add IMO 2005 - 2025
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- IMO/download_script/download.py +75 -0
- IMO/md/en-IMO-2005-notes.md +285 -0
- IMO/md/en-IMO-2006-notes.md +289 -0
- IMO/md/en-IMO-2007-notes.md +281 -0
- IMO/md/en-IMO-2008-notes.md +248 -0
- IMO/md/en-IMO-2009-notes.md +290 -0
- IMO/md/en-IMO-2010-notes.md +317 -0
- IMO/md/en-IMO-2011-notes.md +295 -0
- IMO/md/en-IMO-2012-notes.md +286 -0
- IMO/md/en-IMO-2013-notes.md +0 -0
- IMO/md/en-IMO-2014-notes.md +0 -0
- IMO/md/en-IMO-2015-notes.md +0 -0
- IMO/md/en-IMO-2016-notes.md +303 -0
- IMO/md/en-IMO-2017-notes.md +383 -0
- IMO/md/en-IMO-2018-notes.md +0 -0
- IMO/md/en-IMO-2019-notes.md +0 -0
- IMO/md/en-IMO-2020-notes.md +0 -0
- IMO/md/en-IMO-2021-notes.md +0 -0
- IMO/md/en-IMO-2022-notes.md +316 -0
- IMO/md/en-IMO-2023-notes.md +0 -0
- IMO/md/en-IMO-2024-notes.md +0 -0
- IMO/md/en-IMO-2025-notes.md +0 -0
- IMO/raw/en-IMO-2005-notes.pdf +3 -0
- IMO/raw/en-IMO-2006-notes.pdf +3 -0
- IMO/raw/en-IMO-2007-notes.pdf +3 -0
- IMO/raw/en-IMO-2008-notes.pdf +3 -0
- IMO/raw/en-IMO-2009-notes.pdf +3 -0
- IMO/raw/en-IMO-2010-notes.pdf +3 -0
- IMO/raw/en-IMO-2011-notes.pdf +3 -0
- IMO/raw/en-IMO-2012-notes.pdf +3 -0
- IMO/raw/en-IMO-2013-notes.pdf +3 -0
- IMO/raw/en-IMO-2014-notes.pdf +3 -0
- IMO/raw/en-IMO-2015-notes.pdf +3 -0
- IMO/raw/en-IMO-2016-notes.pdf +3 -0
- IMO/raw/en-IMO-2017-notes.pdf +3 -0
- IMO/raw/en-IMO-2018-notes.pdf +3 -0
- IMO/raw/en-IMO-2019-notes.pdf +3 -0
- IMO/raw/en-IMO-2020-notes.pdf +3 -0
- IMO/raw/en-IMO-2021-notes.pdf +3 -0
- IMO/raw/en-IMO-2022-notes.pdf +3 -0
- IMO/raw/en-IMO-2023-notes.pdf +3 -0
- IMO/raw/en-IMO-2024-notes.pdf +3 -0
- IMO/raw/en-IMO-2025-notes.pdf +3 -0
- IMO/segment_script/segment.py +182 -0
- IMO/segmented/en-IMO-2005-notes.jsonl +6 -0
- IMO/segmented/en-IMO-2006-notes.jsonl +6 -0
- IMO/segmented/en-IMO-2007-notes.jsonl +6 -0
- IMO/segmented/en-IMO-2008-notes.jsonl +6 -0
- IMO/segmented/en-IMO-2009-notes.jsonl +6 -0
- IMO/segmented/en-IMO-2010-notes.jsonl +6 -0
IMO/download_script/download.py
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# -----------------------------------------------------------------------------
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# Author: Jiawei Liu
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# Date: 2025-10-25
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# -----------------------------------------------------------------------------
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'''
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Download script for APMO
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To run:
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`python IMO/download_script/download.py`
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'''
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import re
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import requests
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from bs4 import BeautifulSoup
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from tqdm import tqdm
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from pathlib import Path
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from requests.adapters import HTTPAdapter
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from urllib3.util.retry import Retry
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from urllib.parse import urljoin
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def build_session(
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max_retries: int = 3,
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backoff_factor: int = 2,
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session: requests.Session = None
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) -> requests.Session:
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"""
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Build a requests session with retries
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Args:
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max_retries (int, optional): Number of retries. Defaults to 3.
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backoff_factor (int, optional): Backoff factor. Defaults to 2.
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session (requests.Session, optional): Session object. Defaults to None.
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"""
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session = session or requests.Session()
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adapter = HTTPAdapter(max_retries=Retry(total=max_retries, backoff_factor=backoff_factor))
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session.mount("http://", adapter)
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session.mount("https://", adapter)
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session.headers.update({
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"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"
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})
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return session
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def main():
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base_url = "https://web.evanchen.cc/problems.html"
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req_session = build_session()
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output_dir = Path(__file__).parent.parent / "raw"
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output_dir.mkdir(parents=True, exist_ok=True)
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resp = req_session.get(base_url)
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soup = BeautifulSoup(resp.text, 'html.parser')
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imo_exams = soup.find_all("a", href=lambda h: h and re.search(r"IMO\-\d{4}\-notes.pdf$", h))
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imo_pdf_links = [urljoin(base_url, link["href"]) for link in imo_exams]
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for link in tqdm(imo_pdf_links):
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output_file = output_dir / f"en-{Path(link).name}"
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# Check if the file already exists
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if output_file.exists():
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continue
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pdf_resp = req_session.get(link)
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if pdf_resp.status_code != 200:
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print(f"Failed to download {link}")
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continue
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output_file.write_bytes(pdf_resp.content)
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if __name__ == "__main__":
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main()
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IMO/md/en-IMO-2005-notes.md
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# IMO 2005 Solution Notes
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Evan Chen 《陳誼廷》
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18 October 2025
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This is a compilation of solutions for the 2005 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
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These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
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Corrections and comments are welcome!
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## Contents
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0 Problems 2
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1 Solutions to Day 1 3
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1.1 IMO 2005/1, proposed by Bogdan Enescu (ROU) 3
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1.2 IMO 2005/2, proposed by Nicholas de Bruijn (NLD) 4
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1.3 IMO 2005/3, proposed by Hojoo Lee (KOR) 6
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2 Solutions to Day 2 8
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2.1 IMO 2005/4, proposed by Mariusz Skalba (POL) 8
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2.2 IMO 2005/5, proposed by Waldemar Pompe (POL) 9
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2.3 IMO 2005/6, proposed by Radu Gologan, Dan Schwartz (ROU) 10
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## Problems
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1. Six points are chosen on the sides of an equilateral triangle \(A B C\) .. \(A_{1}\) \(A_{2}\) on \(B C\) \(B_{1}\) \(B_{2}\) on \(C A\) and \(C_{1}\) \(C_{2}\) on \(A B\) , such that they are the vertices of a convex hexagon \(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\) with equal side lengths. Prove that the lines \(A_{1}B_{2}\) \(B_{1}C_{2}\) and \(C_{1}A_{2}\) are concurrent.
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2. Let \(a_{1}\) , \(a_{2}\) , ... be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \(n\) the numbers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) leave \(n\) different remainders upon division by \(n\) . Prove that every integer occurs exactly once in the sequence.
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3. Let \(x,y,z > 0\) satisfy \(x y z\geq 1\) . Prove that
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\[\frac{x^{5} - x^{2}}{x^{5} + y^{2} + z^{2}} +\frac{y^{5} - y^{2}}{x^{2} + y^{5} + z^{2}} +\frac{z^{5} - z^{2}}{x^{2} + y^{2} + z^{5}}\geq 0.\]
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4. Determine all positive integers relatively prime to all the terms of the infinite sequence
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\[a_{n} = 2^{n} + 3^{n} + 6^{n} - 1,\quad n\geq 1.\]
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5. Let \(A B C D\) be a fixed convex quadrilateral with \(B C = D A\) and \(\overline{{B C}}\not\parallel\overline{{D A}}\) . Let two variable points \(E\) and \(F\) lie on the sides \(B C\) and \(D A\) , respectively, and satisfy \(B E = D F\) . The lines \(A C\) and \(B D\) meet at \(P\) , the lines \(B D\) and \(E F\) meet at \(Q\) , the lines \(E F\) and \(A C\) meet at \(R\) . Prove that the circumcircles of the triangles \(P Q R\) , as \(E\) and \(F\) vary, have a common point other than \(P\) .
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6. In a mathematical competition 6 problems were posed to the contestants. Each pair of problems was solved by more than \(\frac{2}{5}\) of the contestants. Nobody solved all 6 problems. Show that there were at least 2 contestants who each solved exactly 5 problems.
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## \(\S 1\) Solutions to Day 1
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## \(\S 1.1\) IMO 2005/1, proposed by Bogdan Enescu (ROU)
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Available online at https://aops.com/community/p281571.
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## Problem statement
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Six points are chosen on the sides of an equilateral triangle \(ABC\) : \(A_1\) , \(A_2\) on \(BC\) , \(B_1\) , \(B_2\) on \(CA\) and \(C_1\) , \(C_2\) on \(AB\) , such that they are the vertices of a convex hexagon \(A_1A_2B_1B_2C_1C_2\) with equal side lengths. Prove that the lines \(A_1B_2\) , \(B_1C_2\) and \(C_1A_2\) are concurrent.
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The six sides of the hexagon, when oriented, comprise six vectors with vanishing sum. However note that
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\[\overrightarrow{A_1A_2} +\overrightarrow{B_1B_2} +\overrightarrow{C_1C_2} = 0.\]
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Thus
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\[\overrightarrow{A_2B_1} +\overrightarrow{B_2C_1} +\overrightarrow{C_2A_1} = 0\]
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and since three unit vectors with vanishing sum must be rotations of each other by \(120^{\circ}\) , it follows they must also form an equilateral triangle.
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| 79 |
+
Consequently, triangles \(A_1A_2B_1\) , \(B_1B_2C_1\) , \(C_1C_2A_1\) are congruent, as \(\angle A_2 = \angle B_2 = \angle C_2\) . So triangle \(A_1B_1C_1\) is equilateral and the diagonals are concurrent at the center.
|
| 80 |
+
|
| 81 |
+
|
| 82 |
+
|
| 83 |
+
## \(\S 1.2\) IMO 2005/2, proposed by Nicholas de Bruijn (NLD)
|
| 84 |
+
|
| 85 |
+
Available online at https://aops.com/community/p281572.
|
| 86 |
+
|
| 87 |
+
## Problem statement
|
| 88 |
+
|
| 89 |
+
Let \(a_{1}\) , \(a_{2}\) , ... be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \(n\) the numbers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) leave \(n\) different remainders upon division by \(n\) . Prove that every integer occurs exactly once in the sequence.
|
| 90 |
+
|
| 91 |
+
Obviously every integer appears at most once (otherwise take \(n\) much larger). So we will prove every integer appears at least once.
|
| 92 |
+
|
| 93 |
+
Claim — For any \(i < j\) we have \(|a_{i} - a_{j}| < j\) .
|
| 94 |
+
|
| 95 |
+
Proof. Otherwise, let \(n = |a_{i} - a_{j}| \neq 0\) . Then \(i, j \in [1, n]\) and \(a_{i} \equiv a_{j} \pmod{n}\) , contradiction.
|
| 96 |
+
|
| 97 |
+
Claim — For any \(n\) , the set \(\{a_{1}, \ldots , a_{n}\}\) is of the form \(\{k + 1, \ldots , k + n\}\) for some integer \(k\) .
|
| 98 |
+
|
| 99 |
+
Proof. By induction, with the base case \(n = 1\) being vacuous. For the inductive step, suppose \(\{a_{1}, \ldots , a_{n}\} = \{k + 1, \ldots , k + n\}\) are determined. Then
|
| 100 |
+
|
| 101 |
+
\[a_{n + 1} \equiv k \pmod{n + 1}.\]
|
| 102 |
+
|
| 103 |
+
Moreover by the earlier claim we have
|
| 104 |
+
|
| 105 |
+
\[|a_{n + 1} - a_{1}| < n + 1.\]
|
| 106 |
+
|
| 107 |
+
From this we deduce \(a_{n + 1} \in \{k, k + n + 1\}\) as desired.
|
| 108 |
+
|
| 109 |
+
This gives us actually a complete description of all possible sequences satisfying the hypothesis: choose any value of \(a_{1}\) to start. Then, for the \(n\) th term, the set \(S = \{a_{1}, \ldots , a_{n - 1}\}\) is (in some order) a set of \(n - 1\) consecutive integers. We then let \(a_{n} = \max S + 1\) or \(a_{n} = \min S - 1\) . A picture of six possible starting terms is shown below.
|
| 110 |
+
|
| 111 |
+
|
| 112 |
+
|
| 113 |
+
|
| 114 |
+
|
| 115 |
+
|
| 116 |
+
Finally, we observe that the condition that the sequence has infinitely many positive and negative terms (which we have not used until now) implies it is unbounded above and below. Thus it must contain every integer.
|
| 117 |
+
|
| 118 |
+
|
| 119 |
+
|
| 120 |
+
## \(\S 1.3\) IMO 2005/3, proposed by Hojoo Lee (KOR)
|
| 121 |
+
|
| 122 |
+
Available online at https://aops.com/community/p281573.
|
| 123 |
+
|
| 124 |
+
## Problem statement
|
| 125 |
+
|
| 126 |
+
Let \(x,y,z > 0\) satisfy \(x y z\geq 1\) . Prove that
|
| 127 |
+
|
| 128 |
+
\[\frac{x^{5} - x^{2}}{x^{5} + y^{2} + z^{2}} +\frac{y^{5} - y^{2}}{x^{2} + y^{5} + z^{2}} +\frac{z^{5} - z^{2}}{x^{2} + y^{2} + z^{5}}\geq 0.\]
|
| 129 |
+
|
| 130 |
+
Negating both sides and adding 3 eliminates the minus signs:
|
| 131 |
+
|
| 132 |
+
\[\sum_{\mathrm{cyc}}\frac{1}{x^{5} + y^{2} + z^{2}}\leq \frac{3}{x^{2} + y^{2} + z^{2}}.\]
|
| 133 |
+
|
| 134 |
+
Thus we only need to consider the case \(x y z = 1\)
|
| 135 |
+
|
| 136 |
+
Direct expansion and Muirhead works now! As advertised, once we show it suffices to analyze if \(x y z = 1\) the inequality becomes more economically written as
|
| 137 |
+
|
| 138 |
+
\[S = \sum_{\mathrm{cyc}}x^{2}(x^{2} - y z)(y^{4} + x^{3}z + x z^{3})(z^{4} + x^{3}y + x y^{3})\stackrel {?}{\geq}0.\]
|
| 139 |
+
|
| 140 |
+
So, clearing all the denominators gives
|
| 141 |
+
|
| 142 |
+
\[S = \sum_{\mathrm{cyc}}x^{2}(x^{2} - y z)\left[y^{4}z^{4} + x^{3}y^{5} + x y^{7} + x^{3}z^{5} + x^{6}y z + x^{4}y^{3}z + x z^{7} + x^{4}y z^{3} + x^{2}y^{3}z^{3}\right]\] \[\quad = \sum_{\mathrm{cyc}}\left[x^{4}y^{4}z^{4} + x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{8}y^{3}z + x^{5}z^{7} + x^{8}y z^{3} + x^{6}y^{3}z^{3}\right]\] \[\quad -\sum_{\mathrm{cyc}}\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{3}y^{8}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{3}y z^{8} + x^{6}y^{2}z^{4} + x^{4}y^{4}z^{4}\right]\] \[\quad = \sum_{\mathrm{cyc}}\left[x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{5}z^{7} + x^{6}y^{3}z^{3}\right]\] \[\quad -\sum_{\mathrm{cyc}}\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{2}z^{4}\right]\]
|
| 143 |
+
|
| 144 |
+
In other words we need to show
|
| 145 |
+
|
| 146 |
+
\[\sum_{\mathrm{sym}}\left(2x^{7}y^{5} + \frac{1}{2} x^{10}y z + \frac{1}{2} x^{6}y^{3}z^{3}\right)\geq \sum_{\mathrm{sym}}\left(\frac{1}{2} x^{8}y^{2}z^{2} + \frac{1}{2} x^{5}y^{5}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{5}z\right).\]
|
| 147 |
+
|
| 148 |
+
which follows by summing
|
| 149 |
+
|
| 150 |
+
\[\sum_{\mathrm{sym}}\frac{x^{10}y z + x^{6}y^{3}z^{3}}{2}\geq \sum_{\mathrm{sym}}x^{8}y^{2}z^{2}\] \[\frac{1}{2}\sum_{\mathrm{sym}}x^{8}y^{2}z^{2}\geq \frac{1}{2}\sum_{\mathrm{sym}}x^{6}y^{4}z^{2}\] \[\frac{1}{2}\sum_{\mathrm{sym}}x^{7}y^{5}\geq \frac{1}{2}\sum_{\mathrm{sym}}x^{5}y^{5}z^{2}\]
|
| 151 |
+
|
| 152 |
+
|
| 153 |
+
|
| 154 |
+
\[{\frac{1}{2}}\sum_{\mathrm{sym}}x^{7}y^{5}\geq{\frac{1}{2}}\sum_{\mathrm{sym}}x^{6}y^{4}z^{2}\] \[\sum_{\mathrm{sym}}x^{7}y^{5}\geq\sum_{\mathrm{sym}}x^{6}y^{5}z.\]
|
| 155 |
+
|
| 156 |
+
The first line here comes from AM- GM, the rest come from Muirhead.
|
| 157 |
+
|
| 158 |
+
Remark. More elegant approach is to use Cauchy in the form
|
| 159 |
+
|
| 160 |
+
\[\frac{1}{x^{5} + y^{2} + z^{2}} \leq \frac{x^{-1} + y^{2} + z^{2}}{(x^{2} + y^{2} + z^{2})^{2}}.\]
|
| 161 |
+
|
| 162 |
+
|
| 163 |
+
|
| 164 |
+
## \(\S 2\) Solutions to Day 2
|
| 165 |
+
|
| 166 |
+
## \(\S 2.1\) IMO 2005/4, proposed by Mariusz Skalba (POL)
|
| 167 |
+
|
| 168 |
+
Available online at https://aops.com/community/p282138.
|
| 169 |
+
|
| 170 |
+
## Problem statement
|
| 171 |
+
|
| 172 |
+
Determine all positive integers relatively prime to all the terms of the infinite sequence
|
| 173 |
+
|
| 174 |
+
\[a_{n} = 2^{n} + 3^{n} + 6^{n} - 1,\quad n\geq 1.\]
|
| 175 |
+
|
| 176 |
+
The answer is 1 only (which works).
|
| 177 |
+
|
| 178 |
+
It suffices to show there are no primes. For the primes \(p = 2\) and \(p = 3\) , take \(a_{2} = 48\) . For any prime \(p \geq 5\) notice that
|
| 179 |
+
|
| 180 |
+
\[a_{p - 2} = 2^{p - 2} + 3^{p - 2} + 6^{p - 2} - 1\] \[\equiv \frac{1}{2} +\frac{1}{3} +\frac{1}{6} -1\pmod {p\] \[\equiv 0\pmod {p\]
|
| 181 |
+
|
| 182 |
+
so no other larger prime works.
|
| 183 |
+
|
| 184 |
+
|
| 185 |
+
|
| 186 |
+
## \(\S 2.2\) IMO 2005/5, proposed by Waldemar Pompe (POL)
|
| 187 |
+
|
| 188 |
+
Available online at https://aops.com/community/p282140.
|
| 189 |
+
|
| 190 |
+
## Problem statement
|
| 191 |
+
|
| 192 |
+
Let \(A B C D\) be a fixed convex quadrilateral with \(B C = D A\) and \(\overline{{B C}}\not\parallel\overline{{D A}}\) . Let two variable points \(E\) and \(F\) lie on the sides \(B C\) and \(D A\) , respectively, and satisfy \(B E = D F\) . The lines \(A C\) and \(B D\) meet at \(P\) , the lines \(B D\) and \(E F\) meet at \(Q\) , the lines \(E F\) and \(A C\) meet at \(R\) . Prove that the circumcircles of the triangles \(P Q R\) , as \(E\) and \(F\) vary, have a common point other than \(P\) .
|
| 193 |
+
|
| 194 |
+
Let \(M\) be the Miquel point of complete quadrilateral \(A D B C\) ; in other words, let \(M\) be the second intersection point of the circumcircles of \(\triangle A P D\) and \(\triangle B P C\) . (A good diagram should betray this secret; all the points are given in the picture.) This makes lots of sense since we know \(E\) and \(F\) will be sent to each other under the spiral similarity too.
|
| 195 |
+
|
| 196 |
+
|
| 197 |
+
|
| 198 |
+
|
| 199 |
+
Thus \(M\) is the Miquel point of complete quadrilateral \(F A C E\) . As \(R = \overline{{F E}}\cap \overline{{A C}}\) we deduce \(F A R M\) is a cyclic quadrilateral (among many others, but we'll only need one).
|
| 200 |
+
|
| 201 |
+
Now look at complete quadrilateral \(A F Q P\) . Since \(M\) lies on \((D F Q)\) and \((R A F)\) , it follows that \(M\) is in fact the Miquel point of \(A F Q P\) as well. So \(M\) lies on \((P Q R)\) .
|
| 202 |
+
|
| 203 |
+
Thus \(M\) is the fixed point that we wanted.
|
| 204 |
+
|
| 205 |
+
Remark. Naturally, the congruent length condition can be relaxed to \(D F / D A = B E / B C\) .
|
| 206 |
+
|
| 207 |
+
|
| 208 |
+
|
| 209 |
+
## \(\S 2.3\) IMO 2005/6, proposed by Radu Gologan, Dan Schwartz (ROU)
|
| 210 |
+
|
| 211 |
+
Available online at https://aops.com/community/p282141.
|
| 212 |
+
|
| 213 |
+
## Problem statement
|
| 214 |
+
|
| 215 |
+
In a mathematical competition 6 problems were posed to the contestants. Each pair of problems was solved by more than \(\frac{2}{5}\) of the contestants. Nobody solved all 6 problems. Show that there were at least 2 contestants who each solved exactly 5 problems.
|
| 216 |
+
|
| 217 |
+
Assume not and at most one contestant solved five problems. By adding in solves, we can assume WLOG that one contestant solved problems one through five, and every other contestant solved four of the six problems.
|
| 218 |
+
|
| 219 |
+
We split the remaining contestants based on whether they solved P6. Let \(a_{i}\) denote the number of contestants who solved \(\{1,2,\ldots ,5\} \setminus \{i\}\) (and missed P6). Let \(b_{ij}\) denote the number of contestants who solved \(\{1,2,\ldots ,5,6\} \setminus \{i,j\}\) , for \(1\leq i< j\leq 5\) (thus in particular they solved P6). Thus
|
| 220 |
+
|
| 221 |
+
\[n = 1 + \sum_{1\leq i\leq 5}a_{i} + \sum_{1\leq i< j\leq 5}b_{ij}\]
|
| 222 |
+
|
| 223 |
+
denotes the total number of contestants.
|
| 224 |
+
|
| 225 |
+
Considering contestants who solved P1/P6 we have
|
| 226 |
+
|
| 227 |
+
\[t_{1}:= b_{23} + b_{24} + b_{25} + b_{34} + b_{35} + b_{45}\geq \frac{2}{5} n + \frac{1}{5}\]
|
| 228 |
+
|
| 229 |
+
and we similarly define \(t_{2}\) , \(t_{3}\) , \(t_{4}\) , \(t_{5}\) . (We have written \(\frac{2}{5} n + \frac{1}{5}\) since we know the left- hand side is an integer strictly larger than \(\frac{2}{5} n\) .) Also, by considering contestants who solved P1/P2 we have
|
| 230 |
+
|
| 231 |
+
\[t_{12} = 1 + a_{3} + a_{4} + a_{5} + b_{34} + b_{35} + b_{45}\geq \frac{2}{5} n + \frac{1}{5}\]
|
| 232 |
+
|
| 233 |
+
and we similarly define \(t_{ij}\) for \(1\leq i< j\leq 5\)
|
| 234 |
+
|
| 235 |
+
Claim — The number \(\frac{2n + 1}{5}\) is equal to some integer \(k\) , fourteen of the \(t\) 's are equal to \(k\) , and the last one is equal to \(k + 1\) .
|
| 236 |
+
|
| 237 |
+
Proof. First, summing all fifteen equations gives
|
| 238 |
+
|
| 239 |
+
\[6n + 4 = 10 + 6(n - 1) = 10 + \sum_{1\leq i\leq 5}6a_{i} + \sum_{1\leq i< j\leq 5}6b_{ij}\] \[\qquad = \sum_{1\leq i\leq 5}t_{i} + \sum_{1\leq i< j\leq 5}t_{ij}.\]
|
| 240 |
+
|
| 241 |
+
Thus the sum of the 15 \(t\) 's is \(6n + 4\) . But since all the \(t\) 's are integers at least \(\frac{2n + 1}{5} = \frac{6n + 3}{15}\) , the conclusion follows. \(\square\)
|
| 242 |
+
|
| 243 |
+
However, we will also manipulate the equations to get the following.
|
| 244 |
+
|
| 245 |
+
|
| 246 |
+
|
| 247 |
+
Claim — We have
|
| 248 |
+
|
| 249 |
+
\[t_{45} \equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \pmod {3}.\]
|
| 250 |
+
|
| 251 |
+
Proof. This follows directly by computing the coefficient of the \(a\) 's and \(b\) 's. We will nonetheless write out a derivation of this equation, to motivate it, but the proof stands without it.
|
| 252 |
+
|
| 253 |
+
Let \(B = \sum_{1 \leq i < j \leq 5} b_{ij}\) be the sum of all \(b\) 's. First, note that
|
| 254 |
+
|
| 255 |
+
\[t_{1} + t_{2} = B + b_{34} + b_{45} + b_{35} - b_{12}\] \[\qquad = B + (t_{12} - 1 - a_{3} - a_{4} - a_{5}) - b_{12}\] \[\Rightarrow b_{12} = B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5}).\]
|
| 256 |
+
|
| 257 |
+
This means we have more or less solved for each \(b_{ij}\) in terms of only \(t\) and \(a\) variables. Now
|
| 258 |
+
|
| 259 |
+
\[t_{45} = 1 + a_{1} + a_{2} + a_{3} + b_{12} + b_{23} + b_{31}\] \[\qquad = 1 + a_{1} + a_{2} + a_{3}\] \[\qquad +[B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5})]\] \[\qquad +[B - (t_{2} + t_{3}) + t_{23} - 1 - (a_{1} + a_{4} + a_{5})]\] \[\qquad +[B - (t_{3} + t_{1}) + t_{13} - 1 - (a_{2} + a_{4} + a_{5})]\] \[\qquad \equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \pmod {3}\]
|
| 260 |
+
|
| 261 |
+
as desired.
|
| 262 |
+
|
| 263 |
+
However, we now show the two claims are incompatible (and this is easy, many ways to do this). There are two cases.
|
| 264 |
+
|
| 265 |
+
- Say \(t_{5} = k + 1\) and the others are \(k\) . Then the equation for \(t_{45}\) gives that \(k \equiv 6k + 1 \pmod {3}\) . But now the equation for \(t_{12}\) give \(k \equiv 6k \pmod {3}\) .
|
| 266 |
+
|
| 267 |
+
- Say \(t_{45} = k + 1\) and the others are \(k\) . Then the equation for \(t_{45}\) gives that \(k + 1 \equiv 6k \pmod {3}\) . But now the equation for \(t_{12}\) give \(k \equiv 6k + 1 \pmod {3}\) .
|
| 268 |
+
|
| 269 |
+
Remark. It is significantly easier to prove that there is at least one contestant who solved five problems. One can see it by dropping the \(+10\) in the proof of the claim, and arrives at a contradiction. In this situation it is not even necessary to set up the many \(a\) and \(b\) variables; just note that the expected number of contestants solving any particular pair of problems is \(\frac{\binom{4}{2}n}{\binom{6}{2}} = \frac{2}{5} n\) .
|
| 270 |
+
|
| 271 |
+
The fact that \(\frac{2n + 1}{5}\) should be an integer also follows quickly, since if not one can improve the bound to \(\frac{2n + 5}{5}\) and quickly run into a contradiction. Again one can get here without setting up \(a\) and \(b\) .
|
| 272 |
+
|
| 273 |
+
The main difficulty seems to be the precision required in order to nail down the second 5- problem solve.
|
| 274 |
+
|
| 275 |
+
Remark. The second claim may look miraculous, but the proof shows that it is not too unnatural to consider \(t_{1} + t_{2} - t_{12}\) to isolate \(b_{12}\) in terms of \(a\) 's and \(t\) 's. The main trick is: why mod 3?
|
| 276 |
+
|
| 277 |
+
The reason is that if one looks closely, for a fixed \(k\) we have a system of 15 equations in 15 variables. Unless the determinant \(D\) of that system happens to be zero, this means there will be a rational solution in \(a\) and \(b\) , whose denominators are bounded by \(D\) . However if
|
| 278 |
+
|
| 279 |
+
|
| 280 |
+
|
| 281 |
+
\(p\mid D\) then we may conceivably run into mod \(p\) issues.
|
| 282 |
+
|
| 283 |
+
This motivates the choice \(p = 3\) , since it is easy to see the determinant is divisible by 3, since constant shifts of \(\vec{a}\) and \(\vec{b}\) are also solutions mod 3. (The choice \(p = 2\) is a possible guess as well for this reason, but the problem seems to have better 3- symmetry.)
|
| 284 |
+
|
| 285 |
+
|
IMO/md/en-IMO-2006-notes.md
ADDED
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| 1 |
+
|
| 2 |
+
# IMO 2006 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2006 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2006/1, proposed by Hojoo Lee (KOR) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2006/2, proposed by Dušan Djukić (SRB) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2006/3, proposed by Finbarr Holland (IRL) 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 6
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2006/4, proposed by Zuming Feng (USA) 6
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2006/5, proposed by Dan Schwarz (ROU) 7
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2006/6, proposed by Dušan Djukić (SRB) 8
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Let \(A B C\) be a triangle with incenter \(I\) . A point \(P\) in the interior of the triangle satisfies
|
| 39 |
+
|
| 40 |
+
\[\angle P B A + \angle P C A = \angle P B C + \angle P C B.\]
|
| 41 |
+
|
| 42 |
+
Show that \(A P\geq A I\) and that equality holds if and only if \(P = I\)
|
| 43 |
+
|
| 44 |
+
2. Let \(P\) be a regular 2006-gon. A diagonal is called good if its endpoints divide the boundary of \(P\) into two parts, each composed of an odd number of sides of \(P\) . The sides of \(P\) are also called good. Suppose \(P\) has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of \(P\) . Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.
|
| 45 |
+
|
| 46 |
+
3. Determine the least real number \(M\) such that the inequality
|
| 47 |
+
|
| 48 |
+
\[\left|a b(a^{2} - b^{2}) + b c(b^{2} - c^{2}) + c a(c^{2} - a^{2})\right|\leq M\left(a^{2} + b^{2} + c^{2}\right)^{2}\]
|
| 49 |
+
|
| 50 |
+
holds for all real numbers \(a\) , \(b\) and \(c\) .
|
| 51 |
+
|
| 52 |
+
4. Determine all pairs \((x,y)\) of integers such that
|
| 53 |
+
|
| 54 |
+
\[1 + 2^{x} + 2^{2x + 1} = y^{2}.\]
|
| 55 |
+
|
| 56 |
+
5. Let \(P(x)\) be a polynomial of degree \(n > 1\) with integer coefficients and let \(k\) be a positive integer. Consider the polynomial
|
| 57 |
+
|
| 58 |
+
\[Q(x) = P(P(\ldots P(P(x))\ldots))\]
|
| 59 |
+
|
| 60 |
+
where \(P\) occurs \(k\) times. Prove that there are at most \(n\) integers \(t\) such that \(Q(t) = t\) .
|
| 61 |
+
|
| 62 |
+
6. Assign to each side \(b\) of a convex polygon \(P\) the maximum area of a triangle that has \(b\) as a side and is contained in \(P\) . Show that the sum of the areas assigned to the sides of \(P\) is at least twice the area of \(P\) .
|
| 63 |
+
|
| 64 |
+
|
| 65 |
+
|
| 66 |
+
## \(\S 1\) Solutions to Day 1
|
| 67 |
+
|
| 68 |
+
## \(\S 1.1\) IMO 2006/1, proposed by Hojoo Lee (KOR)
|
| 69 |
+
|
| 70 |
+
Available online at https://aops.com/community/p571966.
|
| 71 |
+
|
| 72 |
+
## Problem statement
|
| 73 |
+
|
| 74 |
+
Let \(ABC\) be a triangle with incenter \(I\) . A point \(P\) in the interior of the triangle satisfies
|
| 75 |
+
|
| 76 |
+
\[\angle PBA + \angle PCA = \angle PBC + \angle PCB.\]
|
| 77 |
+
|
| 78 |
+
Show that \(AP \geq AI\) and that equality holds if and only if \(P = I\) .
|
| 79 |
+
|
| 80 |
+
The condition rewrites as
|
| 81 |
+
|
| 82 |
+
\[\angle PBC + \angle PCB = (\angle B - \angle PBC) + (\angle C - \angle PCB) \Rightarrow \angle PBC + \angle PCB = \frac{\angle B + \angle C}{2}\]
|
| 83 |
+
|
| 84 |
+
which means that
|
| 85 |
+
|
| 86 |
+
\[\angle BPC = 180^{\circ} - \frac{\angle B + \angle C}{2} = 90^{\circ} + \frac{\angle A}{2} = \angle BIC.\]
|
| 87 |
+
|
| 88 |
+
Since \(P\) and \(I\) are both inside \(\triangle ABC\) that implies \(P\) lies on the circumcircle of \(\triangle BIC\) . It's well- known (by "Fact 5") that the circumcenter of \(\triangle BIC\) is the arc midpoint \(M\) of \(\widehat{BC}\) . Therefore
|
| 89 |
+
|
| 90 |
+
\[AI + IM = AM \leq AP + PM \Rightarrow AI \leq AP\]
|
| 91 |
+
|
| 92 |
+
with equality holding iff \(A\) , \(P\) , \(M\) are collinear, or \(P = I\) .
|
| 93 |
+
|
| 94 |
+
|
| 95 |
+
|
| 96 |
+
## \(\S 1.2\) IMO 2006/2, proposed by Dušan Djukić (SRB)
|
| 97 |
+
|
| 98 |
+
Available online at https://aops.com/community/p571973.
|
| 99 |
+
|
| 100 |
+
## Problem statement
|
| 101 |
+
|
| 102 |
+
Let \(P\) be a regular 2006- gon. A diagonal is called good if its endpoints divide the boundary of \(P\) into two parts, each composed of an odd number of sides of \(P\) . The sides of \(P\) are also called good. Suppose \(P\) has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of \(P\) . Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.
|
| 103 |
+
|
| 104 |
+
Call a triangle with the desired property special. We prove the maximum number of special triangles is 1003, achieved by paring up the sides of the polygon.
|
| 105 |
+
|
| 106 |
+
We present two solutions for the upper bound. Both of them rely first on two geometric notes:
|
| 107 |
+
|
| 108 |
+
- In a special triangle, the good sides are congruent (and not congruent to the third side).
|
| 109 |
+
|
| 110 |
+
- No two isosceles triangles share a good side.
|
| 111 |
+
|
| 112 |
+
\(\P\) Solution using bijections. Call a good diagonal special if it's part of a special triangle; special diagonals come in pairs. Consider the minor arc cut out by a special diagonal \(d\) , which has an odd number of sides. Since special diagonals come in pairs, one can associate to \(d\) a side of the polygon not covered by any special diagonals from \(d\) . Hence there are at most 2006 special diagonals, so at most 1003 special triangles.
|
| 113 |
+
|
| 114 |
+
\(\P\) Solution using graph theory. Consider the tree \(T\) formed by the 2004 triangles in the dissection, with obvious adjacency. Let \(F\) be the forest obtained by deleting any edge corresponding to a good diagonal. Then the resulting graph \(F\) has only degrees 1 and 3, with special triangles only occurring at degree 1 vertices.
|
| 115 |
+
|
| 116 |
+
If there are \(k\) good diagonals drawn, then this forest consists of \(k + 1\) trees. A tree with \(n_{i}\) vertices ( \(0 \leq i \leq k\) ) consequently has \(\frac{n_{i} + 2}{2}\) leaves. However by the earlier remark at least \(k\) leaves don't give special triangles (one on each side of a special diagonal); so the number of leaves that do give good triangles is at most
|
| 117 |
+
|
| 118 |
+
\[-k + \sum_{i}\frac{n_{i} + 2}{2} = -k + \frac{2004 + 2(k + 1)}{2} = 1003.\]
|
| 119 |
+
|
| 120 |
+
|
| 121 |
+
|
| 122 |
+
## \(\S 1.3\) IMO 2006/3, proposed by Finbarr Holland (IRL)
|
| 123 |
+
|
| 124 |
+
Available online at https://aops.com/community/p571945.
|
| 125 |
+
|
| 126 |
+
## Problem statement
|
| 127 |
+
|
| 128 |
+
Determine the least real number \(M\) such that the inequality
|
| 129 |
+
|
| 130 |
+
\[\left|a b(a^{2} - b^{2}) + b c(b^{2} - c^{2}) + c a(c^{2} - a^{2})\right|\leq M\left(a^{2} + b^{2} + c^{2}\right)^{2}\]
|
| 131 |
+
|
| 132 |
+
holds for all real numbers \(a\) , \(b\) and \(c\) .
|
| 133 |
+
|
| 134 |
+
It's the same as
|
| 135 |
+
|
| 136 |
+
\[|(a - b)(b - c)(c - a)(a + b + c)|\leq M\left(a^{2} + b^{2} + c^{2}\right)^{2}.\]
|
| 137 |
+
|
| 138 |
+
Let \(x = a - b\) , \(y = b - c\) , \(z = c - a\) , \(s = a + b + c\) . Then we want to have
|
| 139 |
+
|
| 140 |
+
\[|xyz s|\leq \frac{M}{9} (x^{2} + y^{2} + z^{2} + s^{2})^{2}.\]
|
| 141 |
+
|
| 142 |
+
Here \(x + y + z = 0\) .
|
| 143 |
+
|
| 144 |
+
Now if \(x\) and \(y\) have the same sign, we can replace them with the average (this increases the LHS and decreases RHS). So we can have \(x = y\) , \(z = - 2x\) . Now WLOG \(x > 0\) to get
|
| 145 |
+
|
| 146 |
+
\[2x^{3}\cdot s\leq \frac{M}{9}\left(6x^{2} + s^{2}\right)^{2}.\]
|
| 147 |
+
|
| 148 |
+
After this routine calculation gives \(M = \frac{9}{12}\sqrt{2}\) works and is optimal (by \(6x^{2} + s^{2} = 2x^{2} + 2x^{2} + 2x^{2} + s^{2}\) and AM- GM).
|
| 149 |
+
|
| 150 |
+
|
| 151 |
+
|
| 152 |
+
## \(\S 2\) Solutions to Day 2
|
| 153 |
+
|
| 154 |
+
## \(\S 2.1\) IMO 2006/4, proposed by Zuming Feng (USA)
|
| 155 |
+
|
| 156 |
+
Available online at https://aops.com/community/p572815.
|
| 157 |
+
|
| 158 |
+
## Problem statement
|
| 159 |
+
|
| 160 |
+
Determine all pairs \((x,y)\) of integers such that
|
| 161 |
+
|
| 162 |
+
\[1 + 2^{x} + 2^{2x + 1} = y^{2}.\]
|
| 163 |
+
|
| 164 |
+
Answers: \((0,\pm 2)\) , \((4,\pm 23)\) , which work.
|
| 165 |
+
|
| 166 |
+
Assume \(x \geq 4\) .
|
| 167 |
+
|
| 168 |
+
\[2^{x}\left(1 + 2^{x + 1}\right) = 2^{x} + 2^{2x + 1} = y^{2} - 1 = (y - 1)(y + 1).\]
|
| 169 |
+
|
| 170 |
+
So either:
|
| 171 |
+
|
| 172 |
+
- \(y = 2^{x - 1}m + 1\) for some odd \(m\) , so
|
| 173 |
+
|
| 174 |
+
\[1 + 2^{x + 1} = m\left(2^{x - 2}m + 1\right)\Longrightarrow 2^{x} = \frac{4(1 - m)}{m^{2} - 8}.\]
|
| 175 |
+
|
| 176 |
+
- \(y = 2^{x - 1}m - 1\) for some odd \(m\) , so
|
| 177 |
+
|
| 178 |
+
\[1 + 2^{x + 1} = m\left(2^{x - 2}m - 1\right)\Longrightarrow 2^{x} = \frac{4(1 + m)}{m^{2} - 8}.\]
|
| 179 |
+
|
| 180 |
+
In particular we need \(4|1\pm m|\geq 2^{4}|m^{2} - 8|\) , which is enough to imply \(m< 5\) . From here easily recover \(x = 4\) , \(m = 3\) as the last solution (in the second case).
|
| 181 |
+
|
| 182 |
+
|
| 183 |
+
|
| 184 |
+
## \(\S 2.2\) IMO 2006/5, proposed by Dan Schwarz (ROU)
|
| 185 |
+
|
| 186 |
+
Available online at https://aops.com/community/p572821.
|
| 187 |
+
|
| 188 |
+
## Problem statement
|
| 189 |
+
|
| 190 |
+
Let \(P(x)\) be a polynomial of degree \(n > 1\) with integer coefficients and let \(k\) be a positive integer. Consider the polynomial
|
| 191 |
+
|
| 192 |
+
\[Q(x) = P(P(\ldots P(P(x))\ldots))\]
|
| 193 |
+
|
| 194 |
+
where \(P\) occurs \(k\) times. Prove that there are at most \(n\) integers \(t\) such that \(Q(t) = t\) .
|
| 195 |
+
|
| 196 |
+
First, we prove that:
|
| 197 |
+
|
| 198 |
+
Claim (Putnam 2000 et al) — If a number is periodic under \(P\) then in fact it's fixed by \(P \circ P\) .
|
| 199 |
+
|
| 200 |
+
Proof. Let \(x_{1}\) , \(x_{2}\) , ..., \(x_{n}\) be a minimal orbit. Then
|
| 201 |
+
|
| 202 |
+
\[x_{i} - x_{i + 1} \mid P(x_{i}) - P(x_{i + 1}) = x_{i + 1} - x_{i + 2}\]
|
| 203 |
+
|
| 204 |
+
and so on cyclically.
