satisfies the original relation. We show some possible approaches to prove that this is the only possible function.
+
+Solution. First we perform the following substitutions on the original relation:
+
+1. With $a=b=1$, we find that $f(1)+1 \mid f(1)^{2}+1$, which implies $f(1)=1$.
+2. With $a=1$, we find that $b+1 \mid f(b)+1$. In particular, $b \leq f(b)$ for all $b \in \mathbb{Z}^{+}$.
+3. With $b=1$, we find that $f(a)+1 \mid a^{2}+f(a)$, and thus $f(a)+1 \mid a^{2}-1$. In particular, $f(a) \leq a^{2}-2$ for all $a \geq 2$.
+
+Now, let $p$ be any odd prime. Substituting $a=p$ and $b=f(p)$ in the original relation, we find that $2 f(p) \mid p^{2}+f(p) f(f(p))$. Therefore, $f(p) \mid p^{2}$. Hence the possible values of $f(p)$ are $1, p$ and $p^{2}$. By (2) above, $f(p) \geq p$ and by (3) above $f(p) \leq p^{2}-2$. So $f(p)=p$ for all primes $p$.
+
+Substituting $a=p$ into the original relation, we find that $b+p \mid p^{2}+p f(b)$. However, since $(b+p)(f(b)+p-b)=p^{2}-b^{2}+b f(b)+p f(b)$, we have $b+p \mid b f(b)-b^{2}$. Thus, for any fixed $b$ this holds for arbitrarily large primes $p$ and therefore we must have $b f(b)-b^{2}=0$, or $f(b)=b$, as desired.
+
+Solution 2: As above, we have relations (1)-(3). In (2) and (3), for $b=2$ we have $3 \mid f(2)+1$ and $f(2)+1 \mid 3$. These imply $f(2)=2$.
+
+Now, using $a=2$ we get $2+b \mid 4+2 f(b)$. Let $f(b)=x$. We have
+
+$$
+\begin{aligned}
+1+x & \equiv 0 \quad(\bmod b+1) \\
+4+2 x & \equiv 0 \quad(\bmod b+2)
+\end{aligned}
+$$
+
+From the first equation $x \equiv b(\bmod b+1)$ so $x=b+(b+1) t$ for some integer $t \geq 0$. Then
+
+$$
+0 \equiv 4+2 x \equiv 4+2(b+(b+1) t) \equiv 4+2(-2-t) \equiv-2 t \quad(\bmod b+2)
+$$
+
+Also $t \leq b-2$ because $1+x \mid b^{2}-1$ by (3).
+If $b+2$ is odd, then $t \equiv 0(\bmod b+2)$. Then $t=0$, which implies $f(b)=b$.
+If $b+2$ is even, then $t \equiv 0(\bmod (b+2) / 2)$. Then $t=0$ or $t=(b+2) / 2$. But if $t \neq 0$, then by definition $(b+4) / 2=(1+t)=(x+1) /(b+1)$ and since $x+1 \mid b^{2}-1$, then $(b+4) / 2$ divides $b-1$. Therefore $b+4 \mid 10$ and the only possibility is $b=6$. So for even $b, b \neq 6$ we have $f(b)=b$.
+
+Finally, by (2) and (3), for $b=6$ we have $7 \mid f(6)+1$ and $f(6)+1 \mid 35$. This means $f(6)=6$ or $f(6)=34$. The later is discarded as, for $a=5, b=6$, we have by the original equation that $11 \mid 5(5+f(6))$. Therefore $f(n)=n$ for every positive integer $n$.
+
+Solution 3: We proceed by induction. As in Solution 1, we have $f(1)=1$. Suppose that $f(n-1)=n-1$ for some integer $n \geq 2$.
+
+With the substitution $a=n$ and $b=n-1$ in the original relation we obtain that $f(n)+$ $n-1 \mid n^{2}+f(n)(n-1)$. Since $f(n)+n-1 \mid(n-1)(f(n)+n-1)$, then $f(n)+n-1 \mid 2 n-1$.
+
+With the substitution $a=n-1$ and $b=n$ in the original relation we obtain that $2 n-$ $1 \mid(n-1)^{2}+(n-1) f(n)=(n-1)(n-1+f(n))$. Since $(2 n-1, n-1)=1$, we deduce that $2 n-1 \mid f(n)+n-1$.
+
+Therefore, $f(n)+n-1=2 n-1$, which implies the desired $f(n)=n$.
+
+Problem 2. Let $m$ be a fixed positive integer. The infinite sequence $\left\{a_{n}\right\}_{n \geq 1}$ is defined in the following way: $a_{1}$ is a positive integer, and for every integer $n \geq 1$ we have
+
+$$
+a_{n+1}= \begin{cases}a_{n}^{2}+2^{m} & \text { if } a_{n}<2^{m} \\ a_{n} / 2 & \text { if } a_{n} \geq 2^{m}\end{cases}
+$$
+
+For each $m$, determine all possible values of $a_{1}$ such that every term in the sequence is an integer.
+
+Answer: The only value of $m$ for which valid values of $a_{1}$ exist is $m=2$. In that case, the only solutions are $a_{1}=2^{\ell}$ for $\ell \geq 1$.
+
+Solution. Suppose that for integers $m$ and $a_{1}$ all the terms of the sequence are integers. For each $i \geq 1$, write the $i$ th term of the sequence as $a_{i}=b_{i} 2^{c_{i}}$ where $b_{i}$ is the largest odd divisor of $a_{i}$ (the "odd part" of $a_{i}$ ) and $c_{i}$ is a nonnegative integer.
+
+Lemma 1. The sequence $b_{1}, b_{2}, \ldots$ is bounded above by $2^{m}$.
+Proof. Suppose this is not the case and take an index $i$ for which $b_{i}>2^{m}$ and for which $c_{i}$ is minimal. Since $a_{i} \geq b_{i}>2^{m}$, we are in the second case of the recursion. Therefore, $a_{i+1}=a_{i} / 2$ and thus $b_{i+1}=b_{i}>2^{m}$ and $c_{i+1}=c_{i}-1<c_{i}$. This contradicts the minimality of $c_{i}$.
+
+Lemma 2. The sequence $b_{1}, b_{2}, \ldots$ is nondecreasing.
+Proof. If $a_{i} \geq 2^{m}$, then $a_{i+1}=a_{i} / 2$ and thus $b_{i+1}=b_{i}$. On the other hand, if $a_{i}<2^{m}$, then
+
+$$
+a_{i+1}=a_{i}^{2}+2^{m}=b_{i}^{2} 2^{2 c_{i}}+2^{m}
+$$
+
+and we have the following cases:
+
+- If $2 c_{i}>m$, then $a_{i+1}=2^{m}\left(b_{i}^{2} 2^{2 c_{i}-m}+1\right)$, so $b_{i+1}=b_{i}^{2} 2^{2 c_{i}-m}+1>b_{i}$.
+- If $2 c_{i}<m$, then $a_{i+1}=2^{2 c_{i}}\left(b_{i}^{2}+2^{m-2 c_{i}}\right)$, so $b_{i+1}=b_{i}^{2}+2^{m-2 c_{i}}>b_{i}$.
+- If $2 c_{i}=m$, then $a_{i+1}=2^{m+1} \cdot \frac{b_{i}^{2}+1}{2}$, so $b_{i+1}=\left(b_{i}^{2}+1\right) / 2 \geq b_{i}$ since $b_{i}^{2}+1 \equiv 2(\bmod 4)$.
+
+By combining these two lemmas we obtain that the sequence $b_{1}, b_{2}, \ldots$ is eventually constant. Fix an index $j$ such that $b_{k}=b_{j}$ for all $k \geq j$. Since $a_{n}$ descends to $a_{n} / 2$ whenever $a_{n} \geq 2^{m}$, there are infinitely many terms which are smaller than $2^{m}$. Thus, we can choose an $i>j$ such that $a_{i}<2^{m}$. From the proof of Lemma $2, a_{i}<2^{m}$ and $b_{i+1}=b_{i}$ can happen simultaneously only when $2 c_{i}=m$ and $b_{i+1}=b_{i}=1$. By Lemma 2 , the sequence $b_{1}, b_{2}, \ldots$ is constantly 1 and thus $a_{1}, a_{2}, \ldots$ are all powers of two. Tracing the sequence starting from $a_{i}=2^{c_{i}}=2^{m / 2}<2^{m}$,
+
+$$
+2^{m / 2} \rightarrow 2^{m+1} \rightarrow 2^{m} \rightarrow 2^{m-1} \rightarrow 2^{2 m-2}+2^{m}
+$$
+
+Note that this last term is a power of two if and only if $2 m-2=m$. This implies that $m$ must be equal to 2 . When $m=2$ and $a_{1}=2^{\ell}$ for $\ell \geq 1$ the sequence eventually cycles through $2,8,4,2, \ldots$. When $m=2$ and $a_{1}=1$ the sequence fails as the first terms are $1,5,5 / 2$.
+
+Solution 2: Let $m$ be a positive integer and suppose that $\left\{a_{n}\right\}$ consists only of positive integers. Call a number small if it is smaller than $2^{m}$ and large otherwise. By the recursion,
+after a small number we have a large one and after a large one we successively divide by 2 until we get a small one.
+
+First, we note that $\left\{a_{n}\right\}$ is bounded. Indeed, $a_{1}$ turns into a small number after a finite number of steps. After this point, each small number is smaller than $2^{m}$, so each large number is smaller than $2^{2 m}+2^{m}$. Now, since $\left\{a_{n}\right\}$ is bounded and consists only of positive integers, it is eventually periodic. We focus only on the cycle.
+
+Any small number $a_{n}$ in the cycle can be writen as $a / 2$ for $a$ large, so $a_{n} \geq 2^{m-1}$, then $a_{n+1} \geq 2^{2 m-2}+2^{m}=2^{m-2}\left(4+2^{m}\right)$, so we have to divide $a_{n+1}$ at least $m-1$ times by 2 until we get a small number. This means that $a_{n+m}=\left(a_{n}^{2}+2^{m}\right) / 2^{m-1}$, so $2^{m-1} \mid a_{n}^{2}$, and therefore $2^{\lceil(m-1) / 2\rceil} \mid a_{n}$ for any small number $a_{n}$ in the cycle. On the other hand, $a_{n} \leq 2^{m}-1$, so $a_{n+1} \leq 2^{2 m}-2^{m+1}+1+2^{m} \leq 2^{m}\left(2^{m}-1\right)$, so we have to divide $a_{n+1}$ at most $m$ times by two until we get a small number. This means that after $a_{n}$, the next small number is either $N=a_{m+n}=\left(a_{n}^{2} / 2^{m-1}\right)+2$ or $a_{m+n+1}=N / 2$. In any case, $2^{\lceil(m-1) / 2\rceil}$ divides $N$.