|
| 205 |
+
|
| 206 |
+
If any of the quantities are zero we are done. Else, we must eventually have \(x_{i} - x_{i + 1} = -(x_{i + 1} - x_{i + 2})\) , so \(x_{i} = x_{i + 2}\) and we get 2- periodicity.
|
| 207 |
+
|
| 208 |
+
The tricky part is to study the 2- orbits. Suppose there exists a fixed pair \(u \neq v\) with \(P(u) = v\) , \(P(v) = u\) . (If no such pair exists, we are already done.) Let \((a, b)\) be any other pair with \(P(a) = b\) , \(P(b) = a\) , possibly even \(a = b\) , but \(\{a, b\} \cap \{u, v\} = \emptyset\) . Then we should have
|
| 209 |
+
|
| 210 |
+
\[u - a \mid P(u) - P(a) = v - b \mid P(v) - P(b) = u - a\]
|
| 211 |
+
|
| 212 |
+
and so \(u - a\) and \(v - b\) divide each other (and are nonzero). Similarly, \(u - b\) and \(v - a\) divide each other.
|
| 213 |
+
|
| 214 |
+
Hence \(u - a = \pm (v - b)\) and \(u - b = \pm (v - a)\) . We consider all four cases:
|
| 215 |
+
|
| 216 |
+
- If \(u - a = v - b\) and \(u - b = v - a\) then \(u - v = b - a = b\) , contradiction.
|
| 217 |
+
|
| 218 |
+
- If \(u - a = -(v - b)\) and \(u - b = -(v - a)\) then \(u + v = u - v = a + b\) .
|
| 219 |
+
|
| 220 |
+
- If \(u - a = -(v - b)\) and \(u - b = v - a\) , we get \(a + b = u + v\) from the first one (discarding the second).
|
| 221 |
+
|
| 222 |
+
- If \(u - a = v - b\) and \(u - b = -(v - a)\) , we get \(a + b = u + v\) from the second one (discarding the first one).
|
| 223 |
+
|
| 224 |
+
Thus in all possible situations we have
|
| 225 |
+
|
| 226 |
+
\[a + b = c := u + v\]
|
| 227 |
+
|
| 228 |
+
a fixed constant.
|
| 229 |
+
|
| 230 |
+
Therefore, any pair \((a, b)\) with \(P(a) = b\) and \(P(b) = a\) actually satisfies \(P(a) = c - a\) . And since \(\deg P > 1\) , this means there are at most \(n\) roots to \(a + P(a) = c\) , as needed.
|
| 231 |
+
|
| 232 |
+
|
| 233 |
+
|
| 234 |
+
## \(\S 2.3\) IMO 2006/6, proposed by Dušan Djukić (SRB)
|
| 235 |
+
|
| 236 |
+
Available online at https://aops.com/community/p572824.
|
| 237 |
+
|
| 238 |
+
## Problem statement
|
| 239 |
+
|
| 240 |
+
Assign to each side \(b\) of a convex polygon \(P\) the maximum area of a triangle that has \(b\) as a side and is contained in \(P\) . Show that the sum of the areas assigned to the sides of \(P\) is at least twice the area of \(P\) .
|
| 241 |
+
|
| 242 |
+
We say a polygon in almost convex if all its angles are at most \(180^{\circ}\) .
|
| 243 |
+
|
| 244 |
+
Note that given any convex or almost convex polygon, we can take any side \(b\) and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point).
|
| 245 |
+
|
| 246 |
+
## Lemma
|
| 247 |
+
|
| 248 |
+
Let \(N\) be an even integer. Then any almost convex \(N\) - gon with area \(S\) should have an inscribed triangle with area at least \(2S / N\) .
|
| 249 |
+
|
| 250 |
+
The main work is the proof of the lemma.
|
| 251 |
+
|
| 252 |
+
Label the polygon \(P_{0}P_{1}\ldots P_{N - 1}\) . Consider the \(N / 2\) major diagonals of the almost convex \(N\) - gon, \(P_{0}P_{N / 2}\) , \(P_{1}P_{N / 2 + 1}\) , et cetera. A butterfly refers to a self- intersecting quadrilateral \(P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\) . An example of a butterfly is shown below for \(N = 8\) .
|
| 253 |
+
|
| 254 |
+
|
| 255 |
+
|
| 256 |
+
|
| 257 |
+
Claim — Every point \(X\) in the polygon is contained in the wingspan of some butterfly.
|
| 258 |
+
|
| 259 |
+
Proof. Consider a windmill- like process which
|
| 260 |
+
|
| 261 |
+
- starts from some oriented red line \(P_{0}P_{N / 2}\) , oriented to face \(P_{0}P_{N / 2}\)
|
| 262 |
+
|
| 263 |
+
|
| 264 |
+
|
| 265 |
+
- rotates through \(P_{0}P_{N / 2} \cap P_{1}P_{N / 2 + 1}\) to get line \(P_{1}P_{N / 2 + 1}\) ,
|
| 266 |
+
|
| 267 |
+
- rotates through \(P_{1}P_{N / 2 + 1} \cap P_{2}P_{N / 2 + 2}\) to get line \(P_{2}P_{N / 2 + 2}\) ,
|
| 268 |
+
|
| 269 |
+
- ...et cetera, until returning to line \(P_{N / 2}P_{0}\) , but in the reverse orientation.
|
| 270 |
+
|
| 271 |
+
At the end of the process, every point in the plane has switched sides with our moving line. The moment that \(X\) crosses the moving red line, we get it contained in a butterfly, as needed. \(\square\)
|
| 272 |
+
|
| 273 |
+
Claim — If \(A B D C = P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\) is a butterfly, one of the triangles \(A B C\) , \(B C D\) , \(C D A\) , \(D A B\) has area at least that of the butterfly.
|
| 274 |
+
|
| 275 |
+
Proof. Let the diagonals of the butterfly meet at \(O\) , and let \(a = A O\) , \(b = B O\) , \(c = C O\) , \(d = D O\) . If we assume WLOG \(d = \min (a,b,c,d)\) then it follows \([A B C] = [A O B] + [B O C] \geq [A O B] + [C O D]\) , as needed. \(\square\)
|
| 276 |
+
|
| 277 |
+
Now, since the \(N / 2\) butterflies cover an area of \(S\) , it follows that one of the butterflies has area at least \(S / (N / 2) = 2S / N\) , and so that butterfly gives a triangle with area at least \(2S / N\) , completing the proof of the lemma.
|
| 278 |
+
|
| 279 |
+
Main proof. Let \(a_{1}\) , ..., \(a_{n}\) be the numbers assigned to the sides. Assume for contradiction \(a_{1} + \dots +a_{n}< 2S\) . We pick even integers \(m_{1}\) , \(m_{2}\) , ..., \(m_{n}\) such that
|
| 280 |
+
|
| 281 |
+
\[\frac{a_{1}}{S} < \frac{2m_{1}}{m_{1} + \dots +m_{n}}\] \[\frac{a_{2}}{S} < \frac{2m_{2}}{m_{1} + \dots +m_{n}}\] \[\vdots\] \[\frac{a_{n}}{S} < \frac{2m_{n}}{m_{1} + \dots +m_{n}}.\]
|
| 282 |
+
|
| 283 |
+
which is possible by rational approximation, since the right- hand sides sum to 2 and the left- hand sides sum to strictly less than 2.
|
| 284 |
+
|
| 285 |
+
Now we break every side of \(P\) into \(m_{i}\) equal parts to get an almost convex \(N\) - gon, where \(N = m_{1} + \dots +m_{n}\) .
|
| 286 |
+
|
| 287 |
+
The main lemma then gives us a triangle \(\Delta\) of the almost convex \(N\) - gon which has area at least \(\frac{2S}{N}\) . If \(\Delta\) used the \(i\) th side then it then follows the label \(a_{i}\) on that side should be at least \(m_{i} \cdot \frac{2S}{N}\) , contradiction.
|
| 288 |
+
|
| 289 |
+
|
IMO/md/en-IMO-2007-notes.md
ADDED
|
@@ -0,0 +1,281 @@
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|
| 1 |
+
|
| 2 |
+
# IMO 2007 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2007 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2007/1, proposed by Michael Albert (NZL) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2007/2, proposed by Charles Leytem (LUX) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2007/3, proposed by Vasily Astakhov (RUS) 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2007/4, proposed by Marek Pechal (CZE) 7
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2007/5, proposed by Kevin Buzzard, Edward Crane (UNK) 8
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2007/6, proposed by Gerhard Woeginger (NLD) 9
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Real numbers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) are fixed. For each \(1 \leq i \leq n\) we let \(d_{i} = \max \{a_{j} : 1 \leq j \leq i\} - \min \{a_{j} : i \leq j \leq n\}\) and let \(d = \max \{d_{i} : 1 \leq i \leq n\}\) .
|
| 39 |
+
|
| 40 |
+
(a) Prove that for any real numbers \(x_{1} \leq \dots \leq x_{n}\) we have
|
| 41 |
+
|
| 42 |
+
\[\max \left\{|x_{i} - a_{i}|:1\leq i\leq n\right\} \geq \frac{1}{2} d.\]
|
| 43 |
+
|
| 44 |
+
(b) Moreover, show that there exists some choice of \(x_{1} \leq \dots \leq x_{n}\) which achieves equality.
|
| 45 |
+
|
| 46 |
+
2. Consider five points \(A\) , \(B\) , \(C\) , \(D\) and \(E\) such that \(ABCD\) is a parallelogram and \(BCED\) is a cyclic quadrilateral. Let \(\ell\) be a line passing through \(A\) . Suppose that \(\ell\) intersects the interior of the segment \(DC\) at \(F\) and intersects line \(BC\) at \(G\) . Suppose also that \(EF = EG = EC\) . Prove that \(\ell\) is the bisector of angle \(DAB\) .
|
| 47 |
+
|
| 48 |
+
3. In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.
|
| 49 |
+
|
| 50 |
+
4. In triangle \(ABC\) the bisector of \(\angle BCA\) meets the circumcircle again at \(R\) , the perpendicular bisector of \(\overline{BC}\) at \(P\) , and the perpendicular bisector of \(\overline{AC}\) at \(Q\) . The midpoint of \(\overline{BC}\) is \(K\) and the midpoint of \(\overline{AC}\) is \(L\) . Prove that the triangles \(RPK\) and \(RQL\) have the same area.
|
| 51 |
+
|
| 52 |
+
5. Let \(a\) and \(b\) be positive integers. Show that if \(4ab - 1\) divides \((4a^{2} - 1)^{2}\) , then \(a = b\) .
|
| 53 |
+
|
| 54 |
+
6. Let \(n\) be a positive integer. Consider
|
| 55 |
+
|
| 56 |
+
\[S = \{(x,y,z) \mid x,y,z \in \{0,1,\ldots ,n\} , x + y + z > 0\}\]
|
| 57 |
+
|
| 58 |
+
as a set of \((n + 1)^{3} - 1\) points in the three- dimensional space. Determine the smallest possible number of planes, the union of which contains \(S\) but does not include \((0,0,0)\) .
|
| 59 |
+
|
| 60 |
+
|
| 61 |
+
|
| 62 |
+
## \(\S 1\) Solutions to Day 1
|
| 63 |
+
|
| 64 |
+
## \(\S 1.1\) IMO 2007/1, proposed by Michael Albert (NZL)
|
| 65 |
+
|
| 66 |
+
Available online at https://aops.com/community/p893741.
|
| 67 |
+
|
| 68 |
+
## Problem statement
|
| 69 |
+
|
| 70 |
+
Real numbers \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) are fixed. For each \(1 \leq i \leq n\) we let \(d_{i} = \max \{a_{j}: 1 \leq j \leq i\} - \min \{a_{j}: i \leq j \leq n\}\) and let \(d = \max \{d_{i}: 1 \leq i \leq n\}\) .
|
| 71 |
+
|
| 72 |
+
(a) Prove that for any real numbers \(x_{1} \leq \dots \leq x_{n}\) we have
|
| 73 |
+
|
| 74 |
+
\[\max \left\{|x_{i} - a_{i}|: 1 \leq i \leq n\right\} \geq \frac{1}{2} d.\]
|
| 75 |
+
|
| 76 |
+
(b) Moreover, show that there exists some choice of \(x_{1} \leq \dots \leq x_{n}\) which achieves equality.
|
| 77 |
+
|
| 78 |
+
Note that we can dispense of \(d_{i}\) immediately by realizing that the definition of \(d\) just says
|
| 79 |
+
|
| 80 |
+
\[d = \max_{1 \leq i \leq j \leq n} (a_{i} - a_{j}).\]
|
| 81 |
+
|
| 82 |
+
If \(a_{1} \leq \dots \leq a_{n}\) are already nondecreasing then \(d = 0\) and there is nothing to prove (for the equality case, just let \(x_{i} = a_{i}\) ), so we will no longer consider this case.
|
| 83 |
+
|
| 84 |
+
Otherwise, consider any indices \(i < j\) with \(a_{i} > a_{j}\) . We first prove (a) by applying the following claim with \(p = a_{i}\) and \(q = a_{j}\) :
|
| 85 |
+
|
| 86 |
+
Claim — For any \(p \leq q\) , we have either \(|p - a_{i}| \geq \frac{1}{2} (a_{i} - a_{j})\) or \(|q - a_{j}| \geq \frac{1}{2} (a_{i} - a_{j})
|
| 87 |
+
|
| 88 |
+
Proof. Assume for contradiction both are false. Then \(p > a_{i} - \frac{1}{2} (a_{i} - a_{j}) = a_{j} + \frac{1}{2} (a_{i} - a_{j}) > q\) , contradiction. \(\square\)
|
| 89 |
+
|
| 90 |
+
As for (b), we let \(i < j\) be any indices for which \(a_{i} - a_{j} = d > 0\) achieves the maximal difference. We then define \(x_{\bullet}\) in three steps:
|
| 91 |
+
|
| 92 |
+
We set \(x_{k} = \frac{a_{i} + a_{j}}{2}\) for \(k = i, \ldots , j\) . We recursively set \(x_{k} = \max (x_{k - 1}, a_{k})\) for \(k = j + 1, j + 2, \ldots\) . We recursively set \(x_{k} = \min (x_{k + 1}, a_{k})\) for \(k = i - 1, i - 2, \ldots\) .
|
| 93 |
+
|
| 94 |
+
By definition, these \(x_{\bullet}\) are weakly increasing. To prove this satisfies (b) we only need to check that
|
| 95 |
+
|
| 96 |
+
\[|x_{k} - a_{k}| \leq \frac{a_{i} - a_{j}}{2} \qquad (\star)\]
|
| 97 |
+
|
| 98 |
+
for any index \(k\) (as equality holds for \(k = i\) or \(k = j\) ).
|
| 99 |
+
|
| 100 |
+
We note \((\star)\) holds for \(i < k < j\) by construction. For \(k > j\) , note that \(x_{k} \in \{a_{j}, a_{j + 1}, \ldots , a_{k}\}\) by construction, so \((\star)\) follows from our choice of \(i\) and \(j\) giving the largest possible difference; the case \(k < i\) is similar.
|
| 101 |
+
|
| 102 |
+
|
| 103 |
+
|
| 104 |
+
## \(\S 1.2\) IMO 2007/2, proposed by Charles Leytem (LUX)
|
| 105 |
+
|
| 106 |
+
Available online at https://aops.com/community/p893744.
|
| 107 |
+
|
| 108 |
+
## Problem statement
|
| 109 |
+
|
| 110 |
+
Consider five points \(A\) , \(B\) , \(C\) , \(D\) and \(E\) such that \(ABCD\) is a parallelogram and \(BCED\) is a cyclic quadrilateral. Let \(\ell\) be a line passing through \(A\) . Suppose that \(\ell\) intersects the interior of the segment \(DC\) at \(F\) and intersects line \(BC\) at \(G\) . Suppose also that \(EF = EG = EC\) . Prove that \(\ell\) is the bisector of angle \(DAB\) .
|
| 111 |
+
|
| 112 |
+
Let \(M\) , \(N\) , \(P\) denote the midpoints of \(\overline{CF}\) , \(\overline{CG}\) , \(\overline{AC}\) (noting \(P\) is also the midpoint of \(\overline{BD}\) ).
|
| 113 |
+
|
| 114 |
+
By a homothety at \(C\) with ratio \(\frac{1}{2}\) , we find \(\overline{MNP}\) is the image of line \(\ell \equiv \overline{AGF}\) .
|
| 115 |
+
|
| 116 |
+
|
| 117 |
+
|
| 118 |
+
|
| 119 |
+
However, since we also have \(\overline{EM} \perp \overline{CF}\) and \(\overline{EN} \perp \overline{CG}\) (from \(EF = EG = EC\) ) we conclude \(\overline{PMN}\) is the Simson line of \(E\) with respect to \(\triangle BCD\) , which implies \(\overline{EP} \perp \overline{BD}\) . In other words, \(\overline{EP}\) is the perpendicular bisector of \(\overline{BD}\) , so \(E\) is the midpoint of arc \(\overline{BCD}\) .
|
| 120 |
+
|
| 121 |
+
Finally,
|
| 122 |
+
|
| 123 |
+
\[\angle (\overline{AB}, \ell) = \angle (\overline{CD}, \overline{MNP}) = \angle CMN = \angle CEN\] \[\qquad = 90^{\circ} - \angle NCE = 90^{\circ} + \angle ECB\]
|
| 124 |
+
|
| 125 |
+
which means that \(\ell\) is parallel to a bisector of \(\angle BCD\) , and hence to one of \(\angle BAD\) . (Moreover since \(F\) lies on the interior of \(\overline{CD}\) , it is actually the internal bisector.)
|
| 126 |
+
|
| 127 |
+
|
| 128 |
+
|
| 129 |
+
## \(\S 1.3\) IMO 2007/3, proposed by Vasily Astakhov (RUS)
|
| 130 |
+
|
| 131 |
+
Available online at https://aops.com/community/p893746.
|
| 132 |
+
|
| 133 |
+
## Problem statement
|
| 134 |
+
|
| 135 |
+
In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.
|
| 136 |
+
|
| 137 |
+
Take the obvious graph interpretation \(G\) . We paint red any vertices in one of the maximal cliques \(K\) , which we assume has \(2r\) vertices, and paint the remaining vertices green. We let \(\alpha (\bullet)\) denote the clique number.
|
| 138 |
+
|
| 139 |
+
Initially, let the two rooms \(A = K\) , \(B = G - K\) .
|
| 140 |
+
|
| 141 |
+
Claim — We can move at most \(r\) vertices of \(A\) into \(B\) to arrive at \(\alpha (A) \leq \alpha (B) \leq \alpha (A) + 1\) .
|
| 142 |
+
|
| 143 |
+
Proof. This is actually obvious by discrete continuity. We move one vertex at a time, noting \(\alpha (A)\) decreases by one at each step, while \(\alpha (B)\) increases by either zero or one at each step.
|
| 144 |
+
|
| 145 |
+
We stop once \(\alpha (B) \geq \alpha (A)\) , which happens before we have moved \(r\) vertices (since then we have \(\alpha (B) \geq r = \alpha (A)\) ). The conclusion follows. \(\square\)
|
| 146 |
+
|
| 147 |
+
So let's consider the situation
|
| 148 |
+
|
| 149 |
+
\[\alpha (A) = k \geq r \qquad \text{and} \qquad \alpha (B) = k + 1.\]
|
| 150 |
+
|
| 151 |
+
At this point \(A\) is a set of \(k\) red vertices, while \(B\) has the remaining \(2r - k\) red vertices (and all the green ones). An example is shown below with \(k = 4\) and \(2r = 6\) .
|
| 152 |
+
|
| 153 |
+
|
| 154 |
+
|
| 155 |
+
|
| 156 |
+
Now, if we can move any red vertex from \(B\) back to \(A\) without changing the clique number of \(B\) , we do so, and win.
|
| 157 |
+
|
| 158 |
+
Otherwise, it must be the case that every \((k + 1)\) - clique in \(B\) uses every red vertex in \(B\) . For each \((k + 1)\) - clique in \(B\) (in arbitrary order), we do the following procedure.
|
| 159 |
+
|
| 160 |
+
|
| 161 |
+
|
| 162 |
+
- If all \(k + 1\) vertices are still green, pick one and re-color it blue. This is possible since \(k + 1 > 2r - k\) .
|
| 163 |
+
|
| 164 |
+
- Otherwise, do nothing.
|
| 165 |
+
|
| 166 |
+
Then we move all the blue vertices from \(B\) to \(A\) , one at a time, in the same order we re- colored them. This forcibly decreases the clique number of \(B\) to \(k\) , since the clique number is \(k + 1\) just before the last blue vertex is moved, and strictly less than \(k + 1\) (hence equal to \(k\) ) immediately after that.
|
| 167 |
+
|
| 168 |
+
Claim — After this, \(\alpha (A) = k\) still holds.
|
| 169 |
+
|
| 170 |
+
Proof. Assume not, and we have a \((k + 1)\) - clique which uses \(b\) blue vertices and \((k + 1) - b\) red vertices in \(A\) . Together with the \(2r - k\) red vertices already in \(B\) we then get a clique of size
|
| 171 |
+
|
| 172 |
+
\[b + ((k + 1 - b)) + (2r - k) = 2r + 1\]
|
| 173 |
+
|
| 174 |
+
which is a contradiction.
|
| 175 |
+
|
| 176 |
+
Remark. Dragomir Grozev posted the following motivation on his blog:
|
| 177 |
+
|
| 178 |
+
I think, it's a natural idea to place all students in one room and begin moving them one by one into the other one. Then the max size of the cliques in the first and second room increase (resp. decrease) at most with one. So, there would be a moment both sizes are almost the same. At that moment we may adjust something.
|
| 179 |
+
|
| 180 |
+
Trying the idea, I had some difficulties keeping track of the maximal cliques in the both rooms. It seemed easier all the students in one of the rooms to comprise a clique. It could be achieved by moving only the members of the maximal clique. Following this path the remaining obstacles can be overcome naturally.
|
| 181 |
+
|
| 182 |
+
|
| 183 |
+
|
| 184 |
+
## \(\S 2\) Solutions to Day 2
|
| 185 |
+
|
| 186 |
+
## \(\S 2.1\) IMO 2007/4, proposed by Marek Pechal (CZE)
|
| 187 |
+
|
| 188 |
+
Available online at https://aops.com/community/p894655.
|
| 189 |
+
|
| 190 |
+
## Problem statement
|
| 191 |
+
|
| 192 |
+
In triangle \(ABC\) the bisector of \(\angle BCA\) meets the circumcircle again at \(R\) , the perpendicular bisector of \(\overline{BC}\) at \(P\) , and the perpendicular bisector of \(\overline{AC}\) at \(Q\) . The midpoint of \(\overline{BC}\) is \(K\) and the midpoint of \(\overline{AC}\) is \(L\) . Prove that the triangles \(RPK\) and \(RQL\) have the same area.
|
| 193 |
+
|
| 194 |
+
We first begin by proving the following claim.
|
| 195 |
+
|
| 196 |
+
Claim — We have \(CQ = PR\) (equivalently, \(CP = QR\) ).
|
| 197 |
+
|
| 198 |
+
Proof. Let \(O = \overline{LQ} \cap \overline{KP}\) be the circumcenter. Then
|
| 199 |
+
|
| 200 |
+
\[\angle OPQ = \angle KPC = 90^{\circ} - \angle PCK = 90^{\circ} - \angle LCQ = \angle \angle CQL = \angle PQO.\]
|
| 201 |
+
|
| 202 |
+
Thus \(OP = OQ\) . Since \(OC = OR\) as well, we get the conclusion.
|
| 203 |
+
|
| 204 |
+
Denote by \(X\) and \(Y\) the feet from \(R\) to \(\overline{CA}\) and \(\overline{CB}\) , so \(\triangle CXR \cong \triangle CYR\) . Then, let \(t = \frac{CQ}{CR} = 1 - \frac{CP}{CR}\) .
|
| 205 |
+
|
| 206 |
+
|
| 207 |
+
|
| 208 |
+
|
| 209 |
+
Then it follows that
|
| 210 |
+
|
| 211 |
+
\[[RQ L] = [X Q L] = t(1 - t)\cdot [X R C] = t(1 - t)\cdot [Y C R] = [Y K P] = [R K P]\]
|
| 212 |
+
|
| 213 |
+
as needed.
|
| 214 |
+
|
| 215 |
+
Remark. Trigonometric approaches are very possible (and easier to find) as well: both areas work out to be \(\frac{1}{8} ab \tan \frac{1}{2} C\) .
|
| 216 |
+
|
| 217 |
+
|
| 218 |
+
|
| 219 |
+
## \(\S 2.2\) IMO 2007/5, proposed by Kevin Buzzard, Edward Crane (UNK)
|
| 220 |
+
|
| 221 |
+
Available online at https://aops.com/community/p894656.
|
| 222 |
+
|
| 223 |
+
## Problem statement
|
| 224 |
+
|
| 225 |
+
Let \(a\) and \(b\) be positive integers. Show that if \(4ab - 1\) divides \((4a^{2} - 1)^{2}\) , then \(a = b\) .
|
| 226 |
+
|
| 227 |
+
As usual,
|
| 228 |
+
|
| 229 |
+
\[4ab - 1\mid (4a^{2} - 1)^{2}\iff 4ab - 1\mid (4ab\cdot a - b)^{2}\iff 4ab - 1\mid (a - b)^{2}.\]
|
| 230 |
+
|
| 231 |
+
Then we use a typical Vieta jumping argument. Define
|
| 232 |
+
|
| 233 |
+
\[k = \frac{(a - b)^{2}}{4ab - 1}.\]
|
| 234 |
+
|
| 235 |
+
Note that \(k = 0 \iff a = b\) . So we will prove that \(k > 0\) leads to a contradiction.
|
| 236 |
+
|
| 237 |
+
Indeed, suppose \((a,b)\) is a minimal solution with \(a > b\) (we have \(a\neq b\) since \(k\neq 0\) - By Vieta jumping, \((b,\frac{b^{2} + k}{a})\) is also such a solution. But now
|
| 238 |
+
|
| 239 |
+
\[\frac{b^{2} + k}{a}\geq a\Rightarrow k\geq a^{2} - b^{2}\] \[\Rightarrow \frac{(a - b)^{2}}{4ab - 1}\geq a^{2} - b^{2}\] \[\Rightarrow a - b\geq (4ab - 1)(a + b)\]
|
| 240 |
+
|
| 241 |
+
which is absurd for \(a,b\in \mathbb{Z}_{>0}\) . (In the last step we divided by \(a - b > 0\) -
|
| 242 |
+
|
| 243 |
+
|
| 244 |
+
|
| 245 |
+
## \(\S 2.3\) IMO 2007/6, proposed by Gerhard Woeginger (NLD)
|
| 246 |
+
|
| 247 |
+
Available online at https://aops.com/community/p894658.
|
| 248 |
+
|
| 249 |
+
## Problem statement
|
| 250 |
+
|
| 251 |
+
Let \(n\) be a positive integer. Consider
|
| 252 |
+
|
| 253 |
+
\[S = \{(x,y,z)\mid x,y,z\in \{0,1,\ldots ,n\} ,x + y + z > 0\}\]
|
| 254 |
+
|
| 255 |
+
as a set of \((n + 1)^{3} - 1\) points in the three- dimensional space. Determine the smallest possible number of planes, the union of which contains \(S\) but does not include \((0,0,0)\) .
|
| 256 |
+
|
| 257 |
+
The answer is \(3n\) . Here are two examples of constructions with \(3n\) planes:
|
| 258 |
+
|
| 259 |
+
\(\cdot x + y + z = i\) for \(i = 1,\ldots ,3n\)
|
| 260 |
+
|
| 261 |
+
\(\cdot x = i\) \(y = i\) \(z = i\) for \(i = 1,\ldots ,n\)
|
| 262 |
+
|
| 263 |
+
Suppose for contradiction we have \(N< 3n\) planes. Let them be \(a_{i}x + b_{i}y + c_{i}z + 1 = 0\) for \(i = 1,\ldots ,N\) . Define the polynomials
|
| 264 |
+
|
| 265 |
+
\[A(x,y,z) = \prod_{i = 1}^{n}(x - i)\prod_{i = 1}^{n}(y - i)\prod_{i = 1}^{n}(z - i)\] \[B(x,y,z) = \prod_{i = 1}^{N}(a_{i}x + b_{i}y + c_{i}z + 1).\]
|
| 266 |
+
|
| 267 |
+
Note that \(A(0,0,0) = (- 1)^{n}(n!)^{3}\neq 0\) and \(B(0,0,0) = 1\neq 0\) , but \(A(x,y,z) =\) \(B(x,y,z) = 0\) for any \((x,y,z)\in S\) . Also, the coefficient of \(x^{n}y^{n}z^{n}\) in \(A\) is 1, while the coefficient of \(x^{n}y^{n}z^{n}\) in \(B\) is 0.
|
| 268 |
+
|
| 269 |
+
Now, define
|
| 270 |
+
|
| 271 |
+
\[P(x,y,z):= A(x,y,z) - \lambda B(x,y,z).\]
|
| 272 |
+
|
| 273 |
+
where \(\lambda = \frac{A(0,0,0)}{B(0,0,0)} = (- 1)^{n}(n!)^{3}\) . We now have that
|
| 274 |
+
|
| 275 |
+
\(\cdot P(x,y,z) = 0\) for any \(x,y,z\in \{0,1,\ldots ,n\}^{3}\)
|
| 276 |
+
|
| 277 |
+
But the coefficient of \(x^{n}y^{n}z^{n}\) is 1.
|
| 278 |
+
|
| 279 |
+
This is a contradiction to Alon's combinatorial nullstellensatz.
|
| 280 |
+
|
| 281 |
+
|
IMO/md/en-IMO-2008-notes.md
ADDED
|
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|
| 1 |
+
|
| 2 |
+
# IMO 2008 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2008 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2008/1, proposed by Andrey Gavrilyuk (RUS) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2008/2, proposed by Walther Janous (AUT) 5
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2008/3, proposed by Kęstutis Česnavičius (LTU) 6
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2008/4, proposed by Hojoo Lee (KOR) 7
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2008/5, proposed by Bruno Le Floch and Ilia Smilga (FRA) 8
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2008/6, proposed by Vladimir Shmarov (RUS) 9
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Let \(H\) be the orthocenter of an acute-angled triangle \(ABC\) . The circle \(\Gamma_{A}\) centered at the midpoint of \(\overline{BC}\) and passing through \(H\) intersects the sideline \(BC\) at points \(A_{1}\) and \(A_{2}\) . Similarly, define the points \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , and \(C_{2}\) . Prove that six points \(A_{1}\) , \(A_{2}\) , \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , \(C_{2}\) are concyclic.
|
| 39 |
+
|
| 40 |
+
2. Let \(x\) , \(y\) , \(z\) be real numbers with \(xyz = 1\) , all different from 1. Prove that
|
| 41 |
+
|
| 42 |
+
\[\frac{x^{2}}{(x - 1)^{2}} +\frac{y^{2}}{(y - 1)^{2}} +\frac{z^{2}}{(z - 1)^{2}}\geq 1\]
|
| 43 |
+
|
| 44 |
+
and show that equality holds for infinitely many choices of rational numbers \(x\) , \(y\) , \(z\) .
|
| 45 |
+
|
| 46 |
+
3. Prove that there are infinitely many positive integers \(n\) such that \(n^{2} + 1\) has a prime factor greater than \(2n + \sqrt{2n}\) .
|
| 47 |
+
|
| 48 |
+
4. Find all functions \(f\) from the positive reals to the positive reals such that
|
| 49 |
+
|
| 50 |
+
\[\frac{f(w)^{2} + f(x)^{2}}{f(y^{2}) + f(z^{2})} = \frac{w^{2} + x^{2}}{y^{2} + z^{2}}\]
|
| 51 |
+
|
| 52 |
+
for all positive real numbers \(w\) , \(x\) , \(y\) , \(z\) satisfying \(wx = yz\) .
|
| 53 |
+
|
| 54 |
+
5. Let \(n\) and \(k\) be positive integers with \(k \geq n\) and \(k - n\) an even number. There are \(2n\) lamps labelled 1, 2, ..., \(2n\) each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on). Let \(N\) be the number of such sequences consisting of \(k\) steps and resulting in the state where lamps 1 through \(n\) are all on, and lamps \(n + 1\) through \(2n\) are all off. Let \(M\) be number of such sequences consisting of \(k\) steps, resulting in the state where lamps 1 through \(n\) are all on, and lamps \(n + 1\) through \(2n\) are all off, but where none of the lamps \(n + 1\) through \(2n\) is ever switched on. Determine \(\frac{N}{M}\) .
|
| 55 |
+
|
| 56 |
+
6. Let \(ABCD\) be a convex quadrilateral with \(BA \neq BC\) . Denote the incircles of triangles \(ABC\) and \(ADC\) by \(\omega_{1}\) and \(\omega_{2}\) respectively. Suppose that there exists a circle \(\omega\) tangent to ray \(BA\) beyond \(A\) and to the ray \(BC\) beyond \(C\) , which is also tangent to the lines \(AD\) and \(CD\) . Prove that the common external tangents to \(\omega_{1}\) and \(\omega_{2}\) intersect on \(\omega\) .
|
| 57 |
+
|
| 58 |
+
|
| 59 |
+
|
| 60 |
+
## \(\S 1\) Solutions to Day 1
|
| 61 |
+
|
| 62 |
+
## \(\S 1.1\) IMO 2008/1, proposed by Andrey Gavrilyuk (RUS)
|
| 63 |
+
|
| 64 |
+
Available online at https://aops.com/community/p1190553.
|
| 65 |
+
|
| 66 |
+
## Problem statement
|
| 67 |
+
|
| 68 |
+
Let \(H\) be the orthocenter of an acute- angled triangle \(ABC\) . The circle \(\Gamma_{A}\) centered at the midpoint of \(\overline{BC}\) and passing through \(H\) intersects the sideline \(BC\) at points \(A_{1}\) and \(A_{2}\) . Similarly, define the points \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , and \(C_{2}\) . Prove that six points \(A_{1}\) , \(A_{2}\) , \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , \(C_{2}\) are concyclic.
|
| 69 |
+
|
| 70 |
+
We show two solutions.
|
| 71 |
+
|
| 72 |
+
\(\P\) First solution using power of a point. Let \(D\) , \(E\) , \(F\) be the centers of \(\Gamma_{A}\) , \(\Gamma_{B}\) , \(\Gamma_{C}\) (in other words, the midpoints of the sides).
|
| 73 |
+
|
| 74 |
+
We first show that \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , \(C_{2}\) are concyclic. It suffices to prove that \(A\) lies on the radical axis of the circles \(\Gamma_{B}\) and \(\Gamma_{C}\) .
|
| 75 |
+
|
| 76 |
+
|
| 77 |
+
|
| 78 |
+
|
| 79 |
+
Let \(X\) be the second intersection of \(\Gamma_{B}\) and \(\Gamma_{C}\) . Clearly \(\overline{XH}\) is perpendicular to the line joining the centers of the circles, namely \(\overline{EF}\) . But \(\overline{EF} \parallel \overline{BC}\) , so \(\overline{XH} \perp \overline{BC}\) . Since \(\overline{AH} \perp \overline{BC}\) as well, we find that \(A\) , \(X\) , \(H\) are collinear, as needed.
|
| 80 |
+
|
| 81 |
+
Thus, \(B_{1}\) , \(B_{2}\) , \(C_{1}\) , \(C_{2}\) are concyclic. Similarly, \(C_{1}\) , \(C_{2}\) , \(A_{1}\) , \(A_{2}\) are concyclic, as are \(A_{1}\) , \(A_{2}\) , \(B_{1}\) , \(B_{2}\) . Now if any two of these three circles coincide, we are done; else the pairwise radical axii are not concurrent, contradiction. (Alternatively, one can argue directly that \(O\) is the center of all three circles, by taking the perpendicular bisectors.)
|
| 82 |
+
|
| 83 |
+
\(\P\) Second solution using length chase (Ritwin Narra). We claim the circumcenter \(O\) of \(\triangle ABC\) is in fact the center of \((A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\) .
|
| 84 |
+
|
| 85 |
+
Define \(D\) , \(E\) , \(F\) as before. Then since \(\overline{OD} \perp \overline{A_{1}A_{2}}\) and \(DA_{1} = DA_{2}\) , which means \(OA_{1} = OA_{2}\) . Similarly, we have \(OB_{1} = OB_{2}\) and \(OC_{1} = OC_{2}\) .
|
| 86 |
+
|
| 87 |
+
Now since \(DA_{1} = DA_{2} = DH\) , we have \(OA_{1}^{2} = OD^{2} + HD^{2}\) . We seek to show
|
| 88 |
+
|
| 89 |
+
\[OD^{2} + HD^{2} = OE^{2} + HE^{2} = OF^{2} + HF^{2}.\]
|
| 90 |
+
|
| 91 |
+
|
| 92 |
+
|
| 93 |
+
This is clear by Appollonius's Theorem since \(D\) , \(E\) , and \(F\) lie on the nine- point circle, which is centered at the midpoint of \(\overline{OH}\) .
|
| 94 |
+
|
| 95 |
+
|
| 96 |
+
|
| 97 |
+
## \(\S 1.2\) IMO 2008/2, proposed by Walther Janous (AUT)
|
| 98 |
+
|
| 99 |
+
Available online at https://aops.com/community/p1190551.
|
| 100 |
+
|
| 101 |
+
## Problem statement
|
| 102 |
+
|
| 103 |
+
Let \(x\) , \(y\) , \(z\) be real numbers with \(xyz = 1\) , all different from 1. Prove that
|
| 104 |
+
|
| 105 |
+
\[\frac{x^{2}}{(x - 1)^{2}} +\frac{y^{2}}{(y - 1)^{2}} +\frac{z^{2}}{(z - 1)^{2}}\geq 1\]
|
| 106 |
+
|
| 107 |
+
and show that equality holds for infinitely many choices of rational numbers \(x\) , \(y\) , \(z\) .
|
| 108 |
+
|
| 109 |
+
Let \(x = a / b\) , \(y = b / c\) , \(z = c / a\) , so we want to show
|
| 110 |
+
|
| 111 |
+
\[\left(\frac{a}{a - b}\right)^{2} + \left(\frac{b}{b - c}\right)^{2} + \left(\frac{c}{c - a}\right)^{2}\geq 1.\]
|
| 112 |
+
|
| 113 |
+
A boring computation shows this is equivalent to
|
| 114 |
+
|
| 115 |
+
\[\frac{(a^{2}b + b^{2}c + c^{2}a - 3abc)^{2}}{(a - b)^{2}(b - c)^{2}(c - a)^{2}}\geq 0\]
|
| 116 |
+
|
| 117 |
+
which proves the inequality (and it is unsurprising we are in such a situation, given that there is an infinite curve of rationals).
|
| 118 |
+
|
| 119 |
+
For equality, it suffices to show there are infinitely many integer solutions to
|
| 120 |
+
|
| 121 |
+
\[a^{2}b + b^{2}c + c^{2}a = 3abc\iff \frac{a}{c} +\frac{b}{a} +\frac{c}{a} = 3\]
|
| 122 |
+
|
| 123 |
+
or equivalently that there are infinitely many rational solutions to
|
| 124 |
+
|
| 125 |
+
\[u + v + \frac{1}{uv} = 3.\]
|
| 126 |
+
|
| 127 |
+
For any \(0\neq u\in \mathbb{Q}\) the real solution for \(u\) is
|
| 128 |
+
|
| 129 |
+
\[v = \frac{-u + (u - 1)\sqrt{1 - 4 / u} + 3}{2}\]
|
| 130 |
+
|
| 131 |
+
and there are certainly infinitely many rational numbers \(u\) for which \(1 - 4 / u\) is a rational square (say, \(u = \frac{- 4}{q^{2} - 1}\) for \(q\neq \pm 1\) a rational number).
|
| 132 |
+
|
| 133 |
+
|
| 134 |
+
|
| 135 |
+
## \(\S 1.3\) IMO 2008/3, proposed by Kestutis Česnavičius (LTU)
|
| 136 |
+
|
| 137 |
+
Available online at https://aops.com/community/p1190546.
|
| 138 |
+
|
| 139 |
+
## Problem statement
|
| 140 |
+
|
| 141 |
+
Problem statementProve that there are infinitely many positive integers \(n\) such that \(n^{2} + 1\) has a prime factor greater than \(2n + \sqrt{2n}\) .
|
| 142 |
+
|
| 143 |
+
The idea is to pick the prime \(p\) first!
|
| 144 |
+
|
| 145 |
+
Select any large prime \(p \geq 2013\) , and let \(h = \left\lceil \sqrt{p}\right\rceil\) . We will try to find an \(n\) such that
|
| 146 |
+
|
| 147 |
+
\[n \leq \frac{1}{2} (p - h) \quad \text{and} \quad p \mid n^{2} + 1.\]
|
| 148 |
+
|
| 149 |
+
This implies \(p \geq 2n + \sqrt{p}\) which is enough to ensure \(p \geq 2n + \sqrt{2n}\) .
|
| 150 |
+
|
| 151 |
+
Assume \(p \equiv 1\) (mod 8) henceforth. Then there exists some \(\frac{1}{2} p < x < p\) such that \(x^{2} \equiv - 1\) (mod \(p\) ), and we set
|
| 152 |
+
|
| 153 |
+
\[x = \frac{p + 1}{2} +t.\]
|
| 154 |
+
|
| 155 |
+
Claim — We have \(t \geq \frac{h - 1}{2}\) and hence may take \(n = p - x\) .
|
| 156 |
+
|
| 157 |
+
Proof. Assume for contradiction this is false; then
|
| 158 |
+
|
| 159 |
+
\[0\equiv 4(x^{2} + 1)\pmod{p\] \[\quad = (p + 1 + 2t)^{2} + 4\] \[\quad \equiv (2t + 1)^{2} + 4\pmod{p\] \[\quad < h^{2} + 4\]
|
| 160 |
+
|
| 161 |
+
So we have that \((2t + 1)^{2} + 4\) is positive and divisible by \(p\) , yet at most \(\left\lceil \sqrt{p}\right\rceil^{2} + 4 < 2p\) . So it must be the case that \((2t + 1)^{2} + 4 = p\) , but this has no solutions modulo 8. \(\square\)
|
| 162 |
+
|
| 163 |
+
|
| 164 |
+
|
| 165 |
+
## \(\S 2\) Solutions to Day 2
|
| 166 |
+
|
| 167 |
+
## \(\S 2.1\) IMO 2008/4, proposed by Hojoo Lee (KOR)
|
| 168 |
+
|
| 169 |
+
Available online at https://aops.com/community/p1191683.
|
| 170 |
+
|
| 171 |
+
## Problem statement
|
| 172 |
+
|
| 173 |
+
Find all functions \(f\) from the positive reals to the positive reals such that
|
| 174 |
+
|
| 175 |
+
\[\frac{f(w)^{2} + f(x)^{2}}{f(y^{2}) + f(z^{2})} = \frac{w^{2} + x^{2}}{y^{2} + z^{2}}\]
|
| 176 |
+
|
| 177 |
+
for all positive real numbers \(w\) , \(x\) , \(y\) , \(z\) satisfying \(wx = yz\) .
|
| 178 |
+
|
| 179 |
+
The answers are \(f(x) \equiv x\) and \(f(x) \equiv 1 / x\) . These work, so we show they are the only ones.
|
| 180 |
+
|
| 181 |
+
First, setting \((t, t, t, t)\) gives \(f(t^{2}) = f(t)^{2}\) . In particular, \(f(1) = 1\) . Next, setting \((t, 1, \sqrt{t}, \sqrt{t})\) gives
|
| 182 |
+
|
| 183 |
+
\[\frac{f(t)^{2} + 1}{2f(t)} = \frac{t^{2} + 1}{2t}\]
|
| 184 |
+
|
| 185 |
+
which as a quadratic implies \(f(t) \in \{t, 1 / t\}\) .
|
| 186 |
+
|
| 187 |
+
Now assume \(f(a) = a\) and \(f(b) = 1 / b\) . Setting \((\sqrt{a}, \sqrt{b}, 1, \sqrt{ab})\) gives
|
| 188 |
+
|
| 189 |
+
\[\frac{a + 1 / b}{f(ab) + 1} = \frac{a + b}{ab + 1}.\]
|
| 190 |
+
|
| 191 |
+
One can check the two cases on \(f(ab)\) each imply \(a = 1\) and \(b = 1\) respectively. Hence the only answers are those claimed.
|
| 192 |
+
|
| 193 |
+
|
| 194 |
+
|
| 195 |
+
## \(\S 2.2\) IMO 2008/5, proposed by Bruno Le Floch and Ilia Smilga (FRA)
|
| 196 |
+
|
| 197 |
+
Available online at https://aops.com/community/p1191679.
|
| 198 |
+
|
| 199 |
+
## Problem statement
|
| 200 |
+
|
| 201 |
+
Let \(n\) and \(k\) be positive integers with \(k\geq n\) and \(k - n\) an even number. There are \(2n\) lamps labelled 1, 2, ..., \(2n\) each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on). Let \(N\) be the number of such sequences consisting of \(k\) steps and resulting in the state where lamps 1 through \(n\) are all on, and lamps \(n + 1\) through \(2n\) are all off. Let \(M\) be number of such sequences consisting of \(k\) steps, resulting in the state where lamps 1 through \(n\) are all on, and lamps \(n + 1\) through \(2n\) are all off, but where none of the lamps \(n + 1\) through \(2n\) is ever switched on. Determine \(\frac{N}{M}\) .
|
| 202 |
+
|
| 203 |
+
The answer is \(2^{k - n}\) .
|
| 204 |
+
|
| 205 |
+
Consider the following map \(\Psi\) from \(N\) - sequences to \(M\) - sequences:
|
| 206 |
+
|
| 207 |
+
- change every instance of \(n + 1\) to 1;
|
| 208 |
+
|
| 209 |
+
- change every instance of \(n + 2\) to 2;
|
| 210 |
+
|
| 211 |
+
:
|
| 212 |
+
|
| 213 |
+
- change every instance of \(2n\) to \(n\) .
|
| 214 |
+
|
| 215 |
+
(For example, suppose \(k = 9\) , \(n = 3\) ; then \(144225253 \mapsto 111222223\) .)
|
| 216 |
+
|
| 217 |
+
Clearly this is map is well- defined and surjective. So all that remains is:
|
| 218 |
+
|
| 219 |
+
Claim — Every \(M\) - sequence has exactly \(2^{k - n}\) pre- images under \(\Psi\) .
|
| 220 |
+
|
| 221 |
+
Proof. Indeed, suppose that there are \(c_{1}\) instances of lamp 1. Then we want to pick an odd subset of the 1's to change to \(n + 1\) 's, so \(2^{c_{1} - 1}\) ways to do this. And so on. Hence the number of pre- images is
|
| 222 |
+
|
| 223 |
+
\[\prod_{i}2^{c_{i} - 1} = 2^{k - n}.\]
|
| 224 |
+
|
| 225 |
+
|
| 226 |
+
|
| 227 |
+
## \(\S 2.3\) IMO 2008/6, proposed by Vladimir Shmarov (RUS)
|
| 228 |
+
|
| 229 |
+
Available online at https://aops.com/community/p1191671.
|
| 230 |
+
|
| 231 |
+
## Problem statement
|
| 232 |
+
|
| 233 |
+
Let \(A B C D\) be a convex quadrilateral with \(B A \neq B C\) . Denote the incircles of triangles \(A B C\) and \(A D C\) by \(\omega_{1}\) and \(\omega_{2}\) respectively. Suppose that there exists a circle \(\omega\) tangent to ray \(B A\) beyond \(A\) and to the ray \(B C\) beyond \(C\) , which is also tangent to the lines \(A D\) and \(C D\) . Prove that the common external tangents to \(\omega_{1}\) and \(\omega_{2}\) intersect on \(\omega\) .
|
| 234 |
+
|
| 235 |
+
By the external version of Pitot theorem, the existence of \(\omega\) implies that
|
| 236 |
+
|
| 237 |
+
\[B A + A D = C B + C D.\]
|
| 238 |
+
|
| 239 |
+
Let \(\overline{{P Q}}\) and \(\overline{{S T}}\) be diameters of \(\omega_{1}\) and \(\omega_{2}\) with \(P,T\in \overline{{A C}}\) . Then the length relation on \(A B C D\) implies that \(P\) and \(T\) are reflections about the midpoint of \(\overline{{A C}}\)
|
| 240 |
+
|
| 241 |
+
Now orient \(A C\) horizontally and let \(K\) be the "uppermost" point of \(\omega\) , as shown.
|
| 242 |
+
|
| 243 |
+
|
| 244 |
+
|
| 245 |
+
|
| 246 |
+
Consequently, a homothety at \(B\) maps \(Q\) , \(T\) , \(K\) to each other (since \(T\) is the uppermost of the excircle, \(Q\) of the incircle). Similarly, a homothety at \(D\) maps \(P\) , \(S\) , \(K\) to each other. As \(\overline{{P Q}}\) and \(\overline{{S T}}\) are parallel diameters it then follows \(K\) is the exsimilicenter of \(\omega_{1}\) and \(\omega_{2}\) .
|
| 247 |
+
|
| 248 |
+
|
IMO/md/en-IMO-2009-notes.md
ADDED
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| 1 |
+
|
| 2 |
+
# IMO 2009 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2009 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2009/1, proposed by Ross Atkins (AUS) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2009/2, proposed by Sergei Berlov (RUS) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2009/3, proposed by Gabriel Carroll (USA) 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk
|
| 29 |
+
|
| 30 |
+
(BEL) 7
|
| 31 |
+
|
| 32 |
+
2.2 IMO 2009/5, proposed by Bruno Le Floch (FRA) 9
|
| 33 |
+
|
| 34 |
+
2.3 IMO 2009/6, proposed by Dmitry Khramtsov (RUS) 10
|
| 35 |
+
|
| 36 |
+
|
| 37 |
+
|
| 38 |
+
## Problems
|
| 39 |
+
|
| 40 |
+
1. Let \(n,k\geq 2\) be positive integers and let \(a_{1}\) , \(a_{2}\) , \(a_{3}\) , ..., \(a_{k}\) be distinct integers in the set \(\{1,2,\ldots ,n\}\) such that \(n\) divides \(a_{i}(a_{i + 1} - 1)\) for \(i = 1,2,\ldots ,k - 1\) . Prove that \(n\) does not divide \(a_{k}(a_{1} - 1)\) .
|
| 41 |
+
|
| 42 |
+
2. Let \(A B C\) be a triangle with circumcenter \(o\) . The points \(P\) and \(Q\) are interior points of the sides \(C A\) and \(A B\) respectively. Let \(K\) , \(L\) , \(M\) be the midpoints of \(\overline{{B P}}\) \(\overline{{C Q}}\) , \(\overline{{P Q}}\) . Suppose that \(\overline{{P Q}}\) is tangent to the circumcircle of \(\triangle K L M\) . Prove that \(O P = O Q\) .
|
| 43 |
+
|
| 44 |
+
3. Suppose that \(s_{1},s_{2},s_{3},\ldots\) is a strictly increasing sequence of positive integers such that the sub-sequences \(s_{s_{1}}\) , \(s_{s_{2}}\) , \(s_{s_{3}}\) , ...and \(s_{s_{1} + 1}\) , \(s_{s_{2} + 1}\) , \(s_{s_{3} + 1}\) , ...are both arithmetic progressions. Prove that the sequence \(s_{1}\) , \(s_{2}\) , \(s_{3}\) , ...is itself an arithmetic progression.
|
| 45 |
+
|
| 46 |
+
4. Let \(A B C\) be a triangle with \(A B = A C\) . The angle bisectors of \(\angle C A B\) and \(\angle A B C\) meet the sides \(B C\) and \(C A\) at \(D\) and \(E\) , respectively. Let \(K\) be the incenter of triangle \(A D C\) . Suppose that \(\angle B E K = 45^{\circ}\) . Find all possible values of \(\angle C A B\) .
|
| 47 |
+
|
| 48 |
+
5. Find all functions \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that for positive integers \(a\) and \(b\) , the numbers
|
| 49 |
+
|
| 50 |
+
\[a, f(b), f(b + f(a) - 1)\]
|
| 51 |
+
|
| 52 |
+
are the sides of a non-degenerate triangle.
|
| 53 |
+
|
| 54 |
+
6. Let \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) be distinct positive integers and let \(M\) be a set of \(n - 1\) positive integers not containing \(s = a_{1} + \dots +a_{n}\) . A grasshopper is to jump along the real axis, starting at the point 0 and making \(n\) jumps to the right with lengths \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in \(M\) .
|
| 55 |
+
|
| 56 |
+
|
| 57 |
+
|
| 58 |
+
## \(\S 1\) Solutions to Day 1
|
| 59 |
+
|
| 60 |
+
## \(\S 1.1\) IMO 2009/1, proposed by Ross Atkins (AUS)
|
| 61 |
+
|
| 62 |
+
Available online at https://aops.com/community/p1561571.
|
| 63 |
+
|
| 64 |
+
## Problem statement
|
| 65 |
+
|
| 66 |
+
Let \(n,k\geq 2\) be positive integers and let \(a_{1}\) , \(a_{2}\) , \(a_{3}\) , ..., \(a_{k}\) be distinct integers in the set \(\{1,2,\ldots ,n\}\) such that \(n\) divides \(a_{i}(a_{i + 1} - 1)\) for \(i = 1,2,\ldots ,k - 1\) . Prove that \(n\) does not divide \(a_{k}(a_{1} - 1)\) .
|
| 67 |
+
|
| 68 |
+
We proceed indirectly and assume that
|
| 69 |
+
|
| 70 |
+
\[a_{i}(a_{i + 1} - 1)\equiv 0\pmod {n}\]
|
| 71 |
+
|
| 72 |
+
for \(i = 1,\ldots ,k\) (indices taken modulo \(k\) ). We claim that this implies all the \(a_{i}\) are equal modulo \(n\) .
|
| 73 |
+
|
| 74 |
+
Let \(q = p^{e}\) be any prime power dividing \(n\) . Then, \(a_{1}(a_{2} - 1)\equiv 0\) (mod \(q\) ), so \(p\) divides either \(a_{1}\) or \(a_{2} - 1\) .
|
| 75 |
+
|
| 76 |
+
- If \(p \mid a_{1}\) , then \(p \nmid a_{1} - 1\) . Then
|
| 77 |
+
|
| 78 |
+
\[a_{k}(a_{1} - 1)\equiv 0\pmod {q}\implies a_{k}\equiv 0\pmod {q}.\]
|
| 79 |
+
|
| 80 |
+
In particular, \(p \mid a_{k}\) . So repeating this argument, we get \(a_{k - 1} \equiv 0\) (mod \(q\) ), \(a_{k - 2} \equiv 0\) (mod \(q\) ), and so on.
|
| 81 |
+
|
| 82 |
+
- Similarly, if \(p \mid a_{2} - 1\) then \(p \nmid a_{2}\) , and
|
| 83 |
+
|
| 84 |
+
\[a_{2}(a_{3} - 1)\equiv 0\pmod {q}\implies a_{3}\equiv 1\pmod {q}.\]
|
| 85 |
+
|
| 86 |
+
In particular, \(p \mid a_{3} - 1\) . So repeating this argument, we get \(a_{4} \equiv 0\) (mod \(q\) ), \(a_{5} \equiv 0\) (mod \(q\) ), and so on.
|
| 87 |
+
|
| 88 |
+
Either way, we find \(a_{i}\) (mod \(q\) ) is constant (and either 0 or 1).
|
| 89 |
+
|
| 90 |
+
Since \(q\) was an arbitrary prime power dividing \(n\) , by Chinese remainder theorem we conclude that \(a_{i}\) (mod \(n\) ) is constant as well. But this contradicts the assumption of distinctness.
|
| 91 |
+
|
| 92 |
+
|
| 93 |
+
|
| 94 |
+
## \(\S 1.2\) IMO 2009/2, proposed by Sergei Berlov (RUS)
|
| 95 |
+
|
| 96 |
+
Available online at https://aops.com/community/p1561572.
|
| 97 |
+
|
| 98 |
+
## Problem statement
|
| 99 |
+
|
| 100 |
+
Let \(ABC\) be a triangle with circumcenter \(O\) . The points \(P\) and \(Q\) are interior points of the sides \(CA\) and \(AB\) respectively. Let \(K\) , \(L\) , \(M\) be the midpoints of \(\overline{BP}\) , \(\overline{CQ}\) , \(\overline{PQ}\) . Suppose that \(\overline{PQ}\) is tangent to the circumcircle of \(\triangle KLM\) . Prove that \(OP = OQ\) .
|
| 101 |
+
|
| 102 |
+
By power of a point, we have \(- AQ \cdot QB = OQ^2 - R^2\) and \(- AP \cdot PC = OP^2 - R^2\) . Therefore, it suffices to show \(AQ \cdot QB = AP \cdot PC\) .
|
| 103 |
+
|
| 104 |
+
|
| 105 |
+
|
| 106 |
+
|
| 107 |
+
As \(\overline{ML} \parallel \overline{AC}\) and \(\overline{MK} \parallel \overline{AB}\) we have that
|
| 108 |
+
|
| 109 |
+
\[\angle APQ = \angle LMP = \angle LKM\] \[\angle PQA = \angle KMQ = \angle MLK\]
|
| 110 |
+
|
| 111 |
+
and consequently we have the (opposite orientation) similarity
|
| 112 |
+
|
| 113 |
+
\[\triangle APQ \sim \triangle MKL.\]
|
| 114 |
+
|
| 115 |
+
Therefore
|
| 116 |
+
|
| 117 |
+
\[\frac{AQ}{AP} = \frac{ML}{MK} = \frac{2ML}{2MK} = \frac{PC}{QB}\]
|
| 118 |
+
|
| 119 |
+
id est \(AQ \cdot QB = AP \cdot PC\) , which is what we wanted to prove.
|
| 120 |
+
|
| 121 |
+
|
| 122 |
+
|
| 123 |
+
## \(\S 1.3\) IMO 2009/3, proposed by Gabriel Carroll (USA)
|
| 124 |
+
|
| 125 |
+
Available online at https://aops.com/community/p1561573.
|
| 126 |
+
|
| 127 |
+
## Problem statement
|
| 128 |
+
|
| 129 |
+
Suppose that \(s_{1},s_{2},s_{3},\ldots\) is a strictly increasing sequence of positive integers such that the sub- sequences \(s_{s_{1}},s_{s_{2}},s_{s_{3}},\ldots\) and \(s_{s_{1} + 1},s_{s_{2} + 1},s_{s_{3} + 1},\ldots\) are both arithmetic progressions. Prove that the sequence \(s_{1},s_{2},s_{3},\ldots\) is itself an arithmetic progression.
|
| 130 |
+
|
| 131 |
+
We present two solutions.
|
| 132 |
+
|
| 133 |
+
\(\P\) First solution (Alex Zhai). Let \(s(n):= s_{n}\) and write
|
| 134 |
+
|
| 135 |
+
\[s(s(n)) = Dn + A\] \[s(s(n) + 1) = D^{\prime}n + B.\]
|
| 136 |
+
|
| 137 |
+
In light of the bounds \(s(s(n))\leq s(s(n) + 1)\leq s(s(n + 1))\) we right away recover \(D = D^{\prime}\) and \(A\leq B\)
|
| 138 |
+
|
| 139 |
+
Let \(d_{n} = s(n + 1) - s(n)\) . Note that \(\sup d_{n}< \infty\) since \(d_{n}\) is bounded above by \(A\)
|
| 140 |
+
|
| 141 |
+
Then we let
|
| 142 |
+
|
| 143 |
+
\[m:= \min d_{n},\qquad M:= \max d_{n}.\]
|
| 144 |
+
|
| 145 |
+
Now suppose \(a\) achieves the maximum, meaning \(s(a + 1) - s(a) = M\) . Then
|
| 146 |
+
|
| 147 |
+
\[\underbrace{d_{s(s(a))} + \cdot\cdot\cdot + d_{s(s(a + 1)) - 1}}_{D\mathrm{~terms}} = \underbrace{\left[s(s(s(a + 1))) - s(s(s(a)))\right]}_{D\mathrm{~terms}}\] \[\qquad = (D\cdot s(a + 1) + A) - (D\cdot s(a) + A) = DM.\]
|
| 148 |
+
|
| 149 |
+
Now \(M\) was maximal hence \(M = d_{s(s(a))} = \dots = d_{s(s(a + 1)) - 1}\) . But \(d_{s(s(a))} = B - A\) is a constant. Hence \(M = B - A\) . In the same way \(m = B - A\) as desired.
|
| 150 |
+
|
| 151 |
+
\(\P\) Second solution. We retain the notation \(D\) , \(A\) , \(B\) above, as well as \(m = \min_{n}s(n+\) \(1) - s(n)\geq B - A\) . We do the involution trick first as:
|
| 152 |
+
|
| 153 |
+
\[D = \left[s(s(s(n) + 1)) - s(s(s(n)))\right] = s(Dn + B) - s(Dn + A)\]
|
| 154 |
+
|
| 155 |
+
and hence we recover \(D\geq m(B - A)\)
|
| 156 |
+
|
| 157 |
+
The edge case \(D = B - A\) is easy since then \(m = 1\) and \(D = s(Dn + B) - s(Dn + A)\) forces \(s\) to be a constant shift. So henceforth assume \(D > B - A\)
|
| 158 |
+
|
| 159 |
+
The idea is that right now the \(B\) terms are "too big", so we want to use the involution trick in a way that makes as many " \(A\) minus \(B\) " shape expressions as possible. This motivates considering \(s(s(s(n + 1))) - s(s(s(n) + 1) + 1) > 0\) , since the first expression will have all \(A\) 's and the second expression will have all \(B\) 's. Calculation gives:
|
| 160 |
+
|
| 161 |
+
\[s(D(n + 1) + A) - s(Dn + B + 1) = \left[s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\right]\] \[\qquad = (D s(n + 1) + A) - (D(s(n) + 1) + B)\] \[\qquad = D(s(n + 1) - s(n)) + A - B - D.\]
|
| 162 |
+
|
| 163 |
+
|
| 164 |
+
|
| 165 |
+
Then by picking \(n\) achieving the minimum \(m\) ,
|
| 166 |
+
|
| 167 |
+
\[\underbrace{m(D + A - B - 1)}_{>0}\leq s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\leq Dm + A - B - D\]
|
| 168 |
+
|
| 169 |
+
which becomes
|
| 170 |
+
|
| 171 |
+
\[(D - m(B - A)) + ((B - A) - m)\leq 0.\]
|
| 172 |
+
|
| 173 |
+
Since both of these quantities were supposed to be nonnegative, we conclude \(m = B - A\) and \(D = m^{2}\) . Now the estimate \(D = s(Dn + B) - s(Dn + A)\geq m(B - A)\) is actually sharp, so it follows that \(s(n)\) is arithmetic.
|
| 174 |
+
|
| 175 |
+
|
| 176 |
+
|
| 177 |
+
## \(\S 2\) Solutions to Day 2
|
| 178 |
+
|
| 179 |
+
## \(\S 2.1\) IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk (BEL)
|
| 180 |
+
|
| 181 |
+
Available online at https://aops.com/community/p1562847.
|
| 182 |
+
|
| 183 |
+
## Problem statement
|
| 184 |
+
|
| 185 |
+
Let \(A B C\) be a triangle with \(A B = A C\) . The angle bisectors of \(\angle C A B\) and \(\angle A B C\) meet the sides \(B C\) and \(C A\) at \(D\) and \(E\) , respectively. Let \(K\) be the incenter of triangle \(A D C\) . Suppose that \(\angle B E K = 45^{\circ}\) . Find all possible values of \(\angle C A B\) .
|
| 186 |
+
|
| 187 |
+
Here is the solution presented in my book EGM0.
|
| 188 |
+
|
| 189 |
+
Let \(I\) be the incenter of \(A B C\) , and set \(\angle D A C = 2x\) (so that \(0^{\circ}< x< 45^{\circ}\) ). From \(\angle A I E = \angle D I C\) , it is easy to compute
|
| 190 |
+
|
| 191 |
+
\[\angle K I E = 90^{\circ} - 2x, \angle E C I = 45^{\circ} - x, \angle I E K = 45^{\circ}, \angle K E C = 3x.\]
|
| 192 |
+
|
| 193 |
+
Having chased all the angles we want, we need a relationship. We can find it by considering the side ratio \(\frac{I K}{K C}\) . Using the angle bisector theorem, we can express this in terms of triangle \(I D C\) ; however we can also express it in terms of triangle \(I E C\) .
|
| 194 |
+
|
| 195 |
+
|
| 196 |
+
|
| 197 |
+
|
| 198 |
+
By the law of sines, we obtain
|
| 199 |
+
|
| 200 |
+
\[\frac{I K}{K C} = \frac{\sin 45^{\circ}\cdot\frac{E K}{\sin(90^{\circ} - 2x)}}{\sin(3x)\cdot\frac{E K}{\sin(45^{\circ} - x)}} = \frac{\sin 45^{\circ}\sin(45^{\circ} - x)}{\sin(3x)\sin(90^{\circ} - 2x)}.\]
|
| 201 |
+
|
| 202 |
+
Also, by the angle bisector theorem on \(\triangle I D C\) , we have
|
| 203 |
+
|
| 204 |
+
\[\frac{I K}{K C} = \frac{I D}{D C} = \frac{\sin(45^{\circ} - x)}{\sin(45^{\circ} + x)}.\]
|
| 205 |
+
|
| 206 |
+
|
| 207 |
+
|
| 208 |
+
Equating these and cancelling \(\sin \left(45^{\circ} - x\right) \neq 0\) gives
|
| 209 |
+
|
| 210 |
+
\[\sin 45^{\circ}\sin \left(45^{\circ} + x\right) = \sin 3x\sin \left(90^{\circ} - 2x\right).\]
|
| 211 |
+
|
| 212 |
+
Applying the product- sum formula (again, we are just trying to break down things as much as possible), this just becomes
|
| 213 |
+
|
| 214 |
+
\[\cos \left(x\right) - \cos \left(90^{\circ} + x\right) = \cos \left(5x - 90^{\circ}\right) - \cos \left(90^{\circ} + x\right)\]
|
| 215 |
+
|
| 216 |
+
or \(\cos x = \cos \left(5x - 90^{\circ}\right)\)
|
| 217 |
+
|
| 218 |
+
At this point we are basically done; the rest is making sure we do not miss any solutions and write up the completion nicely. One nice way to do this is by using product- sum in reverse as
|
| 219 |
+
|
| 220 |
+
\[0 = \cos \left(5x - 90^{\circ}\right) - \cos x = 2\sin \left(3x - 45^{\circ}\right)\sin \left(2x - 45^{\circ}\right).\]
|
| 221 |
+
|
| 222 |
+
This way we merely consider the two cases
|
| 223 |
+
|
| 224 |
+
\[\sin \left(3x - 45^{\circ}\right) = 0 \mathrm{and} \sin \left(2x - 45^{\circ}\right) = 0.\]
|
| 225 |
+
|
| 226 |
+
Notice that \(\sin \theta = 0\) if and only \(\theta\) is an integer multiple of \(180^{\circ}\) . Using the bound \(0^{\circ} < x < 45^{\circ}\) , it is easy to see that the permissible values of \(x\) are \(x = 15^{\circ}\) and \(x = \frac{45^{\circ}}{2}\) . As \(\angle A = 4x\) , this corresponds to \(\angle A = 60^{\circ}\) and \(\angle A = 90^{\circ}\) , which can be seen to work.
|
| 227 |
+
|
| 228 |
+
|
| 229 |
+
|
| 230 |
+
## \(\S 2.2\) IMO 2009/5, proposed by Bruno Le Floch (FRA)
|
| 231 |
+
|
| 232 |
+
Available online at https://aops.com/community/p1562848.
|
| 233 |
+
|
| 234 |
+
## Problem statement
|
| 235 |
+
|
| 236 |
+
Find all functions \(f\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that for positive integers \(a\) and \(b\) , the numbers
|
| 237 |
+
|
| 238 |
+
\[a, f(b), f(b + f(a) - 1)\]
|
| 239 |
+
|
| 240 |
+
are the sides of a non- degenerate triangle.
|
| 241 |
+
|
| 242 |
+
The only function is the identity function (which works). We prove it is the only one. Let \(P(a,b)\) denote the given statement.
|
| 243 |
+
|
| 244 |
+
Claim — We have \(f(1) = 1\) , and \(f(f(n)) = n\) . (In particular \(f\) is a bijection.)
|
| 245 |
+
|
| 246 |
+
Proof. Note that
|
| 247 |
+
|
| 248 |
+
\[P(1,b)\Rightarrow f(b) = f(b + f(1) - 1).\]
|
| 249 |
+
|
| 250 |
+
Otherwise, the function \(f\) is periodic modulo \(N = f(1) - 1\geq 1\) . This is impossible since we can fix \(b\) and let \(a\) be arbitrarily large in some residue class modulo \(N\) .
|
| 251 |
+
|
| 252 |
+
Hence \(f(1) = 1\) , so taking \(P(n,1)\) gives \(f(f(n)) = n\) .
|
| 253 |
+
|
| 254 |
+
Claim — Let \(\delta = f(2) - 1 > 0\) . Then for every \(n\) ,
|
| 255 |
+
|
| 256 |
+
\[f(n + 1) = f(n) + \delta \quad \mathrm{or}\quad f(n - 1) = f(n) + \delta\]
|
| 257 |
+
|
| 258 |
+
Proof. Use
|
| 259 |
+
|
| 260 |
+
\[P(2,f(n))\Rightarrow n - 2< f(f(n) + \delta)< n + 2.\]
|
| 261 |
+
|
| 262 |
+
Let \(y = f(f(n) + \delta)\) , hence \(n - 2< y< n + 2\) and \(f(y) = f(n) + \delta\) . But, remark that if \(y = n\) , we get \(\delta = 0\) , contradiction. So \(y\in \{n + 1,n - 1\}\) and that is all. \(\square\)
|
| 263 |
+
|
| 264 |
+
We now show \(f\) is an arithmetic progression with common difference \(+\delta\) . Indeed we already know \(f(1) = 1\) and \(f(2) = 1 + \delta\) . Now suppose \(f(1) = 1,\ldots ,f(n) = 1 + (n - 1)\delta\) . Then by induction for any \(n\geq 2\) , the second case can't hold, so we have \(f(n + 1) = f(n) + \delta\) , as desired.
|
| 265 |
+
|
| 266 |
+
Combined with \(f(f(n)) = n\) , we recover that \(f\) is the identity.
|
| 267 |
+
|
| 268 |
+
|
| 269 |
+
|
| 270 |
+
## \(\S 2.3\) IMO 2009/6, proposed by Dmitry Khramtsov (RUS)
|
| 271 |
+
|
| 272 |
+
Available online at https://aops.com/community/p1562840.
|
| 273 |
+
|
| 274 |
+
## Problem statement
|
| 275 |
+
|
| 276 |
+
Let \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) be distinct positive integers and let \(M\) be a set of \(n - 1\) positive integers not containing \(s = a_{1} + \dots +a_{n}\) . A grasshopper is to jump along the real axis, starting at the point 0 and making \(n\) jumps to the right with lengths \(a_{1}\) , \(a_{2}\) , ..., \(a_{n}\) in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in \(M\) .
|
| 277 |
+
|
| 278 |
+
The proof is by induction on \(n\) . Assume \(a_{1} < \dots < a_{n}\) and call each element of \(M\) a mine. Let \(x = s - a_{n}\) . We consider four cases, based on whether \(x\) has a mine and whether there is a mine past \(x\) .
|
| 279 |
+
|
| 280 |
+
- If \(x\) has no mine, and there is a mine past \(x\) , then at most \(n - 2\) mines in \([0, x]\) and so we use induction to reach \(x\) , then leap from \(x\) to \(s\) and win.
|
| 281 |
+
|
| 282 |
+
- If \(x\) has no mine but there is also no mine to the right of \(x\) , then let \(m\) be the maximal mine. By induction hypothesis on \(M \setminus \{m\}\) , there is a path to \(x\) using \(\{a_{1}, \ldots , a_{n - 1}\}\) which avoids mines except possibly \(m\) . If the path hits the mine \(m\) on the hop of length \(a_{k}\) , we then swap that hop with \(a_{n}\) , and finish.
|
| 283 |
+
|
| 284 |
+
- If \(x\) has a mine, but there are no mines to the right of \(x\) , we can repeat the previous case with \(m = x\) .
|
| 285 |
+
|
| 286 |
+
- Now suppose \(x\) has a mine, and there is a mine past \(x\) . There should exist an integer \(1 \leq i \leq n - 1\) such that \(s - a_{i}\) and \(y = s - a_{i} - a_{n}\) both have no mine. By induction hypothesis, we can then reach \(y\) in \(n - 2\) steps (as there are two mines to the right of \(y\) ); then \(y \to s - a_{i} \to s\) finishes.
|
| 287 |
+
|
| 288 |
+
Remark. It seems much of the difficulty of the problem is realizing induction will actually work. Attempts at induction are, indeed, a total minefield (ha!), and given the position P6 of the problem, it is expected that many contestants will abandon induction after some cursory attempts fail.
|
| 289 |
+
|
| 290 |
+
|
IMO/md/en-IMO-2010-notes.md
ADDED
|
@@ -0,0 +1,317 @@
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|
| 1 |
+
|
| 2 |
+
# IMO 2010 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2010 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2010/1, proposed by Pierre Bornsztein (FRA) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2010/2, proposed by Tai Wai Ming and Wang Chongli (HKG) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2010/3, proposed by Gabriel Carroll (USA) 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 6
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2010/4, proposed by Marcin Kuczma (POL) 6
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2010/5, proposed by Netherlands 8
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2010/6, proposed by Morteza Saghafiyan (IRN) 9
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Find all functions \(f\colon \mathbb{R}\to \mathbb{R}\) such that for all \(x,y\in \mathbb{R}\)
|
| 39 |
+
|
| 40 |
+
\[f(\lfloor x\rfloor y) = f(x)\lfloor f(y)\rfloor .\]
|
| 41 |
+
|
| 42 |
+
2. Let \(I\) be the incenter of a triangle \(A B C\) and let \(\Gamma\) be its circumcircle. Let line \(A I\) intersect \(\Gamma\) again at \(D\) . Let \(E\) be a point on arc \(\overrightarrow{B D C}\) and \(F\) a point on side \(B C\) such that
|
| 43 |
+
|
| 44 |
+
\[\angle B A F = \angle C A E< \frac{1}{2}\angle B A C.\]
|
| 45 |
+
|
| 46 |
+
Finally, let \(G\) be the midpoint of \(\overline{{I F}}\) . Prove that \(\overline{{D G}}\) and \(\overline{{E I}}\) intersect on \(\Gamma\)
|
| 47 |
+
|
| 48 |
+
3. Find all functions \(g\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that
|
| 49 |
+
|
| 50 |
+
\[(g(m) + n)(g(n) + m)\]
|
| 51 |
+
|
| 52 |
+
is always a perfect square.
|
| 53 |
+
|
| 54 |
+
4. Let \(P\) be a point interior to triangle \(A B C\) (with \(C A\neq C B\) ). The lines \(A P\) , \(B P\) and \(C P\) meet again its circumcircle \(\Gamma\) at \(K\) , \(L\) , \(M\) , respectively. The tangent line at \(C\) to \(\Gamma\) meets the line \(A B\) at \(S\) . Show that from \(S C = S P\) follows \(M K = M L\) .
|
| 55 |
+
|
| 56 |
+
5. Each of the six boxes \(B_{1}\) , \(B_{2}\) , \(B_{3}\) , \(B_{4}\) , \(B_{5}\) , \(B_{6}\) initially contains one coin. The following two types of operations are allowed:
|
| 57 |
+
|
| 58 |
+
a) Choose a non-empty box \(B_{j}\) , \(1\leq j\leq 5\) , remove one coin from \(B_{j}\) and add two coins to \(B_{j + 1}\) ;
|
| 59 |
+
b) Choose a non-empty box \(B_{k}\) , \(1\leq k\leq 4\) , remove one coin from \(B_{k}\) and swap the contents (possibly empty) of the boxes \(B_{k + 1}\) and \(B_{k + 2}\) .
|
| 60 |
+
|
| 61 |
+
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes \(B_{1}\) , \(B_{2}\) , \(B_{3}\) , \(B_{4}\) , \(B_{5}\) become empty, while box \(B_{6}\) contains exactly \(2010^{2010^{2010}}\) coins.
|
| 62 |
+
|
| 63 |
+
6. Let \(a_{1},a_{2},a_{3},\ldots\) be a sequence of positive real numbers, and \(s\) be a positive integer, such that
|
| 64 |
+
|
| 65 |
+
\[a_{n} = \max \{a_{k} + a_{n - k}\mid 1\leq k\leq n - 1\} \mathrm{~for~all~}n > s.\]
|
| 66 |
+
|
| 67 |
+
Prove there exist positive integers \(\ell \leq s\) and \(N\) , such that
|
| 68 |
+
|
| 69 |
+
\[a_{n} = a_{\ell} + a_{n - \ell}\mathrm{~for~all~}n\geq N.\]
|
| 70 |
+
|
| 71 |
+
|
| 72 |
+
|
| 73 |
+
## \(\S 1\) Solutions to Day 1
|
| 74 |
+
|
| 75 |
+
## \(\S 1.1\) IMO 2010/1, proposed by Pierre Bornsztein (FRA)
|
| 76 |
+
|
| 77 |
+
Available online at https://aops.com/community/p1935849.
|
| 78 |
+
|
| 79 |
+
## Problem statement
|
| 80 |
+
|
| 81 |
+
Find all functions \(f\colon \mathbb{R}\to \mathbb{R}\) such that for all \(x,y\in \mathbb{R}\)
|
| 82 |
+
|
| 83 |
+
\[f(\lfloor x\rfloor y) = f(x)\lfloor f(y)\rfloor .\]
|
| 84 |
+
|
| 85 |
+
The only solutions are \(f(x)\equiv c\) , where \(c = 0\) or \(1\leq c< 2\) . It's easy to see these work.
|
| 86 |
+
|
| 87 |
+
Plug in \(x = 0\) to get \(f(0) = f(0)\lfloor f(y)\rfloor\) , so either
|
| 88 |
+
|
| 89 |
+
\[1\leq f(y)< 2\quad \forall y\qquad \mathrm{or}\qquad f(0) = 0\]
|
| 90 |
+
|
| 91 |
+
In the first situation, plug in \(y = 0\) to get \(f(x)\lfloor f(0)\rfloor = f(0)\) , thus \(f\) is constant. Thus assume henceforth \(f(0) = 0\)
|
| 92 |
+
|
| 93 |
+
Now set \(x = y = 1\) to get
|
| 94 |
+
|
| 95 |
+
\[f(1) = f(1)\lfloor f(1)\rfloor\]
|
| 96 |
+
|
| 97 |
+
so either \(f(1) = 0\) or \(1\leq f(1)< 2\) . We split into cases:
|
| 98 |
+
|
| 99 |
+
- If \(f(1) = 0\) , pick \(x = 1\) to get \(f(y)\equiv 0\) .
|
| 100 |
+
|
| 101 |
+
- If \(1\leq f(1)< 2\) , then \(y = 1\) gives
|
| 102 |
+
|
| 103 |
+
\[f(\lfloor x\rfloor) = f(x)\]
|
| 104 |
+
|
| 105 |
+
from \(y = 1\) , in particular \(f(x) = 0\) for \(0\leq x< 1\) . Choose \((x,y) = \left(2,\frac{1}{2}\right)\) to get \(f(1) = f(2)\left\lfloor f\left(\frac{1}{2}\right)\right\rfloor = 0\) .
|
| 106 |
+
|
| 107 |
+
|
| 108 |
+
|
| 109 |
+
## §1.2 IMO 2010/2, proposed by Tai Wai Ming and Wang Chongli (HKG)
|
| 110 |
+
|
| 111 |
+
Available online at https://aops.com/community/p1935927.
|
| 112 |
+
|
| 113 |
+
## Problem statement
|
| 114 |
+
|
| 115 |
+
Let \(I\) be the incenter of a triangle \(A B C\) and let \(\Gamma\) be its circumcircle. Let line \(A I\) intersect \(\Gamma\) again at \(D\) . Let \(E\) be a point on arc \(\overrightarrow{B D C}\) and \(F\) a point on side \(B C\) such that
|
| 116 |
+
|
| 117 |
+
\[\angle B A F = \angle C A E< \frac{1}{2}\angle B A C.\]
|
| 118 |
+
|
| 119 |
+
Finally, let \(G\) be the midpoint of \(\overline{{I F}}\) . Prove that \(\overline{{D G}}\) and \(\overline{{E I}}\) intersect on \(\Gamma\)
|
| 120 |
+
|
| 121 |
+
Let \(\overline{{E I}}\) meet \(\Gamma\) again at \(K\) . Then it suffices to show that \(\overline{{K D}}\) bisects \(\overline{{I F}}\) . Let \(\overline{{A F}}\) meet \(\Gamma\) again at \(H\) , so \(\overline{{H E}}\parallel \overline{{B C}}\) . By Pascal theorem on
|
| 122 |
+
|
| 123 |
+
\[A H E K D D\]
|
| 124 |
+
|
| 125 |
+
we then obtain that \(P = \overline{{A H}}\cap \overline{{K D}}\) lies on a line through \(I\) parallel to \(\overline{{B C}}\) .
|
| 126 |
+
|
| 127 |
+
Let \(I_{A}\) be the \(A\) - excenter, and set \(Q = \overline{{I_{A}F}}\cap \overline{{I P}}\) , and \(T = \overline{{A I D I_{A}}}\cap \overline{{B F C}}\) . Then
|
| 128 |
+
|
| 129 |
+
\[-1 = (A I;T I_{A})\stackrel {E}{=}(I Q;\infty P)\]
|
| 130 |
+
|
| 131 |
+
where \(\infty\) is the point at infinity along \(\overline{{I P Q}}\) . Thus \(P\) is the midpoint of \(\overline{{I Q}}\) . Since \(D\) is the midpoint of \(\overline{{I I_{A}}}\) by "Fact 5", it follows that \(\overline{{D P}}\) bisects \(\overline{{I F}}\) .
|
| 132 |
+
|
| 133 |
+
|
| 134 |
+
|
| 135 |
+
|
| 136 |
+
|
| 137 |
+
|
| 138 |
+
## \(\S 1.3\) IMO 2010/3, proposed by Gabriel Carroll (USA)
|
| 139 |
+
|
| 140 |
+
Available online at https://aops.com/community/p1935854.
|
| 141 |
+
|
| 142 |
+
## Problem statement
|
| 143 |
+
|
| 144 |
+
Find all functions \(g\colon \mathbb{Z}_{>0}\to \mathbb{Z}_{>0}\) such that
|
| 145 |
+
|
| 146 |
+
\[(g(m) + n)(g(n) + m)\]
|
| 147 |
+
|
| 148 |
+
is always a perfect square.
|
| 149 |
+
|
| 150 |
+
For \(c\geq 0\) , the function \(g(n) = n + c\) works; we prove this is the only possibility.
|
| 151 |
+
|
| 152 |
+
First, the main point of the problem is that:
|
| 153 |
+
|
| 154 |
+
Claim — We have \(g(n)\equiv g(n^{\prime})\) (mod \(p\) ) \(\Longrightarrow n\equiv n^{\prime}\) (mod \(p\) ).
|
| 155 |
+
|
| 156 |
+
Proof. Pick a large integer \(M\) such that
|
| 157 |
+
|
| 158 |
+
\[\nu_{p}(M + g(n)),\quad \nu_{p}(M + g(n^{\prime}))\quad \mathrm{are~both~odd}.\]
|
| 159 |
+
|
| 160 |
+
(It's not hard to see this is always possible.) Now, since each of
|
| 161 |
+
|
| 162 |
+
\[(M + g(n))(n + g(M))\] \[(M + g(n^{\prime}))(n^{\prime} + g(M))\]
|
| 163 |
+
|
| 164 |
+
is a square, we get \(n\equiv n^{\prime}\equiv - g(M)\) (mod \(p\) ).
|
| 165 |
+
|
| 166 |
+
This claim implies that
|
| 167 |
+
|
| 168 |
+
- The numbers \(g(n)\) and \(g(n + 1)\) differ by \(\pm 1\) for any \(n\) , and
|
| 169 |
+
|
| 170 |
+
- The function \(g\) is injective.
|
| 171 |
+
|
| 172 |
+
It follows \(g\) is a linear function with slope \(\pm 1\) , hence done.
|
| 173 |
+
|
| 174 |
+
|
| 175 |
+
|
| 176 |
+
## \(\S 2\) Solutions to Day 2
|
| 177 |
+
|
| 178 |
+
## \(\S 2.1\) IMO 2010/4, proposed by Marcin Kuczma (POL)
|
| 179 |
+
|
| 180 |
+
Available online at https://aops.com/community/p1936916.