+
+If $m$ is odd, then $x^{2} \equiv-2\left(\bmod 2^{\lceil(m-1) / 2\rceil}\right)$ has a solution $x=a_{n} / 2^{(m-1) / 2}$. If $(m-1) / 2 \geq$ $2 \Longleftrightarrow m \geq 5$ then $x^{2} \equiv-2(\bmod 4)$, which has no solution. So if $m$ is odd, then $m \leq 3$.
+
+If $m$ is even, then $2^{m-1}\left|a_{n}^{2} \Longrightarrow 2^{\lceil(m-1) / 2\rceil}\right| a_{n} \Longleftrightarrow 2^{m / 2} \mid a_{n}$. Then if $a_{n}=2^{m / 2} x$, $2 x^{2} \equiv-2\left(\bmod 2^{m / 2}\right) \Longleftrightarrow x^{2} \equiv-1\left(\bmod 2^{(m / 2)-1}\right)$, which is not possible for $m \geq 6$. So if $m$ is even, then $m \leq 4$.
+
+The cases $m=1,2,3,4$ are handed manually, checking the possible small numbers in the cycle, which have to be in the interval $\left[2^{m-1}, 2^{m}\right)$ and be divisible by $2^{\lceil(m-1) / 2\rceil}$ :
+
+- For $m=1$, the only small number is 1 , which leads to 5 , then $5 / 2$.
+- For $m=2$, the only eligible small number is 2 , which gives the cycle $(2,8,4)$. The only way to get to 2 is by dividing 4 by 2 , so the starting numbers greater than 2 are all numbers that lead to 4 , which are the powers of 2 .
+- For $m=3$, the eligible small numbers are 4 and 6 ; we then obtain $4,24,12,6,44,22,11,11 / 2$.
+- For $m=4$, the eligible small numbers are 8 and 12 ; we then obtain $8,80,40,20,10, \ldots$ or $12,160,80,40,20,10, \ldots$, but in either case 10 is not an elegible small number.
+
+
+## Problem 3. Let $A B C$
+
+be a scalene triangle with circumcircle $\Gamma$. Let $M$ be the midpoint of $B C$. A variable point $P$ is selected in the line segment $A M$. The circumcircles of triangles $B P M$ and $C P M$ intersect $\Gamma$ again at points $D$ and $E$, respectively. The lines $D P$ and $E P$ intersect (a second time) the circumcircles to triangles $C P M$ and $B P M$ at $X$ and $Y$, respectively. Prove that as $P$ varies, the circumcircle of $\triangle A X Y$ passes through a fixed point $T$ distinct from $A$.
+
+Solution. Let $N$ be the radical center of the circumcircles of triangles $A B C, B M P$ and $C M P$. The pairwise radical axes of these circles are $B D, C E$ and $P M$, and hence they concur at $N$. Now, note that in directed angles:
+
+$$
+\angle M C E=\angle M P E=\angle M P Y=\angle M B Y .
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_62de8cd12a71a76ef11bg-4.jpg?height=1320&width=1320&top_left_y=168&top_left_x=374)
+
+It follows that $B Y$ is parallel to $C E$, and analogously that $C X$ is parallel to $B D$. Then, if $L$ is the intersection of $B Y$ and $C X$, it follows that $B N C L$ is a parallelogram. Since $B M=M C$ we deduce that $L$ is the reflection of $N$ with respect to $M$, and therefore $L \in A M$. Using power of a point from $L$ to the circumcircles of triangles $B P M$ and $C P M$, we have
+
+$$
+L Y \cdot L B=L P \cdot L M=L X \cdot L C
+$$
+
+Hence, $B Y X C$ is cyclic. Using the cyclic quadrilateral we find in directed angles:
+
+$$
+\angle L X Y=\angle L B C=\angle B C N=\angle N D E .
+$$
+
+Since $C X \| B N$, it follows that $X Y \| D E$.
+Let $Q$ and $R$ be two points in $\Gamma$ such that $C Q, B R$, and $A M$ are all parallel. Then in directed angles:
+
+$$
+\angle Q D B=\angle Q C B=\angle A M B=\angle P M B=\angle P D B .
+$$
+
+Then $D, P, Q$ are collinear. Analogously $E, P, R$ are collinear. From here we get $\angle P R Q=$ $\angle P D E=\angle P X Y$, since $X Y$ and $D E$ are parallel. Therefore $Q R Y X$ is cyclic. Let $S$ be the radical center of the circumcircle of triangle $A B C$ and the circles $B C Y X$ and $Q R Y X$. This point lies in the lines $B C, Q R$ and $X Y$ because these are the radical axes of the circles. Let $T$ be the second intersection of $A S$ with $\Gamma$. By power of a point from $S$ to the circumcircle of $A B C$ and the circle $B C X Y$ we have
+
+$$
+S X \cdot S Y=S B \cdot S C=S T \cdot S A
+$$
+
+Therefore $T$ is in the circumcircle of triangle $A X Y$. Since $Q$ and $R$ are fixed regardless of the choice of $P$, then $S$ is also fixed, since it is the intersection of $Q R$ and $B C$. This implies $T$ is also fixed, and therefore, the circumcircle of triangle $A X Y$ goes through $T \neq A$ for any choice of $P$.
+
+Now we show an alternative way to prove that $B C X Y$ and $Q R X T$ are cyclic.
+Solution 2. Let the lines $D P$ and $E P$ meet the circumcircle of $A B C$ again at $Q$ and $R$, respectively. Then $\angle D Q C \angle D B C=\angle D P M$, so $Q C \| P M$. Similarly, $R B \| P M$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_62de8cd12a71a76ef11bg-5.jpg?height=643&width=992&top_left_y=638&top_left_x=535)
+
+Now, $\angle Q C B=\angle P M B=\angle P X C=\angle(Q X, C X)$, which is half of the arc $Q C$ in the circumcircle $\omega_{C}$ of $Q X C$. So $\omega_{C}$ is tangent to $B S$; analogously, $\omega_{B}$, the circumcicle of $R Y B$, is also tangent to $B C$. Since $B R \| C Q$, the inscribed trapezoid $B R Q C$ is isosceles, and by symmetry $Q R$ is also tangent to both circles, and the common perpendicular bisector of $B R$ and $C Q$ passes through the centers of $\omega_{B}$ and $\omega_{C}$. Since $M B=M C$ and $P M\|B R\| C Q$, the line $P M$ is the radical axis of $\omega_{B}$ and $\omega_{C}$.
+
+However, $P M$ is also the radical axis of the circumcircles $\gamma_{B}$ of $P M B$ and $\gamma_{C}$ of $P M C$. Let $C X$ and $P M$ meet at $Z$. Let $p(K, \omega)$ denote the power of a point $K$ with respect to a circumference $\omega$. We have
+
+$$
+p\left(Z, \gamma_{B}\right)=p\left(Z, \gamma_{C}\right)=Z X \cdot Z C=p\left(Z, \omega_{B}\right)=p\left(Z, \omega_{C}\right)
+$$
+
+Point $Z$ is thus the radical center of $\gamma_{B}, \gamma_{C}, \omega_{B}, \omega_{C}$. Thus, the radical axes $B Y, C X, P M$ meet at $Z$. From here,
+
+$$
+\begin{aligned}
+& Z Y \cdot Z B=Z C \cdot Z X \Rightarrow B C X Y \text { cyclic } \\
+& P Y \cdot P R=P X \cdot P Q \Rightarrow Q R X T \text { cyclic. }
+\end{aligned}
+$$
+
+We may now finish as in Solution 1.
+
+Problem 4. Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the number in each chosen unit square is replaced by the corresponding average. Is it always possible to make the numbers in all squares become the same after finitely many turns?
+
+Answer: No
+
+Solution. Let $n$ be a positive integer relatively prime to 2 and 3 . We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all the numbers equal. Our aim is to choose $n$ and an initial configuration modulo $n$ for which no process modulo $n$ reaches a board with all numbers equal modulo $n$. We split this goal into two lemmas.
+
+Lemma 1. There is a $2 \times 3$ board that stays constant modulo 5 and whose entries are not all equal.
+
+Proof. Here is one such a board:
+![](https://cdn.mathpix.com/cropped/2024_11_22_62de8cd12a71a76ef11bg-6.jpg?height=217&width=315&top_left_y=694&top_left_x=870)
+
+The fact that the board remains constant regardless of the choice of squares can be checked square by square.
+
+Lemma 2. If there is an $r \times s$ board with $r \geq 2, s \geq 2$, that stays constant modulo 5 , then there is also a $k r \times l s$ board with the same property.
+
+Proof. We prove by a case by case analysis that repeateadly reflecting the $r \times s$ with respect to an edge preserves the property:
+
+- If a cell had 4 neighbors, after reflections it still has the same neighbors.
+- If a cell with $a$ had 3 neighbors $b, c, d$, we have by hypothesis that $a \equiv 3^{-1}(b+c+d) \equiv$ $2(b+c+d)(\bmod 5)$. A reflection may add $a$ as a neighbor of the cell and now
+
+$$
+4^{-1}(a+b+c+d) \equiv 4(a+b+c+d) \equiv 4 a+2 a \equiv a \quad(\bmod 5)
+$$
+
+- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c) \equiv 3(b+c)$ $(\bmod 5)$. If the reflections add one $a$ as neighbor, now
+
+$$
+3^{-1}(a+b+c) \equiv 2(3(b+c)+b+c) \equiv 8(b+c) \equiv 3(b+c) \equiv a \quad(\bmod 5)
+$$
+
+- If a cell with $a$ had 2 neighbors $b, c$, we have by hypothesis that $a \equiv 2^{-1}(b+c)(\bmod 5)$. If the reflections add two $a$ 's as neighbors, now
+
+$$
+4^{-1}(2 a+b+c) \equiv\left(2^{-1} a+2^{-1} a\right) \equiv a \quad(\bmod 5)
+$$
+
+In the three cases, any cell is still preserved modulo 5 after an operation. Hence we can fill in the $k r \times l s$ board by $k \times l$ copies by reflection.