|
| 181 |
+
|
| 182 |
+
## Problem statement
|
| 183 |
+
|
| 184 |
+
Let \(P\) be a point interior to triangle \(ABC\) (with \(CA \neq CB\) ). The lines \(AP\) , \(BP\) and \(CP\) meet again its circumcircle \(\Gamma\) at \(K\) , \(L\) , \(M\) , respectively. The tangent line at \(C\) to \(\Gamma\) meets the line \(AB\) at \(S\) . Show that from \(SC = SP\) follows \(MK = ML\) .
|
| 185 |
+
|
| 186 |
+
We present two solutions using harmonic bundles.
|
| 187 |
+
|
| 188 |
+
\(\P\) First solution (Evan Chen). Let \(N\) be the antipode of \(M\) , and let \(NP\) meet \(\Gamma\) again at \(D\) . Focus only on \(CDMN\) for now (ignoring the condition). Then \(C\) and \(D\) are feet of altitudes in \(\triangle MNP\) ; it is well- known that the circumcircle of \(\triangle CDP\) is orthogonal to \(\Gamma\) (passing through the orthocenter of \(\triangle MPN\) ).
|
| 189 |
+
|
| 190 |
+
|
| 191 |
+
|
| 192 |
+
|
| 193 |
+
Now, we are given that point \(S\) is such that \(\overline{SC}\) is tangent to \(\Gamma\) , and \(SC = SP\) . It follows that \(S\) is the circumcenter of \(\triangle CDP\) , and hence \(\overline{SC}\) and \(\overline{SD}\) are tangents to \(\Gamma\) .
|
| 194 |
+
|
| 195 |
+
Then \(- 1 = (AB;CD) \stackrel{P}{=} (KL;MN)\) . Since \(\overline{MN}\) is a diameter, this implies \(MK = ML\) .
|
| 196 |
+
|
| 197 |
+
Remark. I think it's more natural to come up with this solution in reverse. Namely, suppose we define the points the other way: let \(\overline{SD}\) be the other tangent, so \((AB;CD) = - 1\) . Then project through \(P\) to get \((KL;MN) = - 1\) , where \(N\) is the second intersection of \(\overline{DP}\) . However, if \(ML = MK\) then \(KMLN\) must be a kite. Thus one can recover the solution in reverse.
|
| 198 |
+
|
| 199 |
+
## \(\P\) Second solution (Sebastian Jeon). We have
|
| 200 |
+
|
| 201 |
+
\[SP^{2} = SC^{2} = SA\cdot SB\Rightarrow \angle SPA = \angle PBA = \angle LBA = \angle LKA = \angle LKP\]
|
| 202 |
+
|
| 203 |
+
(the latter half is Reim's theorem). Therefore \(\overline{SP}\) and \(\overline{LK}\) are parallel.
|
| 204 |
+
|
| 205 |
+
|
| 206 |
+
|
| 207 |
+
Now, let \(\overline{SP}\) meet \(\Gamma\) again at \(X\) and \(Y\) , and let \(Q\) be the antipode of \(P\) on \((S)\) . Then
|
| 208 |
+
|
| 209 |
+
\[SP^{2} = SQ^{2} = SX\cdot SY\Longrightarrow (PQ;XY) = -1\Longrightarrow \angle QCP = 90^{\circ}\]
|
| 210 |
+
|
| 211 |
+
that \(\overline{CP}\) bisects \(\angle XCY\) . Since \(\overline{XY} \parallel \overline{KL}\) , it follows \(\overline{CP}\) bisects to \(\angle LCK\) too.
|
| 212 |
+
|
| 213 |
+
|
| 214 |
+
|
| 215 |
+
## \(\S 2.2\) IMO 2010/5, proposed by Netherlands
|
| 216 |
+
|
| 217 |
+
Available online at https://aops.com/community/p1936917.
|
| 218 |
+
|
| 219 |
+
## Problem statement
|
| 220 |
+
|
| 221 |
+
Each of the six boxes \(B_{1}\) , \(B_{2}\) , \(B_{3}\) , \(B_{4}\) , \(B_{5}\) , \(B_{6}\) initially contains one coin. The following two types of operations are allowed:
|
| 222 |
+
|
| 223 |
+
1. Choose a non-empty box \(B_{j}\) , \(1 \leq j \leq 5\) , remove one coin from \(B_{j}\) and add two coins to \(B_{j + 1}\) ;
|
| 224 |
+
|
| 225 |
+
2. Choose a non-empty box \(B_{k}\) , \(1 \leq k \leq 4\) , remove one coin from \(B_{k}\) and swap the contents (possibly empty) of the boxes \(B_{k + 1}\) and \(B_{k + 2}\) .
|
| 226 |
+
|
| 227 |
+
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes \(B_{1}\) , \(B_{2}\) , \(B_{3}\) , \(B_{4}\) , \(B_{5}\) become empty, while box \(B_{6}\) contains exactly \(2010^{2010^{2010}}\) coins.
|
| 228 |
+
|
| 229 |
+
First,
|
| 230 |
+
|
| 231 |
+
\[(1,1,1,1,1,1) \to (0,3,1,0,3,1) \to (0,0,7,0,0,7)\] \[\to (0,0,6,2,0,7) \to (0,0,6,1,2,7) \to (0,0,6,1,0,11)\] \[\to (0,0,6,0,11,0) \to (0,0,5,11,0,0).\]
|
| 232 |
+
|
| 233 |
+
and henceforth we ignore boxes \(B_{1}\) and \(B_{2}\) , looking at just the last four boxes; so we write the current position as \((5,11,0,0)\) .
|
| 234 |
+
|
| 235 |
+
We prove a lemma:
|
| 236 |
+
|
| 237 |
+
Claim — Let \(k \geq 0\) and \(n > 0\) . From \((k, n, 0, 0)\) we may reach \((k - 1, 2^{n}, 0, 0)\) .
|
| 238 |
+
|
| 239 |
+
Proof. Working with only the last three boxes for now,
|
| 240 |
+
|
| 241 |
+
\[(n,0,0)\to (n - 1,2,0)\to (n - 1,0,4)\] \[\to (n - 2,4,0)\to (n - 2,0,8)\] \[\to (n - 3,8,0)\to (n - 3,0,16)\] \[\to \dots \to (1,2^{n - 1},0)\to (1,0,2^{n})\to (0,2^{n},0).\]
|
| 242 |
+
|
| 243 |
+
Finally we have \((k, n, 0, 0) \to (k, 0, 2^{n}, 0) \to (k - 1, 2^{n}, 0, 0)\) .
|
| 244 |
+
|
| 245 |
+
Now from \((5,11,0,0)\) we go as follows:
|
| 246 |
+
|
| 247 |
+
\[(5,11,0,0)\to (4,2^{11},0,0)\to \left(3,2^{21},0,0\right)\to \left(2,2^{22^{11}},0,0\right)\] \[\to \left(1,2^{22^{11}},0,0\right)\to \left(0,2^{22^{21^{11}}},0,0\right).\]
|
| 248 |
+
|
| 249 |
+
Let \(A = 2^{22^{22^{11}}} > 2010^{2010^{2010}} = B\) . Then by using move 2 repeatedly on the fourth box (i.e., throwing away several coins by swapping the empty \(B_{5}\) and \(B_{6}\) ), we go from \((0, A, 0, 0)\) to \((0, B / 4, 0, 0)\) . From there we reach \((0, 0, 0, B)\) .
|
| 250 |
+
|
| 251 |
+
|
| 252 |
+
|
| 253 |
+
## \(\S 2.3\) IMO 2010/6, proposed by Morteza Saghafiyan (IRN)
|
| 254 |
+
|
| 255 |
+
Available online at https://aops.com/community/p1936918.
|
| 256 |
+
|
| 257 |
+
## Problem statement
|
| 258 |
+
|
| 259 |
+
Let \(a_{1},a_{2},a_{3},\ldots\) be a sequence of positive real numbers, and \(s\) be a positive integer, such that
|
| 260 |
+
|
| 261 |
+
\[a_{n} = \max \{a_{k} + a_{n - k}\mid 1\leq k\leq n - 1\} \mathrm{~for~all~}n > s.\]
|
| 262 |
+
|
| 263 |
+
Prove there exist positive integers \(\ell \leq s\) and \(N\) , such that
|
| 264 |
+
|
| 265 |
+
\[a_{n} = a_{\ell} + a_{n - \ell}\mathrm{~for~all~}n\geq N.\]
|
| 266 |
+
|
| 267 |
+
Let
|
| 268 |
+
|
| 269 |
+
\[w_{1} = \frac{a_{1}}{1},\quad w_{2} = \frac{a_{2}}{2},\quad \ldots ,\quad w_{s} = \frac{a_{s}}{s}.\]
|
| 270 |
+
|
| 271 |
+
(The choice of the letter \(w\) is for "weight".) We claim the right choice of \(\ell\) is the one maximizing \(w_{\ell}\) .
|
| 272 |
+
|
| 273 |
+
Our plan is to view each \(a_{n}\) as a linear combination of the weights \(w_{1},\ldots ,w_{s}\) and track their coefficients.
|
| 274 |
+
|
| 275 |
+
To this end, let's define an \(n\) - type to be a vector \(T = \langle t_{1},\ldots ,t_{s}\rangle\) of nonnegative integers such that
|
| 276 |
+
|
| 277 |
+
\(n = t_{1} + \dots +t_{s}\) ; and
|
| 278 |
+
|
| 279 |
+
\(t_{i}\) is divisible by \(i\) for every \(i\) .
|
| 280 |
+
|
| 281 |
+
We then define its valuation as \(v(T) = \sum_{i = 1}^{s}w_{i}t_{i}\) .
|
| 282 |
+
|
| 283 |
+
Now we define a \(n\) - type to be valid according to the following recursive rule. For \(1\leq n\leq s\) the only valid \(n\) - types are
|
| 284 |
+
|
| 285 |
+
\[T_{1} = \langle 1,0,0,\ldots ,0\rangle\] \[T_{2} = \langle 0,2,0,\ldots ,0\rangle\] \[T_{3} = \langle 0,0,3,\ldots ,0\rangle\] \[\vdots\] \[T_{s} = \langle 0,0,0,\ldots ,s\rangle\]
|
| 286 |
+
|
| 287 |
+
for \(n = 1,\ldots ,s\) , respectively. Then for any \(n > s\) , an \(n\) - type is valid if it can be written as the sum of a valid \(k\) - type and a valid \((n - k)\) - type, componentwise. These represent the linear combinations possible in the recursion; in other words the recursion in the problem is phrased as
|
| 288 |
+
|
| 289 |
+
\[a_{n} = \max_{T\mathrm{~is~a~valid~}n\mathrm{-type}}v(T).\]
|
| 290 |
+
|
| 291 |
+
In fact, we have the following description of valid \(n\) - types:
|
| 292 |
+
|
| 293 |
+
Claim — Assume \(n > s\) . Then an \(n\) - type \(\langle t_{1},\ldots ,t_{s}\rangle\) is valid if and only if either
|
| 294 |
+
|
| 295 |
+
there exist indices \(i< j\) with \(i + j > s\) , \(t_{i}\geq i\) and \(t_{j}\geq j\) ; or
|
| 296 |
+
|
| 297 |
+
there exists an index \(i > s / 2\) with \(t_{i}\geq 2i\) .
|
| 298 |
+
|
| 299 |
+
|
| 300 |
+
|
| 301 |
+
Proof. Immediate by forwards induction on \(n > s\) that all \(n\) - types have this property.
|
| 302 |
+
|
| 303 |
+
The reverse direction is by downwards induction on \(n\) . Indeed if \(\sum_{i} \frac{t_{i}}{i} > 2\) , then we may subtract off on of \(\{T_{1}, \ldots , T_{s}\}\) while preserving the condition; and the case \(\sum_{i} \frac{t_{i}}{i} = 2\) is essentially by definition. \(\square\)
|
| 304 |
+
|
| 305 |
+
Remark. The claim is a bit confusingly stated in its two cases; really the latter case should be thought of as the situation \(i = j\) but requiring that \(t_{i} / i\) is counted with multiplicity.
|
| 306 |
+
|
| 307 |
+
Now, for each \(n > s\) we pick a valid \(n\) - type \(T_{n}\) with \(a_{n} = v(T_{n})\) ; if there are ties, we pick one for which the \(\ell\) th entry is as large as possible.
|
| 308 |
+
|
| 309 |
+
Claim — For any \(n > s\) and index \(i \neq \ell\) , the \(i\) th entry of \(T_{n}\) is at most \(2s + \ell i\) .
|
| 310 |
+
|
| 311 |
+
Proof. If not, we can go back \(i\ell\) steps to get a valid \((n - i\ell)\) - type \(T\) achieved by decreasing the \(i\) th entry of \(T_{n}\) by \(i\ell\) . But then we can add \(\ell\) to the \(\ell\) th entry \(i\) times to get another \(n\) - type \(T^{\prime}\) which obviously has valuation at least as large, but with larger \(\ell\) th entry. \(\square\)
|
| 312 |
+
|
| 313 |
+
Now since all other entries in \(T_{n}\) are bounded, eventually the sequence \((T_{n})_{n > s}\) just consists of repeatedly adding 1 to the \(\ell\) th entry, as required.
|
| 314 |
+
|
| 315 |
+
Remark. One big step is to consider \(w_{k} = a_{k} / k\) . You can get this using wishful thinking or by examining small cases. (In addition this normalization makes it easier to see why the largest \(w\) plays an important role, since then in the definition of type, the \(n\) - types all have a sum of \(n\) . Unfortunately, it makes the characterization of valid \(n\) - types somewhat clumsier too.)
|
| 316 |
+
|
| 317 |
+
|
IMO/md/en-IMO-2011-notes.md
ADDED
|
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|
| 1 |
+
|
| 2 |
+
# IMO 2011 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2011 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2011/1, proposed by Fernando Campos (MEX) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2011/2, proposed by Geoff Smith (UNK) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2011/3, proposed by Igor Voronovich, BLR 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2011/4, proposed by Morteza Saghafian (IRN) 7
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2011/5, proposed by Mahyar Sefidgaran (IRN) 9
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2011/6, proposed by Japan 10
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Given any set \(A = \{a_{1},a_{2},a_{3},a_{4}\}\) of four distinct positive integers, we denote the sum \(a_{1} + a_{2} + a_{3} + a_{4}\) by \(s_{A}\) . Let \(n_{A}\) denote the number of pairs \((i,j)\) with \(1\leq i< j\leq 4\) for which \(a_{i} + a_{j}\) divides \(s_{A}\) . Find all sets \(A\) of four distinct positive integers which achieve the largest possible value of \(n_{A}\) .
|
| 39 |
+
|
| 40 |
+
2. Let \(S\) be a finite set of at least two points in the plane. Assume that no three points of \(S\) are collinear. A windmill is a process that starts with a line \(\ell\) going through a single point \(P\in S\) . The line rotates clockwise about the pivot \(P\) until the first time that the line meets some other point belonging to \(S\) . This point, \(Q\) , takes over as the new pivot, and the line now rotates clockwise about \(Q\) , until it next meets a point of \(S\) . This process continues indefinitely.
|
| 41 |
+
|
| 42 |
+
Show that we can choose a point \(P\) in \(S\) and a line \(\ell\) going through \(P\) such that the resulting windmill uses each point of \(S\) as a pivot infinitely many times.
|
| 43 |
+
|
| 44 |
+
3. Let \(f\colon \mathbb{R}\to \mathbb{R}\) be a real-valued function defined on the set of real numbers that satisfies
|
| 45 |
+
|
| 46 |
+
\[f(x + y)\leq yf(x) + f(f(x))\]
|
| 47 |
+
|
| 48 |
+
for all real numbers \(x\) and \(y\) . Prove that \(f(x) = 0\) for all \(x\leq 0\)
|
| 49 |
+
|
| 50 |
+
4. Let \(n > 0\) be an integer. We are given a balance and \(n\) weights of weight \(2^{0}\) , \(2^{1}\) , ..., \(2^{n - 1}\) . We are to place each of the \(n\) weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed. Determine the number of ways in which this can be done.
|
| 51 |
+
|
| 52 |
+
5. Let \(f\colon \mathbb{Z}\to \mathbb{Z}_{>0}\) be a function such that \(f(m - n)\mid f(m) - f(n)\) for \(m,n\in \mathbb{Z}\) Prove that if \(m,n\in \mathbb{Z}\) satisfy \(f(m)\leq f(n)\) then \(f(m)\mid f(n)\)
|
| 53 |
+
|
| 54 |
+
6. Let \(A B C\) be an acute triangle with circumcircle \(\Gamma\) . Let \(\ell\) be a tangent line to \(\Gamma\) and let \(\ell_{a}\) , \(\ell_{b}\) , \(\ell_{c}\) be the lines obtained by reflecting \(\ell\) in the lines \(B C\) , \(C A\) , and \(A B\) , respectively. Show that the circumcircle of the triangle determined by the lines \(\ell_{a}\) , \(\ell_{b}\) , and \(\ell_{c}\) is tangent to the circle \(\Gamma\) .
|
| 55 |
+
|
| 56 |
+
|
| 57 |
+
|
| 58 |
+
## \(\S 1\) Solutions to Day 1
|
| 59 |
+
|
| 60 |
+
## \(\S 1.1\) IMO 2011/1, proposed by Fernando Campos (MEX)
|
| 61 |
+
|
| 62 |
+
Available online at https://aops.com/community/p2363530.
|
| 63 |
+
|
| 64 |
+
## Problem statement
|
| 65 |
+
|
| 66 |
+
Given any set \(A = \{a_{1},a_{2},a_{3},a_{4}\}\) of four distinct positive integers, we denote the sum \(a_{1} + a_{2} + a_{3} + a_{4}\) by \(s_{A}\) . Let \(n_{A}\) denote the number of pairs \((i,j)\) with \(1\leq i< j\leq 4\) for which \(a_{i} + a_{j}\) divides \(s_{A}\) . Find all sets \(A\) of four distinct positive integers which achieve the largest possible value of \(n_{A}\) .
|
| 67 |
+
|
| 68 |
+
There are two curves of solutions, namely \(\{x,5x,7x,11x\}\) and \(\{x,11x,19x,29x\}\) , for any positive integer \(x\) , achieving \(n_{A} = 4\) (easy to check). We'll show that \(n_{A}\leq 4\) and equality holds only in one of the curves.
|
| 69 |
+
|
| 70 |
+
Let \(A = \{a< b< c< d\}\) .
|
| 71 |
+
|
| 72 |
+
Claim — We have \(n_{A}\leq 4\) with equality iff
|
| 73 |
+
|
| 74 |
+
\[a + b\mid c + d,\quad a + c\mid b + d,\quad a + d = b + c.\]
|
| 75 |
+
|
| 76 |
+
Proof. Note \(a + b\mid s_{A}\iff a + b\mid c + d\) etc. Now \(c + d\nmid a + b\) and \(b + d\nmid a + c\) for size reasons, so we already have \(n_{A}\leq 4\) ; moreover \(a + d\mid b + c\) and \(b + c\mid a + d\) if and only if \(a + d = b + c\) . \(\square\)
|
| 77 |
+
|
| 78 |
+
We now show the equality curve is the one above.
|
| 79 |
+
|
| 80 |
+
\[a + c\mid b + d\iff a + c\mid -a + 2b + c\iff a + c\mid 2(b - a).\]
|
| 81 |
+
|
| 82 |
+
Since \(a + c > |b - a|\) , so we must have \(a + c = 2(b - a)\) . So we now have
|
| 83 |
+
|
| 84 |
+
\[c = 2b - 3a\] \[d = b + c - a = 3b + c - 4a.\]
|
| 85 |
+
|
| 86 |
+
The last condition is
|
| 87 |
+
|
| 88 |
+
\[a + b\mid c + d = 5b - 7a\iff a + b\mid 12a.\]
|
| 89 |
+
|
| 90 |
+
Now, let \(x = \gcd (a,b)\) . The expressions for \(c\) and \(d\) above imply that \(x\mid c,d\) so we may scale down so that \(x = 1\) . Then \(\gcd (a + b,a) = \gcd (a,b) = 1\) and so \(a + b\mid 12\)
|
| 91 |
+
|
| 92 |
+
We have \(c > b\) , so \(3a< b\) . The only pairs \((a,b)\) with \(3a< 2b\) , \(\gcd (a,b) = 1\) and \(a + b\mid 12\) are \((a,b)\in \{(1,5),(1,11)\}\) which give the solutions earlier.
|
| 93 |
+
|
| 94 |
+
|
| 95 |
+
|
| 96 |
+
## \(\S 1.2\) IMO 2011/2, proposed by Geoff Smith (UNK)
|
| 97 |
+
|
| 98 |
+
Available online at https://aops.com/community/p2363537.
|
| 99 |
+
|
| 100 |
+
## Problem statement
|
| 101 |
+
|
| 102 |
+
Let \(S\) be a finite set of at least two points in the plane. Assume that no three points of \(S\) are collinear. A windmill is a process that starts with a line \(\ell\) going through a single point \(P\in S\) . The line rotates clockwise about the pivot \(P\) until the first time that the line meets some other point belonging to \(S\) . This point, \(Q\) , takes over as the new pivot, and the line now rotates clockwise about \(Q\) , until it next meets a point of \(S\) . This process continues indefinitely.
|
| 103 |
+
|
| 104 |
+
Show that we can choose a point \(P\) in \(S\) and a line \(\ell\) going through \(P\) such that the resulting windmill uses each point of \(S\) as a pivot infinitely many times.
|
| 105 |
+
|
| 106 |
+
Orient \(\ell\) in some direction, and color the plane such that its left half is red and right half is blue. The critical observation is that:
|
| 107 |
+
|
| 108 |
+
Claim — The number of points on the red side of \(\ell\) does not change, nor does the number of points on the blue side (except at a moment when \(\ell\) contains two points).
|
| 109 |
+
|
| 110 |
+
Thus, if \(|S| = n + 1\) , it suffices to pick the initial configuration so that there are \(\lfloor n / 2\rfloor\) red and \(\lceil n / 2\rceil\) blue points. Then when the line \(\ell\) does a full \(180^{\circ}\) rotation, the red and blue sides "switch", so the windmill has passed through every point.
|
| 111 |
+
|
| 112 |
+
(See official shortlist for verbose write- up; this is deliberately short to make a point.)
|
| 113 |
+
|
| 114 |
+
|
| 115 |
+
|
| 116 |
+
## \(\S 1.3\) IMO 2011/3, proposed by Igor Voronovich, BLR
|
| 117 |
+
|
| 118 |
+
Available online at https://aops.com/community/p2363539.
|
| 119 |
+
|
| 120 |
+
## Problem statement
|
| 121 |
+
|
| 122 |
+
Let \(f\colon \mathbb{R}\to \mathbb{R}\) be a real- valued function defined on the set of real numbers that satisfies
|
| 123 |
+
|
| 124 |
+
\[f(x + y)\leq yf(x) + f(f(x))\]
|
| 125 |
+
|
| 126 |
+
for all real numbers \(x\) and \(y\) . Prove that \(f(x) = 0\) for all \(x\leq 0\)
|
| 127 |
+
|
| 128 |
+
We begin by rewriting the given as
|
| 129 |
+
|
| 130 |
+
\[f(z)\leq (z - x)f(x) + f(f(x))\quad \forall x,z\in \mathbb{R}\qquad (\heartsuit)\]
|
| 131 |
+
|
| 132 |
+
(which is better anyways since control over inputs to \(f\) is more valuable). We start by eliminating the double \(f\) : let \(z = f(w)\) to get
|
| 133 |
+
|
| 134 |
+
\[f(f(w))\leq (f(w) - x)f(x) + f(f(x))\]
|
| 135 |
+
|
| 136 |
+
and then use the symmetry trick to write
|
| 137 |
+
|
| 138 |
+
\[f(f(x))\leq (f(x) - w)f(w) + f(f(w))\]
|
| 139 |
+
|
| 140 |
+
so that when we sum we get
|
| 141 |
+
|
| 142 |
+
\[w f(w) + x f(x)\leq 2f(x)f(w).\]
|
| 143 |
+
|
| 144 |
+
Next we use cancellation trick: set \(w = 2f(x)\) in the above to get
|
| 145 |
+
|
| 146 |
+
\[x f(x)\leq 0\quad \forall x\in \mathbb{R}.\qquad (\spadesuit)\]
|
| 147 |
+
|
| 148 |
+
Claim — For every \(p\in \mathbb{R}\) , we have \(f(p)\leq 0\)
|
| 149 |
+
|
| 150 |
+
Proof. Assume \(f(p) > 0\) for some \(p\in \mathbb{R}\) . Then for any negative number \(z\)
|
| 151 |
+
|
| 152 |
+
\[0\leq f(z)\leq \stackrel {(\heartsuit)}{(z - p)}f(p) + f(f(p)).\]
|
| 153 |
+
|
| 154 |
+
which is false if we let \(z\to - \infty\)
|
| 155 |
+
|
| 156 |
+
Together with \((\spadesuit)\) we derive \(f(x) = 0\) for \(x< 0\) . Finally, letting \(x\) and \(z\) be any negative numbers in \((\heartsuit)\) , we get \(f(0)\geq 0\) , so \(f(0) = 0\) too.
|
| 157 |
+
|
| 158 |
+
Remark. As another corollary of the claim, \(f(f(x)) = 0\) for all \(x\) .
|
| 159 |
+
|
| 160 |
+
Remark. A nontrivial example of a working \(f\) is to take
|
| 161 |
+
|
| 162 |
+
|
| 163 |
+
|
| 164 |
+
or some other negative function growing rapidly in absolute value for \(x > 0\) .
|
| 165 |
+
|
| 166 |
+
|
| 167 |
+
|
| 168 |
+
■
|
| 169 |
+
|
| 170 |
+
|
| 171 |
+
|
| 172 |
+
## \(\S 2\) Solutions to Day 2
|
| 173 |
+
|
| 174 |
+
## \(\S 2.1\) IMO 2011/4, proposed by Morteza Saghafian (IRN)
|
| 175 |
+
|
| 176 |
+
Available online at https://aops.com/community/p2365036.
|
| 177 |
+
|
| 178 |
+
## Problem statement
|
| 179 |
+
|
| 180 |
+
Let \(n > 0\) be an integer. We are given a balance and \(n\) weights of weight \(2^{0}\) , \(2^{1}\) , ..., \(2^{n - 1}\) . We are to place each of the \(n\) weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed. Determine the number of ways in which this can be done.
|
| 181 |
+
|
| 182 |
+
The answer is \(a_{n} = (2n - 1)!!\)
|
| 183 |
+
|
| 184 |
+
We refer to what we're counting as a valid \(n\) - sequence: an order of which weights to place, and whether to place them on the left or right pan.
|
| 185 |
+
|
| 186 |
+
We use induction, with \(n = 1\) being obvious. Now consider the weight \(2^{0} = 1\)
|
| 187 |
+
|
| 188 |
+
- If we delete it from any valid \(n\) -sequence, we get a valid \((n - 1)\) -sequence with all weights doubled.
|
| 189 |
+
|
| 190 |
+
- Given a valid \((n - 1)\) -sequence with all weights doubled, we may insert \(2^{0} = 1\) it into \(2n - 1\) ways. Indeed, we may insert it anywhere, and designate it either left or right, except we may not designate right if we choose to insert \(2^{0} = 1\) at the very beginning.
|
| 191 |
+
|
| 192 |
+
Consequently, we have that
|
| 193 |
+
|
| 194 |
+
\[a_{n} = (2n - 1)\cdot a_{n - 1}.\]
|
| 195 |
+
|
| 196 |
+
Since \(a_{1} = 1\) , the conclusion follows.
|
| 197 |
+
|
| 198 |
+
Remark (Gabriel Levin). An alternate approach can be done by considering the heaviest weight \(2^{n - 1}\) instead of the lightest weight \(2^{0} = 1\) . This gives a more complicated recursion:
|
| 199 |
+
|
| 200 |
+
\[a_{n} = \sum_{k = 0}^{n - 1}\binom{n - 1}{k}2^{n - k - 1}(n - k - 1)!a_{k}\]
|
| 201 |
+
|
| 202 |
+
by summing over the index \(k\) corresponding to the number of weights placed before the heaviest weight \(2^{n - 1}\) is placed.
|
| 203 |
+
|
| 204 |
+
This recursion can then be rewritten as
|
| 205 |
+
|
| 206 |
+
\[a_{n} = \sum_{k = 0}^{n - 1}\binom{n - 1}{k}2^{n - k - 1}(n - k - 1)!a_{k}\] \[\qquad = (n - 1)!\sum_{k = 0}^{n - 1}\frac{2^{n - k - 1}a_{k}}{k!}\] \[\qquad = 2(n - 1)!\sum_{k = 0}^{n - 1}\frac{2^{(n - 1) - k - 1}a_{k}}{k!}\]
|
| 207 |
+
|
| 208 |
+
|
| 209 |
+
|
| 210 |
+
\[= 2(n - 1)(n - 2)!\left(\sum_{k = 0}^{(n - 1) - 1}\frac{2^{(n - 1) - k - 1}a_k}{k!}\right) + 2(n - 1)!\frac{\frac{1}{2}a_{n - 1}}{(n - 1)!}\] \[= 2(n - 1)a_{n - 1} + a_{n - 1}\] \[= (2n - 1)a_{n - 1}\]
|
| 211 |
+
|
| 212 |
+
and since \(a_{1} = 1\) , we have \(a_{n} = (2n - 1)!!\) by induction on \(n\) .
|
| 213 |
+
|
| 214 |
+
|
| 215 |
+
|
| 216 |
+
## \(\S 2.2\) IMO 2011/5, proposed by Mahyar Sefidgaran (IRN)
|
| 217 |
+
|
| 218 |
+
Available online at https://aops.com/community/p2365041.
|
| 219 |
+
|
| 220 |
+
## Problem statement
|
| 221 |
+
|
| 222 |
+
Let \(f\colon \mathbb{Z}\to \mathbb{Z}_{>0}\) be a function such that \(f(m - n)\mid f(m) - f(n)\) for \(m,n\in \mathbb{Z}\) . Prove that if \(m,n\in \mathbb{Z}\) satisfy \(f(m)\leq f(n)\) then \(f(m)\mid f(n)\)
|
| 223 |
+
|
| 224 |
+
Let \(P(m,n)\) denote the given assertion. First, we claim \(f\) is even. This is straight calculation:
|
| 225 |
+
|
| 226 |
+
\[\bullet P(x,0)\Rightarrow f(x)\mid f(x) - f(0)\Rightarrow f(x)\mid M:= f(0).\] \[\bullet P(0,x)\Rightarrow f(-x)\mid M - f(x)\Rightarrow f(-x)\mid f(x).\mathrm{Analogously},f(x)\mid f(-x).\mathrm{So\] \[f(x) = f(-x)\mathrm{and} f\mathrm{~is~even.}\]
|
| 227 |
+
|
| 228 |
+
Claim — Let \(x,y,z\) be integers with \(x + y + z = 0\) . Then among \(f(x),f(y),f(z)\) two of them are equal and divide the third.
|
| 229 |
+
|
| 230 |
+
Proof. Let \(a = f(\pm x)\) , \(b = f(\pm y)\) , \(c = f(\pm z)\) be positive integers. Note that
|
| 231 |
+
|
| 232 |
+
\[a\mid b - c\] \[b\mid c - a\] \[c\mid a - b\]
|
| 233 |
+
|
| 234 |
+
from \(P(y, - z)\) and similarly. WLOG \(c = \max (a,b,c)\) ; then \(c > |a - b|\) so \(a = b\) . Thus \(a = b\mid c\) from the first two. \(\square\)
|
| 235 |
+
|
| 236 |
+
This implies the problem, by taking \(x\) and \(y\) in the previous claim to be the integers \(m\) and \(n\) .
|
| 237 |
+
|
| 238 |
+
Remark. At https://aops.com/community/c6h418981p2381909, Davi Medeiros gives the following characterization of functions \(f\) satisfying the hypothesis.
|
| 239 |
+
|
| 240 |
+
Pick \(f(0)\) , \(k\) positive integers, a chain \(d_{1}\mid d_{2}\mid \dots \mid d_{k}\) of divisors of \(f(0)\) , and positive integers \(a_{1},a_{2},\ldots ,a_{k - 1}\) , greater than 1 (if \(k = 1\) , \(a_{i}\) doesn't exist, for every \(i\) ). We'll define \(f\) as follows:
|
| 241 |
+
|
| 242 |
+
\(f(n) = d_{1}\) , for every integer \(n\) that is not divisible by \(a_{1}\) ; \(f(a_{1}n) = d_{2}\) , for every integer \(n\) that is not divisible by \(a_{2}\) ; \(f(a_{1}a_{2}n) = d_{3}\) , for every integer \(n\) that is not divisible by \(a_{3}\) ; \(f(a_{1}a_{2}a_{3}n) = d_{4}\) , for every integer \(n\) that is not divisible by \(a_{4}\) ; \(\dots\) \(f(a_{1}a_{2}\ldots a_{k - 1}n) = d_{k}\) , for every integer \(n\) ;
|
| 243 |
+
|
| 244 |
+
|
| 245 |
+
|
| 246 |
+
## \(\S 2.3\) IMO 2011/6, proposed by Japan
|
| 247 |
+
|
| 248 |
+
Available online at https://aops.com/community/p2365045.
|
| 249 |
+
|
| 250 |
+
## Problem statement
|
| 251 |
+
|
| 252 |
+
Let \(A B C\) be an acute triangle with circumcircle \(\Gamma\) . Let \(\ell\) be a tangent line to \(\Gamma\) and let \(\ell_{a}\) , \(\ell_{b}\) , \(\ell_{c}\) be the lines obtained by reflecting \(\ell\) in the lines \(B C\) , \(C A\) , and \(A B\) , respectively. Show that the circumcircle of the triangle determined by the lines \(\ell_{a}\) , \(\ell_{b}\) , and \(\ell_{c}\) is tangent to the circle \(\Gamma\) .
|
| 253 |
+
|
| 254 |
+
This is a hard problem with many beautiful solutions. The following solution is not very beautiful but not too hard to find during an olympiad, as the only major insight it requires is the construction of \(A_{2}\) , \(B_{2}\) , and \(C_{2}\) .
|
| 255 |
+
|
| 256 |
+
|
| 257 |
+
|
| 258 |
+
|
| 259 |
+
We apply complex numbers with \(\omega\) the unit circle and \(p = 1\) . Let \(A_{1} = \ell_{B} \cap \ell_{C}\) , and let \(a_{2} = a^{2}\) (in other words, \(A_{2}\) is the reflection of \(P\) across the diameter of \(\omega\) through \(A\) ). Define the points \(B_{1}\) , \(C_{1}\) , \(B_{2}\) , \(C_{2}\) similarly.
|
| 260 |
+
|
| 261 |
+
We claim that \(\overline{A_{1}A_{2}}\) , \(\overline{B_{1}B_{2}}\) , \(\overline{C_{1}C_{2}}\) concur at a point on \(\Gamma\) .
|
| 262 |
+
|
| 263 |
+
We begin by finding \(A_{1}\) . If we reflect the points \(1 + i\) and \(1 - i\) over \(\overline{AB}\) , then we get two points \(Z_{1}\) , \(Z_{2}\) with
|
| 264 |
+
|
| 265 |
+
\[z_{1} = a + b - a b(1 - i) = a + b - a b + a b i\] \[z_{2} = a + b - a b(1 + i) = a + b - a b - a b i.\]
|
| 266 |
+
|
| 267 |
+
Therefore,
|
| 268 |
+
|
| 269 |
+
\[z_{1} - z_{2} = 2a b i\] \[\overline{{z_{1}z_{2}}} -\overline{{z_{2}}} z_{1} = -2i\left(a + b + \frac{1}{a} +\frac{1}{b} -2\right).\]
|
| 270 |
+
|
| 271 |
+
|
| 272 |
+
|
| 273 |
+
Now \(\ell_{C}\) is the line \(\overline{Z_{1}Z_{2}}\) , so with the analogous equation \(\ell_{B}\) we obtain:
|
| 274 |
+
|
| 275 |
+
\[a_{1} = \frac{-2i\left(a + b + \frac{1}{a} +\frac{1}{b} - 2\right)(2a c i) + 2i\left(a + c + \frac{1}{a} +\frac{1}{c} - 2\right)(2a b i)}{\left(-\frac{2}{a b} i\right)(2a c i) - \left(-\frac{2}{a c} i\right)(2a b i)}\] \[\quad = \frac{\left[c - b\right]a^{2} + \left[\frac{c}{b} -\frac{b}{c} -2c + 2b\right]a + (c - b)}{\frac{c}{b} -\frac{b}{c}}\] \[\quad = a + \frac{(c - b)\left[a^{2} - 2a + 1\right]}{(c - b)(c + b) / b c}\] \[\quad = a + \frac{b c}{b + c} (a - 1)^{2}.\]
|
| 276 |
+
|
| 277 |
+
Then the second intersection of \(\overline{A_{1}A_{2}}\) with \(\omega\) is given by
|
| 278 |
+
|
| 279 |
+
\[\frac{a_{1} - a_{2}}{1 - a_{2}\overline{a_{1}}} = \frac{a + \frac{b c}{b + c}(a - 1)^{2} - a^{2}}{1 - a - a^{2}\cdot\frac{(1 - 1 / a)^{2}}{b + c}}\] \[\qquad = \frac{a + \frac{b c}{b + c}(1 - a)}{1 - \frac{1}{b + c}(1 - a)}\] \[\qquad = \frac{a b + b c + c a - a b c}{a + b + c - 1}.\]
|
| 280 |
+
|
| 281 |
+
Thus, the claim is proved.
|
| 282 |
+
|
| 283 |
+
Finally, it suffices to show \(\overline{A_{1}B_{1}}\parallel \overline{A_{2}B_{2}}\) . One can also do this with complex numbers; it amounts to showing \(a^{2} - b^{2}\) , \(a - b\) , \(i\) (corresponding to \(\overline{A_{2}B_{2}}\) , \(\overline{A_{1}B_{1}}\) , \(\overline{PP}\) ) have their arguments an arithmetic progression, equivalently
|
| 284 |
+
|
| 285 |
+
\[\frac{(a - b)^{2}}{i(a^{2} - b^{2})}\in \mathbb{R}\iff \frac{(a - b)^{2}}{i(a^{2} - b^{2})} = \frac{\left(\frac{1}{a} - \frac{1}{b}\right)^{2}}{\frac{1}{i}\left(\frac{1}{a^{2}} - \frac{1}{b^{2}}\right)}\]
|
| 286 |
+
|
| 287 |
+
which is obvious.
|
| 288 |
+
|
| 289 |
+
Remark. One can use directed angle chasing for this last part too. Let \(\overline{BC}\) meet \(\ell\) at \(K\) and \(\overline{B_{2}C_{2}}\) meet \(\ell\) at \(L\) . Evidently
|
| 290 |
+
|
| 291 |
+
\[-\angle B_{2}L P = \angle L P B_{2} + \angle P B_{2}L\] \[\qquad = 2\angle K P B + \angle P B_{2}C_{2}\] \[\qquad = 2\angle K P B + 2\angle P B C\] \[\qquad = -2\angle P K B\] \[\qquad = \angle P K B_{1}\]
|
| 292 |
+
|
| 293 |
+
as required.
|
| 294 |
+
|
| 295 |
+
|
IMO/md/en-IMO-2012-notes.md
ADDED
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@@ -0,0 +1,286 @@
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|
| 1 |
+
|
| 2 |
+
# IMO 2012 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2012 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2012/1, proposed by Evangelos Psychas (HEL) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2012/2, proposed by Angelo di Pasquale (AUS) 4
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2012/3, proposed by David Arthur (CAN) 5
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2012/4, proposed by Liam Baker (SAF) 7
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2012/5, proposed by Josef Tkadlec (CZE) 8
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2012/6, proposed by Dusan Djukic (SRB) 9
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. Let \(A B C\) be a triangle and \(J\) the center of the \(A\) -excircle. This excircle is tangent to the side \(B C\) at \(M\) , and to the lines \(A B\) and \(A C\) at \(K\) and \(L\) , respectively. The lines \(L M\) and \(B J\) meet at \(F\) , and the lines \(K M\) and \(C J\) meet at \(G\) . Let \(S\) be the point of intersection of the lines \(A F\) and \(B C\) , and let \(T\) be the point of intersection of the lines \(A G\) and \(B C\) . Prove that \(M\) is the midpoint of \(\overline{S T}\) .
|
| 39 |
+
|
| 40 |
+
2. Let \(a_{2}\) , \(a_{3}\) , ..., \(a_{n}\) be positive reals with product 1, where \(n \geq 3\) . Show that
|
| 41 |
+
|
| 42 |
+
\[(1 + a_{2})^{2}(1 + a_{3})^{3}\dots (1 + a_{n})^{n} > n^{n}.\]
|
| 43 |
+
|
| 44 |
+
3. The liar's guessing game is a game played between two players \(A\) and \(B\) . The rules of the game depend on two fixed positive integers \(k\) and \(n\) which are known to both players.
|
| 45 |
+
|
| 46 |
+
At the start of the game \(A\) chooses integers \(x\) and \(N\) with \(1 \leq x \leq N\) . Player \(A\) keeps \(x\) secret, and truthfully tells \(N\) to player \(B\) . Player \(B\) now tries to obtain information about \(x\) by asking player \(A\) questions as follows: each question consists of \(B\) specifying an arbitrary set \(S\) of positive integers (possibly one specified in some previous question), and asking \(A\) whether \(x\) belongs to \(S\) . Player \(B\) may ask as many questions as he wishes. After each question, player \(A\) must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any \(k + 1\) consecutive answers, at least one answer must be truthful.
|
| 47 |
+
|
| 48 |
+
After \(B\) has asked as many questions as he wants, he must specify a set \(X\) of at most \(n\) positive integers. If \(x\) belongs to \(X\) , then \(B\) wins; otherwise, he loses. Prove that:
|
| 49 |
+
|
| 50 |
+
(a) If \(n \geq 2^{k}\) , then \(B\) can guarantee a win.
|
| 51 |
+
(b) For all sufficiently large \(k\) , there exists an integer \(n \geq (1.99)^{k}\) such that \(B\) cannot guarantee a win.