+
+Since 2|2018 and 3|2019, we can get through reflections the following board:
+![](https://cdn.mathpix.com/cropped/2024_11_22_62de8cd12a71a76ef11bg-6.jpg?height=406&width=629&top_left_y=2327&top_left_x=719)
+
+By the lemmas above, the board is invariant modulo 5, so the answer is no.
+
+Problem 5. Determine all the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
+
+$$
+f\left(x^{2}+f(y)\right)=f(f(x))+f\left(y^{2}\right)+2 f(x y)
+$$
+
+for all real number $x$ and $y$.
+Answer: The possible functions are $f(x)=0$ for all $x$ and $f(x)=x^{2}$ for all $x$.
+Solution. By substituting $x=y=0$ in the given equation of the problem, we obtain that $f(0)=0$. Also, by substituting $y=0$, we get $f\left(x^{2}\right)=f(f(x))$ for any $x$.
+
+Furthermore, by letting $y=1$ and simplifying, we get
+
+$$
+2 f(x)=f\left(x^{2}+f(1)\right)-f\left(x^{2}\right)-f(1)
+$$
+
+from which it follows that $f(-x)=f(x)$ must hold for every $x$.
+Suppose now that $f(a)=f(b)$ holds for some pair of numbers $a, b$. Then, by letting $y=a$ and $y=b$ in the given equation, comparing the two resulting identities and using the fact that $f\left(a^{2}\right)=f(f(a))=f(f(b))=f\left(b^{2}\right)$ also holds under the assumption, we get the fact that
+
+$$
+f(a)=f(b) \Rightarrow f(a x)=f(b x) \quad \text { for any real number } x
+$$
+
+Consequently, if for some $a \neq 0, f(a)=0$, then we see that, for any $x, f(x)=f\left(a \cdot \frac{x}{a}\right)=$ $f\left(0 \cdot \frac{x}{a}\right)=f(0)=0$, which gives a trivial solution to the problem.
+
+In the sequel, we shall try to find a non-trivial solution for the problem. So, let us assume from now on that if $a \neq 0$ then $f(a) \neq 0$ must hold. We first note that since $f(f(x))=f\left(x^{2}\right)$ for all $x$, the right-hand side of the given equation equals $f\left(x^{2}\right)+f\left(y^{2}\right)+2 f(x y)$, which is invariant if we interchange $x$ and $y$. Therefore, we have
+
+$$
+f\left(x^{2}\right)+f\left(y^{2}\right)+2 f(x y)=f\left(x^{2}+f(y)\right)=f\left(y^{2}+f(x)\right) \quad \text { for every pair } x, y
+$$
+
+Next, let us show that for any $x, f(x) \geq 0$ must hold. Suppose, on the contrary, $f(s)=-t^{2}$ holds for some pair $s, t$ of non-zero real numbers. By setting $x=s, y=t$ in the right hand side of (2), we get $f\left(s^{2}+f(t)\right)=f\left(t^{2}+f(s)\right)=f(0)=0$, so $f(t)=-s^{2}$. We also have $f\left(t^{2}\right)=f\left(-t^{2}\right)=f(f(s))=f\left(s^{2}\right)$. By applying (2) with $x=\sqrt{s^{2}+t^{2}}$ and $y=s$, we obtain
+
+$$
+f\left(s^{2}+t^{2}\right)+2 f\left(s \cdot \sqrt{s^{2}+t^{2}}\right)=0
+$$
+
+and similarly, by applying (2) with $x=\sqrt{s^{2}+t^{2}}$ and $y=t$, we obtain
+
+$$
+f\left(s^{2}+t^{2}\right)+2 f\left(t \cdot \sqrt{s^{2}+t^{2}}\right)=0
+$$
+
+Consequently, we obtain
+
+$$
+f\left(s \cdot \sqrt{s^{2}+t^{2}}\right)=f\left(t \cdot \sqrt{s^{2}+t^{2}}\right)
+$$
+
+By applying (1) with $a=s \sqrt{s^{2}+t^{2}}, b=t \sqrt{s^{2}+t^{2}}$ and $x=1 / \sqrt{s^{2}+t^{2}}$, we obtain $f(s)=$ $f(t)=-s^{2}$, from which it follows that
+
+$$
+0=f\left(s^{2}+f(s)\right)=f\left(s^{2}\right)+f\left(s^{2}\right)+2 f\left(s^{2}\right)=4 f\left(s^{2}\right)
+$$
+
+a contradiction to the fact $s^{2}>0$. Thus we conclude that for all $x \neq 0, f(x)>0$ must be satisfied.
+
+Now, we show the following fact
+
+$$
+k>0, f(k)=1 \Leftrightarrow k=1
+$$
+
+Let $k>0$ for which $f(k)=1$. We have $f\left(k^{2}\right)=f(f(k))=f(1)$, so by $(1), f(1 / k)=f(k)=$ 1 , so we may assume $k \geq 1$. By applying (2) with $x=\sqrt{k^{2}-1}$ and $y=k$, and using $f(x) \geq 0$, we get
+
+$$
+f\left(k^{2}-1+f(k)\right)=f\left(k^{2}-1\right)+f\left(k^{2}\right)+2 f\left(k \sqrt{k^{2}-1}\right) \geq f\left(k^{2}-1\right)+f\left(k^{2}\right) .
+$$
+
+This simplifies to $0 \geq f\left(k^{2}-1\right) \geq 0$, so $k^{2}-1=0$ and thus $k=1$.
+Next we focus on showing $f(1)=1$. If $f(1)=m \leq 1$, then we may proceed as above by setting $x=\sqrt{1-m}$ and $y=1$ to get $m=1$. If $f(1)=m \geq 1$, now we note that $f(m)=f(f(1))=f\left(1^{2}\right)=f(1)=m \leq m^{2}$. We may then proceed as above with $x=\sqrt{m^{2}-m}$ and $y=1$ to show $m^{2}=m$ and thus $m=1$.
+
+We are now ready to finish. Let $x>0$ and $m=f(x)$. Since $f(f(x))=f\left(x^{2}\right)$, then $f\left(x^{2}\right)=$ $f(m)$. But by (1), $f\left(m / x^{2}\right)=1$. Therefore $m=x^{2}$. For $x<0$, we have $f(x)=f(-x)=f\left(x^{2}\right)$ as well. Therefore, for all $x, f(x)=x^{2}$.
+
+Solution 2 After proving that $f(x)>0$ for $x \neq 0$ as in the previous solution, we may also proceed as follows. We claim that $f$ is injective on the positive real numbers. Suppose that $a>b>0$ satisfy $f(a)=f(b)$. Then by setting $x=1 / b$ in (1) we have $f(a / b)=f(1)$. Now, by induction on $n$ and iteratively setting $x=a / b$ in (1) we get $f\left((a / b)^{n}\right)=1$ for any positive integer $n$.
+
+Now, let $m=f(1)$ and $n$ be a positive integer such that $(a / b)^{n}>m$. By setting $x=$ $\sqrt{(a / b)^{n}-m}$ and $y=1$ in (2) we obtain that
+$f\left((a / b)^{n}-m+f(1)\right)=f\left((a / b)^{n}-m\right)+f\left(1^{2}\right)+2 f\left(\sqrt{\left.(a / b)^{n}-m\right)}\right) \geq f\left((a / b)^{n}-m\right)+f(1)$.
+Since $f\left((a / b)^{n}\right)=f(1)$, this last equation simplifies to $f\left((a / b)^{n}-m\right) \leq 0$ and thus $m=$ $(a / b)^{n}$. But this is impossible since $m$ is constant and $a / b>1$. Thus, $f$ is injective on the positive real numbers. Since $f(f(x))=f\left(x^{2}\right)$, we obtain that $f(x)=x^{2}$ for any real value $x$.
+

APMO/md/en-apmo2020_sol.md

@@ -0,0 +1,234 @@
+# 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.
+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)
+
+Let $P$ be the intersection of $A C$ and $B F$. Then we have
+
+$$
+\angle P A E=\angle C B A=\angle B A C=\angle B F C .
+$$
+
+This implies $A, P, F, E$ are concyclic. It follows that
+
+$$
+\angle F P E=\angle F A E=\angle F B A,
+$$
+
+and hence $A B$ and $E P$ are parallel. So $E, P, D$ are collinear, and the result follows.
+Solution 2
+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.
+![](https://cdn.mathpix.com/cropped/2024_11_22_207f52578b01f443eb43g-2.jpg?height=714&width=673&top_left_y=215&top_left_x=753)
+
+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:
+
+If a sequence $a_{1}, a_{2}, \ldots$ of positive integers fulfills the inequalities
+
+$$
+a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+r a_{n+1}}
+$$
+
+for every positive integer $n$, then there exists a positive integer $M$ such that $a_{n+2}=a_{n}$ for every $n \geq M$.
+
+Solution 1. First, let us assume that $r>2$, and take a positive integer $a \geq 1 /(r-2)$.
+Then, if we let $a_{n}=a+\lfloor n / 2\rfloor$ for $n=1,2, \ldots$, the sequence $a_{n}$ satisfies the inequalities
+
+$$
+\sqrt{a_{n}^{2}+r a_{n+1}} \geq \sqrt{a_{n}^{2}+r a_{n}} \geq \sqrt{a_{n}^{2}+\left(2+\frac{1}{a}\right) a_{n}} \geq a_{n}+1=a_{n+2}
+$$
+
+but since $a_{n+2}>a_{n}$ for any $n$, we see that $r$ does not satisfy the condition given in the problem.
+Now 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.
+By induction we prove the following assertion:
+
+$$
+a_{m+2 k} \leq a_{m+2 k-1}=a_{m+1} \text { holds for every positive integer } k
+$$
+
+The truth of $(\dagger)$ for $k=1$ follows from the inequalities below
+
+$$
+2 a_{m+2}-1=a_{m+2}^{2}-\left(a_{m+2}-1\right)^{2} \leq a_{m}^{2}+2 a_{m+1}-\left(a_{m+2}-1\right)^{2} \leq 2 a_{m+1}
+$$
+
+Let us assume that $(\dagger)$ holds for some positive integer $k$. From
+
+$$
+a_{m+1}^{2} \leq a_{m+2 k+1}^{2} \leq a_{m+2 k-1}^{2}+2 a_{m+2 k} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
+$$
+
+it follows that $a_{m+2 k+1}=a_{m+1}$ must hold. Furthermore, since $a_{m+2 k} \leq a_{m+1}$, we have
+
+$$
+a_{m+2 k+2}^{2} \leq a_{m+2 k}^{2}+2 a_{m+2 k+1} \leq a_{m+1}^{2}+2 a_{m+1}<\left(a_{m+1}+1\right)^{2}
+$$
+
+from which it follows that $a_{m+2 k+2} \leq a_{m+1}$, which proves the assertion $(\dagger)$.
+We 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 $k \geq K$.
+
+## Solution 2
+
+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:
+(a) If $a_{n+1} \leq a_{n}$ for some $n \geq 1$, then
+
+$$
+a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n}^{2}+2 a_{n}+1}=a_{n}+1,
+$$
+
+hence $a_{n}=a_{n+2}$.
+(b) If $a_{n} \leq a_{n+1}$ for some $n \geq 1$, then
+
+$$
+a_{n} \leq a_{n+2} \leq \sqrt{a_{n}^{2}+2 a_{n+1}}<\sqrt{a_{n+1}^{2}+2 a_{n+1}+1}=a_{n+1}+1
+$$
+
+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.
+
+## Solution:
+
+We claim that $k=2^{a}$ for all $a \geq 0$.
+Let $A=\{1,2,4,8, \ldots\}$ 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 $s(T)=0$.)
+We 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.
+Next, 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.
+Lemma: For all sufficiently large $x \in S$, the smallest element of $S$ larger than $x$ is $2 x$.
+Proof 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.
+We 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
+a 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.
+From 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.
+Solution 2. We give an alternative proof of the first half of the lemma in the Solution 1 above.
+Let $s_{1}<s_{2}<\cdots$ 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 $k-1$.
+Fix a $t$ such that $s_{t}>2(m+1)$. Write
+
+$$
+A_{t-1}(x)=u(x)+k\left(x^{m+1}+\cdots+x^{s_{t}-1}\right)+x^{s_{t}} v(x)
+$$
+
+for some $u(x), v(x)$ where $u(x)$ is of degree at most $m$.
+Note that
+
+$$
+A_{t+1}(x)=A_{t-1}(x)+x^{s_{t}} A_{t-1}(x)+x^{s_{t+1}} A_{t-1}(x)+x^{s_{t}+s_{t+1}} A_{t-1}(x)
+$$
+
+If $s_{t+1}+m+1<2 s_{t}$, 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 $s_{t+1} \geq 2 s_{t}-(m+1)>$ $s_{t}+m+1$.
+Now
+
+$$
+A_{t}(x)=A_{t-1}(x)+x^{s_{t}} u(x)+k\left(x^{s_{t}+m+1}+\cdots x^{2 s_{t}-1}\right)+x^{2 s_{t}} v(x)
+$$
+
+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:
+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:
+
+Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
+Suppose $P(x)=c x+d$, 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 $s_{j}-s_{i} \equiv d$ $(\bmod c)$.
+Consider $c+1$ 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 $j=n$.
+Part 2: All polynomials with $\operatorname{deg} P \neq 1$ do not satisfy the given property.
+Lemma: 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 $[y-A, y+B]$.
+Proof 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
+loss 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.
+Suppose $\operatorname{deg} P \neq 1$. 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 $P$.
+Alternate 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 $d(\bmod c)$.
+Alternate Construction when $P(x)$ is constant or of even degree
+If $P(x)$ 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
+
+$$
+0,1,-1,2,3,-2,4,5,-3, \cdots
+$$
+
+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 $\frac{t}{s}$.
+
+## 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
+
+$$
+q=0 \cdot a_{1}+1 \cdot a_{2}+\cdots+(n-1) a_{n}
+$$
+
+Since each number $a_{i}$ is at least 1 , we always have
+
+$$
+q \leq(n-1) p-(1+\cdots+(n-1))=(n-1) p-\frac{n(n-1)}{2}=(n-1) s+\frac{n(n-1)}{2}
+$$
+
+Also, 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
+
+$$
+(j-1) a_{i}-(i-1)\left(a_{i}-1\right) \geq i a_{i}-(i-1)\left(a_{i}-1\right) \geq a_{i}
+$$
+
+Hence $q-t$ 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
+$t=1$. So after that move, we always have
+
+$$
+\begin{aligned}
+t & \leq q+1-\frac{(n+2)(n-1)}{2} \\
+& \leq(n-1) s+\frac{n(n-1)}{2}-\frac{(n+2)(n-1)}{2}+1 \\
+& =(n-1) s-(n-2)<(n-1) s
+\end{aligned}
+$$
+
+Hence, $\frac{t}{s}<n-1$. So $\frac{t}{s}$ must be a rational number in $[1, n-1)$.
+
+After 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
+
+$$
+t=(a-1) a^{n-2}+a \cdot a^{n-3}(a-1)(n-2)=a^{n-2}(a-1)(n-1)
+$$
+
+This completes the induction.
+Now 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
+
+$$
+\frac{t}{s}=\frac{a^{n-2}(a-1)(n-1)+b}{a^{n-1}-1+b}
+$$
+
+So 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
+
+$$
+\frac{p}{q}=\frac{a^{n-2}(a-1)(n-1)+b}{a^{n-1}-1+b}
+$$
+
+Rearranging, we see that this happens if and only if
+
+$$
+b=\frac{q a^{n-2}(a-1)(n-1)-p\left(a^{n-1}-1\right)}{p-q} .
+$$
+
+If 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 $a$.
+
+$$
+\begin{aligned}
+q a^{n-2}(a-1)(n-1)-p\left(a^{n-1}-1\right) & >q a^{n-2}(a-1)(n-1)-p a^{n-1} \\
+& =a^{n-2}(a(q(n-1)-p)-(n-1))
+\end{aligned}
+$$
+
+which is positive for sufficiently large $a$ since $q(n-1)-p>0$.
+
+Alternative 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
+red. 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
+
+$$
+\begin{aligned}
+(a-1) w+(b-1)(w-1)+\operatorname{gcd}(a, b) & \leq(a-1) w+(b-1)(w-1)+b \\
+& =(p-1) w+1-w \\
+& \leq(p-1) w .
+\end{aligned}
+$$
+
+This completes the induction.
+Now, 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, $\frac{t}{s}<n-1$.
+

APMO/md/en-apmo2021_sol.md

@@ -0,0 +1,185 @@
+# 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 $x^{2}=r\lfloor x\rfloor$.
+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 $n$.
+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:
+
+- $P(x)=x+d$, for some integer $d \geq-2022$.
+- $P(x)=-x+d$, for some integer $d \leq 2022$.
+
+Suppose $P$ 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 $x \geq M$.
+
+Lemma 1. For any positive integer $n$, the integers $P(1), P(2), \ldots, P(n)$ leave pairwise distinct remainders upon division by $n$.
+
+Proof. 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
+
+$$
+r, n+r, 2 n+r, \ldots,(k-1) n+r
+$$
+
+upon 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.
+
+Next, 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.
+We can now write $P(x)=c x+d$ with $c>0$. We prove that $c=1$ by two ways.
+Solution 1 If $c \geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the Lemma. Hence $c=1$.
+
+Solution 2 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, $c=1$.
+
+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 $L M$.
+
+## Solution 1
+
+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 $\angle A C B=\gamma$.
+We have the following:
+
+$$
+\begin{array}{r}
+\angle C B L=\frac{1}{2} \angle A B C=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \gamma \text { and } \angle B C L=90^{\circ}-\frac{1}{2} \alpha-\frac{1}{2} \beta, \\
+\angle C B N=90^{\circ}-\frac{1}{2} \beta \text { and } \angle B C N=90^{\circ}-\frac{1}{2} \gamma, \\
+\angle M B L=\angle M B C+\angle C B L=90^{\circ}-\frac{1}{2} \gamma \text { and } \angle M C L=90^{\circ}-\frac{1}{2} \beta, \\
+\angle L C N=\angle L B N=180^{\circ}-\frac{1}{2}(\alpha+\beta+\gamma) .
+\end{array}
+$$
+
+Applying the sine rule to $\triangle M B L$ and $\triangle M C L$ we obtain
+
+$$
+\frac{M B}{M L}=\frac{M C}{M L}=\frac{\sin \angle B L M}{\sin \angle M B L}=\frac{\sin \angle C L M}{\sin \angle M C L}
+$$
+
+It follows that
+
+$$
+\frac{\sin \angle B L M}{\sin \angle C L M}=\frac{\sin \angle M B L}{\sin \angle M C L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}
+$$
+
+Now
+
+$$
+\frac{\sin \angle B L M}{\sin \angle M L C} \cdot \frac{\sin \angle L C N}{\sin \angle N C B} \cdot \frac{\sin \angle N B C}{\sin \angle N B L}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)} \cdot \frac{\sin \left(90^{\circ}-\frac{1}{2} \beta\right)}{\sin \left(90^{\circ}-\frac{1}{2} \gamma\right)}=1
+$$
+
+Hence $L M, B N, C N$ are concurrent and therefore $L, M, N$ are collinear.
+
+## Alternative proof
+
+We proceed similarly as above until the equation (1).
+We use the following lemma.
+Lemma: 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 $\beta=\delta$.
+Proof of Lemma: Let $\theta=\alpha+\beta=\gamma+\delta$. Then $\frac{\sin (\theta-\beta)}{\sin \beta}=\frac{\sin (\theta-\delta)}{\sin \delta}$.