|
| 52 |
+
|
| 53 |
+
4. Find all functions \(f: \mathbb{Z} \to \mathbb{Z}\) such that, for all integers \(a\) , \(b\) , \(c\) that satisfy \(a + b + c = 0\) , the following equality holds:
|
| 54 |
+
|
| 55 |
+
\[f(a)^{2} + f(b)^{2} + f(c)^{2} = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\]
|
| 56 |
+
|
| 57 |
+
5. Let \(A B C\) be a triangle with \(\angle B C A = 90^{\circ}\) , and let \(D\) be the foot of the altitude from \(C\) . Let \(X\) be a point in the interior of the segment \(C D\) . Let \(K\) be the point on the segment \(A X\) such that \(B K = B C\) . Similarly, let \(L\) be the point on the segment \(B X\) such that \(A L = A C\) . Let \(M = \overline{A L} \cap \overline{B K}\) . Prove that \(M K = M L\) .
|
| 58 |
+
|
| 59 |
+
6. Find all positive integers \(n\) for which there exist non-negative integers \(a_{1}, a_{2}, \ldots , a_{n}\) such that
|
| 60 |
+
|
| 61 |
+
\[\frac{1}{2^{a_{1}}} + \frac{1}{2^{a_{2}}} + \dots + \frac{1}{2^{a_{n}}} = \frac{1}{3^{a_{1}}} + \frac{2}{3^{a_{2}}} + \dots + \frac{n}{3^{a_{n}}} = 1.\]
|
| 62 |
+
|
| 63 |
+
|
| 64 |
+
|
| 65 |
+
## \(\S 1\) Solutions to Day 1
|
| 66 |
+
|
| 67 |
+
## \(\S 1.1\) IMO 2012/1, proposed by Evangelos Psychas (HEL)
|
| 68 |
+
|
| 69 |
+
Available online at https://aops.com/community/p2736397.
|
| 70 |
+
|
| 71 |
+
## Problem statement
|
| 72 |
+
|
| 73 |
+
Let \(A B C\) be a triangle and \(J\) the center of the \(A\) - excircle. This excircle is tangent to the side \(B C\) at \(M\) , and to the lines \(A B\) and \(A C\) at \(K\) and \(L\) , respectively. The lines \(L M\) and \(B J\) meet at \(F\) , and the lines \(K M\) and \(C J\) meet at \(G\) . Let \(S\) be the point of intersection of the lines \(A F\) and \(B C\) , and let \(T\) be the point of intersection of the lines \(A G\) and \(B C\) . Prove that \(M\) is the midpoint of \(\overline{S T}\) .
|
| 74 |
+
|
| 75 |
+
We employ barycentric coordinates with reference \(\triangle A B C\) . As usual \(a = B C\) , \(b = C A\) , \(c = A B\) , \(s = \frac{1}{2} (a + b + c)\) .
|
| 76 |
+
|
| 77 |
+
It's obvious that \(K = (- (s - c):s:0)\) , \(M = (0:s - b:s - c)\) . Also, \(J = (- a:b:c)\) . We then obtain
|
| 78 |
+
|
| 79 |
+
\[G = \left(-a:b: - \frac{as + (s - c)b}{s - b}\right).\]
|
| 80 |
+
|
| 81 |
+
It follows that
|
| 82 |
+
|
| 83 |
+
\[T = \left(0:b: - \frac{-as + (s - c)}{s - b}\right) = (0:b(s - b):b(s - c) - as).\]
|
| 84 |
+
|
| 85 |
+
Normalizing, we see that \(T = \left(0, - \frac{b}{a},1 + \frac{b}{a}\right)\) , from which we quickly obtain \(M T = s\) . Similarly, \(M S = s\) , so we're done.
|
| 86 |
+
|
| 87 |
+
|
| 88 |
+
|
| 89 |
+
## \(\S 1.2\) IMO 2012/2, proposed by Angelo di Pasquale (AUS)
|
| 90 |
+
|
| 91 |
+
Available online at https://aops.com/community/p2736375.
|
| 92 |
+
|
| 93 |
+
## Problem statement
|
| 94 |
+
|
| 95 |
+
Let \(a_{2}\) , \(a_{3}\) , ..., \(a_{n}\) be positive reals with product 1, where \(n \geq 3\) . Show that
|
| 96 |
+
|
| 97 |
+
\[(1 + a_{2})^{2}(1 + a_{3})^{3}\ldots (1 + a_{n})^{n} > n^{n}.\]
|
| 98 |
+
|
| 99 |
+
Try the dumbest thing possible: by AM- GM,
|
| 100 |
+
|
| 101 |
+
\[(1 + a_{2})^{2}\geq 2^{2}a_{2}\] \[(1 + a_{3})^{3} = \left(\frac{1}{2} +\frac{1}{2} +a_{3}\right)^{3}\geq \frac{3^{3}}{2^{2}}a_{3}\] \[(1 + a_{4})^{4} = \left(\frac{1}{3} +\frac{1}{3} +\frac{1}{3} +a_{4}\right)^{4}\geq \frac{4^{4}}{3^{3}}a_{4}\] \[\qquad \vdots\]
|
| 102 |
+
|
| 103 |
+
and so on. Multiplying these all gives the result. The inequality is strict since it's not possible that \(a_{2} = 1\) , \(a_{3} = \frac{1}{2}\) , et cetera.
|
| 104 |
+
|
| 105 |
+
|
| 106 |
+
|
| 107 |
+
## \(\S 1.3\) IMO 2012/3, proposed by David Arthur (CAN)
|
| 108 |
+
|
| 109 |
+
Available online at https://aops.com/community/p2736406.
|
| 110 |
+
|
| 111 |
+
## Problem statement
|
| 112 |
+
|
| 113 |
+
The liar's guessing game is a game played between two players \(A\) and \(B\) . The rules of the game depend on two fixed positive integers \(k\) and \(n\) which are known to both players.
|
| 114 |
+
|
| 115 |
+
At the start of the game \(A\) chooses integers \(x\) and \(N\) with \(1\leq x\leq N\) . Player \(A\) keeps \(x\) secret, and truthfully tells \(N\) to player \(B\) . Player \(B\) now tries to obtain information about \(x\) by asking player \(A\) questions as follows: each question consists of \(B\) specifying an arbitrary set \(S\) of positive integers (possibly one specified in some previous question), and asking \(A\) whether \(x\) belongs to \(S\) . Player \(B\) may ask as many questions as he wishes. After each question, player \(A\) must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any \(k + 1\) consecutive answers, at least one answer must be truthful.
|
| 116 |
+
|
| 117 |
+
After \(B\) has asked as many questions as he wants, he must specify a set \(X\) of at most \(n\) positive integers. If \(x\) belongs to \(X\) , then \(B\) wins; otherwise, he loses. Prove that:
|
| 118 |
+
|
| 119 |
+
(a) If \(n\geq 2^{k}\) , then \(B\) can guarantee a win.
|
| 120 |
+
|
| 121 |
+
(b) For all sufficiently large \(k\) , there exists an integer \(n\geq (1.99)^{k}\) such that \(B\) cannot guarantee a win.
|
| 122 |
+
|
| 123 |
+
Call the players Alice and Bob.
|
| 124 |
+
|
| 125 |
+
Part (a): We prove the following.
|
| 126 |
+
|
| 127 |
+
Claim — If \(N\geq 2^{k} + 1\) , then in \(2k + 1\) questions, Bob can rule out some number in \(\{1,\ldots ,2^{k} + 1\}\) form being equal to \(x\) .
|
| 128 |
+
|
| 129 |
+
Proof. First, Bob asks the question \(S_{0} = \{2^{k} + 1\}\) until Alice answers "yes" or until Bob has asked \(k + 1\) questions. If Alice answers "no" to all of these then Bob rules out \(2^{k} + 1\) . So let's assume Alice just said "yes".
|
| 130 |
+
|
| 131 |
+
Now let \(T = \{1,\ldots ,2^{k}\}\) . Then, he asks \(k\) - follow up questions \(S_{1}\) , ..., \(S_{k}\) defined as follows:
|
| 132 |
+
|
| 133 |
+
- \(S_{1} = \{1,3,5,7,\ldots ,2^{k} - 1\}\) consists of all numbers in \(T\) whose least significant digit in binary is 1.
|
| 134 |
+
- \(S_{2} = \{2,3,6,7,\ldots ,2^{k} - 2,2^{k} - 1\}\) consists of all numbers in \(T\) whose second least significant digit in binary is 1.
|
| 135 |
+
- More generally \(S_{i}\) consists of all numbers in \(T\) whose \(i\)th least significant digit in binary is 1.
|
| 136 |
+
|
| 137 |
+
WLOG Alice answers these all as "yes" (the other cases are similar). Among the last \(k + 1\) answers at least one must be truthful, and the number \(2^{k}\) (having zeros in all relevant digits) does not appear in any of \(S_{0}\) , ..., \(S_{k}\) and is ruled out. \(\square\)
|
| 138 |
+
|
| 139 |
+
|
| 140 |
+
|
| 141 |
+
Thus in this way Bob can repeatedly find non- possibilities for \(x\) (and then relabel the remaining candidates \(1, \ldots , N - 1\) ) until he arrives at a set of at most \(2^{k}\) numbers.
|
| 142 |
+
|
| 143 |
+
Part (b): It suffices to consider \(n = \lceil 1.99^{k}\rceil\) and \(N = n + 1\) for large \(k\) . At the \(t\) th step, Bob asks some question \(S_{t}\) ; we phrase each of Alice's answers in the form " \(x \notin B_{t}\) ", where \(B_{t}\) is either \(S_{t}\) or its complement. (You may think of these as "bad sets"; the idea is to show we can avoid having any number appear in \(k + 1\) consecutive bad sets, preventing Bob from ruling out any numbers.)
|
| 144 |
+
|
| 145 |
+
Main idea: for every number \(1 \leq x \leq N\) , at time step \(t\) we define its weight to be
|
| 146 |
+
|
| 147 |
+
\[w(x) = 1.998^{e}\]
|
| 148 |
+
|
| 149 |
+
where \(e\) is the largest number such that \(x \in B_{t - 1} \cap B_{t - 2} \cap \dots \cap B_{t - e}\) .
|
| 150 |
+
|
| 151 |
+
Claim ��� Alice can ensure the total weight never exceeds \(1.998^{k + 1}\) for large \(k\) .
|
| 152 |
+
|
| 153 |
+
Proof. Let \(W_{t}\) denote the sum of weights after the \(t\) th question. We have \(W_{0} = N < 1000n\) . We will prove inductively that \(W_{t} < 1000n\) always.
|
| 154 |
+
|
| 155 |
+
At time \(t\) , Bob specifies a question \(S_{t}\) . We have Alice choose \(B_{t}\) as whichever of \(S_{t}\) or \(\overline{S_{t}}\) has lesser total weight, hence at most \(W_{t} / 2\) . The weights of for \(B_{t}\) increase by a factor of 1.998, while the weights for \(\overline{B_{t}}\) all reset to 1. So the new total weight after time \(t\) is
|
| 156 |
+
|
| 157 |
+
\[W_{t + 1} \leq 1.998 \cdot \frac{W_{t}}{2} + \# \overline{B_{t}} \leq 0.999W_{t} + n.\]
|
| 158 |
+
|
| 159 |
+
Thus if \(W_{t} < 1000n\) then \(W_{t + 1} < 1000n\) .
|
| 160 |
+
|
| 161 |
+
To finish, note that \(1000n < 1000 \left(1.99^{k} + 1\right) < 1.998^{k + 1}\) for \(k\) large.
|
| 162 |
+
|
| 163 |
+
In particular, no individual number can have weight \(1.998^{k + 1}\) . Thus for every time step \(t\) we have
|
| 164 |
+
|
| 165 |
+
\[B_{t} \cap B_{t + 1} \cap \dots \cap B_{t + k} = \emptyset .\]
|
| 166 |
+
|
| 167 |
+
Then once Bob stops, if he declares a set of \(n\) positive integers, and \(x\) is an integer Bob did not choose, then Alice's question history is consistent with \(x\) being Alice's number, as among any \(k + 1\) consecutive answers she claimed that \(x \in \overline{B_{t}}\) for some \(t\) in that range.
|
| 168 |
+
|
| 169 |
+
Remark (Motivation). In our \(B_{t}\) setup, let's think backwards. The problem is equivalent to avoiding \(e = k + 1\) at any time step \(t\) , for any number \(x\) . That means
|
| 170 |
+
|
| 171 |
+
- have at most two elements with \(e = k\) at time \(t - 1\) ,- thus have at most four elements with \(e = k - 1\) at time \(t - 2\) ,- thus have at most eight elements with \(e = k - 2\) at time \(t - 3\) ,- and so on.
|
| 172 |
+
|
| 173 |
+
We already exploited this in solving part (a). In any case it's now natural to try letting \(w(x) = 2^{e}\) , so that all the cases above sum to "equally bad" situations: since \(8 \cdot 2^{k - 2} = 4 \cdot 2^{k - 1} = 2 \cdot 2^{k}\) , say.
|
| 174 |
+
|
| 175 |
+
However, we then get \(W_{t + 1} \leq \frac{1}{2} (2W_{t}) + n\) , which can increase without bound due to contributions from numbers resetting to zero. The way to fix this is to change the weight to \(w(x) = 1.998^{e}\) , taking advantage of the little extra space we have due to having \(n \geq 1.99^{k}\) rather than \(n \geq 2^{k}\) .
|
| 176 |
+
|
| 177 |
+
|
| 178 |
+
|
| 179 |
+
## \(\S 2\) Solutions to Day 2
|
| 180 |
+
|
| 181 |
+
## \(\S 2.1\) IMO 2012/4, proposed by Liam Baker (SAF)
|
| 182 |
+
|
| 183 |
+
Available online at https://aops.com/community/p2737336.
|
| 184 |
+
|
| 185 |
+
## Problem statement
|
| 186 |
+
|
| 187 |
+
Find all functions \(f:\mathbb{Z}\to \mathbb{Z}\) such that, for all integers \(a,b,c\) that satisfy \(a + b + c = 0\) the following equality holds:
|
| 188 |
+
|
| 189 |
+
\[f(a)^{2} + f(b)^{2} + f(c)^{2} = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\]
|
| 190 |
+
|
| 191 |
+
Answer: for arbitrary \(k\in \mathbb{Z}\) , we have
|
| 192 |
+
|
| 193 |
+
(i) \(f(x) = k x^{2}\)
|
| 194 |
+
|
| 195 |
+
(ii) \(f(x) = 0\) for even \(x\) , and \(f(x) = k\) for odd \(x\) , and
|
| 196 |
+
|
| 197 |
+
(iii) \(f(x) = 0\) for \(x\equiv 0\) (mod 4), \(f(x) = k\) for odd \(x\) , and \(f(x) = 4k\) for \(x\equiv 2\) (mod 4).
|
| 198 |
+
|
| 199 |
+
These can be painfully seen to work. (It's more natural to think of these as \(f(x) = x^{2}\) , \(f(x) = x^{2}\) (mod 4), \(f(x) = x^{2}\) (mod 8), and multiples thereof.)
|
| 200 |
+
|
| 201 |
+
Set \(a = b = c = 0\) to get \(f(0) = 0\) . Then set \(c = 0\) to get \(f(a) = f(- a)\) , so \(f\) is even. Now
|
| 202 |
+
|
| 203 |
+
\[f(a)^{2} + f(b)^{2} + f(a + b)^{2} = 2f(a + b)\left(f(a) + f(b)\right) + 2f(a)f(b)\]
|
| 204 |
+
|
| 205 |
+
or
|
| 206 |
+
|
| 207 |
+
\[(f(a + b) - (f(a) + f(b)))^{2} = 4f(a)f(b).\]
|
| 208 |
+
|
| 209 |
+
Hence \(f(a)f(b)\) is a perfect square for all \(a,b\in \mathbb{Z}\) . So there exists a \(\lambda\) such that \(f(n) = \lambda g(n)^{2}\) , where \(g(n)\geq 0\) . From here we recover
|
| 210 |
+
|
| 211 |
+
\[\boxed{g(a + b) = \pm g(a)\pm g(b).}\]
|
| 212 |
+
|
| 213 |
+
Also \(g(0) = 0\)
|
| 214 |
+
|
| 215 |
+
Let \(k = g(1)\neq 0\) . We now split into cases on \(g(2)\)
|
| 216 |
+
|
| 217 |
+
- \(g(2) = 0\) . Put \(b = 2\) in original to get \(g(a + 2) = \pm g(a) = +g(a)\)
|
| 218 |
+
|
| 219 |
+
- \(g(2) = 2k\) . Cases on \(g(4)\) :
|
| 220 |
+
|
| 221 |
+
- \(g(4) = 0\) , then we get \((g(n))_{n\geq 0} = (0,1,2,1,0,1,2,1,\ldots)\) . This works.
|
| 222 |
+
|
| 223 |
+
- \(g(4) = 4k\) . This only happens when \(g(1) = k\) , \(g(2) = 2k\) , \(g(3) = 3k\) , \(g(4) = 4k\) . Then
|
| 224 |
+
|
| 225 |
+
\[*g(5) = \pm 3k\pm 2k = \pm 4k\pm k.\]
|
| 226 |
+
|
| 227 |
+
\[*g(6) = \pm 4k\pm 2k = \pm 5k\pm k.\]
|
| 228 |
+
|
| 229 |
+
\\*...
|
| 230 |
+
|
| 231 |
+
and so by induction \(g(n) = nk\)
|
| 232 |
+
|
| 233 |
+
|
| 234 |
+
|
| 235 |
+
## \(\S 2.2\) IMO 2012/5, proposed by Josef Tkadlec (CZE)
|
| 236 |
+
|
| 237 |
+
Available online at https://aops.com/community/p2737425.
|
| 238 |
+
|
| 239 |
+
## Problem statement
|
| 240 |
+
|
| 241 |
+
Let \(A B C\) be a triangle with \(\angle B C A = 90^{\circ}\) , and let \(D\) be the foot of the altitude from \(C\) . Let \(X\) be a point in the interior of the segment \(C D\) . Let \(K\) be the point on the segment \(A X\) such that \(B K = B C\) . Similarly, let \(L\) be the point on the segment \(B X\) such that \(A L = A C\) . Let \(M = \overline{A L} \cap \overline{B K}\) . Prove that \(M K = M L\) .
|
| 242 |
+
|
| 243 |
+
Let \(\omega_{A}\) and \(\omega_{B}\) be the circles through \(C\) centered at \(A\) and \(B\) ; extend rays \(A K\) and \(B L\) to hit \(\omega_{B}\) and \(\omega_{A}\) again at \(K^{*}\) , \(L^{*}\) . By radical center \(X\) , we have \(K L K^{*}L^{*}\) is cyclic, say with circumcircle \(\omega\) .
|
| 244 |
+
|
| 245 |
+
|
| 246 |
+
|
| 247 |
+
|
| 248 |
+
By orthogonality of \((A)\) and \((B)\) we find that \(\overline{A L}\) , \(\overline{A L^{*}}\) , \(\overline{B K}\) , \(\overline{B K^{*}}\) are tangents to \(\omega\) (in particular, \(K L K^{*}L^{*}\) is harmonic). In particular \(\overline{M K}\) and \(\overline{M L}\) are tangents to \(\omega\) , so \(M K = M L\) .
|
| 249 |
+
|
| 250 |
+
|
| 251 |
+
|
| 252 |
+
## \(\S 2.3\) IMO 2012/6, proposed by Dusan Djukic (SRB)
|
| 253 |
+
|
| 254 |
+
Available online at https://aops.com/community/p2737435.
|
| 255 |
+
|
| 256 |
+
## Problem statement
|
| 257 |
+
|
| 258 |
+
Find all positive integers \(n\) for which there exist non- negative integers \(a_{1},a_{2},\ldots ,a_{n}\) such that
|
| 259 |
+
|
| 260 |
+
\[\frac{1}{2^{a_{1}}} +\frac{1}{2^{a_{2}}} +\dots +\frac{1}{2^{a_{n}}} = \frac{1}{3^{a_{1}}} +\frac{2}{3^{a_{2}}} +\dots +\frac{n}{3^{a_{n}}} = 1.\]
|
| 261 |
+
|
| 262 |
+
The answer is \(n\equiv 1,2\) (mod 4). To see these are necessary, note that taking the latter equation modulo 2 gives
|
| 263 |
+
|
| 264 |
+
\[1 = \frac{1}{3^{a_{1}}} +\frac{2}{3^{a_{2}}} +\dots +\frac{n}{3^{a_{n}}}\equiv 1 + 2 + \ldots +n\pmod {2}.\]
|
| 265 |
+
|
| 266 |
+
Now we prove these are sufficient. The following nice construction was posted on AOPs by the user cfheolipixin.
|
| 267 |
+
|
| 268 |
+
Claim — If \(n = 2k - 1\) works then so does \(n = 2k\)
|
| 269 |
+
|
| 270 |
+
Proof. Replace
|
| 271 |
+
|
| 272 |
+
\[\frac{k}{3^{r}} = \frac{k}{3^{r + 1}} +\frac{2k}{3^{r + 1}}. \quad (*)\]
|
| 273 |
+
|
| 274 |
+
Claim — If \(n = 4k + 2\) works then so does \(n = 4k + 13\)
|
| 275 |
+
|
| 276 |
+
Proof. First use the identity
|
| 277 |
+
|
| 278 |
+
\[\frac{k + 2}{3^{r}} = \frac{k + 2}{3^{r + 2}} +\frac{4k + 3}{3^{r + 3}} +\frac{4k + 5}{3^{r + 3}} +\frac{4k + 7}{3^{r + 3}} +\frac{4k + 9}{3^{r + 3}} +\frac{4k + 11}{3^{r + 3}} +\frac{4k + 13}{3^{r + 3}}\]
|
| 279 |
+
|
| 280 |
+
to fill in the odd numbers. The even numbers can then be instantiated with \((\ast)\) too.
|
| 281 |
+
|
| 282 |
+
Thus it suffices to construct base cases for \(n = 1\) , \(n = 5\) , \(n = 9\) . They are
|
| 283 |
+
|
| 284 |
+
\[1 = \frac{1}{3^{0}}\] \[\quad = \frac{1}{3^{2}} +\frac{2}{3^{2}} +\frac{3}{3^{2}} +\frac{4}{3^{3}} +\frac{5}{3^{3}}\] \[\quad = \frac{1}{3^{2}} +\frac{2}{3^{3}} +\frac{3}{3^{3}} +\frac{4}{3^{3}} +\frac{5}{3^{3}} +\frac{6}{3^{4}} +\frac{7}{3^{4}} +\frac{8}{3^{4}} +\frac{9}{3^{4}}.\]
|
| 285 |
+
|
| 286 |
+
|
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|
| 1 |
+
|
| 2 |
+
# IMO 2016 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2016 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2016/1, proposed by Art Waterschoot (BEL) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2016/2, proposed by Trevor Tao (AUS) 5
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2016/3, proposed by Aleksandr Gaifullin (RUS) 7
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 8
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2016/4, proposed by Gerhard Woeginger (LUX) 8
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2016/5, proposed by Russia 9
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2016/6, proposed by Josef Tkadlec (CZE) 11
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. In convex pentagon \(A B C D E\) with \(\angle B > 90^{\circ}\) , let \(F\) be a point on \(\overline{{A C}}\) such that \(\angle F B C = 90^{\circ}\) . It is given that \(F A = F B\) \(D A = D C\) \(E A = E D\) , and rays \(\overline{{A C}}\) and \(\overline{{A D}}\) trisect \(\angle B A E\) . Let \(M\) be the midpoint of \(\overline{{C F}}\) . Let \(X\) be the point such that \(A M X E\) is a parallelogram. Show that \(\overline{{F X}}\) \(\overline{{E M}}\) \(\overline{{B D}}\) are concurrent.
|
| 39 |
+
|
| 40 |
+
2. Find all integers \(n\) for which each cell of \(n\times n\) table can be filled with one of the letters \(I\) \(M\) and \(o\) in such a way that:
|
| 41 |
+
|
| 42 |
+
In each row and column, one third of the entries are \(I\) , one third are \(M\) and one third are \(o\) ; and in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are \(I\) , one third are \(M\) and one third are \(o\) .
|
| 43 |
+
|
| 44 |
+
Note that an \(n\times n\) table has \(4n - 2\) diagonals.
|
| 45 |
+
|
| 46 |
+
3. Let \(P = A_{1}A_{2}\ldots A_{k}\) be a convex polygon in the plane. The vertices \(A_{1}\) \(A_{2}\) ,..., \(A_{k}\) have integral coordinates and lie on a circle. Let \(S\) be the area of \(P\) . An odd positive integer \(n\) is given such that the squares of the side lengths of \(P\) are integers divisible by \(n\) . Prove that \(2S\) is an integer divisible by \(n\) .
|
| 47 |
+
|
| 48 |
+
4. A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let \(P(n) = n^{2} + n + 1\) . What is the smallest possible positive integer value of \(b\) such that there exists a non-negative integer \(a\) for which the set
|
| 49 |
+
|
| 50 |
+
\[\{P(a + 1),P(a + 2),\ldots ,P(a + b)\}\]
|
| 51 |
+
|
| 52 |
+
is fragrant?
|
| 53 |
+
|
| 54 |
+
5. The equation
|
| 55 |
+
|
| 56 |
+
\[(x - 1)(x - 2)\ldots (x - 2016) = (x - 1)(x - 2)\ldots (x - 2016)\]
|
| 57 |
+
|
| 58 |
+
is written on the board, with 2016 linear factors on each side. What is the least possible value of \(k\) for which it is possible to erase exactly \(k\) of these 4032 linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
|
| 59 |
+
|
| 60 |
+
6. There are \(n\geq 2\) line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands \(n - 1\) times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.
|
| 61 |
+
|
| 62 |
+
(a) Prove that Geoff can always fulfill his wish if \(n\) is odd.
|
| 63 |
+
|
| 64 |
+
(b) Prove that Geoff can never fulfill his wish if \(n\) is even.
|
| 65 |
+
|
| 66 |
+
|
| 67 |
+
|
| 68 |
+
## \(\S 1\) Solutions to Day 1
|
| 69 |
+
|
| 70 |
+
## \(\S 1.1\) IMO 2016/1, proposed by Art Waeterschoot (BEL)
|
| 71 |
+
|
| 72 |
+
Available online at https://aops.com/community/p6637656.
|
| 73 |
+
|
| 74 |
+
## Problem statement
|
| 75 |
+
|
| 76 |
+
In convex pentagon \(ABCDE\) with \(\angle B > 90^{\circ}\) , let \(F\) be a point on \(\overline{AC}\) such that \(\angle FBC = 90^{\circ}\) . It is given that \(FA = FB\) , \(DA = DC\) , \(EA = ED\) , and rays \(\overline{AC}\) and \(\overline{AD}\) trisect \(\angle BAE\) . Let \(M\) be the midpoint of \(\overline{CF}\) . Let \(X\) be the point such that \(AMXE\) is a parallelogram. Show that \(\overline{FX}\) , \(\overline{EM}\) , \(\overline{BD}\) are concurrent.
|
| 77 |
+
|
| 78 |
+
Here is a "long" solution which I think shows where the "power" in the configuration comes from (it should be possible to come up with shorter solutions by cutting more directly to the desired conclusion). Throughout the proof, we let
|
| 79 |
+
|
| 80 |
+
\[\theta = \angle FAB = \angle FBA = \angle DAC = \angle DCA = \angle EAD = \angle EDA.\]
|
| 81 |
+
|
| 82 |
+
We begin by focusing just on \(ABCD\) with point \(F\) , ignoring for now the points \(E\) and \(X\) (and to some extent even point \(M\) ). It turns out this is a very familiar configuration.
|
| 83 |
+
|
| 84 |
+
## Lemma (Central lemma)
|
| 85 |
+
|
| 86 |
+
The points \(F\) and \(C\) are the incenter and \(A\) - excenter of \(\triangle DAB\) . Moreover, \(\triangle DAB\) is isosceles with \(DA = DB\) .
|
| 87 |
+
|
| 88 |
+
Proof. The proof uses three observations:
|
| 89 |
+
|
| 90 |
+
- We already know that \(\overline{FAC}\) is the angle bisector of \(\angle ABD\) .
|
| 91 |
+
|
| 92 |
+
- We were given \(\angle FBC = 90^{\circ}\) .
|
| 93 |
+
|
| 94 |
+
- Next, note that \(\triangle AFB \sim \triangle ADC\) (they are similar isosceles triangles). From this it follows that \(AF \cdot AC = AB \cdot AD\) .
|
| 95 |
+
|
| 96 |
+
These three facts, together with \(F\) lying inside \(\triangle ABD\) , are enough to imply the result. \(\square\)
|
| 97 |
+
|
| 98 |
+
|
| 99 |
+
|
| 100 |
+
|
| 101 |
+
|
| 102 |
+
|
| 103 |
+
## Corollary
|
| 104 |
+
|
| 105 |
+
The point \(M\) is the midpoint of arc \(\overline{BD}\) of \((DAB)\) , and the center of cyclic quadrilateral \(FDCB\) .
|
| 106 |
+
|
| 107 |
+
Proof. Fact 5.
|
| 108 |
+
|
| 109 |
+
Using these observations as the anchor for everything that follows, we now prove several claims about \(X\) and \(E\) in succession.
|
| 110 |
+
|
| 111 |
+
|
| 112 |
+
|
| 113 |
+
|
| 114 |
+
Claim — Point \(E\) is the midpoint of arc \(\overline{AD}\) in \((ABMD)\) , and hence lies on ray \(BF\) .
|
| 115 |
+
|
| 116 |
+
Proof. This follows from \(\angle EDA = \theta = \angle EBA\) .
|
| 117 |
+
|
| 118 |
+
Claim — Points \(X\) is the second intersection of ray \(\overline{ED}\) with \((BFDC)\) .
|
| 119 |
+
|
| 120 |
+
Proof. First, \(\overline{ED} \parallel \overline{AC}\) already since \(\angle AED = 180^\circ - 2\theta\) and \(\angle CAE = 2\theta\) . Now since \(DB = DA\) , we get \(MB = MD = ED = EA\) . Thus, \(MX = AE = MB\) , so \(X\) also lies on the circle \((BFDC)\) centered at \(M\) .
|
| 121 |
+
|
| 122 |
+
Claim — The quadrilateral \(EXMF\) is an isosceles trapezoid.
|
| 123 |
+
|
| 124 |
+
Proof. We already know \(\overline{EX} \parallel \overline{FM}\) . Since \(\angle EFA = 180^\circ - \angle AFB = 2\theta = \angle FAE\) , we have \(EF = EA\) as well (and \(F \neq A\) ). As \(EXMA\) was a parallelogram, it follows \(EXMF\) is an isosceles trapezoid.
|
| 125 |
+
|
| 126 |
+
The problem then follows by radical axis theorem on the three circles \((AEDMB)\) , \((BFDXC)\) and \((EXMF)\) .
|
| 127 |
+
|
| 128 |
+
|
| 129 |
+
|
| 130 |
+
## \(\S 1.2\) IMO 2016/2, proposed by Trevor Tao (AUS)
|
| 131 |
+
|
| 132 |
+
Available online at https://aops.com/community/p6637677.
|
| 133 |
+
|
| 134 |
+
## Problem statement
|
| 135 |
+
|
| 136 |
+
Find all integers \(n\) for which each cell of \(n\times n\) table can be filled with one of the letters \(I\) \(M\) and \(o\) in such a way that:
|
| 137 |
+
|
| 138 |
+
In each row and column, one third of the entries are \(I\) , one third are \(M\) and one third are \(o\) ; and
|
| 139 |
+
|
| 140 |
+
in any diagonal, if the number of entries on the diagonal is a multiple of three, then one third of the entries are \(I\) , one third are \(M\) and one third are \(o\) .
|
| 141 |
+
|
| 142 |
+
Note that an \(n\times n\) table has \(4n - 2\) diagonals.
|
| 143 |
+
|
| 144 |
+
The answer is \(n\) divisible by 9.
|
| 145 |
+
|
| 146 |
+
First we construct \(n = 9\) and by extension every multiple of 9.
|
| 147 |
+
|
| 148 |
+
<table><tr><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td></tr><tr><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td></tr><tr><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td></tr><tr><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td></tr><tr><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td></tr><tr><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td></tr><tr><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td></tr><tr><td>M</td><td>M</td><td>M</td><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td></tr><tr><td>O</td><td>O</td><td>O</td><td>I</td><td>I</td><td>I</td><td>M</td><td>M</td><td>M</td></tr></table>
|
| 149 |
+
|
| 150 |
+
We now prove \(9\mid n\) is necessary.
|
| 151 |
+
|
| 152 |
+
Let \(n = 3k\) , which divides the given grid into \(k^{2}\) sub- boxes (of size \(3\times 3\) each). We say a multiset of squares \(S\) is clean if the letters distribute equally among them; note that unions of clean multisets are clean.
|
| 153 |
+
|
| 154 |
+
Consider the following clean sets (given to us by problem statement):
|
| 155 |
+
|
| 156 |
+
All columns indexed 2 (mod 3), All rows indexed 2 (mod 3), and All \(4k - 2\) diagonals mentioned in the problem.
|
| 157 |
+
|
| 158 |
+
Take their union. This covers the center of each box four times, and every other cell exactly once. We conclude the set of \(k^{2}\) center squares are clean, hence \(3\mid k^{2}\) and so \(9\mid n\) , as desired.
|
| 159 |
+
|
| 160 |
+
|
| 161 |
+
|
| 162 |
+
## \(\S 1.3\) IMO 2016/3, proposed by Aleksandr Gaifullin (RUS)
|
| 163 |
+
|
| 164 |
+
Available online at https://aops.com/community/p6637660.
|
| 165 |
+
|
| 166 |
+
## Problem statement
|
| 167 |
+
|
| 168 |
+
Let \(P = A_{1}A_{2}\ldots A_{k}\) be a convex polygon in the plane. The vertices \(A_{1}\) , \(A_{2}\) , ..., \(A_{k}\) have integral coordinates and lie on a circle. Let \(S\) be the area of \(P\) . An odd positive integer \(n\) is given such that the squares of the side lengths of \(P\) are integers divisible by \(n\) . Prove that \(2S\) is an integer divisible by \(n\) .
|
| 169 |
+
|
| 170 |
+
Solution by Jeck Lim: We will prove the result just for \(n = p^{e}\) where \(p\) is an odd prime and \(e \geq 1\) . The case \(k = 3\) is resolved by Heron's formula directly: we have \(S = \frac{1}{4} \sqrt{2(a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2}) - a^{4} - b^{4} - c^{4}}\) , so if \(p^{e} \mid \gcd (a^{2}, b^{2}, c^{2})\) then \(p^{2e} \mid S^{2}\) .
|
| 171 |
+
|
| 172 |
+
Now we show we can pick a diagonal and induct down on \(k\) by using inversion.
|
| 173 |
+
|
| 174 |
+
Let the polygon be \(A_{1}A_{2}\ldots A_{k + 1}\) and suppose for contradiction that all sides are divisible by \(p^{e}\) but no diagonals are. Let \(O = A_{k + 1}\) for notational convenience. By applying inversion around \(O\) with radius 1, we get the "generalized Ptolemy theorem"
|
| 175 |
+
|
| 176 |
+
\[\frac{A_{1}A_{2}}{OA_{1}\cdot OA_{2}} +\frac{A_{2}A_{3}}{OA_{2}\cdot OA_{3}} +\dots +\frac{A_{k - 1}A_{k}}{OA_{k - 1}\cdot OA_{k}} = \frac{A_{1}A_{k}}{OA_{1}\cdot OA_{k}}\]
|
| 177 |
+
|
| 178 |
+
or, making use of square roots,
|
| 179 |
+
|
| 180 |
+
\[\sqrt{\frac{A_{1}A_{2}^{2}}{OA_{1}^{2}\cdot OA_{2}^{2}}} +\sqrt{\frac{A_{2}A_{3}^{2}}{OA_{2}^{2}\cdot OA_{3}^{2}}} +\dots +\sqrt{\frac{A_{k - 1}A_{k}^{2}}{OA_{k - 1}^{2}\cdot OA_{k}^{2}}} = \sqrt{\frac{A_{1}A_{k}^{2}}{OA_{1}^{2}\cdot OA_{k}^{2}}}\]
|
| 181 |
+
|
| 182 |
+
Suppose \(\nu_{p}\) of all diagonals is strictly less than \(e\) . Then the relation becomes
|
| 183 |
+
|
| 184 |
+
\[\sqrt{q_{1}} +\dots +\sqrt{q_{k - 1}} = \sqrt{q}\]
|
| 185 |
+
|
| 186 |
+
where \(q_{i}\) are positive rational numbers. Since there are no nontrivial relations between square roots (see this link) there is a positive rational number \(b\) such that \(r_{i} = \sqrt{q_{i} / b}\) and \(r = \sqrt{q / b}\) are all rational numbers. Then
|
| 187 |
+
|
| 188 |
+
\[\sum r_{i} = r.\]
|
| 189 |
+
|
| 190 |
+
However, the condition implies that \(\nu_{p}(q_{i}^{2}) > \nu_{p}(q^{2})\) for all \(i\) (check this for \(i = 1\) , \(i = k - 1\) and \(2 \leq i \leq k - 2\) ), and hence \(\nu_{p}(r_{i}) > \nu_{p}(r)\) . This is absurd.
|
| 191 |
+
|
| 192 |
+
Remark. I think you basically have to use some Ptolemy- like geometric property, and also all correct solutions I know of for \(n = p^{e}\) depend on finding a diagonal and inducting down. (Actually, the case \(k = 4\) is pretty motivating; Ptolemy implies one can cut in two.)
|
| 193 |
+
|
| 194 |
+
|
| 195 |
+
|
| 196 |
+
## \(\S 2\) Solutions to Day 2
|
| 197 |
+
|
| 198 |
+
## \(\S 2.1\) IMO 2016/4, proposed by Gerhard Woeginger (LUX)
|
| 199 |
+
|
| 200 |
+
Available online at https://aops.com/community/p6642559.
|
| 201 |
+
|
| 202 |
+
## Problem statement
|
| 203 |
+
|
| 204 |
+
A set of positive integers is called fragrant if it contains at least two elements and each of its elements has a prime factor in common with at least one of the other elements. Let \(P(n) = n^{2} + n + 1\) . What is the smallest possible positive integer value of \(b\) such that there exists a non- negative integer \(a\) for which the set
|
| 205 |
+
|
| 206 |
+
\[\{P(a + 1),P(a + 2),\ldots ,P(a + b)\}\]
|
| 207 |
+
|
| 208 |
+
is fragrant?
|
| 209 |
+
|
| 210 |
+
The answer is \(b = 6\)
|
| 211 |
+
|
| 212 |
+
First, we prove \(b\geq 6\) must hold. It is not hard to prove the following divisibilities by Euclid:
|
| 213 |
+
|
| 214 |
+
\[\gcd (P(n),P(n + 1))\mid 1\] \[\gcd (P(n),P(n + 2))\mid 7\] \[\gcd (P(n),P(n + 3))\mid 3\] \[\gcd (P(n),P(n + 4))\mid 19.\]
|
| 215 |
+
|
| 216 |
+
Now assume for contradiction \(b\leq 5\) . Then any GCD's among \(P(a + 1)\) , ..., \(P(a + b)\) must be among \(\{3,7,19\}\) . Consider a multi- graph on \(\{a + 1,\ldots ,a + b\}\) where we join two elements with nontrivial GCD and label the edge with the corresponding prime. Then we readily see there is at most one edge each of \(\{3,7,19\}\) : id est at most one edge of gap 2, 3, 4 (and no edges of gap 1). (By the gap of an edge \(e = \{u,v\}\) we mean \(|u - v|\) .) But one can see that it's now impossible for every vertex to have nonzero degree, contradiction.
|
| 217 |
+
|
| 218 |
+
To construct \(b = 6\) we use the Chinese remainder theorem: select \(a\) with
|
| 219 |
+
|
| 220 |
+
\[a + 1\equiv 7\pmod{19\] \[a + 5\equiv 11\pmod{19\] \[a + 2\equiv 2\pmod{7\] \[a + 4\equiv 4\pmod{7\] \[a + 3\equiv 1\pmod{3\] \[a + 6\equiv 1\pmod{3\]
|
| 221 |
+
|
| 222 |
+
|
| 223 |
+
|
| 224 |
+
## \(\S 2.2\) IMO 2016/5, proposed by Russia
|
| 225 |
+
|
| 226 |
+
Available online at https://aops.com/community/p6642565.
|
| 227 |
+
|
| 228 |
+
## Problem statement
|
| 229 |
+
|
| 230 |
+
The equation
|
| 231 |
+
|
| 232 |
+
\[(x - 1)(x - 2)\ldots (x - 2016) = (x - 1)(x - 2)\ldots (x - 2016)\]
|
| 233 |
+
|
| 234 |
+
is written on the board, with 2016 linear factors on each side. What is the least possible value of \(k\) for which it is possible to erase exactly \(k\) of these 4032 linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
|
| 235 |
+
|
| 236 |
+
The answer is 2016. Obviously this is necessary in order to delete duplicated factors. We now prove it suffices to deleted 2 (mod 4) and 3 (mod 4) guys from the left- hand side, and 0 (mod 4), 1 (mod 4) from the right- hand side.
|
| 237 |
+
|
| 238 |
+
Consider the 1008 inequalities
|
| 239 |
+
|
| 240 |
+
\[(x - 1)(x - 4)< (x - 2)(x - 3)\] \[(x - 5)(x - 8)< (x - 6)(x - 7)\] \[(x - 9)(x - 12)< (x - 10)(x - 11)\] \[\vdots\] \[(x - 2013)(x - 2016)< (x - 2014)(x - 2015).\]
|
| 241 |
+
|
| 242 |
+
Notice that in all these inequalities, at most one of them has non- positive numbers in it, and we never have both zero. If there is exactly one negative term among the \(1008\cdot 2 = 2016\) sides, it is on the left and we can multiply all together. Thus the only case that remains is if \(x\in (4m - 2,4m - 1)\) for some \(m\) , say the \(m\) th inequality. In that case, the two sides of that inequality differ by a factor of at least 9.