+
+$$
+\begin{gathered}
+\Longleftrightarrow \sin (\theta-\beta) \sin \delta=\sin (\theta-\delta) \sin \beta \\
+\Longleftrightarrow(\sin \theta \cos \beta-\sin \beta \cos \theta) \sin \delta=(\sin \theta \cos \delta-\sin \delta \cos \theta) \sin \beta \\
+\Longleftrightarrow \sin \theta \cos \beta \sin \delta=\sin \theta \cos \delta \sin \beta \\
+\Longleftrightarrow \sin \theta \sin (\beta-\delta)=0
+\end{gathered}
+$$
+
+Since $0<\theta<\pi$, then $\sin \theta \neq 0$. Therefore, $\sin (\beta-\delta)=0$, and we must have $\beta=\delta$.
+Applying the sine rule to $\triangle N B L$ and $\triangle N C L$ we obtain
+
+$$
+\begin{aligned}
+& \frac{N B}{N L}=\frac{\sin \angle B L N}{\sin \angle L B N} \\
+& \frac{N C}{N L}=\frac{\sin \angle C L N}{\sin \angle L C N}
+\end{aligned}
+$$
+
+Since $\angle L B N=\angle L C N$, it follows that
+
+$$
+\frac{\sin \angle B L N}{\sin \angle C L N}=\frac{N B}{N C}=\frac{\sin \angle B C N}{\sin \angle C B N}=\frac{\cos (\gamma / 2)}{\cos (\beta / 2)}=\frac{\sin \angle B L M}{\sin \angle C L M}
+$$
+
+By 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.
+
+## Solution 2
+
+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
+
+$$
+\begin{aligned}
+\angle M B L & =\angle M B C+\angle C B L=\frac{1}{2}(\angle M B C+\angle M C B+\angle A B C) \\
+& =\frac{1}{2}(\angle M B A+\angle M C B)=90^{\circ}-\frac{\angle B C E}{2}=\angle B C N .
+\end{aligned}
+$$
+
+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:
+(a) No good subset consists of 888 cells.
+(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.
+(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.
+(a) Basic tiles
+(b) Inductive construction
+(c) $16 \times 16$
+![](https://cdn.mathpix.com/cropped/2024_11_22_5e137eeb9eefa8d28dceg-4.jpg?height=373&width=1633&top_left_y=1078&top_left_x=249)
+
+Note 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.
+For 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.
+
+## 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 $b$.
+
+## Solution 1.
+
+There are two families of functions which satisfy the condition:
+(1) $f(n)= \begin{cases}0 & \text { if } n \text { is even, and } \\ \text { any perfect square } & \text { if } n \text { is odd }\end{cases}$
+(2) $f(n)=n^{2}$, for every integer $n$.
+
+It 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).
+Claim 1. $f(0)=0$ and $f(n)$ is a perfect square for every integer $n$.
+Proof. 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.
+Suppose, 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 $a \in \mathbb{Z}$.
+Plugging $(a, b) \rightarrow(0,3)$ 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 .
+For every integer $n$, plugging $(a, b) \rightarrow(0,-n)$ in $\left(^{*}\right)$ shows that $f(n)$ is a square.
+Replacing $b$ with $f(a)-b$, we find that for all integers $a$ and $b$,
+
+$$
+f(b)+(f(a)-b) f(2 a) \text { is a square. }
+$$
+
+Now, let $S$ be the set of all integers $n$ such that $f(n)=0$. We have two cases:
+
+- Case 1: $S$ is unbounded from above.
+
+We 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.
+
+- Case 2: $S$ is bounded from above.
+
+Let $T$ 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.
+Fix 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
+
+$$
+(k+m)(k-m)=k^{2}-m^{2}=p \Longrightarrow k+m=p, k-m=1 \Longrightarrow k=n, m=n-1
+$$
+
+Thus, $f(n)=k^{2}=n^{2}$, giving us $n=\frac{p+1}{2} \in T$.
+Next, 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 $n$.
+Finally, 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.
+
+## Another approach of Case 1.
+
+Claim 2. One of the following is true.
+(i) For every integer $n, f(2 n)=0$.
+(ii) There exists an integer $K>0$ such that for every integer $n \geq K, f(n)>0$.
+
+Proof. 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 $f(n)>0$.
+For 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.
+If $f$ belongs to Case (i), Claim 1 shows that $f$ belongs to Family (1).
+If $f$ belongs to Case (ii), then $S$ is bounded from above. From Case 2 we get $f(n)=n^{2}$.
+

APMO/md/en-apmo2022_sol.md

@@ -0,0 +1,345 @@
+# 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
+
+## Solution 1.1
+
+By 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.
+
+- 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.
+- 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 $a^{3}=b^{2} c$.
+Note that $a \equiv b \equiv 1$ modulo $a-1$. So we have
+
+$$
+1 \equiv a^{3}=b^{2} c \equiv c \quad(\bmod a-1) .
+$$
+
+If $c<a$, 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.
+If $c \geq a$, 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.
+Therefore, the solutions described above are the only solutions.
+
+## Solution 1.2
+
+We 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.
+Now, it remains to find all positive integers $x, c, d$ satisfying
+
+$$
+x^{2} c d-1 \mid x^{3} c-1
+$$
+
+That 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 $d-x \geq x^{2} c d-1$.
+
+- If $x=d$ then $b=a$.
+- 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 $a=b=1$.
+- 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 $b=1$.
+Hence 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.
+
+
+## Solution 1.3
+
+All answers are $(n, n)$ and $(n, 1)$ where $n$ is any positive integer. They all clearly work.
+To 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 $\operatorname{gcd}(a, b) \left\lvert\, \frac{b-1}{a-1}-1\right.$.
+The 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,
+
+$$
+\operatorname{gcd}(a, b)^{3}=\operatorname{gcd}\left(a^{3}, b^{3}\right) \geq \operatorname{gcd}\left(b^{2}, b^{3}\right)=b^{2} .
+$$
+
+Combining these two inequalities, we get that
+
+$$
+b^{\frac{2}{3}}<\frac{b-1}{a-1}<2 \frac{b}{a}
+$$
+
+This implies $a<2 b^{\frac{1}{3}}$. However, $b^{2} \mid a^{3}$ gives $b \leq a^{\frac{3}{2}}$. This forces
+
+$$
+a<2\left(a^{\frac{3}{2}}\right)^{\frac{1}{3}}=2 \sqrt{a} \Longrightarrow a<4 .
+$$
+
+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
+
+## Solution 2.1
+
+Let 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.
+First, notice that $\triangle B E D$ is isosceles with $E B=E D$. This implies $\angle E B C=\angle E D P$.
+Then, $\angle D A G=\angle D F G=\angle E B C=\angle E D P$ which implies $A G \| D C$. Hence, $A G C D$ is an isosceles trapezoid.
+Also, $A G \| D C$ and $A E=E D$. This implies $\triangle A E G \cong \triangle D E P$ and $A G=D P$.
+Since $B$ 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.
+
+## Solution 2.2
+
+Set 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
+
+$$
+x^{2}+y^{2}+2 f x+2 g y+h=0
+$$
+
+Substituting 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
+
+$$
+x^{2}+y^{2}+(d-c) x+\left(\frac{c d}{a}-a\right) y-c d=0
+$$
+
+Similarly, the equation of the circumcircle of $\triangle B D E$ is
+
+$$
+x^{2}+y^{2}+d x+\left(\frac{d^{2}}{2 a}-\frac{a}{2}\right) y=0
+$$
+
+Then (3)-(2) gives the equation of the line $D F$ which is
+
+$$
+c x+\frac{a^{2}+d^{2}-2 c d}{2 a} y+c d=0
+$$
+
+Solving (3) and (4) simultaneously, we get
+
+$$
+F=\left(\frac{c\left(d^{2}-a^{2}-2 c d\right)}{a^{2}+(d-2 c)^{2}}, \frac{2 a c(c-d)}{a^{2}+(d-2 c)^{2}}\right)
+$$
+
+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
+
+$$
+\left\{\frac{n}{202}\right\}+\left\{\frac{2 n}{202}\right\}+\cdots+\left\{\frac{k n}{202}\right\}=\frac{k}{2}
+$$
+
+where $\{x\}$ denote the fractional part of $x$.
+Note: $\{x\}$ 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 $0 \leq n<202$.
+
+If $n=0$, 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 $101 \nmid n$.
+For 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
+
+$$
+n(1+2+\ldots+k)-202\left(a_{1}+a_{2}+\ldots+a_{k}\right)=101 k
+$$
+
+Equivalently, letting $z=a_{1}+a_{2}+\ldots+a_{k}$,
+
+$$
+n k(k+1)-404 z=202 k
+$$
+
+Since $n$ 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.
+
+- 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 .
+- 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.
+- 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 $n$.
+Determine all pairs of integers $(n, k)$ such that Cathy can win this game.
+
+## Solution
+
+We claim Cathy can win if and only if $n \leq 2^{k-1}$.
+
+First, 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.
+
+Next, 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
+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.
+
+## Solution
+
+The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
+
+$$
+\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
+$$
+
+Cyclic 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.
+
+## Solution 5.1
+
+Since the expression is cyclic, we could WLOG $a=\max \{a, b, c, d\}$. Let
+
+$$
+S(a, b, c, d)=(a-b)(b-c)(c-d)(d-a)
+$$
+
+Note 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 $S(a, b, c, d)<0$.
+
+- Exactly 1 of $a-b, b-c, c-d, d-a$ is negative.
+
+Since $a=\max \{a, b, c, d\}$, then we must have $d-a<0$. This forces $a>b>c>d$. Now, let us write
+
+$$
+S(a, b, c, d)=-(a-b)(b-c)(c-d)(a-d)
+$$
+
+Write $a-b=y, b-c=x, c-d=w$ for some positive reals $w, x, y>0$. Plugging to the original condition, we have
+
+$$
+(d+w+x+y)^{2}+(d+w+x)^{2}+(d+w)^{2}+d^{2}-1=0(*)
+$$
+
+and we want to prove that $w x y(w+x+y) \leq \frac{1}{8}$. Consider the expression $(*)$ as a quadratic in $d$ :
+
+$$
+4 d^{2}+d(6 w+4 x+2 y)+\left((w+x+y)^{2}+(w+x)^{2}+w^{2}-1\right)=0
+$$
+
+Since $d$ is a real number, then the discriminant of the given equation has to be non-negative, i.e. we must have
+
+$$
+\begin{aligned}
+4 & \geq 4\left((w+x+y)^{2}+(w+x)^{2}+w^{2}\right)-(3 w+2 x+y)^{2} \\
+& =\left(3 w^{2}+2 w y+3 y^{2}\right)+4 x(w+x+y) \\
+& \geq 8 w y+4 x(w+x+y) \\
+& =4(x(w+x+y)+2 w y)
+\end{aligned}
+$$
+
+However, AM-GM gives us
+
+$$
+w x y(w+x+y) \leq \frac{1}{2}\left(\frac{x(w+x+y)+2 w y}{2}\right)^{2} \leq \frac{1}{8}
+$$
+
+This 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
+
+$$
+(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}\right)
+$$
+
+Thus, any cyclic permutation of the above solution will achieve the minimum equality.
+
+- 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
+
+$$
+\begin{aligned}
+S(a, b, c, d) & =(a-b)(b-c)(c-d)(d-a) \\
+& =(a-d)(d-c)(c-b)(b-a) \\
+& =S(a, d, c, b)
+\end{aligned}
+$$
+
+Now, note that $a>d>c>b$. By the previous case, $S(a, d, c, b) \geq-\frac{1}{8}$, which implies that
+
+$$
+S(a, b, c, d)=S(a, d, c, b) \geq-\frac{1}{8}
+$$
+
+as well. Equality holds if and only if
+
+$$
+(a, b, c, d)=\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right)
+$$
+
+and its cyclic permutation.
+
+## Solution 5.2
+
+The minimum value is $-\frac{1}{8}$. There are eight equality cases in total. The first one is
+
+$$
+\left(\frac{1}{4}+\frac{\sqrt{3}}{4},-\frac{1}{4}-\frac{\sqrt{3}}{4}, \frac{1}{4}-\frac{\sqrt{3}}{4},-\frac{1}{4}+\frac{\sqrt{3}}{4}\right) .
+$$
+
+Cyclic 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:
+
+Claim 1. In order to get the minimum value, we must have $a+b+c+d=0$.
+
+Proof. Assume not, let $\delta=\frac{a+b+c+d}{4}$ and note that
+
+$$
+(a-\delta)^{2}+(b-\delta)^{2}+(c-\delta)^{2}+(d-\delta)^{2}<a^{2}+b^{2}+c^{2}+d^{2}
+$$
+
+so by shifting by $\delta$ and scaling, we get an even smaller value of $(a-b)(b-c)(c-d)(d-a)$.
+
+The key idea is to substitute the variables
+
+$$
+\begin{aligned}
+& x=a c+b d \\
+& y=a b+c d \\
+& z=a d+b c
+\end{aligned}
+$$
+
+so that the original expression is just $(x-y)(x-z)$. We also have the conditions $x, y, z \geq-0.5$ because of:
+
+$$
+2 x+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a+c)^{2}+(b+d)^{2} \geq 0
+$$
+
+Moreover, notice that
+
+$$
+0=(a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(x+y+z) \Longrightarrow x+y+z=\frac{-1}{2}
+$$
+
+Now, we reduce to the following optimization problem.
+Claim 2. Let $x, y, z \geq-0.5$ such that $x+y+z=-0.5$. Then, the minimum value of
+
+$$
+(x-y)(x-z)
+$$
+
+is $-1 / 8$. Moreover, the equality case occurs when $x=-1 / 4$ and $\{y, z\}=\{1 / 4,-1 / 2\}$.
+Proof. We notice that
+
+$$
+\begin{aligned}
+(x-y)(x-z)+\frac{1}{8} & =\left(2 y+z+\frac{1}{2}\right)\left(2 z+y+\frac{1}{2}\right)+\frac{1}{8} \\
+& =\frac{1}{8}(4 y+4 z+1)^{2}+\left(y+\frac{1}{2}\right)\left(z+\frac{1}{2}\right) \geq 0
+\end{aligned}
+$$
+
+The last inequality is true since both $y+\frac{1}{2}$ and $z+\frac{1}{2}$ are not less than zero.
+The 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.
+
+Note: 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
+
+$$
+-2 x+\left(a^{2}+b^{2}+c^{2}+d^{2}\right)=(a-c)^{2}+(b-d)^{2} \geq 0
+$$
+
+We 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:
+
+- 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.
+- If $x=0.5$, then $y=z=-0.5$, so $(x-y)(x-z)=1$, not the minimum.
+- 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.
+- If $y=0.5$, then $x=z=-0.5$, so $(x-y)(x-z)=0$, not the minimum.
+- If $y=-0.5$, then $z=-x$, so
+
+$$
+(x-y)(x-z)=2 x(x+0.5)
+$$
+
+which obtain the minimum at $x=-1 / 4$. This gives the desired equality case.
+
+- If the global minimum lies in the interior $(-0.5,0.5)^{3}$, then we apply Lagrange multiplier:
+
+$$
+\begin{aligned}
+& \frac{\partial}{\partial x}(x-y)(x-z)=\lambda \frac{\partial}{\partial x}(x+y+z) \\
+& \frac{\partial}{\partial y}(x-y)(x-z)=\lambda \frac{\partial}{\partial y}(x+y+z) \\
+& \frac{\partial}{\partial z}(x-y)(x-z) \Longrightarrow z-x=\lambda \frac{\partial}{\partial z}(x+y+z) \\
+& \Longrightarrow y-x=\lambda .
+\end{aligned}
+$$
+
+Adding the last two equations gives $\lambda=0$, or $x=y=z$. This gives $(x-y)(x-z)=0$, not the minimum.
+
+Having exhausted all cases, we are done.
+

APMO/md/en-apmo2023_sol.md

@@ -0,0 +1,243 @@
+# APMO 2023 - Problems and Solutions 
+
+## Problem 1
+
+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.
+Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.
+
+## Solution 1
+
+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 $B$.
+String 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:
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-01.jpg?height=592&width=1346&top_left_y=1195&top_left_x=321)
+
+Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-2)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+2\right) \sqrt{2}+\frac{(n+(n-1)) \sqrt{2}}{2}$, this case is done.
+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 $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-01.jpg?height=621&width=1337&top_left_y=2094&top_left_x=328)
+
+Since $\left(a_{1}+a_{2}+\cdots+a_{k}\right) \sqrt{2}+\frac{((n-3)+(n-1)) \sqrt{2}}{2}=\left(b_{1}+b_{2}+\cdots+b_{\ell}+1\right) \sqrt{2}+\frac{(n+(n-2)) \sqrt{2}}{2}$, this case is also done.
+In 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.
+Finally, 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:
+
+- 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.
+- 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 $B$;
+- 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 );
+- 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 );
+- 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 ).
+- 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 $B$.
+
+
+## Solution 2
+
+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:
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-02.jpg?height=815&width=1051&top_left_y=1948&top_left_x=471)
+
+By the induction hypothesis, one can arrange the remaining $n-6$ squares away from the six larger squares, so we are done.
+
+## Problem 2
+
+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 $n$.
+
+Answer: $n=6$.
+
+## 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
+$p_{k}-1=\frac{\sigma(n)}{n}=\prod_{i=1}^{k}\left(1+\frac{1}{p_{i}}+\cdots+\frac{1}{p_{i}^{\alpha_{i}}}\right)<\prod_{i=1}^{k} \frac{1}{1-\frac{1}{p_{i}}}=\prod_{i=1}^{k}\left(1+\frac{1}{p_{i}-1}\right) \leq \prod_{i=1}^{k}\left(1+\frac{1}{i}\right)=k+1$,
+that 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 $p_{k} \leq 3$.
+If $k=1$ 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 $\beta>1$,
+
+$$
+\frac{\sigma(n)}{n}>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=2 .
+$$
+
+Therefore $\alpha=\beta=1$ and the only answer is $n=6$.
+Comment: 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
+
+$$
+\left\{\begin{array} { l } 
+{ 2 ^ { \alpha + 1 } - 1 = 3 ^ { \beta } } \\
+{ 3 ^ { \beta + 1 } - 1 = 2 ^ { \alpha + 2 } }
+\end{array} \Longleftrightarrow \left\{\begin{array} { c } 
+{ 2 ^ { \alpha + 1 } - 1 = 3 ^ { \beta } } \\
+{ 3 \cdot ( 2 ^ { \alpha + 1 } - 1 ) - 1 = 2 \cdot 2 ^ { \alpha + 1 } }
+\end{array} \Longleftrightarrow \left\{\begin{array}{r}
+2^{\alpha+1}=4 \\
+3^{\beta}=3
+\end{array}\right.\right.\right.
+$$
+
+and $n=2^{\alpha} 3^{\beta}=6$.
+
+## Problem 3
+
+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
+
+$$
+r_{2}-r_{1}=I_{1} I_{2} \sin \theta=I_{3} I_{4} \sin \theta=r_{4}-r_{3}
+$$
+
+which 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 $r_{2}=r_{4}$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-05.jpg?height=509&width=812&top_left_y=842&top_left_x=591)
+
+Now 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
+
+$$
+C Y>A W \Longrightarrow B W>D Y \Longrightarrow D Z>B X \Longrightarrow C X>A Z
+$$
+
+which is a contradiction. Therefore $A Z=C X \Longrightarrow A W=C Y$ and $W X Y Z$ is a parallelogram.
+Comment: There are several ways to prove that $r_{1}=r_{3}$ and $r_{2}=r_{4}$. The proposer shows the following three alternative approaches:
+Using 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 $\ell_{1}$.
+Similarly, $P$ 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 $r_{2}=r_{4}$.
+Using 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 $r_{2}=r_{4}$.
+Using 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).
+
+Since $A I_{1} \| C I_{3}$ 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.
+Hence $A B E I_{1} I_{2}$ 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 4
+
+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
+
+$$
+f((c+1) x+f(y))=f(x+2 y)+2 c x \quad \text { for all } x, y \in \mathbb{R}_{>0}
+$$
+
+Answer: $f(x)=2 x$ for all $x>0$.
+
+## Solution 1
+
+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,
+
+$$
+(c+1) x+f(y)=x+2 y \Longleftrightarrow x=\frac{2 y-f(y)}{c}
+$$
+
+Notice 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 $y>0$.