|
| 243 |
+
|
| 244 |
+
Claim — We have
|
| 245 |
+
|
| 246 |
+
\[\prod_{k\geq 0}\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)} < e.\]
|
| 247 |
+
|
| 248 |
+
Proof of claim using logarithms. It's equivalent to prove
|
| 249 |
+
|
| 250 |
+
\[\sum_{k\geq 0}\log \left(1 + \frac{2}{(4k + 1)(4k + 4)}\right)< 1.\]
|
| 251 |
+
|
| 252 |
+
To this end, we use the deep fact that \(\log (1 + t)\leq t\) , and thus it follows from \(\begin{array}{r}{\sum_{k\geq 0}\frac{1}{(4k + 1)(4k + 4)} < \frac{1}{2}} \end{array}\) , which one can obtain for example by noticing it's less than \(\frac{1}{4}\frac{\pi^{2}}{6}\) . \(\square\)
|
| 253 |
+
|
| 254 |
+
Remark (Elementary proof of claim, given by Espen Slettnes). To avoid calculus as above,
|
| 255 |
+
|
| 256 |
+
|
| 257 |
+
|
| 258 |
+
for each \(N \geq 0\) , note the partial product is bounded by
|
| 259 |
+
|
| 260 |
+
\[\prod_{k = 0}^{N}\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)} = \frac{2}{1}\cdot \left(\frac{3}{4}\cdot \frac{6}{5}\right)\cdot \left(\frac{7}{8}\cdot \frac{10}{9}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot\] \[\qquad < 2\cdot 1\cdot 1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \frac{4N + 3}{4N + 4} < 2< e.\]
|
| 261 |
+
|
| 262 |
+
This solves the problem, because then the factors being multiplied on by the positive inequalities before the \(m\) th one are both less than \(e\) , and \(e^{2} < 9\) . In symbols, for \(4m - 2 < x < 4m - 1\) we should have
|
| 263 |
+
|
| 264 |
+
\[\frac{(x - (4m - 6))(x - (4m - 5))}{(x - (4m - 7))(x - (4m - 4))} \times \dots \times \frac{(x - 2)(x - 3)}{(x - 1)(x - 4)} < e\]
|
| 265 |
+
|
| 266 |
+
and
|
| 267 |
+
|
| 268 |
+
\[\frac{(x - (4m + 2))(x - (4m + 3))}{(x - (4m + 1))(x - (4m + 4))} \times \dots \times \frac{(x - 2014)(x - 2015)}{(x - 2013)(x - 2016)} < e\]
|
| 269 |
+
|
| 270 |
+
because the \((k + 1)\) st term of each left- hand side is at most \(\frac{(4k + 2)(4k + 3)}{(4k + 1)(4k + 4)}\) , for \(k \geq 0\) . As \(e^{2} < 9\) , we're okay.
|
| 271 |
+
|
| 272 |
+
|
| 273 |
+
|
| 274 |
+
## \(\S 2.3\) IMO 2016/6, proposed by Josef Tkadlec (CZE)
|
| 275 |
+
|
| 276 |
+
Available online at https://aops.com/community/p6642576.
|
| 277 |
+
|
| 278 |
+
## Problem statement
|
| 279 |
+
|
| 280 |
+
There are \(n\geq 2\) line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands \(n - 1\) times. Every time he claps, each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time.
|
| 281 |
+
|
| 282 |
+
(a) Prove that Geoff can always fulfill his wish if \(n\) is odd.
|
| 283 |
+
|
| 284 |
+
(b) Prove that Geoff can never fulfill his wish if \(n\) is even.
|
| 285 |
+
|
| 286 |
+
The following solution was communicated to me by Yang Liu.
|
| 287 |
+
|
| 288 |
+
Imagine taking a larger circle \(\omega\) encasing all \(\binom{n}{2}\) intersection points. Denote by \(P_{1}\) , \(P_{2}\) , ..., \(P_{2n}\) the order of the points on \(\omega\) in clockwise order; we imagine placing the frogs on \(P_{i}\) instead. Observe that, in order for every pair of segments to meet, each line segment must be of the form \(P_{i}P_{i + n}\) .
|
| 289 |
+
|
| 290 |
+
|
| 291 |
+
|
| 292 |
+
|
| 293 |
+
Then:
|
| 294 |
+
|
| 295 |
+
(a) Place the frogs on \(P_{1}\) , \(P_{3}\) , ..., \(P_{2n - 1}\) . A simple parity arguments shows this works.
|
| 296 |
+
|
| 297 |
+
(b) Observe that we cannot place frogs on consecutive \(P_{i}\) , so the \(n\) frogs must be placed on alternating points. But since we also are supposed to not place frogs on diametrically opposite points, for even \(n\) we immediately get a contradiction.
|
| 298 |
+
|
| 299 |
+
|
| 300 |
+
|
| 301 |
+
Remark. Yang says: this is easy to guess if you just do a few small cases and notice that the pairs of "violating points" just forms a large cycle around.
|
| 302 |
+
|
| 303 |
+
|
IMO/md/en-IMO-2017-notes.md
ADDED
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@@ -0,0 +1,383 @@
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|
| 1 |
+
|
| 2 |
+
# IMO 2017 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
18 October 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2017 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2017/1, proposed by Stephan Wagner (SAF) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2017/2, proposed by Dorlir Ahmeti (ALB) 5
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2017/3, proposed by Gerhard Woeginger (AUT) 7
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 9
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2017/4, proposed by Charles Leytem (LUX) 9
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2017/5, proposed by Grigory Chelnokov (RUS) 11
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2017/6, proposed by John Berman (USA) 12
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. For each integer \(a_{0} > 1\) , define the sequence \(a_{0}\) , \(a_{1}\) , \(a_{2}\) , ..., by
|
| 39 |
+
|
| 40 |
+
\[a_{n + 1} = \left\{ \begin{array}{ll}\sqrt{a_{n}} & \mathrm{if~}\sqrt{a_{n}} \mathrm{is~an~integer,}\\ a_{n} + 3 & \mathrm{otherwise} \end{array} \right.\]
|
| 41 |
+
|
| 42 |
+
for each \(n \geq 0\) . Determine all values of \(a_{0}\) for which there is a number \(A\) such that \(a_{n} = A\) for infinitely many values of \(n\) .
|
| 43 |
+
|
| 44 |
+
2. Solve over \(\mathbb{R}\) the functional equation
|
| 45 |
+
|
| 46 |
+
\[f(f(x)f(y)) + f(x + y) = f(xy).\]
|
| 47 |
+
|
| 48 |
+
3. A hunter and an invisible rabbit play a game in the plane. The rabbit and hunter start at points \(A_{0} = B_{0}\) . In the \(n\) th round of the game ( \(n \geq 1\) ), three things occur in order:
|
| 49 |
+
|
| 50 |
+
(i) The rabbit moves invisibly from \(A_{n - 1}\) to a point \(A_{n}\) such that \(A_{n - 1}A_{n} = 1\) .
|
| 51 |
+
|
| 52 |
+
(ii) The hunter has a tracking device (e.g. dog) which reports an approximate location \(P_{n}\) of the rabbit, such that \(P_{n}A_{n} \leq 1\) .
|
| 53 |
+
|
| 54 |
+
(iii) The hunter moves visibly from \(B_{n - 1}\) to a point \(B_{n}\) such that \(B_{n - 1}B_{n} = 1\) .
|
| 55 |
+
|
| 56 |
+
Let \(N = 10^{9}\) . Can the hunter guarantee that \(A_{N}B_{N} < 100\) ?
|
| 57 |
+
|
| 58 |
+
4. Let \(R\) and \(S\) be different points on a circle \(\Omega\) such that \(\overline{RS}\) is not a diameter. Let \(\ell\) be the tangent line to \(\Omega\) at \(R\) . Point \(T\) is such that \(S\) is the midpoint of \(\overline{RT}\) . Point \(J\) is chosen on minor arc \(RS\) of \(\Omega\) so that the circumcircle \(\Gamma\) of triangle \(JST\) intersects \(\ell\) at two distinct points. Let \(A\) be the common point of \(\Gamma\) and \(\ell\) closer to \(R\) . Line \(AJ\) meets \(\Omega\) again at \(K\) . Prove that line \(KT\) is tangent to \(\Gamma\) .
|
| 59 |
+
|
| 60 |
+
5. Fix \(N \geq 1\) . A collection of \(N(N + 1)\) soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove \(N(N - 1)\) players from this row to obtain a new row of \(2N\) players in which the following \(N\) conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, ..., no one stands between the two shortest players. Prove that this is possible.
|
| 61 |
+
|
| 62 |
+
6. An irreducible lattice point is an ordered pair of integers \((x, y)\) satisfying \(\gcd (x, y) = 1\) . Prove that if \(S\) is a finite set of irreducible lattice points then there exists a nonconstant homogeneous polynomial \(f(x, y)\) with integer coefficients such that \(f(x, y) = 1\) for each \((x, y) \in S\) .
|
| 63 |
+
|
| 64 |
+
|
| 65 |
+
|
| 66 |
+
## \(\S 1\) Solutions to Day 1
|
| 67 |
+
|
| 68 |
+
## \(\S 1.1\) IMO 2017/1, proposed by Stephan Wagner (SAF)
|
| 69 |
+
|
| 70 |
+
Available online at https://aops.com/community/p8633268.
|
| 71 |
+
|
| 72 |
+
## Problem statement
|
| 73 |
+
|
| 74 |
+
For each integer \(a_{0} > 1\) , define the sequence \(a_{0}\) , \(a_{1}\) , \(a_{2}\) , ..., by
|
| 75 |
+
|
| 76 |
+
\[a_{n + 1} = \left\{ \begin{array}{ll}\sqrt{a_{n}} & \mathrm{if} \sqrt{a_{n}} \mathrm{is~an~integer,} \\ a_{n} + 3 & \mathrm{otherwise} \end{array} \right.\]
|
| 77 |
+
|
| 78 |
+
for each \(n \geq 0\) . Determine all values of \(a_{0}\) for which there is a number \(A\) such that \(a_{n} = A\) for infinitely many values of \(n\) .
|
| 79 |
+
|
| 80 |
+
The answer is \(a_{0} \equiv 0\) (mod 3) only.
|
| 81 |
+
|
| 82 |
+
\(\P\) First solution. We first compute the minimal term of any sequence, periodic or not.
|
| 83 |
+
|
| 84 |
+
## Lemma
|
| 85 |
+
|
| 86 |
+
Let \(c\) be the smallest term in \(a_{n}\) . Then either \(c \equiv 2\) (mod 3) or \(c = 3\) .
|
| 87 |
+
|
| 88 |
+
Proof. Clearly \(c \neq 1,4\) . Assume \(c \neq 2\) (mod 3). As \(c\) is not itself a square, the next perfect square after \(c\) in the sequence is one of \((\lfloor \sqrt{c} \rfloor +1)^{2}\) , \((\lfloor \sqrt{c} \rfloor +2)^{2}\) , or \((\lfloor \sqrt{c} \rfloor +3)^{2}\) . So by minimality we require
|
| 89 |
+
|
| 90 |
+
\[c \leq \lfloor \sqrt{c} \rfloor +3 \leq \sqrt{c} +3\]
|
| 91 |
+
|
| 92 |
+
which requires \(c \leq 5\) . Since \(c \neq 1,2,4,5\) we conclude \(c = 3\) .
|
| 93 |
+
|
| 94 |
+
Now we split the problem into two cases:
|
| 95 |
+
|
| 96 |
+
- If \(a_{0} \equiv 0\) (mod 3), then all terms of the sequence are 0 (mod 3). The smallest term of the sequence is thus 3 by the lemma and we have
|
| 97 |
+
|
| 98 |
+
\[3 \rightarrow 6 \rightarrow 9 \rightarrow 3\]
|
| 99 |
+
|
| 100 |
+
so \(A = 3\) works fine.
|
| 101 |
+
|
| 102 |
+
- If \(a_{0} \neq 0\) (mod 3), then no term of the sequence is 0 (mod 3), and so in particular 3 does not appear in the sequence. So the smallest term of the sequence is 2 (mod 3) by lemma. But since no squares are 2 (mod 3), the sequence \(a_{k}\) grows without bound forever after, so no such \(A\) can exist.
|
| 103 |
+
|
| 104 |
+
Hence the answer is \(a_{0} \equiv 0\) (mod 3) only.
|
| 105 |
+
|
| 106 |
+
|
| 107 |
+
|
| 108 |
+
Second solution. We clean up the argument by proving the following lemma.
|
| 109 |
+
|
| 110 |
+
## Lemma
|
| 111 |
+
|
| 112 |
+
If \(a_{n}\) is constant modulo 3 and not 2 (mod 3), then \(a_{n}\) must eventually cycle in the form \((m, m + 3, m + 6, \ldots , m^{2})\) , with no squares inside the cycle except \(m^{2}\) .
|
| 113 |
+
|
| 114 |
+
Proof. Observe that \(a_{n}\) must eventually hit a square, say \(a_{k} = c^{2}\) ; the next term is \(a_{k + 1} = c\) . Then it is forever impossible to exceed \(c^{2}\) again, by what is essentially discrete intermediate value theorem. Indeed, suppose \(a_{\ell} > c^{2}\) and take \(\ell > k\) minimal (in particular \(a_{\ell} \neq \sqrt{a_{\ell - 1}}\) ). Thus \(a_{\ell - 1} \in \{c^{2} - 2, c^{2} - 1, c^{2}\}\) and thus for modulo 3 reasons we have \(a_{\ell - 1} = c^{2}\) . But that should imply \(a_{\ell} = c < c^{2}\) , contradiction.
|
| 115 |
+
|
| 116 |
+
We therefore conclude \(\sup \{a_{n}, a_{n + 1}, \ldots \}\) is a decreasing integer sequence in \(n\) . It must eventually stabilize, say at \(m^{2}\) . Now we can't hit a square between \(m\) and \(m^{2}\) , and so we are done. \(\square\)
|
| 117 |
+
|
| 118 |
+
Now, we contend that all \(a_{0} \equiv 0\) (mod 3) work. Indeed, for such \(a_{0}\) we have \(a_{n} \equiv 0\) (mod 3) for all \(n\) , so the lemma implies that the problem statement is valid.
|
| 119 |
+
|
| 120 |
+
Next, we observe that if \(a_{i} \equiv 2\) (mod 3), then the sequence grows without bound afterwards since no squares are 2 (mod 3). In particular, if \(a_{0} \equiv 2\) (mod 3) the answer is no.
|
| 121 |
+
|
| 122 |
+
Finally, we claim that if \(a_{0} \equiv 1\) (mod 3), then eventually some term is 2 (mod 3). Assume for contradiction this is not so; then \(a_{n} \equiv 1\) (mod 3) must hold forever, and the lemma applies to give us a cycle of the form \((m, m + 3, \ldots , m^{2})\) where \(m \equiv 1\) (mod 3). In particular \(m \geq 4\) and
|
| 123 |
+
|
| 124 |
+
\[m \leq (m - 2)^{2} < m^{2}\]
|
| 125 |
+
|
| 126 |
+
but \((m - 2)^{2} \equiv 1\) (mod 3) which is a contradiction.
|
| 127 |
+
|
| 128 |
+
|
| 129 |
+
|
| 130 |
+
## \(\S 1.2\) IMO 2017/2, proposed by Dorlir Ahmeti (ALB)
|
| 131 |
+
|
| 132 |
+
Available online at https://aops.com/community/p8633190.
|
| 133 |
+
|
| 134 |
+
## Problem statement
|
| 135 |
+
|
| 136 |
+
Solve over \(\mathbb{R}\) the functional equation
|
| 137 |
+
|
| 138 |
+
\[f(f(x)f(y)) + f(x + y) = f(xy).\]
|
| 139 |
+
|
| 140 |
+
The only solutions are \(f(x) = 0\) , \(f(x) = x - 1\) and \(f(x) = 1 - x\) , which clearly work. Note that
|
| 141 |
+
|
| 142 |
+
- If \(f\) is a solution, so is \(-f\) .
|
| 143 |
+
|
| 144 |
+
- Moreover, if \(f(0) = 0\) then setting \(y = 0\) gives \(f \equiv 0\) . So henceforth we assume \(f(0) > 0\) .
|
| 145 |
+
|
| 146 |
+
Claim — We have \(f(z) = 0 \iff z = 1\) . Also, \(f(0) = 1\) and \(f(1) = 0\) .
|
| 147 |
+
|
| 148 |
+
Proof. For the forwards direction, if \(f(z) = 0\) and \(z \neq 1\) one may put \((x, y) = (z, z(z - 1)^{- 1})\) (so that \(x + y = xy\) ) we deduce \(f(0) = 0\) which is a contradiction.
|
| 149 |
+
|
| 150 |
+
For the reverse, \(f(f(0)^2) = 0\) by setting \(x = y = 0\) , and use the previous part. We also conclude \(f(1) = 0\) , \(f(0) = 1\) . \(\square\)
|
| 151 |
+
|
| 152 |
+
Claim — If \(f\) is injective, we are done.
|
| 153 |
+
|
| 154 |
+
Proof. Setting \(y = 0\) in the original equation gives \(f(f(x)) = 1 - f(x)\) . We apply this three times on the expression \(f^3 (x)\) :
|
| 155 |
+
|
| 156 |
+
\[f(1 - f(x)) = f(f(f(x))) = 1 - f(f(x)) = f(x).\]
|
| 157 |
+
|
| 158 |
+
Hence \(1 - f(x) = x\) or \(f(x) = 1 - x\) . \(\square\)
|
| 159 |
+
|
| 160 |
+
Remark. The result \(f(f(x)) + f(x) = 1\) also implies that surjectivity would solve the problem.
|
| 161 |
+
|
| 162 |
+
Claim — \(f\) is injective.
|
| 163 |
+
|
| 164 |
+
Proof. Setting \(y = 1\) in the original equation gives \(f(x + 1) = f(x) - 1\) , and by induction
|
| 165 |
+
|
| 166 |
+
\[f(x + n) = f(x) - n. \quad (1)\]
|
| 167 |
+
|
| 168 |
+
Assume now \(f(a) = f(b)\) . By using (1) we may shift \(a\) and \(b\) to be large enough that we may find \(x\) and \(y\) obeying \(x + y = a + 1\) , \(xy = b\) . Setting these gives
|
| 169 |
+
|
| 170 |
+
\[f(f(x)f(y)) = f(xy) - f(x + y) = f(b) - f(a + 1)\] \[= f(b) + 1 - f(a) = 1\]
|
| 171 |
+
|
| 172 |
+
|
| 173 |
+
|
| 174 |
+
from which we conclude
|
| 175 |
+
|
| 176 |
+
\[f(f(x)f(y) + 1) = 0.\]
|
| 177 |
+
|
| 178 |
+
Hence by the first claim we have \(f(x)f(y) + 1 = 1\) , so \(f(x)f(y) = 0\) . Applying the first claim again gives \(1 \in \{x, y\}\) . But that implies \(a = b\) . \(\square\)
|
| 179 |
+
|
| 180 |
+
Remark. Jessica Wan points out that for any \(a \neq b\) , at least one of \(a^2 > 4(b - 1)\) and \(b^2 > 4(a - 1)\) is true. So shifting via (1) is actually unnecessary for this proof.
|
| 181 |
+
|
| 182 |
+
Remark. One can solve the problem over \(\mathbb{Q}\) using only (1) and the easy parts. Indeed, that already implies \(f(n) = 1 - n\) for all \(n\) . Now we induct to show \(f(p / q) = 1 - p / q\) for all \(0 < p < q\) (on \(q\) ). By choosing \(x = 1 + p / q\) , \(y = 1 + q / p\) , we cause \(xy = x + y\) , and hence \(0 = f(f(1 + p / q)f(1 + q / p))\) or \(1 = f(1 + p / q)f(1 + q / p)\) .
|
| 183 |
+
|
| 184 |
+
By induction we compute \(f(1 + q / p)\) and this gives \(f(p / q + 1) = f(p / q) - 1\) .
|
| 185 |
+
|
| 186 |
+
|
| 187 |
+
|
| 188 |
+
## \(\S 1.3\) IMO 2017/3, proposed by Gerhard Woeginger (AUT)
|
| 189 |
+
|
| 190 |
+
Available online at https://aops.com/community/p8633324.
|
| 191 |
+
|
| 192 |
+
## Problem statement
|
| 193 |
+
|
| 194 |
+
A hunter and an invisible rabbit play a game in the plane. The rabbit and hunter start at points \(A_{0} = B_{0}\) . In the \(n\) th round of the game ( \(n \geq 1\) ), three things occur in order:
|
| 195 |
+
|
| 196 |
+
(i) The rabbit moves invisibly from \(A_{n - 1}\) to a point \(A_{n}\) such that \(A_{n - 1}A_{n} = 1\) .
|
| 197 |
+
|
| 198 |
+
(ii) The hunter has a tracking device (e.g. dog) which reports an approximate location \(P_{n}\) of the rabbit, such that \(P_{n}A_{n} \leq 1\) .
|
| 199 |
+
|
| 200 |
+
(iii) The hunter moves visibly from \(B_{n - 1}\) to a point \(B_{n}\) such that \(B_{n - 1}B_{n} = 1\) .
|
| 201 |
+
|
| 202 |
+
Let \(N = 10^{9}\) . Can the hunter guarantee that \(A_{N}B_{N} < 100\) ?
|
| 203 |
+
|
| 204 |
+
No, the hunter cannot. We will show how to increase the distance in the following way:
|
| 205 |
+
|
| 206 |
+
Claim — Suppose the rabbit is at a distance \(d \geq 1\) from the hunter at some point in time. Then it can increase its distance to at least \(\sqrt{d^{2} + \frac{1}{2}}\) in \(4d\) steps regardless of what the hunter already knows about the rabbit.
|
| 207 |
+
|
| 208 |
+
Proof. Consider a positive integer \(n > d\) , to be chosen later. Let the hunter start at \(B\) and the rabbit at \(A\) , as shown. Let \(\ell\) denote line \(AB\) .
|
| 209 |
+
|
| 210 |
+
Now, we may assume the rabbit reveals its location \(A\) , so that all previous information becomes irrelevant.
|
| 211 |
+
|
| 212 |
+
The rabbit chooses two points \(X\) and \(Y\) symmetric about \(\ell\) such that \(XY = 2\) and \(AX = AY = n\) , as shown. The rabbit can then hop to either \(X\) or \(Y\) , pinging the point \(P_{n}\) on the \(\ell\) each time. This takes \(n\) hops.
|
| 213 |
+
|
| 214 |
+
|
| 215 |
+
|
| 216 |
+
|
| 217 |
+
Now among all points \(H\) the hunter can go to, \(\min \max \{HX, HY\}\) is clearly minimized with \(H \in \ell\) by symmetry. So the hunter moves to a point \(H\) such that \(BH = n\) as well. In that case the new distance is \(HX = HY\) .
|
| 218 |
+
|
| 219 |
+
We now compute
|
| 220 |
+
|
| 221 |
+
\[H X^{2} = 1 + H M^{2} = 1 + \left(\sqrt{A X^{2} - 1} -A H\right)^{2}\] \[\qquad = 1 + \left(\sqrt{n^{2} - 1} - (n - d)\right)^{2}\] \[\qquad \geq 1 + \left(\left(n - \frac{1}{n}\right) - (n - d)\right)^{2}\] \[\qquad = 1 + (d - 1 / n)^{2}\]
|
| 222 |
+
|
| 223 |
+
which exceeds \(d^{2} + \frac{1}{2}\) whenever \(n \geq 4d\) .
|
| 224 |
+
|
| 225 |
+
|
| 226 |
+
|
| 227 |
+
In particular we can always take \(n = 400\) even very crudely; applying the lemma \(2 \cdot 100^{2}\) times, this gives a bound of \(400 \cdot 2 \cdot 100^{2} < 10^{9}\) , as desired.
|
| 228 |
+
|
| 229 |
+
Remark. The step of revealing the location of the rabbit seems critical because as far as I am aware it is basically impossible to keep track of ping locations in the problem.
|
| 230 |
+
|
| 231 |
+
Remark. Reasons to believe the answer is "no": the \(10^{9}\) constant, and also that "follow the last ping" is losing for the hunter.
|
| 232 |
+
|
| 233 |
+
Remark. I think there are roughly two ways you can approach the problem once you recognize the answer.
|
| 234 |
+
|
| 235 |
+
(i) Try and control the location of the pings
|
| 236 |
+
|
| 237 |
+
(ii) Abandon the notion of controlling possible locations, and try to increase the distance by a little bit, say from \(d\) to \(\sqrt{d^{2} + \epsilon}\) . This involves revealing the location of the rabbit before each iteration of several jumps.
|
| 238 |
+
|
| 239 |
+
I think it's clear that the difficulty of my approach is realizing that (ii) is possible; once you do, the two- point approach is more or less the only one possible.
|
| 240 |
+
|
| 241 |
+
My opinion is that (ii) is not that magical; as I said it was the first idea I had. But I am biased, because when I test- solved the problem at the IMO it was called "C5" and not "IMO3"; this effectively told me it was unlikely that the official solution was along the lines of (i), because otherwise it would have been placed much later in the shortlist.
|
| 242 |
+
|
| 243 |
+
|
| 244 |
+
|
| 245 |
+
## \(\S 2\) Solutions to Day 2
|
| 246 |
+
|
| 247 |
+
## \(\S 2.1\) IMO 2017/4, proposed by Charles Leytem (LUX)
|
| 248 |
+
|
| 249 |
+
Available online at https://aops.com/community/p8639236.
|
| 250 |
+
|
| 251 |
+
## Problem statement
|
| 252 |
+
|
| 253 |
+
Let \(R\) and \(S\) be different points on a circle \(\Omega\) such that \(\overline{RS}\) is not a diameter. Let \(\ell\) be the tangent line to \(\Omega\) at \(R\) . Point \(T\) is such that \(S\) is the midpoint of \(\overline{RT}\) . Point \(J\) is chosen on minor arc \(RS\) of \(\Omega\) so that the circumcircle \(\Gamma\) of triangle \(JST\) intersects \(\ell\) at two distinct points. Let \(A\) be the common point of \(\Gamma\) and \(\ell\) closer to \(R\) . Line \(AJ\) meets \(\Omega\) again at \(K\) . Prove that line \(KT\) is tangent to \(\Gamma\) .
|
| 254 |
+
|
| 255 |
+
First solution (elementary). First, note
|
| 256 |
+
|
| 257 |
+
\[\angle RKA = \angle RKJ = \angle RSJ = \angle TSJ = \angle TAJ = \angle TAK\]
|
| 258 |
+
|
| 259 |
+
so \(\overline{RK} \parallel \overline{AT}\) . Now,
|
| 260 |
+
|
| 261 |
+
\(\overline{RA}\) is tangent at \(R\) iff \(\triangle KRS \sim \triangle RTA\) (oppositely), because both equate to \(- \angle RKS = \angle SKR = \angle SRA = \angle TRA\) .
|
| 262 |
+
|
| 263 |
+
Similarly, \(\overline{TK}\) is tangent at \(T\) iff \(\triangle ATS \sim \triangle TRK\) .
|
| 264 |
+
|
| 265 |
+
The two similarities are equivalent because \(RS = ST\) the SAS gives \(KR \cdot TA = RS \cdot RT = TS \cdot TR\) .
|
| 266 |
+
|
| 267 |
+
|
| 268 |
+
|
| 269 |
+
|
| 270 |
+
Remark. The problem is actually symmetric with respect to two circles; \(\overline{RA}\) is tangent at \(R\) if and only if \(\overline{TK}\) at \(T\) .
|
| 271 |
+
|
| 272 |
+
Second solution (inversion). Consider an inversion at \(R\) fixing the circumcircle \(\Gamma\) of \(TSJA\) . Then:
|
| 273 |
+
|
| 274 |
+
- \(T\) and \(S\) swap,
|
| 275 |
+
|
| 276 |
+
|
| 277 |
+
|
| 278 |
+
- \(A\) and \(B\) swap, where \(B\) is the second intersection of \(\ell\) with \(\Gamma\) .
|
| 279 |
+
|
| 280 |
+
- Circle \(\Omega\) inverts to the line through \(T\) parallel to \(\overline{RAB}\) , call it \(\ell\) .
|
| 281 |
+
|
| 282 |
+
- \(J^{*}\) is the second intersection of \(\ell\) with \(\Gamma\) .
|
| 283 |
+
|
| 284 |
+
- \(K^{*}\) is the intersection of \(\ell\) with the circumcircle of \(RBJ^{*}\) ; this implies \(RK^{*}J^{*}B\) is an isosceles trapezoid. In particular, one reads \(\overline{RK^{*}} \parallel \overline{AT}\) from this, hence \(RK^{*}TA\) is a parallelogram.
|
| 285 |
+
|
| 286 |
+
Thus we wish to show the circumcircle of \(RSK^{*}\) is tangent to \(\Gamma\) . But that follows from the final parallelogram observed: \(S\) is the center of the parallelogram since it is the midpoint of the diagonal.
|
| 287 |
+
|
| 288 |
+
Remark. This also implies \(RKTB\) is cyclic, from \(\overline{K^{*}SA}\) collinear. Moreover, quadrilateral \(KK^{*}TS\) is cyclic (by power of a point); this leads to the second official solution to the problem.
|
| 289 |
+
|
| 290 |
+
|
| 291 |
+
|
| 292 |
+
## \(\S 2.2\) IMO 2017/5, proposed by Grigory Chelnokov (RUS)
|
| 293 |
+
|
| 294 |
+
Available online at https://aops.com/community/p8639240.
|
| 295 |
+
|
| 296 |
+
## Problem statement
|
| 297 |
+
|
| 298 |
+
Fix \(N \geq 1\) . A collection of \(N(N + 1)\) soccer players of distinct heights stand in a row. Sir Alex Song wishes to remove \(N(N - 1)\) players from this row to obtain a new row of \(2N\) players in which the following \(N\) conditions hold: no one stands between the two tallest players, no one stands between the third and fourth tallest players, ..., no one stands between the two shortest players. Prove that this is possible.
|
| 299 |
+
|
| 300 |
+
Some opening remarks: location and height are symmetric to each other, if one thinks about this problem as permutation pattern avoidance. So while officially there are multiple solutions, they are basically isomorphic to one another, and I am not aware of any solution otherwise.
|
| 301 |
+
|
| 302 |
+
|
| 303 |
+
|
| 304 |
+
|
| 305 |
+
Take a partition of \(N\) groups in order by height: \(G_{1} = \{1,\ldots ,N + 1\}\) , \(G_{2} = \{N+\) \(2,\ldots ,2N + 2\}\) , and so on. We will pick two people from each group \(G_{k}\)
|
| 306 |
+
|
| 307 |
+
Scan from the left until we find two people in the same group \(G_{k}\) . Delete all people scanned and also everyone in \(G_{k}\) . All the groups still have at least \(N\) people left, so we can induct down with the non- deleted people; the chosen pair is to the far left anyways.
|
| 308 |
+
|
| 309 |
+
Remark. The important bit is to scan by position but group by height, and moreover not change the groups as we scan. Dually, one can have a solution which scans by height but groups by position.
|
| 310 |
+
|
| 311 |
+
|
| 312 |
+
|
| 313 |
+
## \(\S 2.3\) IMO 2017/6, proposed by John Berman (USA)
|
| 314 |
+
|
| 315 |
+
Available online at https://aops.com/community/p8639242.
|
| 316 |
+
|
| 317 |
+
## Problem statement
|
| 318 |
+
|
| 319 |
+
An irreducible lattice point is an ordered pair of integers \((x,y)\) satisfying \(\gcd (x,y) =\) 1. Prove that if \(S\) is a finite set of irreducible lattice points then there exists a nonconstant homogeneous polynomial \(f(x,y)\) with integer coefficients such that \(f(x,y) = 1\) for each \((x,y)\in S\)
|
| 320 |
+
|
| 321 |
+
We present two solutions.
|
| 322 |
+
|
| 323 |
+
\(\P\) First solution (Dan Carmon, Israel). We prove the result by induction on \(|S|\) , with the base case being Bezout's Lemma ( \(n = 1\) ). For the inductive step, suppose we want to add a given pair \((a_{m + 1},b_{m + 1})\) to \(\{(a_{1},\ldots ,a_{m}),(b_{1},\ldots ,b_{m})\}\) .
|
| 324 |
+
|
| 325 |
+
Claim (Standard) — By a suitable linear transformation we may assume
|
| 326 |
+
|
| 327 |
+
\[(a_{m + 1},b_{m + 1}) = (1,0).\]
|
| 328 |
+
|
| 329 |
+
Outline of proof. It would be sufficient to show there exists a \(2\times 2\) matrix \(T = \left[ \begin{array}{ll}u & v\\ s & t \end{array} \right]\) with integer entries such that \(\operatorname *{det}T = 1\) and \(T\cdot \left[ \begin{array}{l}a_{m + 1}\\ b_{m + 1} \end{array} \right] = \left[ \begin{array}{l}1\\ 0 \end{array} \right]\) . Then we could apply \(T\) to all the ordered pairs in \(S\) (viewed as column vectors); if \(f\) was a polynomial that works on the transformed ordered pairs, then \(f(ux + vy,sx + ty)\) works for the original ordered pairs. (Here, the condition that \(\operatorname *{det}T = 1\) ensures that \(T^{- 1}\) has integer entries, and hence that \(T\) maps irreducible lattice points to irreducible lattice points.)
|
| 330 |
+
|
| 331 |
+
To generate such a matrix \(T\) , choose \(T = \left[ \begin{array}{ll}u & v\\ - b_{m + 1} & a_{m + 1} \end{array} \right]\) where \(u\) and \(v\) are chosen via Bezout lemma so that \(ua_{m + 1} + vb_{m + 1} = 1\) . This matrix \(T\) is rigged so that \(\operatorname *{det}T = 1\) and the leftmost column of \(T^{- 1}\) is \(\left[ \begin{array}{l}a_{m + 1}\\ b_{m + 1} \end{array} \right]\) . \(\square\)
|
| 332 |
+
|
| 333 |
+
Remark. This transformation claim is not necessary to proceed; the solution below can be rewritten to avoid it with only cosmetic edits. However, it makes things a bit easier to read.
|
| 334 |
+
|
| 335 |
+
Let \(g(x,y)\) be a polynomial which works on the latter set. We claim we can choose the new polynomial \(f\) of the form
|
| 336 |
+
|
| 337 |
+
\[f(x,y) = g(x,y)^{M} - Cx^{\deg g\cdot M - m}\prod_{i = 1}^{m}(b_{i}x - a_{i}y).\]
|
| 338 |
+
|
| 339 |
+
where \(C\) and \(M\) are integer parameters we may adjust.
|
| 340 |
+
|
| 341 |
+
Since \(f(a_{i},b_{i}) = 1\) by construction we just need
|
| 342 |
+
|
| 343 |
+
\[1 = f(1,0) = g(1,0)^{M} - C\prod b_{i}.\]
|
| 344 |
+
|
| 345 |
+
If \(\prod b_{i} = 0\) we are done, since \(b_{i} = 0 \implies a_{i} = \pm 1\) in that case and so \(g(1,0) = \pm 1\) , thus take \(M = 2\) . So it suffices to prove:
|
| 346 |
+
|
| 347 |
+
|
| 348 |
+
|
| 349 |
+
Claim — We have \(\gcd (g(1,0),b_{i}) = 1\) when \(b_{i}\neq 0\) .
|
| 350 |
+
|
| 351 |
+
Proof. Fix \(i\) . If \(b_{i} = 0\) then \(a_{i} = \pm 1\) and \(g(\pm 1,0) = \pm 1\) . Otherwise know
|
| 352 |
+
|
| 353 |
+
\[1 = g(a_{i},b_{i})\equiv g(a_{i},0)\pmod {b_{i}}\]
|
| 354 |
+
|
| 355 |
+
and since the polynomial is homogeneous with \(\gcd (a_{i},b_{i}) = 1\) it follows \(g(1,0)\neq 0\) (mod \(b_{i}\) ) as well. \(\square\)
|
| 356 |
+
|
| 357 |
+
Then take \(M\) a large multiple of \(\phi (\prod |b_{i}|)\) and we're done.
|
| 358 |
+
|
| 359 |
+
Second solution (Lagrange). The main claim is that:
|
| 360 |
+
|
| 361 |
+
Claim — For every positive integer \(N\) , there is a homogeneous polynomial \(P(x,y)\) such that \(P(x,y)\equiv 1\) (mod \(N\) ) whenever \(\gcd (x,y) = 1\) .
|
| 362 |
+
|
| 363 |
+
(This claim is actually implied by the problem.)
|
| 364 |
+
|
| 365 |
+
Proof. For \(N = p^{e}\) a prime take \((x^{p - 1} + y^{p - 1})^{\phi (N)}\) when \(p\) is odd, and \((x^{2} + xy + y^{2})^{\phi (N)}\) for \(p = 2\) .
|
| 366 |
+
|
| 367 |
+
Now, if \(N\) is a product of primes, we can collate coefficient by coefficient using the Chinese remainder theorem. \(\square\)
|
| 368 |
+
|
| 369 |
+
Let \(S = \{(a_{i},b_{i}) \mid i = 1,\ldots ,m\}\) . We have the natural homogeneous "Lagrange polynomials"
|
| 370 |
+
|
| 371 |
+
\[L_{k}(x,y):= \prod_{i\neq k}(b_{i}x - a_{i}y)\]
|
| 372 |
+
|
| 373 |
+
Now let
|
| 374 |
+
|
| 375 |
+
\[N:= \prod_{k}L_{k}(x_{k},y_{k})\]
|
| 376 |
+
|
| 377 |
+
and take \(P\) as in the claim. Then we can take a large power of \(P\) , and for each \(i\) subtract an appropriate multiple of \(L_{i}(x,y)\) ; that is, choose
|
| 378 |
+
|
| 379 |
+
\[f(x,y) = P(x,y)^{C} - \sum_{i}L_{i}(x,y)\cdot Q_{i}(x,y)\]
|
| 380 |
+
|
| 381 |
+
where \(C\) is large enough that \(C\deg P > \max_{i}\deg L_{i}\) , and \(Q_{i}(x,y)\) is any homogeneous polynomial of degree \(C\deg P - \deg L_{i}\) such that \(L_{k}(x_{k},y_{k})Q_{k}(x_{k},y_{k}) = \frac{P(x_{k},y_{k})^{C - 1}}{N}\) . \(L_{k}(x_{k},y_{k})\) (which is an integer).
|
| 382 |
+
|
| 383 |
+
|
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| 1 |
+
|
| 2 |
+
# IMO 2022 Solution Notes
|
| 3 |
+
|
| 4 |
+
Evan Chen 《陳誼廷》
|
| 5 |
+
|
| 6 |
+
23 July 2025
|
| 7 |
+
|
| 8 |
+
This is a compilation of solutions for the 2022 IMO. The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.
|
| 9 |
+
|
| 10 |
+
These notes will tend to be a bit more advanced and terse than the "official" solutions from the organizers. In particular, if a theorem or technique is not known to beginners but is still considered "standard", then I often prefer to use this theory anyways, rather than try to work around or conceal it. For example, in geometry problems I typically use directed angles without further comment, rather than awkwardly work around configuration issues. Similarly, sentences like "let \(\mathbb{R}\) denote the set of real numbers" are typically omitted entirely.
|
| 11 |
+
|
| 12 |
+
Corrections and comments are welcome!
|
| 13 |
+
|
| 14 |
+
## Contents
|
| 15 |
+
|
| 16 |
+
0 Problems 2
|
| 17 |
+
|
| 18 |
+
1 Solutions to Day 1 3
|
| 19 |
+
|
| 20 |
+
1.1 IMO 2022/1, proposed by Baptiste Serraille (FRA) 3
|
| 21 |
+
|
| 22 |
+
1.2 IMO 2022/2, proposed by Merlijn Staps (NLD) 5
|
| 23 |
+
|
| 24 |
+
1.3 IMO 2022/3, proposed by Ankan Bhattacharya (USA) 6
|
| 25 |
+
|
| 26 |
+
2 Solutions to Day 2 7
|
| 27 |
+
|
| 28 |
+
2.1 IMO 2022/4, proposed by Patrik Bak (SVK) 7
|
| 29 |
+
|
| 30 |
+
2.2 IMO 2022/5, proposed by Tijs Buggenhout (BEL) 9
|
| 31 |
+
|
| 32 |
+
2.3 IMO 2022/6, proposed by Nikola Petrovic (SRB) 10
|
| 33 |
+
|
| 34 |
+
|
| 35 |
+
|
| 36 |
+
## Problems
|
| 37 |
+
|
| 38 |
+
1. The Bank of Oslo issues two types of coin: aluminum (denoted \(A\) ) and bronze (denoted \(B\) ). Marianne has \(n\) aluminum coins and \(n\) bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer \(k \leq 2n\) , Gilberty repeatedly performs the following operation: he identifies the longest chain containing the \(k^{\mathrm{th}}\) coin from the left and moves all coins in that chain to the left end of the row. For example, if \(n = 4\) and \(k = 4\) , the process starting from the ordering \(AABBBA\) would be \(AABBBA\) \(A\) \(BBBA\) \(A\) \(AAABBB\) \(A\) \(BBBBAA\) \(A\) \(BBBBAA\) \(A\) \(\dots\)
|
| 39 |
+
|
| 40 |
+
Find all pairs \((n,k)\) with \(1\leq k\leq 2n\) such that for every initial ordering, at some moment during the process, the leftmost \(n\) coins will all be of the same type.
|
| 41 |
+
|
| 42 |
+
2. Find all functions \(f\colon \mathbb{R}^{+}\to \mathbb{R}^{+}\) such that for each \(x\in \mathbb{R}^{+}\) , there is exactly one \(y\in \mathbb{R}^{+}\) satisfying
|
| 43 |
+
|
| 44 |
+
\[x f(y) + y f(x)\leq 2.\]
|
| 45 |
+
|
| 46 |
+
3. Let \(k\) be a positive integer and let \(S\) be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of \(S\) around the circle such that the product of any two neighbors is of the form \(x^{2} + x + k\) for some positive integer \(x\) .
|
| 47 |
+
|
| 48 |
+
4. Let \(A B C D E\) be a convex pentagon such that \(B C = D E\) . Assume that there is a point \(T\) inside \(A B C D E\) with \(T B = T D\) \(T C = T E\) and \(\angle A B T = \angle T E A\) . Let line \(A B\) intersect lines \(C D\) and \(C T\) at points \(P\) and \(Q\) , respectively. Assume that the points \(P\) , \(B\) , \(A\) , \(Q\) occur on their line in that order. Let line \(A E\) intersect \(C D\) and \(D T\) at points \(R\) and \(S\) , respectively. Assume that the points \(R\) , \(E\) , \(A\) , \(S\) occur on their line in that order. Prove that the points \(P\) , \(S\) , \(Q\) , \(R\) lie on a circle.