+Now 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
+
+$$
+\left\{\begin{array}{r}
+(c+1) x+f(y)=a_{n} \\
+x+2 y=a_{n-1}
+\end{array} \Longleftrightarrow x=a_{n-1}-2 y \quad \text { and } \quad a_{n}=(c+1)\left(a_{n-1}-2 y\right)+f(y) .\right.
+$$
+
+If $x=a_{n-1}-2 y>0$ then $a_{n}>f(y)>2 y$, so inductively all the substitutions make sense.
+For 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
+
+$$
+f\left(a_{n}\right)=f\left(a_{0}\right)+2 c \sum_{i=0}^{n-1} b_{i} .
+$$
+
+One 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
+
+$$
+\begin{aligned}
+f\left(a_{n}\right) & =f\left(a_{0}\right)+2 c \sum_{i=0}^{n-1}\left(\left(b_{0}+\frac{f(y)-2 y}{c}\right)(c+1)^{i}-\frac{f(y)-2 y}{c}\right) \\
+& =f\left(a_{0}\right)+2\left(b_{0}+\frac{f(y)-2 y}{c}\right)\left((c+1)^{n}-1\right)-2 n(f(y)-2 y)
+\end{aligned}
+$$
+
+Since $f\left(a_{n}\right) \geq 2 a_{n}=2 b_{n}+4 y$,
+
+$$
+\begin{aligned}
+& f\left(a_{0}\right)+2\left(b_{0}+\frac{f(y)-2 y}{c}\right)\left((c+1)^{n}-1\right)-2 n(f(y)-2 y) \geq 2 b_{n}+4 y \\
+= & 2\left(b_{0}+\frac{f(y)-2 y}{c}\right)(c+1)^{n}-2 \frac{f(y)-2 y}{c}
+\end{aligned}
+$$
+
+which implies
+
+$$
+f\left(a_{0}\right)+2 \frac{f(y)-2 y}{c} \geq 2\left(b_{0}+\frac{f(y)-2 y}{c}\right)+2 n(f(y)-2 y)
+$$
+
+which is not true for sufficiently large $n$.
+A contradiction is reached, and thus $f(y)=2 y$ for all $y>0$. It is immediate that this function satisfies the functional equation.
+
+## Solution 2
+
+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
+
+$$
+\begin{aligned}
+& g((c+1) x+g(y)+2 y)+2((c+1) x+g(y)+2 y)=g(x+2 y)+2(x+2 y)+2 c x \\
+\Longleftrightarrow & g((c+1) x+g(y)+2 y)+2 g(y)=g(x+2 y) .
+\end{aligned}
+$$
+
+This 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 $z=x+2 y$.
+Now 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 (??),
+
+$$
+g(x+2 y) \geq 2 m \cdot g(y)+2 g(y)=2(m+1) g(y)
+$$
+
+and we are done by plugging $z=x+2 y$ again.
+The 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 5
+
+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:
+First, 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:
+
+- If it is on a line segment, it moves towards the sink.
+- When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink.
+
+If 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 1
+
+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.
+First part: at most $n$ friends can receive a present.
+The 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 $\ell$.
+Now 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.
+First 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 $k+n, 1 \leq k \leq n$.
+We 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
+$k-j$ 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:
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=541&width=552&top_left_y=289&top_left_x=706)
+
+The 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.
+Reintroduce 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 $k+i, i=1,2, \ldots, n-1$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=612&width=732&top_left_y=1259&top_left_x=248)
+
+Paths without chord $k \rightarrow k+n$
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-10.jpg?height=612&width=732&top_left_y=1256&top_left_x=1025)
+
+Corrected paths with chord $k \rightarrow k+n$
+
+Then 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.
+
+## Solution 2
+
+First part: at most n friends can receive a present.
+Similarly to the first solution, consider a circle that encompasses all line segments, extend the lines, and use the endpoints of the chords instead of the line segments, and prove that each chord connects vertices $k$ and $k+n$. We also consider, even in the first part, $n$ presents leaving from $n$ outcoming vertices.
+First we prove that a present always goes to a sink. If it does not, then it loops; let it first enter the loop at point $P$ after turning from chord $a$ to chord $b$. Therefore after it loops once,
+
+[^0]it must turn to chord $b$ at $P$. But $P$ is the intersection of $a$ and $b$, so the present should turn from chord $a$ to chord $b$, which can only be done in one way - the same way it came in first. This means that some part of chord $a$ before the present enters the loop at $P$ is part of the loop, which contradicts the fact that $P$ is the first point in the loop. So no present enters a loop, and every present goes to a sink.
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-11.jpg?height=210&width=264&top_left_y=426&top_left_x=579)
+
+There are no loops
+![](https://cdn.mathpix.com/cropped/2024_11_22_da5040e5bee0df04b59eg-11.jpg?height=210&width=255&top_left_y=426&top_left_x=1135)
+
+No two paths cross
+
+The present paths also do not cross: in fact, every time two paths share a point $P$, intersection of chords $a$ and $b$, one path comes from $a$ to $b$ and the other path comes from $b$ to $a$, and they touch at $P$. This implies the following sequence of facts:
+
+- Every path divides the circle into two regions with paths connecting vertices within each region.
+- All $n$ presents will be delivered to $n$ different persons; that is, all sinks receive a present. This implies that every vertex is an endpoint of a path.
+- The number of chord endpoints inside each region is even, because they are connected within their own region.
+
+Now consider the path starting at vertex 1 , with Tony. It divides the circle into two regions with an even number of vertices in their interior. Then there is an even number of vertices between Tony and the recipient of his present, that is, their vertex is an even numbered one. Second part: all even-numbered vertices can receive a present.
+The construction is the same as the in the previous solution: 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 $k+1, k+2, \ldots, k+n$. Then, since the paths do not cross, $k$ will send a present to $k+1, k-1$ will send a present to $k+2$, and so on, until 1 sends a present to $(k+1)+(k-1)=2 k$.
+
+
+[^0]:    ${ }^{1}$ The paths do not depend uniquely on $k$ and $n$; different chord configurations and vertex labelings may change the paths.
+

APMO/md/en-apmo2024_sol.md

@@ -0,0 +1,292 @@
+# APMO 2024 - Problems and Solutions 
+
+## Problem 1
+
+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 1
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=723&width=1003&top_left_y=678&top_left_x=492)
+
+Let $\ell$ 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
+
+$$
+\frac{B Z}{Z C}=\frac{D Z^{\prime}}{Z^{\prime} E}=\frac{P Z}{Z Q},
+$$
+
+thus $B Z \cdot Q Z=C Z \cdot P Z$, which implies that $Z$ is on $\ell$.
+
+## Solution 2
+
+![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-01.jpg?height=721&width=998&top_left_y=1995&top_left_x=492)
+
+Let 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.
+
+## Solution 3
+
+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 $E$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-02.jpg?height=732&width=1008&top_left_y=682&top_left_x=484)
+
+The 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,
+$\angle D Y^{\prime} E=\angle B Y C=\angle B Y X+\angle X Y C=\angle X Q P+\angle X P Q=180^{\circ}-\angle P X Q=180^{\circ}-\angle D X E$,
+which implies that $D Y^{\prime} E X$ is cyclic. Therefore
+
+$$
+\angle D Y^{\prime} X=\angle D E X=\angle P Q X=\angle B Y X
+$$
+
+which, 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 2
+
+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.
+
+Answer: $L(k)=\left\{\begin{array}{ll}100^{2}-(2 k-100)^{2} & \text { if } k \text { is even } \\ \frac{100^{2}-(2 k-100)^{2}}{2} & \text { if } k \text { is odd }\end{array}\right.$.
+
+## 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 $(x, y) \rightarrow(x \pm 1, y-k) \rightarrow(x \pm 2, y)$.
+Hence 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 3
+
+Let $n$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers. Prove that
+
+$$
+\sum_{i=1}^{n} \frac{1}{2^{i}}\left(\frac{2}{1+a_{i}}\right)^{2^{i}} \geq \frac{2}{1+a_{1} a_{2} \ldots a_{n}}-\frac{1}{2^{n}}
+$$
+
+## Solution
+
+We first prove the following lemma:
+Lemma 1. For $k$ positive integer and $x, y>0$,
+
+$$
+\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}} \geq 2\left(\frac{2}{1+x y}\right)^{2^{k-1}}
+$$
+
+The proof goes by induction. For $k=1$, we have
+
+$$
+\left(\frac{2}{1+x}\right)^{2}+\left(\frac{2}{1+y}\right)^{2} \geq 2\left(\frac{2}{1+x y}\right)
+$$
+
+which reduces to
+
+$$
+x y(x-y)^{2}+(x y-1)^{2} \geq 0 .
+$$
+
+For $k>1$, 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
+
+$$
+\begin{aligned}
+2\left(\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}}\right) & \geq\left(\left(\frac{2}{1+x}\right)^{2^{k-1}}+\left(\frac{2}{1+y}\right)^{2^{k-1}}\right)^{2} \\
+& \geq\left(2\left(\frac{2}{1+x y}\right)^{2^{k-2}}\right)^{2}=4\left(\frac{2}{1+x y}\right)^{2^{k-1}}
+\end{aligned}
+$$
+
+from which the lemma follows.