|
| 49 |
+
|
| 50 |
+
5. Find all triples \((a,b,p)\) of positive integers with \(p\) prime and
|
| 51 |
+
|
| 52 |
+
\[a^{p} = b! + p.\]
|
| 53 |
+
|
| 54 |
+
6. Let \(n\) be a positive integer. A Nordic square is an \(n\times n\) board containing all the integers from 1 to \(n^{2}\) so that each cell contains exactly one number. An uphill path is a sequence of one or more cells such that:
|
| 55 |
+
|
| 56 |
+
a) the first cell in the sequence is a valley, meaning the number written is less than all its orthogonal neighbors,
|
| 57 |
+
b) each subsequent cell in the sequence is orthogonally adjacent to the previous cell, and
|
| 58 |
+
c) the numbers written in the cells in the sequence are in increasing order.
|
| 59 |
+
|
| 60 |
+
Find, as a function of \(n\) , the smallest possible total number of uphill paths in a Nordic square.
|
| 61 |
+
|
| 62 |
+
|
| 63 |
+
|
| 64 |
+
## \(\S 1\) Solutions to Day 1
|
| 65 |
+
|
| 66 |
+
## \(\S 1.1\) IMO 2022/1, proposed by Baptiste Serraille (FRA)
|
| 67 |
+
|
| 68 |
+
Available online at https://aops.com/community/p25635135.
|
| 69 |
+
|
| 70 |
+
## Problem statement
|
| 71 |
+
|
| 72 |
+
The Bank of Oslo issues two types of coin: aluminum (denoted \(A\) ) and bronze (denoted \(B\) ). Marianne has \(n\) aluminum coins and \(n\) bronze coins arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer \(k \leq 2n\) , Gilberty repeatedly performs the following operation: he identifies the longest chain containing the \(k^{\mathrm{th}}\) coin from the left and moves all coins in that chain to the left end of the row. For example, if \(n = 4\) and \(k = 4\) , the process starting from the ordering \(AABBBA\) would be \(AABBBA\) \(A\) \(BB\) \(BA\) \(AA\) \(BB\) \(BA\) \(BB\) \(BA\) \(AA\) \(BB\) \(BA\) \(BB\) \(\dots\)
|
| 73 |
+
|
| 74 |
+
Find all pairs \((n,k)\) with \(1\leq k\leq 2n\) such that for every initial ordering, at some moment during the process, the leftmost \(n\) coins will all be of the same type.
|
| 75 |
+
|
| 76 |
+
Answer: \(n\leq k\leq \left\lceil \frac{3}{2} n\right\rceil\)
|
| 77 |
+
|
| 78 |
+
Call a maximal chain a block. Then the line can be described as a sequence of blocks: it's one of:
|
| 79 |
+
|
| 80 |
+
\[\underbrace{A\ldots A}_{e_1}\underbrace{B\ldots B}_{e_2}\underbrace{A\ldots A}_{e_3}\ldots \underbrace{A\ldots A}_{e_m}\mathrm{~for~odd~}m\] \[\underbrace{A\ldots A}_{e_1}\underbrace{B\ldots B}_{e_2}\underbrace{A\ldots A}_{e_3}\ldots \underbrace{B\ldots B}_{e_m}\mathrm{~for~even~}m\]
|
| 81 |
+
|
| 82 |
+
or the same thing with the roles of \(A\) and \(B\) flipped.
|
| 83 |
+
|
| 84 |
+
The main claim is the following:
|
| 85 |
+
|
| 86 |
+
Claim — The number \(m\) of blocks will never increase after an operation. Moreover, it stays the same if and only if
|
| 87 |
+
|
| 88 |
+
- \(k \leq e_1\) ; or
|
| 89 |
+
|
| 90 |
+
- \(m\) is even and \(e_m \geq 2n + 1 - k\) .
|
| 91 |
+
|
| 92 |
+
Proof. This is obvious, just run the operation and see!
|
| 93 |
+
|
| 94 |
+
The problem asks for which values of \(k\) we always reach \(m = 2\) eventually; we already know that it's non- increasing. We consider a few cases:
|
| 95 |
+
|
| 96 |
+
- If \(k < n\) , then any configuration with \(e_1 = n - 1\) will never change.
|
| 97 |
+
|
| 98 |
+
- If \(k > \lceil 3n / 2 \rceil\) , then take \(m = 4\) and \(e_1 = e_2 = \lfloor n / 2 \rfloor\) and \(e_3 = e_4 = \lceil n / 2 \rceil\) . This configuration retains \(m = 4\) always: the blocks simply rotate.
|
| 99 |
+
|
| 100 |
+
- Conversely, suppose \(k \geq n\) has the property that \(m > 2\) stays fixed. If after the first three operations \(m\) hasn't changed, then we must have \(m \geq 4\) even, and \(e_m, e_{m-1}, e_{m-2} \geq 2n + 1 - k\) . Now,
|
| 101 |
+
|
| 102 |
+
\[n \geq e_m + e_{m-2} \geq 2(2n + 1 - k) \implies k \geq \frac{3}{2} n + 1\]
|
| 103 |
+
|
| 104 |
+
|
| 105 |
+
|
| 106 |
+
so this completes the proof.
|
| 107 |
+
|
| 108 |
+
|
| 109 |
+
|
| 110 |
+
## \(\S 1.2\) IMO 2022/2, proposed by Merlijn Staps (NLD)
|
| 111 |
+
|
| 112 |
+
Available online at https://aops.com/community/p25635138.
|
| 113 |
+
|
| 114 |
+
## Problem statement
|
| 115 |
+
|
| 116 |
+
Find all functions \(f\colon \mathbb{R}^{+}\to \mathbb{R}^{+}\) such that for each \(x\in \mathbb{R}^{+}\) , there is exactly one \(y\in \mathbb{R}^{+}\) satisfying
|
| 117 |
+
|
| 118 |
+
\[x f(y) + y f(x)\leq 2.\]
|
| 119 |
+
|
| 120 |
+
The answer is \(f(x)\equiv 1 / x\) which obviously works (here \(y = x\) ).
|
| 121 |
+
|
| 122 |
+
For the converse, assume we have \(f\) such that each \(x\in \mathbb{R}^{+}\) has a friend \(y\) with \(x f(y) + y f(x)\leq 2\) . By symmetry \(y\) is also the friend of \(x\) .
|
| 123 |
+
|
| 124 |
+
Claim — In fact every number is its own friend.
|
| 125 |
+
|
| 126 |
+
Proof. Assume for contradiction \(a\neq b\) are friends. Then we know that \(a f(a) + a f(a)>\) \(2\Longrightarrow f(a) > \frac{1}{a}\) . Analogously, \(f(b) > \frac{1}{b}\) . However, we then get
|
| 127 |
+
|
| 128 |
+
\[2\geq a f(b) + b f(a) > \frac{a}{b} +\frac{b}{a}\stackrel {\mathrm{AMGM}}{\geq}2\]
|
| 129 |
+
|
| 130 |
+
which is impossible.
|
| 131 |
+
|
| 132 |
+
The problem condition now simplifies to saying
|
| 133 |
+
|
| 134 |
+
\[f(x)\leq \frac{1}{x}\mathrm{~for~all~}x,\qquad x f(y) + y f(x) > 2\mathrm{~for~all~}x\neq y.\]
|
| 135 |
+
|
| 136 |
+
In particular, for any \(x > 0\) and \(\epsilon >0\) we have
|
| 137 |
+
|
| 138 |
+
\[2< x f(x + \epsilon) + (x + \epsilon)f(x)\leq \frac{x}{x + \epsilon} +(x + \epsilon)f(x)\] \[\Longrightarrow f(x) > \frac{x + 2\epsilon}{(x + \epsilon)^{2}} = \frac{1}{x + \frac{\epsilon^{2}}{x + 2\epsilon}}.\]
|
| 139 |
+
|
| 140 |
+
Since this holds for all \(\epsilon >0\) this forces \(f(x)\geq \frac{1}{x}\) as well. We're done.
|
| 141 |
+
|
| 142 |
+
Remark. Alternatively, instead of using \(x + \epsilon\) , it also works to consider \(y = \frac{1}{f(x)}\) . For such a \(y\) , we would have
|
| 143 |
+
|
| 144 |
+
\[x f\left(\frac{1}{f(x)}\right) + \frac{1}{f(x)}\cdot f(x) = x f\left(\frac{1}{f(x)}\right) + 1\leq x\cdot f(x) + 1\leq 1 + 1 = 2\]
|
| 145 |
+
|
| 146 |
+
which gives a similar contradiction.
|
| 147 |
+
|
| 148 |
+
|
| 149 |
+
|
| 150 |
+
## \(\S 1.3\) IMO 2022/3, proposed by Ankan Bhattacharya (USA)
|
| 151 |
+
|
| 152 |
+
Available online at https://aops.com/community/p25635143.
|
| 153 |
+
|
| 154 |
+
## Problem statement
|
| 155 |
+
|
| 156 |
+
Let \(k\) be a positive integer and let \(S\) be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of \(S\) around the circle such that the product of any two neighbors is of the form \(x^{2} + x + k\) for some positive integer \(x\) .
|
| 157 |
+
|
| 158 |
+
We replace "positive integer \(x\) " with "nonnegative integer \(x\) ", and say numbers of the form \(x^{2} + x + k\) are good. We could also replace "nonnegative integer \(x\) " with "integer \(x\) " owing to the obvious map \(x \mapsto 1 - x\) .
|
| 159 |
+
|
| 160 |
+
Claim — If \(p\) is an odd prime, there are at most two odd primes \(q\) and \(r\) less than \(p\) for which \(pq = x^{2} + x + k\) and \(pr = y^{2} + y + k\) are good.
|
| 161 |
+
|
| 162 |
+
Moreover, if the above occurs and \(x, y \geq 0\) , then \(x + y + 1 = p\) and \(xy \equiv k \pmod{p}\) .
|
| 163 |
+
|
| 164 |
+
Proof. The equation \(T^{2} + T + k \equiv 0 \pmod{p}\) has at most two solutions modulo \(p\) , i.e. at most two solutions in the interval \([0, p - 1]\) . Because \(0 \leq x, y < p\) from \(p > \max(q, r)\) and \(k > 0\) , the first half follows.
|
| 165 |
+
|
| 166 |
+
For the second half, Vieta also gives \(x + y \equiv - 1 \pmod{p}\) and \(xy \equiv k \pmod{p}\) , and we know \(0 < x + y < 2p\) . \(\square\)
|
| 167 |
+
|
| 168 |
+
Claim — If two such primes do exist as above, then \(qr\) is also good (!).
|
| 169 |
+
|
| 170 |
+
Proof. Let \(pq = x^{2} + x + k\) and \(pr = y^{2} + y + k\) for \(x, y \geq 0\) as before. Fix \(\alpha \in \mathbb{C}\) such that \(\alpha^{2} + \alpha + k = 0\) ; then for any \(n \in \mathbb{Z}\) , we have
|
| 171 |
+
|
| 172 |
+
\[n^{2} + n + k = \mathrm{Norm}(n - \alpha).\]
|
| 173 |
+
|
| 174 |
+
Hence
|
| 175 |
+
|
| 176 |
+
\[pq \cdot pr = \mathrm{Norm} \left((x - \alpha)(y - \alpha)\right) = \mathrm{Norm} \left((xy - k) - (x + y + 1)\alpha\right)\]
|
| 177 |
+
|
| 178 |
+
But \(\mathrm{Norm}(p) = p^{2}\) , so combining with the second half of the previous claim gives
|
| 179 |
+
|
| 180 |
+
\[qr = \mathrm{Norm} \left(\frac{1}{p} (xy - k) - \alpha\right)\]
|
| 181 |
+
|
| 182 |
+
as needed.
|
| 183 |
+
|
| 184 |
+
These two claims imply the claim directly by induction on \(|S|\) , since one can now delete the largest element of \(S\) .
|
| 185 |
+
|
| 186 |
+
Remark. To show that the condition is not vacuous, the author gave a ring of 385 primes for \(k = 41\) ; see https://aops.com/community/p26068963.
|
| 187 |
+
|
| 188 |
+
|
| 189 |
+
|
| 190 |
+
## \(\S 2\) Solutions to Day 2
|
| 191 |
+
|
| 192 |
+
## \(\S 2.1\) IMO 2022/4, proposed by Patrik Bak (SVK)
|
| 193 |
+
|
| 194 |
+
Available online at https://aops.com/community/p25635154.
|
| 195 |
+
|
| 196 |
+
## Problem statement
|
| 197 |
+
|
| 198 |
+
Let \(ABCDE\) be a convex pentagon such that \(BC = DE\) . Assume that there is a point \(T\) inside \(ABCDE\) with \(TB = TD\) , \(TC = TE\) and \(\angle ABT = \angle TEA\) . Let line \(AB\) intersect lines \(CD\) and \(CT\) at points \(P\) and \(Q\) , respectively. Assume that the points \(P\) , \(B\) , \(A\) , \(Q\) occur on their line in that order. Let line \(AE\) intersect \(CD\) and \(DT\) at points \(R\) and \(S\) , respectively. Assume that the points \(R\) , \(E\) , \(A\) , \(S\) occur on their line in that order. Prove that the points \(P\) , \(S\) , \(Q\) , \(R\) lie on a circle.
|
| 199 |
+
|
| 200 |
+
The conditions imply
|
| 201 |
+
|
| 202 |
+
\[\triangle BTC\cong \triangle DTE,\qquad \mathrm{and}\qquad \triangle BTY\cong \triangle ETX.\]
|
| 203 |
+
|
| 204 |
+
Define \(K = \overline{CT} \cap \overline{AE}\) , \(L = \overline{DT} \cap \overline{AB}\) , \(X = \overline{BT} \cap \overline{AE}\) , \(Y = \overline{ET} \cap \overline{BY}\) .
|
| 205 |
+
|
| 206 |
+
|
| 207 |
+
|
| 208 |
+
|
| 209 |
+
Claim (Main claim) — We have
|
| 210 |
+
|
| 211 |
+
\[\triangle BTQ \sim \triangle ETS, \qquad \text{and} \qquad BY: YL: LQ = EX: XK: KS.\]
|
| 212 |
+
|
| 213 |
+
|
| 214 |
+
|
| 215 |
+
In other words, \(TBYLQ \sim TEXKS\) .
|
| 216 |
+
|
| 217 |
+
Proof. We know \(\triangle BTY \sim \triangle ETX\) . Also, \(\triangle BTL = \triangle BTD = \triangle CTE = \triangle KTE\) and \(\triangle BTQ = \triangle BTC = \triangle DTE = \triangle STE\) . \(\square\)
|
| 218 |
+
|
| 219 |
+
It follows from the claim that:
|
| 220 |
+
|
| 221 |
+
\(\cdot \mathit{T L} / \mathit{T Q} = \mathit{T K} / \mathit{T S}\) , ergo \(T L\cdot T S = T K\cdot T Q\) , so \(K L S Q\) is cyclic; and
|
| 222 |
+
|
| 223 |
+
\(\cdot \mathit{T C} / \mathit{T K} = \mathit{T E} / \mathit{T K} = \mathit{T B} / \mathit{T L} = \mathit{T D} / \mathit{T L}\) , so \(\overline{{K L}}\parallel \overline{{P C D R}}\)
|
| 224 |
+
|
| 225 |
+
With these two bullets, we're done by Reim theorem.
|
| 226 |
+
|
| 227 |
+
|
| 228 |
+
|
| 229 |
+
## \(\S 2.2\) IMO 2022/5, proposed by Tijs Buggenhout (BEL)
|
| 230 |
+
|
| 231 |
+
Available online at https://aops.com/community/p25635158.
|
| 232 |
+
|
| 233 |
+
## Problem statement
|
| 234 |
+
|
| 235 |
+
Find all triples \((a,b,p)\) of positive integers with \(p\) prime and
|
| 236 |
+
|
| 237 |
+
\[a^{p} = b! + p.\]
|
| 238 |
+
|
| 239 |
+
The answer is \((2,2,2)\) and \((3,4,3)\) only, which work.
|
| 240 |
+
|
| 241 |
+
In what follows we assume \(a\geq 2\)
|
| 242 |
+
|
| 243 |
+
Claim — We have \(b\leq 2p - 2\) , and hence \(a< p^{2}\)
|
| 244 |
+
|
| 245 |
+
Proof. For the first half, assume first for contradiction that \(b\geq 2p\) . Then \(b! + p\equiv p\) (mod \(p^{2}\) ), so \(\nu_{p}(b! + p) = 1\) , but \(\nu_{p}(a^{p}) = 1\) never occurs.
|
| 246 |
+
|
| 247 |
+
We can also rule out \(b = 2p - 1\) since that would give
|
| 248 |
+
|
| 249 |
+
\[(2p - 1)! + p = p[(p - 1)!(p + 1)(p + 2)\dots (2p - 1) + 1]\]
|
| 250 |
+
|
| 251 |
+
By Wilson theorem the inner bracket is \((- 1)^{2} + 1\equiv 2\) (mod \(p\) ) exactly, contradiction for \(p > 2\) . And when \(p = 2\) , \(3! + 2 = 8\) is not a perfect square.
|
| 252 |
+
|
| 253 |
+
The second half follows as \(a^{p}\leq (2p - 2)! + p< p^{2p}\) . (Here we used the crude estimate \((2p - 2)! = \prod_{k = 1}^{p - 1}k\cdot (2p - 1 - k)< (p(p - 1))^{p - 1})\) . \(\square\)
|
| 254 |
+
|
| 255 |
+
Claim — We have \(a\geq p\) , and hence \(b\geq p\)
|
| 256 |
+
|
| 257 |
+
Proof. For the first half, assume for contradiction that \(p > a\) . Then
|
| 258 |
+
|
| 259 |
+
\[b! + p = a^{p}\geq a^{p - 1} + p\geq a^{a} + p > a! + p\Rightarrow b > a.\]
|
| 260 |
+
|
| 261 |
+
Then taking modulo \(a\) now gives \(0\equiv 0 + p\) (mod \(a\) ), which is obviously impossible.
|
| 262 |
+
|
| 263 |
+
The second half follows from \(b! = a^{p} - p\geq p! - p > (p - 1)!\) . \(\square\)
|
| 264 |
+
|
| 265 |
+
Claim — We have \(a = p\) exactly.
|
| 266 |
+
|
| 267 |
+
Proof. We know \(p\geq b\) hence \(p\mid b! + p\) , so let \(a = pk\) for \(k< p\) . Then \(k\mid b!\) yet \(k\nmid a^{p} - p\) contradiction. \(\square\)
|
| 268 |
+
|
| 269 |
+
Let's get the small \(p\) out of the way:
|
| 270 |
+
|
| 271 |
+
- For \(p = 2\) , checking \(2\leq b\leq 3\) gives \((a,b) = (2,2)\) only.
|
| 272 |
+
|
| 273 |
+
- For \(p = 3\) , checking \(3\leq b\leq 5\) gives \((a,b) = (3,4)\) only.
|
| 274 |
+
|
| 275 |
+
Once \(p\geq 5\) , if \(b! = p^{p} - p = p(p^{p - 1} - 1)\) then applying Zsigmondy gets a prime factor \(q\equiv 1\) (mod \(p - 1\) ) which divides \(p^{p - 1} - 1\) . Yet \(q\leq b\leq 2p - 2\) and \(q\neq p\) , contradiction.
|
| 276 |
+
|
| 277 |
+
|
| 278 |
+
|
| 279 |
+
## \(\S 2.3\) IMO 2022/6, proposed by Nikola Petrovic (SRB)
|
| 280 |
+
|
| 281 |
+
Available online at https://aops.com/community/p25635163.
|
| 282 |
+
|
| 283 |
+
## Problem statement
|
| 284 |
+
|
| 285 |
+
Let \(n\) be a positive integer. A Nordic square is an \(n\times n\) board containing all the integers from 1 to \(n^{2}\) so that each cell contains exactly one number. An uphill path is a sequence of one or more cells such that:
|
| 286 |
+
|
| 287 |
+
1. the first cell in the sequence is a valley, meaning the number written is less than all its orthogonal neighbors,
|
| 288 |
+
|
| 289 |
+
2. each subsequent cell in the sequence is orthogonally adjacent to the previous cell, and
|
| 290 |
+
|
| 291 |
+
3. the numbers written in the cells in the sequence are in increasing order.
|
| 292 |
+
|
| 293 |
+
Find, as a function of \(n\) , the smallest possible total number of uphill paths in a Nordic square.
|
| 294 |
+
|
| 295 |
+
Answer: \(2n^{2} - 2n + 1\)
|
| 296 |
+
|
| 297 |
+
Bound. The lower bound is the "obvious" one:
|
| 298 |
+
|
| 299 |
+
- For any pair of adjacent cells, say \(a > b\) , one can extend it to a downhill path (the reverse of an uphill path) by walking downwards until one reaches a valley. This gives \(2n(n - 1) = 2n^{2} - 2n\) uphill paths of length \(\geq 2\) .
|
| 300 |
+
|
| 301 |
+
- There is always at least one uphill path of length 1, namely the single cell \(\{1\}\) (or indeed any valley).
|
| 302 |
+
|
| 303 |
+
Construction. For the construction, the ideas it build a tree \(T\) on the grid such that no two cells not in \(T\) are adjacent.
|
| 304 |
+
|
| 305 |
+
An example of such a grid is shown below for \(n = 15\) with \(T\) in yellow and cells not in \(T\) in black; it generalizes to any \(3\mid n\) , and then to any \(n\) by deleting the last \(n\) mod 3 rows and either/both of the leftmost/rightmost column.
|
| 306 |
+
|
| 307 |
+
|
| 308 |
+
|
| 309 |
+
|
| 310 |
+
<table><tr><td>1</td><td>2</td><td>3</td><td>32</td><td></td><td>36</td><td>62</td><td>63</td><td>64</td><td>93</td><td></td><td>97</td><td>123</td><td>124</td><td>125</td></tr><tr><td></td><td>4</td><td></td><td>33</td><td>34</td><td>35</td><td></td><td>65</td><td></td><td>94</td><td>95</td><td>96</td><td></td><td>126</td><td></td></tr><tr><td>6</td><td>5</td><td>7</td><td></td><td>37</td><td></td><td>67</td><td>66</td><td>68</td><td></td><td>98</td><td></td><td>128</td><td>127</td><td>129</td></tr><tr><td></td><td>8</td><td></td><td>39</td><td>38</td><td>40</td><td></td><td>69</td><td></td><td>100</td><td>99</td><td>101</td><td></td><td>130</td><td></td></tr><tr><td>10</td><td>9</td><td>11</td><td></td><td>41</td><td></td><td>71</td><td>70</td><td>72</td><td></td><td>102</td><td></td><td>132</td><td>131</td><td>133</td></tr><tr><td></td><td>12</td><td></td><td>43</td><td>42</td><td>44</td><td></td><td>73</td><td></td><td>104</td><td>103</td><td>105</td><td></td><td>134</td><td></td></tr><tr><td>14</td><td>13</td><td>15</td><td></td><td>45</td><td></td><td>75</td><td>74</td><td>76</td><td></td><td>106</td><td></td><td>136</td><td>135</td><td>137</td></tr><tr><td></td><td>16</td><td></td><td>47</td><td>46</td><td>48</td><td></td><td>77</td><td></td><td>108</td><td>107</td><td>109</td><td></td><td>138</td><td></td></tr><tr><td>18</td><td>17</td><td>19</td><td></td><td>49</td><td></td><td>79</td><td>78</td><td>80</td><td></td><td>110</td><td></td><td>140</td><td>139</td><td>141</td></tr><tr><td></td><td>20</td><td></td><td>51</td><td>50</td><td>52</td><td></td><td>81</td><td></td><td>112</td><td>111</td><td>113</td><td></td><td>142</td><td></td></tr><tr><td>22</td><td>21</td><td>23</td><td></td><td>53</td><td></td><td>83</td><td>82</td><td>84</td><td></td><td>114</td><td></td><td>144</td><td>143</td><td>145</td></tr><tr><td></td><td>24</td><td></td><td>55</td><td>54</td><td>56</td><td></td><td>85</td><td></td><td>116</td><td>115</td><td>117</td><td></td><td>146</td><td></td></tr><tr><td>26</td><td>25</td><td>27</td><td></td><td>57</td><td></td><td>87</td><td>86</td><td>88</td><td></td><td>118</td><td></td><td>148</td><td>147</td><td>149</td></tr><tr><td></td><td>28</td><td></td><td>59</td><td>58</td><td>60</td><td></td><td>89</td><td></td><td>120</td><td>119</td><td>121</td><td></td><td>150</td><td></td></tr><tr><td>30</td><td>29</td><td>31</td><td></td><td>61</td><td></td><td>91</td><td>90</td><td>92</td><td></td><td>122</td><td></td><td>152</td><td>151</td><td>153</td></tr></table>
|
| 311 |
+
|
| 312 |
+
Place 1 anywhere in \(T\) and then place all the small numbers at most \(|T|\) adjacent to previously placed numbers (example above). Then place the remaining numbers outside \(T\) arbitrarily.
|
| 313 |
+
|
| 314 |
+
By construction, as 1 is the only valley, any uphill path must start from 1. And by construction, it may only reach a given pair of terminal cells in one way, i.e. the downhill paths we mentioned are the only one. End proof.
|
| 315 |
+
|
| 316 |
+
|
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size 647216
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IMO/raw/en-IMO-2025-notes.pdf
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version https://git-lfs.github.com/spec/v1
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oid sha256:6b2db390d9abb26d7b2aa3fa38c534953c89db64855b5d7d564587d7a178234a
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size 308511
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IMO/segment_script/segment.py
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| 1 |
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# -----------------------------------------------------------------------------
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| 2 |
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# Author: Jiawei Liu
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| 3 |
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# Date: 2025-10-29
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| 4 |
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# -----------------------------------------------------------------------------
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| 5 |
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import json
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| 6 |
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import re
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| 7 |
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from pathlib import Path
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| 8 |
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from rapidfuzz import fuzz
|
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| 11 |
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| 12 |
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def clean_text(text: str):
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| 13 |
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text = re.sub(
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| 14 |
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r"\nAvailable online at.+?\.\s\s$", "", text, flags=re.DOTALL | re.MULTILINE
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| 15 |
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)
|
| 16 |
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return text
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| 17 |
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| 18 |
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def extract_problems(markdown_text: str):
|
| 20 |
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problems = {}
|
| 21 |
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|
| 22 |
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# 1. Find the block of text containing the problems
|
| 23 |
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# Block starts with '## Problems'
|
| 24 |
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# and ends before '## \(S 1\) Solutions to Day 1'
|
| 25 |
+
problems_section = re.search(
|
| 26 |
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r"^## Problems\s*$(.*?)^##.*Solutions to Day 1",
|
| 27 |
+
markdown_text,
|
| 28 |
+
re.DOTALL | re.MULTILINE,
|
| 29 |
+
)
|
| 30 |
+
|
| 31 |
+
if not problems_section:
|
| 32 |
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print(" - Not found '## Problems' section.")
|
| 33 |
+
return problems
|
| 34 |
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|
| 35 |
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problems_block = problems_section.group(1)
|
| 36 |
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|
| 37 |
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# 2. Split the block into individual problems
|
| 38 |
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# Each problem starts with a number followed by a dot and space (e.g., '1. ')
|
| 39 |
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matchs = list(re.finditer(r"^(\d+)\.\s+", problems_block, flags=re.MULTILINE))
|
| 40 |
+
|
| 41 |
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# Split the text into parts using the problem numbers as delimiters
|
| 42 |
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for i, m in enumerate(matchs):
|
| 43 |
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problem_label = m.group(1)
|
| 44 |
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problem = problems_block[
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| 45 |
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m.end() : matchs[i + 1].start()
|
| 46 |
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if i + 1 < len(matchs)
|
| 47 |
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else len(problems_block)
|
| 48 |
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]
|
| 49 |
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|
| 50 |
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problems[problem_label] = (problem.strip(), m.group())
|
| 51 |
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|
| 52 |
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print(f" - Extracted {len(problems)} problems.")
|
| 53 |
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|
| 54 |
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return problems
|
| 55 |
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|
| 56 |
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|
| 57 |
+
def extract_solutions(markdown_text):
|
| 58 |
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solutions = {}
|
| 59 |
+
|
| 60 |
+
# 1. Find the block of text containing the solutions
|
| 61 |
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# Block starts with '## \(S 1\) Solutions to Day 1' and end of document
|
| 62 |
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solutions_section = re.search(
|
| 63 |
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r"^##.*?Solutions to Day 1\s*$(.*)",
|
| 64 |
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markdown_text,
|
| 65 |
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re.DOTALL | re.MULTILINE,
|
| 66 |
+
)
|
| 67 |
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|
| 68 |
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if not solutions_section:
|
| 69 |
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print(" - Not found Solutions section.")
|
| 70 |
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return solutions
|
| 71 |
+
|
| 72 |
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solutions_block = solutions_section.group(1)
|
| 73 |
+
|
| 74 |
+
## \(\S 1.1\) IMO 2005/1, proposed by Bogdan Enescu (ROU)
|
| 75 |
+
# 2. Split the block into individual solutions
|
| 76 |
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matchs = list(
|
| 77 |
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re.finditer(r"^##.*?IMO\s\d+\/(\d+).*?\n$", solutions_block, flags=re.MULTILINE)
|
| 78 |
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)
|
| 79 |
+
|
| 80 |
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# Split the text into parts using the solution numbers as delimiters
|
| 81 |
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for i, m in enumerate(matchs):
|
| 82 |
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problem_label = m.group(1)
|
| 83 |
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solution = solutions_block[
|
| 84 |
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m.end() : matchs[i + 1].start()
|
| 85 |
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if i + 1 < len(matchs)
|
| 86 |
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else len(solutions_block)
|
| 87 |
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]
|
| 88 |
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| 89 |
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solutions[problem_label] = (solution.strip(), m.group())
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| 90 |
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|
| 91 |
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print(f" - Extracted {len(solutions)} solutions.")
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| 92 |
+
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| 93 |
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return solutions
|
| 94 |
+
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| 95 |
+
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| 96 |
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def join(problems: dict, solutions: dict):
|
| 97 |
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pairs = []
|
| 98 |
+
|
| 99 |
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for problem_label, (problem, p_match) in problems.items():
|
| 100 |
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solution, s_match = solutions.get(problem_label)
|
| 101 |
+
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| 102 |
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# Clean solution by removing the part that overlaps with the problem statement
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| 103 |
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problem_align = fuzz.partial_ratio_alignment(solution, problem)
|
| 104 |
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solution = solution.replace(
|
| 105 |
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solution[problem_align.src_start : problem_align.src_end], ""
|
| 106 |
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)
|
| 107 |
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solution = re.sub(
|
| 108 |
+
r"^\s*## Problem statement", "", solution, flags=re.IGNORECASE
|
| 109 |
+
).strip()
|
| 110 |
+
|
| 111 |
+
if not solution:
|
| 112 |
+
print(f" - Warning: No solution found for problem {problem_label}.")
|
| 113 |
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pairs.append((problem, solution, problem_label, p_match, s_match))
|
| 114 |
+
|
| 115 |
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return pairs
|
| 116 |
+
|
| 117 |
+
|
| 118 |
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def write_pairs(output_file: Path, pairs: list, year: str, project_root: Path):
|
| 119 |
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output_jsonl_text = ""
|
| 120 |
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for problem, solution, problem_label, p_match, s_match in pairs:
|
| 121 |
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output_jsonl_text += (
|
| 122 |
+
json.dumps(
|
| 123 |
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{
|
| 124 |
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"year": year,
|
| 125 |
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"tier": "T0",
|
| 126 |
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"problem_label": problem_label,
|
| 127 |
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"problem_type": None,
|
| 128 |
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"exam": "IMO",
|
| 129 |
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"problem": problem,
|
| 130 |
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"solution": solution,
|
| 131 |
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"metadata": {
|
| 132 |
+
"resource_path": output_file.relative_to(
|
| 133 |
+
project_root
|
| 134 |
+
).as_posix(),
|
| 135 |
+
"problem_match": p_match,
|
| 136 |
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"solution_match": s_match,
|
| 137 |
+
},
|
| 138 |
+
},
|
| 139 |
+
ensure_ascii=False,
|
| 140 |
+
)
|
| 141 |
+
+ "\n"
|
| 142 |
+
)
|
| 143 |
+
|
| 144 |
+
output_file.write_text(output_jsonl_text, encoding="utf-8")
|
| 145 |
+
|
| 146 |
+
|
| 147 |
+
if __name__ == "__main__":
|
| 148 |
+
compet_base_path = Path(__file__).resolve().parent.parent
|
| 149 |
+
compet_md_path = compet_base_path / "md"
|
| 150 |
+
seg_output_path = compet_base_path / "segmented"
|
| 151 |
+
project_root = compet_base_path.parent
|
| 152 |
+
|
| 153 |
+
num_problems = 0
|
| 154 |
+
num_solutions = 0
|
| 155 |
+
|
| 156 |
+
for md_file in list(compet_md_path.glob("**/*notes.md")):
|
| 157 |
+
print(f"Processing {md_file}...")