+The problem now can be deduced from summing the following applications of the lemma, multiplied by the appropriate factor:
+
+$$
+\begin{aligned}
+\frac{1}{2^{n}}\left(\frac{2}{1+a_{n}}\right)^{2^{n}}+\frac{1}{2^{n}}\left(\frac{2}{1+1}\right)^{2^{n}} & \geq \frac{1}{2^{n-1}}\left(\frac{2}{1+a_{n} \cdot 1}\right)^{2^{n-1}} \\
+\frac{1}{2^{n-1}}\left(\frac{2}{1+a_{n-1}}\right)^{2^{n-1}}+\frac{1}{2^{n-1}}\left(\frac{2}{1+a_{n}}\right)^{2^{n-1}} & \geq \frac{1}{2^{n-2}}\left(\frac{2}{1+a_{n-1} a_{n}}\right)^{2^{n-2}} \\
+\frac{1}{2^{n-2}}\left(\frac{2}{1+a_{n-2}}\right)^{2^{n-2}}+\frac{1}{2^{n-2}}\left(\frac{2}{1+a_{n-1} a_{n}}\right)^{2^{n-2}} & \geq \frac{1}{2^{n-3}}\left(\frac{2}{1+a_{n-2} a_{n-1} a_{n}}\right)^{2^{n-3}} \\
+\ldots & )^{2^{k}} \\
+\frac{1}{2^{k}}\left(\frac{2}{1+a_{k}}\right)^{2^{k}}+\frac{1}{2^{k}}\left(\frac{2}{1+a_{k+1} \ldots a_{n-1} a_{n}}\right)^{2^{k-1}} & \geq \frac{1}{2^{k-1}}\left(\frac{2}{1+a_{k} \ldots a_{n-2} a_{n-1} a_{n}}\right)^{2} \\
+\frac{1}{2}\left(\frac{2}{1+a_{1}}\right)^{2}+\frac{1}{2}\left(\frac{2}{1+a_{2} \ldots a_{n-1} a_{n}}\right)^{2} & \geq \frac{2}{1+a_{1} \ldots a_{n-2} a_{n-1} a_{n}}
+\end{aligned}
+$$
+
+Comment: Equality occurs if and only if $a_{1}=a_{2}=\cdots=a_{n}=1$.
+
+Comment: The main motivation for the lemma is trying to "telescope" the sum
+
+$$
+\frac{1}{2^{n}}+\sum_{i=1}^{n} \frac{1}{2^{i}}\left(\frac{2}{1+a_{i}}\right)^{2^{i}}
+$$
+
+that is,
+
+$$
+\frac{1}{2}\left(\frac{2}{1+a_{1}}\right)^{2}+\cdots+\frac{1}{2^{n-1}}\left(\frac{2}{1+a_{n-1}}\right)^{2^{n-1}}+\frac{1}{2^{n}}\left(\frac{2}{1+a_{n}}\right)^{2^{n}}+\frac{1}{2^{n}}\left(\frac{2}{1+1}\right)^{2^{n}}
+$$
+
+to obtain an expression larger than or equal to
+
+$$
+\frac{2}{1+a_{1} a_{2} \ldots a_{n}}
+$$
+
+It 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:
+
+$$
+\frac{1}{2^{k}}\left(\frac{2}{1+x}\right)^{2^{k}}+\frac{1}{2^{k}}\left(\frac{2}{1+y}\right)^{2^{k}} \geq \frac{1}{2^{k-1}}\left(\frac{2}{1+x y}\right)^{2^{i-1}}
+$$
+
+or
+
+$$
+\left(\frac{2}{1+x}\right)^{2^{k}}+\left(\frac{2}{1+y}\right)^{2^{k}} \geq 2\left(\frac{2}{1+x y}\right)^{2^{k-1}}
+$$
+
+## Problem 4
+
+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, $|S(k)|<|S(n)|$.
+We start with a lemma that guides us how the permutation should be set.
+
+## Lemma 1.
+
+$$
+\sum_{i=0}^{t-1}|S(t+i)|=t+\sum_{i=0}^{t-1}|S(2 i)|
+$$
+
+The 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
+
+$$
+\begin{aligned}
+\sum_{i=0}^{t-1}|S(t+i)| & =\sum_{i=0}^{2 t-1}|S(i)|-\sum_{i=0}^{t-1}|S(i)|=\sum_{i=0}^{t-1}|S(2 i)|+\sum_{i=0}^{t-1}|S(2 i+1)|-\sum_{i=0}^{t-1}|S(i)| \\
+& =\sum_{i=0}^{t-1}|S(i)|+\sum_{i=0}^{t-1}(1+|S(i)|)-\sum_{i=0}^{t-1}|S(i)|=t+\sum_{i=0}^{t-1}|S(i)|=t+\sum_{i=0}^{t-1}|S(2 i)|
+\end{aligned}
+$$
+
+The 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, $\left\{k_{i}\right\}=S(t+i) \backslash S\left(2 a_{i}\right)$.)
+In 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.
+Say 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.
+The 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 $y \in A_{t}, 0 \leq y<t-2^{a}$.
+Now 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,
+
+$$
+0 \leq x \leq\left\lfloor\frac{2^{a+1}-t}{2}\right\rfloor-1=2^{a}-\left\lceil\frac{t}{2}\right\rceil-1 \Longleftrightarrow\left\lceil\frac{t}{2}\right\rceil \leq 2^{a}-1-x \leq 2^{a}-1
+$$
+
+and
+
+$$
+\left\lceil\frac{2^{a+1}-t}{2}\right\rceil=2^{a}-\left\lfloor\frac{t}{2}\right\rfloor \leq y \leq 2^{a+1}-t-1 \Longleftrightarrow t-2^{a} \leq 2^{a}-1-y \leq\left\lfloor\frac{t}{2}\right\rfloor-1
+$$
+
+Moreover, $S\left(2^{a}-1-x\right)=S \backslash S(x)$ 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.
+We illustrate the bijection by showing the case $t=23$ :
+
+$$
+A_{23}=\{0,1,2, \ldots, 10\}, \quad B_{23}=\{12,13,14, \ldots, 22\}
+$$
+
+The pairing is
+
+$$
+\left(\begin{array}{ccccccccccc}
+12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\
+8 & 9 & 10 & 7 & 0 & 1 & 2 & 3 & 4 & 5 & 6
+\end{array}\right)
+$$
+
+in which the bijection is between
+
+$$
+\left(\begin{array}{cccc}
+12 & 13 & 14 & 15 \\
+8 & 9 & 10 & 7
+\end{array}\right) \quad \text { and } \quad\left(\begin{array}{llll}
+3 & 2 & 1 & 0 \\
+7 & 6 & 5 & 8
+\end{array}\right) \rightarrow\left(\begin{array}{llll}
+5 & 6 & 7 & 8 \\
+1 & 2 & 3 & 0
+\end{array}\right) .
+$$
+
+## Problem 5
+
+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 1
+
+We start with the following lemma.
+Lemma 1. Points $M, N, P, Q$ are concyclic.
+Point $M$ 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 $L R S$.
+Then, since quadrilaterals $Q N R C, P M A S$, and $A B C D$ are all cyclic, using directed angles (modulo $180^{\circ}$ )
+
+$$
+\begin{aligned}
+\measuredangle N M P & =\measuredangle N M S+\measuredangle S M P=\measuredangle N R S+\measuredangle S A P=\measuredangle N R Q+\measuredangle D A B=\measuredangle N R Q+\measuredangle D C B \\
+& =\measuredangle N R Q+\measuredangle Q C R=\measuredangle N R Q+\measuredangle Q N R=\measuredangle N Q R=\measuredangle N Q P,
+\end{aligned}
+$$
+
+which implies that $M N Q P$ is a cyclic quadrilateral.
+![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-08.jpg?height=975&width=1115&top_left_y=1306&top_left_x=439)
+
+Let $E$ 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 $M N Q P$,
+
+$$
+\measuredangle T E M=\measuredangle L E M=\measuredangle L A M=\measuredangle S A M=\measuredangle S P M=\measuredangle Q P M=\measuredangle Q N M=\measuredangle T N M,
+$$
+
+that 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 $\omega$.
+
+Now 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 $\omega$.
+Therefore, 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 $t$.
+
+## Solution 2
+
+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.
+
+Lemma 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
+
+$$
+\left(\operatorname{pow}_{\Gamma_{1}} A-\operatorname{pow}_{\Gamma_{2}} A\right) x+\left(\operatorname{pow}_{\Gamma_{1}} B-\operatorname{pow}_{\Gamma_{2}} B\right) y+\left(\operatorname{pow}_{\Gamma_{1}} C-\operatorname{pow}_{\Gamma_{2}} C\right) z=0
+$$
+
+Proof: 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 $\operatorname{pow}_{\Gamma_{i}} C=t_{i}$.
+Finally, the radical axis is
+
+$$
+\begin{aligned}
+& \operatorname{pow}_{\Gamma_{1}} P=\operatorname{pow}_{\Gamma_{2}} P \\
+\Longleftrightarrow & \Gamma_{1}(x, y, z)=\Gamma_{2}(x, y, z) \\
+\Longleftrightarrow & r_{1} x+s_{1} y+t_{1} z=r_{2} x+s_{2} y+t_{2} z \\
+\Longleftrightarrow & \left(\operatorname{pow}_{\Gamma_{1}} A-\operatorname{pow}_{\Gamma_{2}} A\right) x+\left(\operatorname{pow}_{\Gamma_{1}} B-\operatorname{pow}_{\Gamma_{2}} B\right) y+\left(\operatorname{pow}_{\Gamma_{1}} C-\operatorname{pow}_{\Gamma_{2}} C\right) z=0 .
+\end{aligned}
+$$
+
+We 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 $E$ :
+
+- $M P$ is the radical axis of the circumcircles of $P A S$ and $P B R$;
+- $N Q$ is the radical axis of the circumcircles of $Q C R$ and $Q D S$;
+- $E K$ is the radical axis of the circumcircles of $K B C$ and $K A D$.
+
+Looking 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 $P Q$.
+![](https://cdn.mathpix.com/cropped/2024_11_22_0826db6e0d83838635cfg-09.jpg?height=515&width=1043&top_left_y=2050&top_left_x=478)
+
+Associate $P$ 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
+
+- MP: $(K A \cdot K P-K B \cdot K P) x+(Q S \cdot Q P-Q R \cdot Q P) y=0$
+- $N Q:(K C \cdot K Q-K D \cdot K Q) x+(P R \cdot P Q-P S \cdot P Q) z=0$
+- MP: $(-Q C \cdot Q K+Q D \cdot Q K) y+(P B \cdot P K-P A \cdot P K) z=0$
+
+These equations simplify to
+
+- $M P:(A B \cdot K P) x+(P Q \cdot R S) y=0$
+- $N Q:(-C D \cdot K Q) x+(P Q \cdot R S) z=0$
+- $M P:(C D \cdot K Q) y+(A B \cdot K P) z=0$
+
+Now, 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
+
+$$
+\left|\begin{array}{ccc}
+u & v & 0 \\
+-w & 0 & v \\
+0 & w & u
+\end{array}\right|=0
+$$
+
+which is a straightforward computation.
+

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