|
| 158 |
+
output_file = seg_output_path / md_file.relative_to(compet_md_path).with_suffix(
|
| 159 |
+
".jsonl"
|
| 160 |
+
)
|
| 161 |
+
output_file.parent.mkdir(parents=True, exist_ok=True)
|
| 162 |
+
|
| 163 |
+
# Read the markdown file
|
| 164 |
+
markdown_text = "\n" + md_file.read_text(encoding="utf-8")
|
| 165 |
+
markdown_text = clean_text(markdown_text)
|
| 166 |
+
|
| 167 |
+
problems = extract_problems(markdown_text)
|
| 168 |
+
solutions = extract_solutions(markdown_text)
|
| 169 |
+
|
| 170 |
+
num_problems += len(problems)
|
| 171 |
+
num_solutions += len(solutions)
|
| 172 |
+
|
| 173 |
+
pairs = join(problems, solutions)
|
| 174 |
+
|
| 175 |
+
year = re.search(r"\d{4}", output_file.stem).group()
|
| 176 |
+
|
| 177 |
+
write_pairs(output_file, pairs, year, project_root)
|
| 178 |
+
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| 179 |
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print()
|
| 180 |
+
|
| 181 |
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print(f"Total problems extracted: {num_problems}")
|
| 182 |
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print(f"Total solutions extracted: {num_solutions}")
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IMO/segmented/en-IMO-2005-notes.jsonl
ADDED
|
@@ -0,0 +1,6 @@
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{"year": "2005", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Six points are chosen on the sides of an equilateral triangle \\(A B C\\) .. \\(A_{1}\\) \\(A_{2}\\) on \\(B C\\) \\(B_{1}\\) \\(B_{2}\\) on \\(C A\\) and \\(C_{1}\\) \\(C_{2}\\) on \\(A B\\) , such that they are the vertices of a convex hexagon \\(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\\) with equal side lengths. Prove that the lines \\(A_{1}B_{2}\\) \\(B_{1}C_{2}\\) and \\(C_{1}A_{2}\\) are concurrent.", "solution": "The six sides of the hexagon, when oriented, comprise six vectors with vanishing sum. However note that \n\n\\[\\overrightarrow{A_1A_2} +\\overrightarrow{B_1B_2} +\\overrightarrow{C_1C_2} = 0.\\] \n\nThus \n\n\\[\\overrightarrow{A_2B_1} +\\overrightarrow{B_2C_1} +\\overrightarrow{C_2A_1} = 0\\] \n\nand since three unit vectors with vanishing sum must be rotations of each other by \\(120^{\\circ}\\) , it follows they must also form an equilateral triangle. \n\n\n \n\nConsequently, triangles \\(A_1A_2B_1\\) , \\(B_1B_2C_1\\) , \\(C_1C_2A_1\\) are congruent, as \\(\\angle A_2 = \\angle B_2 = \\angle C_2\\) . So triangle \\(A_1B_1C_1\\) is equilateral and the diagonals are concurrent at the center.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2005/1, proposed by Bogdan Enescu (ROU) \n"}}
|
| 2 |
+
{"year": "2005", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ... be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer \\(n\\) the numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) leave \\(n\\) different remainders upon division by \\(n\\) . Prove that every integer occurs exactly once in the sequence.", "solution": "Obviously every integer appears at most once (otherwise take \\(n\\) much larger). So we will prove every integer appears at least once. \n\nClaim — For any \\(i < j\\) we have \\(|a_{i} - a_{j}| < j\\) . \n\nProof. Otherwise, let \\(n = |a_{i} - a_{j}| \\neq 0\\) . Then \\(i, j \\in [1, n]\\) and \\(a_{i} \\equiv a_{j} \\pmod{n}\\) , contradiction. \n\nClaim — For any \\(n\\) , the set \\(\\{a_{1}, \\ldots , a_{n}\\}\\) is of the form \\(\\{k + 1, \\ldots , k + n\\}\\) for some integer \\(k\\) . \n\nProof. By induction, with the base case \\(n = 1\\) being vacuous. For the inductive step, suppose \\(\\{a_{1}, \\ldots , a_{n}\\} = \\{k + 1, \\ldots , k + n\\}\\) are determined. Then \n\n\\[a_{n + 1} \\equiv k \\pmod{n + 1}.\\] \n\nMoreover by the earlier claim we have \n\n\\[|a_{n + 1} - a_{1}| < n + 1.\\] \n\nFrom this we deduce \\(a_{n + 1} \\in \\{k, k + n + 1\\}\\) as desired. \n\nThis gives us actually a complete description of all possible sequences satisfying the hypothesis: choose any value of \\(a_{1}\\) to start. Then, for the \\(n\\) th term, the set \\(S = \\{a_{1}, \\ldots , a_{n - 1}\\}\\) is (in some order) a set of \\(n - 1\\) consecutive integers. We then let \\(a_{n} = \\max S + 1\\) or \\(a_{n} = \\min S - 1\\) . A picture of six possible starting terms is shown below. \n\n\n\n\n\n\nFinally, we observe that the condition that the sequence has infinitely many positive and negative terms (which we have not used until now) implies it is unbounded above and below. Thus it must contain every integer.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2005/2, proposed by Nicholas de Bruijn (NLD) \n"}}
|
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{"year": "2005", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Let \\(x,y,z > 0\\) satisfy \\(x y z\\geq 1\\) . Prove that \n\n\\[\\frac{x^{5} - x^{2}}{x^{5} + y^{2} + z^{2}} +\\frac{y^{5} - y^{2}}{x^{2} + y^{5} + z^{2}} +\\frac{z^{5} - z^{2}}{x^{2} + y^{2} + z^{5}}\\geq 0.\\]", "solution": "Negating both sides and adding 3 eliminates the minus signs: \n\n\\[\\sum_{\\mathrm{cyc}}\\frac{1}{x^{5} + y^{2} + z^{2}}\\leq \\frac{3}{x^{2} + y^{2} + z^{2}}.\\] \n\nThus we only need to consider the case \\(x y z = 1\\) \n\nDirect expansion and Muirhead works now! As advertised, once we show it suffices to analyze if \\(x y z = 1\\) the inequality becomes more economically written as \n\n\\[S = \\sum_{\\mathrm{cyc}}x^{2}(x^{2} - y z)(y^{4} + x^{3}z + x z^{3})(z^{4} + x^{3}y + x y^{3})\\stackrel {?}{\\geq}0.\\] \n\nSo, clearing all the denominators gives \n\n\\[S = \\sum_{\\mathrm{cyc}}x^{2}(x^{2} - y z)\\left[y^{4}z^{4} + x^{3}y^{5} + x y^{7} + x^{3}z^{5} + x^{6}y z + x^{4}y^{3}z + x z^{7} + x^{4}y z^{3} + x^{2}y^{3}z^{3}\\right]\\] \\[\\quad = \\sum_{\\mathrm{cyc}}\\left[x^{4}y^{4}z^{4} + x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{8}y^{3}z + x^{5}z^{7} + x^{8}y z^{3} + x^{6}y^{3}z^{3}\\right]\\] \\[\\quad -\\sum_{\\mathrm{cyc}}\\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{3}y^{8}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{3}y z^{8} + x^{6}y^{2}z^{4} + x^{4}y^{4}z^{4}\\right]\\] \\[\\quad = \\sum_{\\mathrm{cyc}}\\left[x^{7}y^{5} + x^{5}y^{7} + x^{7}z^{5} + x^{10}y z + x^{5}z^{7} + x^{6}y^{3}z^{3}\\right]\\] \\[\\quad -\\sum_{\\mathrm{cyc}}\\left[x^{2}y^{5}z^{5} + x^{5}y^{6}z + x^{5}y z^{6} + x^{8}y^{2}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{2}z^{4}\\right]\\] \n\nIn other words we need to show \n\n\\[\\sum_{\\mathrm{sym}}\\left(2x^{7}y^{5} + \\frac{1}{2} x^{10}y z + \\frac{1}{2} x^{6}y^{3}z^{3}\\right)\\geq \\sum_{\\mathrm{sym}}\\left(\\frac{1}{2} x^{8}y^{2}z^{2} + \\frac{1}{2} x^{5}y^{5}z^{2} + x^{6}y^{4}z^{2} + x^{6}y^{5}z\\right).\\] \n\nwhich follows by summing \n\n\\[\\sum_{\\mathrm{sym}}\\frac{x^{10}y z + x^{6}y^{3}z^{3}}{2}\\geq \\sum_{\\mathrm{sym}}x^{8}y^{2}z^{2}\\] \\[\\frac{1}{2}\\sum_{\\mathrm{sym}}x^{8}y^{2}z^{2}\\geq \\frac{1}{2}\\sum_{\\mathrm{sym}}x^{6}y^{4}z^{2}\\] \\[\\frac{1}{2}\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq \\frac{1}{2}\\sum_{\\mathrm{sym}}x^{5}y^{5}z^{2}\\]\n\n\n\n\\[{\\frac{1}{2}}\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq{\\frac{1}{2}}\\sum_{\\mathrm{sym}}x^{6}y^{4}z^{2}\\] \\[\\sum_{\\mathrm{sym}}x^{7}y^{5}\\geq\\sum_{\\mathrm{sym}}x^{6}y^{5}z.\\] \n\nThe first line here comes from AM- GM, the rest come from Muirhead. \n\nRemark. More elegant approach is to use Cauchy in the form \n\n\\[\\frac{1}{x^{5} + y^{2} + z^{2}} \\leq \\frac{x^{-1} + y^{2} + z^{2}}{(x^{2} + y^{2} + z^{2})^{2}}.\\]\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2005/3, proposed by Hojoo Lee (KOR) \n"}}
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{"year": "2005", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Determine all positive integers relatively prime to all the terms of the infinite sequence \n\n\\[a_{n} = 2^{n} + 3^{n} + 6^{n} - 1,\\quad n\\geq 1.\\]", "solution": "The answer is 1 only (which works). \n\nIt suffices to show there are no primes. For the primes \\(p = 2\\) and \\(p = 3\\) , take \\(a_{2} = 48\\) . For any prime \\(p \\geq 5\\) notice that \n\n\\[a_{p - 2} = 2^{p - 2} + 3^{p - 2} + 6^{p - 2} - 1\\] \\[\\equiv \\frac{1}{2} +\\frac{1}{3} +\\frac{1}{6} -1\\pmod {p\\] \\[\\equiv 0\\pmod {p\\] \n\nso no other larger prime works.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2005/4, proposed by Mariusz Skalba (POL) \n"}}
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{"year": "2005", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C D\\) be a fixed convex quadrilateral with \\(B C = D A\\) and \\(\\overline{{B C}}\\not\\parallel\\overline{{D A}}\\) . Let two variable points \\(E\\) and \\(F\\) lie on the sides \\(B C\\) and \\(D A\\) , respectively, and satisfy \\(B E = D F\\) . The lines \\(A C\\) and \\(B D\\) meet at \\(P\\) , the lines \\(B D\\) and \\(E F\\) meet at \\(Q\\) , the lines \\(E F\\) and \\(A C\\) meet at \\(R\\) . Prove that the circumcircles of the triangles \\(P Q R\\) , as \\(E\\) and \\(F\\) vary, have a common point other than \\(P\\) .", "solution": "Let \\(M\\) be the Miquel point of complete quadrilateral \\(A D B C\\) ; in other words, let \\(M\\) be the second intersection point of the circumcircles of \\(\\triangle A P D\\) and \\(\\triangle B P C\\) . (A good diagram should betray this secret; all the points are given in the picture.) This makes lots of sense since we know \\(E\\) and \\(F\\) will be sent to each other under the spiral similarity too. \n\n\n \n\nThus \\(M\\) is the Miquel point of complete quadrilateral \\(F A C E\\) . As \\(R = \\overline{{F E}}\\cap \\overline{{A C}}\\) we deduce \\(F A R M\\) is a cyclic quadrilateral (among many others, but we'll only need one). \n\nNow look at complete quadrilateral \\(A F Q P\\) . Since \\(M\\) lies on \\((D F Q)\\) and \\((R A F)\\) , it follows that \\(M\\) is in fact the Miquel point of \\(A F Q P\\) as well. So \\(M\\) lies on \\((P Q R)\\) . \n\nThus \\(M\\) is the fixed point that we wanted. \n\nRemark. Naturally, the congruent length condition can be relaxed to \\(D F / D A = B E / B C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2005/5, proposed by Waldemar Pompe (POL) \n"}}
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{"year": "2005", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition 6 problems were posed to the contestants. Each pair of problems was solved by more than \\(\\frac{2}{5}\\) of the contestants. Nobody solved all 6 problems. Show that there were at least 2 contestants who each solved exactly 5 problems.", "solution": "Assume not and at most one contestant solved five problems. By adding in solves, we can assume WLOG that one contestant solved problems one through five, and every other contestant solved four of the six problems. \n\nWe split the remaining contestants based on whether they solved P6. Let \\(a_{i}\\) denote the number of contestants who solved \\(\\{1,2,\\ldots ,5\\} \\setminus \\{i\\}\\) (and missed P6). Let \\(b_{ij}\\) denote the number of contestants who solved \\(\\{1,2,\\ldots ,5,6\\} \\setminus \\{i,j\\}\\) , for \\(1\\leq i< j\\leq 5\\) (thus in particular they solved P6). Thus \n\n\\[n = 1 + \\sum_{1\\leq i\\leq 5}a_{i} + \\sum_{1\\leq i< j\\leq 5}b_{ij}\\] \n\ndenotes the total number of contestants. \n\nConsidering contestants who solved P1/P6 we have \n\n\\[t_{1}:= b_{23} + b_{24} + b_{25} + b_{34} + b_{35} + b_{45}\\geq \\frac{2}{5} n + \\frac{1}{5}\\] \n\nand we similarly define \\(t_{2}\\) , \\(t_{3}\\) , \\(t_{4}\\) , \\(t_{5}\\) . (We have written \\(\\frac{2}{5} n + \\frac{1}{5}\\) since we know the left- hand side is an integer strictly larger than \\(\\frac{2}{5} n\\) .) Also, by considering contestants who solved P1/P2 we have \n\n\\[t_{12} = 1 + a_{3} + a_{4} + a_{5} + b_{34} + b_{35} + b_{45}\\geq \\frac{2}{5} n + \\frac{1}{5}\\] \n\nand we similarly define \\(t_{ij}\\) for \\(1\\leq i< j\\leq 5\\) \n\nClaim — The number \\(\\frac{2n + 1}{5}\\) is equal to some integer \\(k\\) , fourteen of the \\(t\\) 's are equal to \\(k\\) , and the last one is equal to \\(k + 1\\) . \n\nProof. First, summing all fifteen equations gives \n\n\\[6n + 4 = 10 + 6(n - 1) = 10 + \\sum_{1\\leq i\\leq 5}6a_{i} + \\sum_{1\\leq i< j\\leq 5}6b_{ij}\\] \\[\\qquad = \\sum_{1\\leq i\\leq 5}t_{i} + \\sum_{1\\leq i< j\\leq 5}t_{ij}.\\] \n\nThus the sum of the 15 \\(t\\) 's is \\(6n + 4\\) . But since all the \\(t\\) 's are integers at least \\(\\frac{2n + 1}{5} = \\frac{6n + 3}{15}\\) , the conclusion follows. \\(\\square\\) \n\nHowever, we will also manipulate the equations to get the following.\n\n\n\nClaim — We have \n\n\\[t_{45} \\equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \\pmod {3}.\\] \n\nProof. This follows directly by computing the coefficient of the \\(a\\) 's and \\(b\\) 's. We will nonetheless write out a derivation of this equation, to motivate it, but the proof stands without it. \n\nLet \\(B = \\sum_{1 \\leq i < j \\leq 5} b_{ij}\\) be the sum of all \\(b\\) 's. First, note that \n\n\\[t_{1} + t_{2} = B + b_{34} + b_{45} + b_{35} - b_{12}\\] \\[\\qquad = B + (t_{12} - 1 - a_{3} - a_{4} - a_{5}) - b_{12}\\] \\[\\Rightarrow b_{12} = B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5}).\\] \n\nThis means we have more or less solved for each \\(b_{ij}\\) in terms of only \\(t\\) and \\(a\\) variables. Now \n\n\\[t_{45} = 1 + a_{1} + a_{2} + a_{3} + b_{12} + b_{23} + b_{31}\\] \\[\\qquad = 1 + a_{1} + a_{2} + a_{3}\\] \\[\\qquad +[B - (t_{1} + t_{2}) + t_{12} - 1 - (a_{3} + a_{4} + a_{5})]\\] \\[\\qquad +[B - (t_{2} + t_{3}) + t_{23} - 1 - (a_{1} + a_{4} + a_{5})]\\] \\[\\qquad +[B - (t_{3} + t_{1}) + t_{13} - 1 - (a_{2} + a_{4} + a_{5})]\\] \\[\\qquad \\equiv 1 + t_{1} + t_{2} + t_{3} + t_{12} + t_{23} + t_{31} \\pmod {3}\\] \n\nas desired. \n\nHowever, we now show the two claims are incompatible (and this is easy, many ways to do this). There are two cases. \n\n- Say \\(t_{5} = k + 1\\) and the others are \\(k\\) . Then the equation for \\(t_{45}\\) gives that \\(k \\equiv 6k + 1 \\pmod {3}\\) . But now the equation for \\(t_{12}\\) give \\(k \\equiv 6k \\pmod {3}\\) . \n\n- Say \\(t_{45} = k + 1\\) and the others are \\(k\\) . Then the equation for \\(t_{45}\\) gives that \\(k + 1 \\equiv 6k \\pmod {3}\\) . But now the equation for \\(t_{12}\\) give \\(k \\equiv 6k + 1 \\pmod {3}\\) . \n\nRemark. It is significantly easier to prove that there is at least one contestant who solved five problems. One can see it by dropping the \\(+10\\) in the proof of the claim, and arrives at a contradiction. In this situation it is not even necessary to set up the many \\(a\\) and \\(b\\) variables; just note that the expected number of contestants solving any particular pair of problems is \\(\\frac{\\binom{4}{2}n}{\\binom{6}{2}} = \\frac{2}{5} n\\) . \n\nThe fact that \\(\\frac{2n + 1}{5}\\) should be an integer also follows quickly, since if not one can improve the bound to \\(\\frac{2n + 5}{5}\\) and quickly run into a contradiction. Again one can get here without setting up \\(a\\) and \\(b\\) . \n\nThe main difficulty seems to be the precision required in order to nail down the second 5- problem solve. \n\nRemark. The second claim may look miraculous, but the proof shows that it is not too unnatural to consider \\(t_{1} + t_{2} - t_{12}\\) to isolate \\(b_{12}\\) in terms of \\(a\\) 's and \\(t\\) 's. The main trick is: why mod 3? \n\nThe reason is that if one looks closely, for a fixed \\(k\\) we have a system of 15 equations in 15 variables. Unless the determinant \\(D\\) of that system happens to be zero, this means there will be a rational solution in \\(a\\) and \\(b\\) , whose denominators are bounded by \\(D\\) . However if\n\n\n\n\\(p\\mid D\\) then we may conceivably run into mod \\(p\\) issues. \n\nThis motivates the choice \\(p = 3\\) , since it is easy to see the determinant is divisible by 3, since constant shifts of \\(\\vec{a}\\) and \\(\\vec{b}\\) are also solutions mod 3. (The choice \\(p = 2\\) is a possible guess as well for this reason, but the problem seems to have better 3- symmetry.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2005-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2005/6, proposed by Radu Gologan, Dan Schwartz (ROU) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with incenter \\(I\\) . A point \\(P\\) in the interior of the triangle satisfies \n\n\\[\\angle P B A + \\angle P C A = \\angle P B C + \\angle P C B.\\] \n\nShow that \\(A P\\geq A I\\) and that equality holds if and only if \\(P = I\\)", "solution": "condition rewrites as \n\n\\[\\angle PBC + \\angle PCB = (\\angle B - \\angle PBC) + (\\angle C - \\angle PCB) \\Rightarrow \\angle PBC + \\angle PCB = \\frac{\\angle B + \\angle C}{2}\\] \n\nwhich means that \n\n\\[\\angle BPC = 180^{\\circ} - \\frac{\\angle B + \\angle C}{2} = 90^{\\circ} + \\frac{\\angle A}{2} = \\angle BIC.\\] \n\nSince \\(P\\) and \\(I\\) are both inside \\(\\triangle ABC\\) that implies \\(P\\) lies on the circumcircle of \\(\\triangle BIC\\) . It's well- known (by \"Fact 5\") that the circumcenter of \\(\\triangle BIC\\) is the arc midpoint \\(M\\) of \\(\\widehat{BC}\\) . Therefore \n\n\\[AI + IM = AM \\leq AP + PM \\Rightarrow AI \\leq AP\\] \n\nwith equality holding iff \\(A\\) , \\(P\\) , \\(M\\) are collinear, or \\(P = I\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2006/1, proposed by Hojoo Lee (KOR) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) be a regular 2006-gon. A diagonal is called good if its endpoints divide the boundary of \\(P\\) into two parts, each composed of an odd number of sides of \\(P\\) . The sides of \\(P\\) are also called good. Suppose \\(P\\) has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of \\(P\\) . Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.", "solution": ". \n\nCall a triangle with the desired property special. We prove the maximum number of special triangles is 1003, achieved by paring up the sides of the polygon. \n\nWe present two solutions for the upper bound. Both of them rely first on two geometric notes: \n\n- In a special triangle, the good sides are congruent (and not congruent to the third side). \n\n- No two isosceles triangles share a good side. \n\n\\(\\P\\) Solution using bijections. Call a good diagonal special if it's part of a special triangle; special diagonals come in pairs. Consider the minor arc cut out by a special diagonal \\(d\\) , which has an odd number of sides. Since special diagonals come in pairs, one can associate to \\(d\\) a side of the polygon not covered by any special diagonals from \\(d\\) . Hence there are at most 2006 special diagonals, so at most 1003 special triangles. \n\n\\(\\P\\) Solution using graph theory. Consider the tree \\(T\\) formed by the 2004 triangles in the dissection, with obvious adjacency. Let \\(F\\) be the forest obtained by deleting any edge corresponding to a good diagonal. Then the resulting graph \\(F\\) has only degrees 1 and 3, with special triangles only occurring at degree 1 vertices. \n\nIf there are \\(k\\) good diagonals drawn, then this forest consists of \\(k + 1\\) trees. A tree with \\(n_{i}\\) vertices ( \\(0 \\leq i \\leq k\\) ) consequently has \\(\\frac{n_{i} + 2}{2}\\) leaves. However by the earlier remark at least \\(k\\) leaves don't give special triangles (one on each side of a special diagonal); so the number of leaves that do give good triangles is at most \n\n\\[-k + \\sum_{i}\\frac{n_{i} + 2}{2} = -k + \\frac{2004 + 2(k + 1)}{2} = 1003.\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2006/2, proposed by Dušan Djukić (SRB) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Determine the least real number \\(M\\) such that the inequality \n\n\\[\\left|a b(a^{2} - b^{2}) + b c(b^{2} - c^{2}) + c a(c^{2} - a^{2})\\right|\\leq M\\left(a^{2} + b^{2} + c^{2}\\right)^{2}\\] \n\nholds for all real numbers \\(a\\) , \\(b\\) and \\(c\\) .", "solution": "It's the same as \n\n\\[|(a - b)(b - c)(c - a)(a + b + c)|\\leq M\\left(a^{2} + b^{2} + c^{2}\\right)^{2}.\\] \n\nLet \\(x = a - b\\) , \\(y = b - c\\) , \\(z = c - a\\) , \\(s = a + b + c\\) . Then we want to have \n\n\\[|xyz s|\\leq \\frac{M}{9} (x^{2} + y^{2} + z^{2} + s^{2})^{2}.\\] \n\nHere \\(x + y + z = 0\\) . \n\nNow if \\(x\\) and \\(y\\) have the same sign, we can replace them with the average (this increases the LHS and decreases RHS). So we can have \\(x = y\\) , \\(z = - 2x\\) . Now WLOG \\(x > 0\\) to get \n\n\\[2x^{3}\\cdot s\\leq \\frac{M}{9}\\left(6x^{2} + s^{2}\\right)^{2}.\\] \n\nAfter this routine calculation gives \\(M = \\frac{9}{12}\\sqrt{2}\\) works and is optimal (by \\(6x^{2} + s^{2} = 2x^{2} + 2x^{2} + 2x^{2} + s^{2}\\) and AM- GM).\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2006/3, proposed by Finbarr Holland (IRL) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Determine all pairs \\((x,y)\\) of integers such that \n\n\\[1 + 2^{x} + 2^{2x + 1} = y^{2}.\\]", "solution": "Answers: \\((0,\\pm 2)\\) , \\((4,\\pm 23)\\) , which work. \n\nAssume \\(x \\geq 4\\) . \n\n\\[2^{x}\\left(1 + 2^{x + 1}\\right) = 2^{x} + 2^{2x + 1} = y^{2} - 1 = (y - 1)(y + 1).\\] \n\nSo either: \n\n- \\(y = 2^{x - 1}m + 1\\) for some odd \\(m\\) , so \n\n\\[1 + 2^{x + 1} = m\\left(2^{x - 2}m + 1\\right)\\Longrightarrow 2^{x} = \\frac{4(1 - m)}{m^{2} - 8}.\\] \n\n- \\(y = 2^{x - 1}m - 1\\) for some odd \\(m\\) , so \n\n\\[1 + 2^{x + 1} = m\\left(2^{x - 2}m - 1\\right)\\Longrightarrow 2^{x} = \\frac{4(1 + m)}{m^{2} - 8}.\\] \n\nIn particular we need \\(4|1\\pm m|\\geq 2^{4}|m^{2} - 8|\\) , which is enough to imply \\(m< 5\\) . From here easily recover \\(x = 4\\) , \\(m = 3\\) as the last solution (in the second case).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2006/4, proposed by Zuming Feng (USA) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(P(x)\\) be a polynomial of degree \\(n > 1\\) with integer coefficients and let \\(k\\) be a positive integer. Consider the polynomial \n\n\\[Q(x) = P(P(\\ldots P(P(x))\\ldots))\\] \n\nwhere \\(P\\) occurs \\(k\\) times. Prove that there are at most \\(n\\) integers \\(t\\) such that \\(Q(t) = t\\) .", "solution": "First, we prove that: \n\nClaim (Putnam 2000 et al) — If a number is periodic under \\(P\\) then in fact it's fixed by \\(P \\circ P\\) . \n\nProof. Let \\(x_{1}\\) , \\(x_{2}\\) , ..., \\(x_{n}\\) be a minimal orbit. Then \n\n\\[x_{i} - x_{i + 1} \\mid P(x_{i}) - P(x_{i + 1}) = x_{i + 1} - x_{i + 2}\\] \n\nand so on cyclically. \n\nIf any of the quantities are zero we are done. Else, we must eventually have \\(x_{i} - x_{i + 1} = -(x_{i + 1} - x_{i + 2})\\) , so \\(x_{i} = x_{i + 2}\\) and we get 2- periodicity. \n\nThe tricky part is to study the 2- orbits. Suppose there exists a fixed pair \\(u \\neq v\\) with \\(P(u) = v\\) , \\(P(v) = u\\) . (If no such pair exists, we are already done.) Let \\((a, b)\\) be any other pair with \\(P(a) = b\\) , \\(P(b) = a\\) , possibly even \\(a = b\\) , but \\(\\{a, b\\} \\cap \\{u, v\\} = \\emptyset\\) . Then we should have \n\n\\[u - a \\mid P(u) - P(a) = v - b \\mid P(v) - P(b) = u - a\\] \n\nand so \\(u - a\\) and \\(v - b\\) divide each other (and are nonzero). Similarly, \\(u - b\\) and \\(v - a\\) divide each other. \n\nHence \\(u - a = \\pm (v - b)\\) and \\(u - b = \\pm (v - a)\\) . We consider all four cases: \n\n- If \\(u - a = v - b\\) and \\(u - b = v - a\\) then \\(u - v = b - a = b\\) , contradiction. \n\n- If \\(u - a = -(v - b)\\) and \\(u - b = -(v - a)\\) then \\(u + v = u - v = a + b\\) . \n\n- If \\(u - a = -(v - b)\\) and \\(u - b = v - a\\) , we get \\(a + b = u + v\\) from the first one (discarding the second). \n\n- If \\(u - a = v - b\\) and \\(u - b = -(v - a)\\) , we get \\(a + b = u + v\\) from the second one (discarding the first one). \n\nThus in all possible situations we have \n\n\\[a + b = c := u + v\\] \n\na fixed constant. \n\nTherefore, any pair \\((a, b)\\) with \\(P(a) = b\\) and \\(P(b) = a\\) actually satisfies \\(P(a) = c - a\\) . And since \\(\\deg P > 1\\) , this means there are at most \\(n\\) roots to \\(a + P(a) = c\\) , as needed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2006/5, proposed by Dan Schwarz (ROU) \n"}}
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{"year": "2006", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Assign to each side \\(b\\) of a convex polygon \\(P\\) the maximum area of a triangle that has \\(b\\) as a side and is contained in \\(P\\) . Show that the sum of the areas assigned to the sides of \\(P\\) is at least twice the area of \\(P\\) .", "solution": "We say a polygon in almost convex if all its angles are at most \\(180^{\\circ}\\) . \n\nNote that given any convex or almost convex polygon, we can take any side \\(b\\) and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point). \n\n## Lemma \n\nLet \\(N\\) be an even integer. Then any almost convex \\(N\\) - gon with area \\(S\\) should have an inscribed triangle with area at least \\(2S / N\\) . \n\nThe main work is the proof of the lemma. \n\nLabel the polygon \\(P_{0}P_{1}\\ldots P_{N - 1}\\) . Consider the \\(N / 2\\) major diagonals of the almost convex \\(N\\) - gon, \\(P_{0}P_{N / 2}\\) , \\(P_{1}P_{N / 2 + 1}\\) , et cetera. A butterfly refers to a self- intersecting quadrilateral \\(P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) . An example of a butterfly is shown below for \\(N = 8\\) . \n\n\n \n\nClaim — Every point \\(X\\) in the polygon is contained in the wingspan of some butterfly. \n\nProof. Consider a windmill- like process which \n\n- starts from some oriented red line \\(P_{0}P_{N / 2}\\) , oriented to face \\(P_{0}P_{N / 2}\\)\n\n\n\n- rotates through \\(P_{0}P_{N / 2} \\cap P_{1}P_{N / 2 + 1}\\) to get line \\(P_{1}P_{N / 2 + 1}\\) , \n\n- rotates through \\(P_{1}P_{N / 2 + 1} \\cap P_{2}P_{N / 2 + 2}\\) to get line \\(P_{2}P_{N / 2 + 2}\\) , \n\n- ...et cetera, until returning to line \\(P_{N / 2}P_{0}\\) , but in the reverse orientation. \n\nAt the end of the process, every point in the plane has switched sides with our moving line. The moment that \\(X\\) crosses the moving red line, we get it contained in a butterfly, as needed. \\(\\square\\) \n\nClaim — If \\(A B D C = P_{i}P_{i + 1}P_{i + 1 + N / 2}P_{i + N / 2}\\) is a butterfly, one of the triangles \\(A B C\\) , \\(B C D\\) , \\(C D A\\) , \\(D A B\\) has area at least that of the butterfly. \n\nProof. Let the diagonals of the butterfly meet at \\(O\\) , and let \\(a = A O\\) , \\(b = B O\\) , \\(c = C O\\) , \\(d = D O\\) . If we assume WLOG \\(d = \\min (a,b,c,d)\\) then it follows \\([A B C] = [A O B] + [B O C] \\geq [A O B] + [C O D]\\) , as needed. \\(\\square\\) \n\nNow, since the \\(N / 2\\) butterflies cover an area of \\(S\\) , it follows that one of the butterflies has area at least \\(S / (N / 2) = 2S / N\\) , and so that butterfly gives a triangle with area at least \\(2S / N\\) , completing the proof of the lemma. \n\nMain proof. Let \\(a_{1}\\) , ..., \\(a_{n}\\) be the numbers assigned to the sides. Assume for contradiction \\(a_{1} + \\dots +a_{n}< 2S\\) . We pick even integers \\(m_{1}\\) , \\(m_{2}\\) , ..., \\(m_{n}\\) such that \n\n\\[\\frac{a_{1}}{S} < \\frac{2m_{1}}{m_{1} + \\dots +m_{n}}\\] \\[\\frac{a_{2}}{S} < \\frac{2m_{2}}{m_{1} + \\dots +m_{n}}\\] \\[\\vdots\\] \\[\\frac{a_{n}}{S} < \\frac{2m_{n}}{m_{1} + \\dots +m_{n}}.\\] \n\nwhich is possible by rational approximation, since the right- hand sides sum to 2 and the left- hand sides sum to strictly less than 2. \n\nNow we break every side of \\(P\\) into \\(m_{i}\\) equal parts to get an almost convex \\(N\\) - gon, where \\(N = m_{1} + \\dots +m_{n}\\) . \n\nThe main lemma then gives us a triangle \\(\\Delta\\) of the almost convex \\(N\\) - gon which has area at least \\(\\frac{2S}{N}\\) . If \\(\\Delta\\) used the \\(i\\) th side then it then follows the label \\(a_{i}\\) on that side should be at least \\(m_{i} \\cdot \\frac{2S}{N}\\) , contradiction.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2006-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2006/6, proposed by Dušan Djukić (SRB) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Real numbers \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) are fixed. For each \\(1 \\leq i \\leq n\\) we let \\(d_{i} = \\max \\{a_{j} : 1 \\leq j \\leq i\\} - \\min \\{a_{j} : i \\leq j \\leq n\\}\\) and let \\(d = \\max \\{d_{i} : 1 \\leq i \\leq n\\}\\) . \n\n(a) Prove that for any real numbers \\(x_{1} \\leq \\dots \\leq x_{n}\\) we have \n\n\\[\\max \\left\\{|x_{i} - a_{i}|:1\\leq i\\leq n\\right\\} \\geq \\frac{1}{2} d.\\] \n\n(b) Moreover, show that there exists some choice of \\(x_{1} \\leq \\dots \\leq x_{n}\\) which achieves equality.", "solution": "Note that we can dispense of \\(d_{i}\\) immediately by realizing that the definition of \\(d\\) just says \n\n\\[d = \\max_{1 \\leq i \\leq j \\leq n} (a_{i} - a_{j}).\\] \n\nIf \\(a_{1} \\leq \\dots \\leq a_{n}\\) are already nondecreasing then \\(d = 0\\) and there is nothing to prove (for the equality case, just let \\(x_{i} = a_{i}\\) ), so we will no longer consider this case. \n\nOtherwise, consider any indices \\(i < j\\) with \\(a_{i} > a_{j}\\) . We first prove (a) by applying the following claim with \\(p = a_{i}\\) and \\(q = a_{j}\\) : \n\nClaim — For any \\(p \\leq q\\) , we have either \\(|p - a_{i}| \\geq \\frac{1}{2} (a_{i} - a_{j})\\) or \\(|q - a_{j}| \\geq \\frac{1}{2} (a_{i} - a_{j}) \n\nProof. Assume for contradiction both are false. Then \\(p > a_{i} - \\frac{1}{2} (a_{i} - a_{j}) = a_{j} + \\frac{1}{2} (a_{i} - a_{j}) > q\\) , contradiction. \\(\\square\\) \n\nAs for (b), we let \\(i < j\\) be any indices for which \\(a_{i} - a_{j} = d > 0\\) achieves the maximal difference. We then define \\(x_{\\bullet}\\) in three steps: \n\nWe set \\(x_{k} = \\frac{a_{i} + a_{j}}{2}\\) for \\(k = i, \\ldots , j\\) . We recursively set \\(x_{k} = \\max (x_{k - 1}, a_{k})\\) for \\(k = j + 1, j + 2, \\ldots\\) . We recursively set \\(x_{k} = \\min (x_{k + 1}, a_{k})\\) for \\(k = i - 1, i - 2, \\ldots\\) . \n\nBy definition, these \\(x_{\\bullet}\\) are weakly increasing. To prove this satisfies (b) we only need to check that \n\n\\[|x_{k} - a_{k}| \\leq \\frac{a_{i} - a_{j}}{2} \\qquad (\\star)\\] \n\nfor any index \\(k\\) (as equality holds for \\(k = i\\) or \\(k = j\\) ). \n\nWe note \\((\\star)\\) holds for \\(i < k < j\\) by construction. For \\(k > j\\) , note that \\(x_{k} \\in \\{a_{j}, a_{j + 1}, \\ldots , a_{k}\\}\\) by construction, so \\((\\star)\\) follows from our choice of \\(i\\) and \\(j\\) giving the largest possible difference; the case \\(k < i\\) is similar.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2007/1, proposed by Michael Albert (NZL) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Consider five points \\(A\\) , \\(B\\) , \\(C\\) , \\(D\\) and \\(E\\) such that \\(ABCD\\) is a parallelogram and \\(BCED\\) is a cyclic quadrilateral. Let \\(\\ell\\) be a line passing through \\(A\\) . Suppose that \\(\\ell\\) intersects the interior of the segment \\(DC\\) at \\(F\\) and intersects line \\(BC\\) at \\(G\\) . Suppose also that \\(EF = EG = EC\\) . Prove that \\(\\ell\\) is the bisector of angle \\(DAB\\) .", "solution": "Let \\(M\\) , \\(N\\) , \\(P\\) denote the midpoints of \\(\\overline{CF}\\) , \\(\\overline{CG}\\) , \\(\\overline{AC}\\) (noting \\(P\\) is also the midpoint of \\(\\overline{BD}\\) ). \n\nBy a homothety at \\(C\\) with ratio \\(\\frac{1}{2}\\) , we find \\(\\overline{MNP}\\) is the image of line \\(\\ell \\equiv \\overline{AGF}\\) . \n\n\n \n\nHowever, since we also have \\(\\overline{EM} \\perp \\overline{CF}\\) and \\(\\overline{EN} \\perp \\overline{CG}\\) (from \\(EF = EG = EC\\) ) we conclude \\(\\overline{PMN}\\) is the Simson line of \\(E\\) with respect to \\(\\triangle BCD\\) , which implies \\(\\overline{EP} \\perp \\overline{BD}\\) . In other words, \\(\\overline{EP}\\) is the perpendicular bisector of \\(\\overline{BD}\\) , so \\(E\\) is the midpoint of arc \\(\\overline{BCD}\\) . \n\nFinally, \n\n\\[\\angle (\\overline{AB}, \\ell) = \\angle (\\overline{CD}, \\overline{MNP}) = \\angle CMN = \\angle CEN\\] \\[\\qquad = 90^{\\circ} - \\angle NCE = 90^{\\circ} + \\angle ECB\\] \n\nwhich means that \\(\\ell\\) is parallel to a bisector of \\(\\angle BCD\\) , and hence to one of \\(\\angle BAD\\) . (Moreover since \\(F\\) lies on the interior of \\(\\overline{CD}\\) , it is actually the internal bisector.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2007/2, proposed by Charles Leytem (LUX) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "In a mathematical competition some competitors are (mutual) friends. Call a group of competitors a clique if each two of them are friends. Given that the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.", "solution": "Take the obvious graph interpretation \\(G\\) . We paint red any vertices in one of the maximal cliques \\(K\\) , which we assume has \\(2r\\) vertices, and paint the remaining vertices green. We let \\(\\alpha (\\bullet)\\) denote the clique number. \n\nInitially, let the two rooms \\(A = K\\) , \\(B = G - K\\) . \n\nClaim — We can move at most \\(r\\) vertices of \\(A\\) into \\(B\\) to arrive at \\(\\alpha (A) \\leq \\alpha (B) \\leq \\alpha (A) + 1\\) . \n\nProof. This is actually obvious by discrete continuity. We move one vertex at a time, noting \\(\\alpha (A)\\) decreases by one at each step, while \\(\\alpha (B)\\) increases by either zero or one at each step. \n\nWe stop once \\(\\alpha (B) \\geq \\alpha (A)\\) , which happens before we have moved \\(r\\) vertices (since then we have \\(\\alpha (B) \\geq r = \\alpha (A)\\) ). The conclusion follows. \\(\\square\\) \n\nSo let's consider the situation \n\n\\[\\alpha (A) = k \\geq r \\qquad \\text{and} \\qquad \\alpha (B) = k + 1.\\] \n\nAt this point \\(A\\) is a set of \\(k\\) red vertices, while \\(B\\) has the remaining \\(2r - k\\) red vertices (and all the green ones). An example is shown below with \\(k = 4\\) and \\(2r = 6\\) . \n\n\n \n\nNow, if we can move any red vertex from \\(B\\) back to \\(A\\) without changing the clique number of \\(B\\) , we do so, and win. \n\nOtherwise, it must be the case that every \\((k + 1)\\) - clique in \\(B\\) uses every red vertex in \\(B\\) . For each \\((k + 1)\\) - clique in \\(B\\) (in arbitrary order), we do the following procedure.\n\n\n\n- If all \\(k + 1\\) vertices are still green, pick one and re-color it blue. This is possible since \\(k + 1 > 2r - k\\) . \n\n- Otherwise, do nothing. \n\nThen we move all the blue vertices from \\(B\\) to \\(A\\) , one at a time, in the same order we re- colored them. This forcibly decreases the clique number of \\(B\\) to \\(k\\) , since the clique number is \\(k + 1\\) just before the last blue vertex is moved, and strictly less than \\(k + 1\\) (hence equal to \\(k\\) ) immediately after that. \n\nClaim — After this, \\(\\alpha (A) = k\\) still holds. \n\nProof. Assume not, and we have a \\((k + 1)\\) - clique which uses \\(b\\) blue vertices and \\((k + 1) - b\\) red vertices in \\(A\\) . Together with the \\(2r - k\\) red vertices already in \\(B\\) we then get a clique of size \n\n\\[b + ((k + 1 - b)) + (2r - k) = 2r + 1\\] \n\nwhich is a contradiction. \n\nRemark. Dragomir Grozev posted the following motivation on his blog: \n\nI think, it's a natural idea to place all students in one room and begin moving them one by one into the other one. Then the max size of the cliques in the first and second room increase (resp. decrease) at most with one. So, there would be a moment both sizes are almost the same. At that moment we may adjust something. \n\nTrying the idea, I had some difficulties keeping track of the maximal cliques in the both rooms. It seemed easier all the students in one of the rooms to comprise a clique. It could be achieved by moving only the members of the maximal clique. Following this path the remaining obstacles can be overcome naturally.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2007/3, proposed by Vasily Astakhov (RUS) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "In triangle \\(ABC\\) the bisector of \\(\\angle BCA\\) meets the circumcircle again at \\(R\\) , the perpendicular bisector of \\(\\overline{BC}\\) at \\(P\\) , and the perpendicular bisector of \\(\\overline{AC}\\) at \\(Q\\) . The midpoint of \\(\\overline{BC}\\) is \\(K\\) and the midpoint of \\(\\overline{AC}\\) is \\(L\\) . Prove that the triangles \\(RPK\\) and \\(RQL\\) have the same area.", "solution": "We first begin by proving the following claim. \n\nClaim — We have \\(CQ = PR\\) (equivalently, \\(CP = QR\\) ). \n\nProof. Let \\(O = \\overline{LQ} \\cap \\overline{KP}\\) be the circumcenter. Then \n\n\\[\\angle OPQ = \\angle KPC = 90^{\\circ} - \\angle PCK = 90^{\\circ} - \\angle LCQ = \\angle \\angle CQL = \\angle PQO.\\] \n\nThus \\(OP = OQ\\) . Since \\(OC = OR\\) as well, we get the conclusion. \n\nDenote by \\(X\\) and \\(Y\\) the feet from \\(R\\) to \\(\\overline{CA}\\) and \\(\\overline{CB}\\) , so \\(\\triangle CXR \\cong \\triangle CYR\\) . Then, let \\(t = \\frac{CQ}{CR} = 1 - \\frac{CP}{CR}\\) . \n\n\n \n\nThen it follows that \n\n\\[[RQ L] = [X Q L] = t(1 - t)\\cdot [X R C] = t(1 - t)\\cdot [Y C R] = [Y K P] = [R K P]\\] \n\nas needed. \n\nRemark. Trigonometric approaches are very possible (and easier to find) as well: both areas work out to be \\(\\frac{1}{8} ab \\tan \\frac{1}{2} C\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2007/4, proposed by Marek Pechal (CZE) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(a\\) and \\(b\\) be positive integers. Show that if \\(4ab - 1\\) divides \\((4a^{2} - 1)^{2}\\) , then \\(a = b\\) .", "solution": "As usual, \n\n\\[4ab - 1\\mid (4a^{2} - 1)^{2}\\iff 4ab - 1\\mid (4ab\\cdot a - b)^{2}\\iff 4ab - 1\\mid (a - b)^{2}.\\] \n\nThen we use a typical Vieta jumping argument. Define \n\n\\[k = \\frac{(a - b)^{2}}{4ab - 1}.\\] \n\nNote that \\(k = 0 \\iff a = b\\) . So we will prove that \\(k > 0\\) leads to a contradiction. \n\nIndeed, suppose \\((a,b)\\) is a minimal solution with \\(a > b\\) (we have \\(a\\neq b\\) since \\(k\\neq 0\\) - By Vieta jumping, \\((b,\\frac{b^{2} + k}{a})\\) is also such a solution. But now \n\n\\[\\frac{b^{2} + k}{a}\\geq a\\Rightarrow k\\geq a^{2} - b^{2}\\] \\[\\Rightarrow \\frac{(a - b)^{2}}{4ab - 1}\\geq a^{2} - b^{2}\\] \\[\\Rightarrow a - b\\geq (4ab - 1)(a + b)\\] \n\nwhich is absurd for \\(a,b\\in \\mathbb{Z}_{>0}\\) . (In the last step we divided by \\(a - b > 0\\) -", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2007/5, proposed by Kevin Buzzard, Edward Crane (UNK) \n"}}
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{"year": "2007", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) be a positive integer. Consider \n\n\\[S = \\{(x,y,z) \\mid x,y,z \\in \\{0,1,\\ldots ,n\\} , x + y + z > 0\\}\\] \n\nas a set of \\((n + 1)^{3} - 1\\) points in the three- dimensional space. Determine the smallest possible number of planes, the union of which contains \\(S\\) but does not include \\((0,0,0)\\) .", "solution": "The answer is \\(3n\\) . Here are two examples of constructions with \\(3n\\) planes: \n\n\\(\\cdot x + y + z = i\\) for \\(i = 1,\\ldots ,3n\\) \n\n\\(\\cdot x = i\\) \\(y = i\\) \\(z = i\\) for \\(i = 1,\\ldots ,n\\) \n\nSuppose for contradiction we have \\(N< 3n\\) planes. Let them be \\(a_{i}x + b_{i}y + c_{i}z + 1 = 0\\) for \\(i = 1,\\ldots ,N\\) . Define the polynomials \n\n\\[A(x,y,z) = \\prod_{i = 1}^{n}(x - i)\\prod_{i = 1}^{n}(y - i)\\prod_{i = 1}^{n}(z - i)\\] \\[B(x,y,z) = \\prod_{i = 1}^{N}(a_{i}x + b_{i}y + c_{i}z + 1).\\] \n\nNote that \\(A(0,0,0) = (- 1)^{n}(n!)^{3}\\neq 0\\) and \\(B(0,0,0) = 1\\neq 0\\) , but \\(A(x,y,z) =\\) \\(B(x,y,z) = 0\\) for any \\((x,y,z)\\in S\\) . Also, the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(A\\) is 1, while the coefficient of \\(x^{n}y^{n}z^{n}\\) in \\(B\\) is 0. \n\nNow, define \n\n\\[P(x,y,z):= A(x,y,z) - \\lambda B(x,y,z).\\] \n\nwhere \\(\\lambda = \\frac{A(0,0,0)}{B(0,0,0)} = (- 1)^{n}(n!)^{3}\\) . We now have that \n\n\\(\\cdot P(x,y,z) = 0\\) for any \\(x,y,z\\in \\{0,1,\\ldots ,n\\}^{3}\\) \n\nBut the coefficient of \\(x^{n}y^{n}z^{n}\\) is 1. \n\nThis is a contradiction to Alon's combinatorial nullstellensatz.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2007-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2007/6, proposed by Gerhard Woeginger (NLD) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(H\\) be the orthocenter of an acute-angled triangle \\(ABC\\) . The circle \\(\\Gamma_{A}\\) centered at the midpoint of \\(\\overline{BC}\\) and passing through \\(H\\) intersects the sideline \\(BC\\) at points \\(A_{1}\\) and \\(A_{2}\\) . Similarly, define the points \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , and \\(C_{2}\\) . Prove that six points \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic.", "solution": ". \n\nWe show two solutions. \n\n\\(\\P\\) First solution using power of a point. Let \\(D\\) , \\(E\\) , \\(F\\) be the centers of \\(\\Gamma_{A}\\) , \\(\\Gamma_{B}\\) , \\(\\Gamma_{C}\\) (in other words, the midpoints of the sides). \n\nWe first show that \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. It suffices to prove that \\(A\\) lies on the radical axis of the circles \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . \n\n\n \n\nLet \\(X\\) be the second intersection of \\(\\Gamma_{B}\\) and \\(\\Gamma_{C}\\) . Clearly \\(\\overline{XH}\\) is perpendicular to the line joining the centers of the circles, namely \\(\\overline{EF}\\) . But \\(\\overline{EF} \\parallel \\overline{BC}\\) , so \\(\\overline{XH} \\perp \\overline{BC}\\) . Since \\(\\overline{AH} \\perp \\overline{BC}\\) as well, we find that \\(A\\) , \\(X\\) , \\(H\\) are collinear, as needed. \n\nThus, \\(B_{1}\\) , \\(B_{2}\\) , \\(C_{1}\\) , \\(C_{2}\\) are concyclic. Similarly, \\(C_{1}\\) , \\(C_{2}\\) , \\(A_{1}\\) , \\(A_{2}\\) are concyclic, as are \\(A_{1}\\) , \\(A_{2}\\) , \\(B_{1}\\) , \\(B_{2}\\) . Now if any two of these three circles coincide, we are done; else the pairwise radical axii are not concurrent, contradiction. (Alternatively, one can argue directly that \\(O\\) is the center of all three circles, by taking the perpendicular bisectors.) \n\n\\(\\P\\) Second solution using length chase (Ritwin Narra). We claim the circumcenter \\(O\\) of \\(\\triangle ABC\\) is in fact the center of \\((A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\\) . \n\nDefine \\(D\\) , \\(E\\) , \\(F\\) as before. Then since \\(\\overline{OD} \\perp \\overline{A_{1}A_{2}}\\) and \\(DA_{1} = DA_{2}\\) , which means \\(OA_{1} = OA_{2}\\) . Similarly, we have \\(OB_{1} = OB_{2}\\) and \\(OC_{1} = OC_{2}\\) . \n\nNow since \\(DA_{1} = DA_{2} = DH\\) , we have \\(OA_{1}^{2} = OD^{2} + HD^{2}\\) . We seek to show \n\n\\[OD^{2} + HD^{2} = OE^{2} + HE^{2} = OF^{2} + HF^{2}.\\]\n\n\n\nThis is clear by Appollonius's Theorem since \\(D\\) , \\(E\\) , and \\(F\\) lie on the nine- point circle, which is centered at the midpoint of \\(\\overline{OH}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2008/1, proposed by Andrey Gavrilyuk (RUS) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(x\\) , \\(y\\) , \\(z\\) be real numbers with \\(xyz = 1\\) , all different from 1. Prove that \n\n\\[\\frac{x^{2}}{(x - 1)^{2}} +\\frac{y^{2}}{(y - 1)^{2}} +\\frac{z^{2}}{(z - 1)^{2}}\\geq 1\\] \n\nand show that equality holds for infinitely many choices of rational numbers \\(x\\) , \\(y\\) , \\(z\\) .", "solution": "Let \\(x = a / b\\) , \\(y = b / c\\) , \\(z = c / a\\) , so we want to show \n\n\\[\\left(\\frac{a}{a - b}\\right)^{2} + \\left(\\frac{b}{b - c}\\right)^{2} + \\left(\\frac{c}{c - a}\\right)^{2}\\geq 1.\\] \n\nA boring computation shows this is equivalent to \n\n\\[\\frac{(a^{2}b + b^{2}c + c^{2}a - 3abc)^{2}}{(a - b)^{2}(b - c)^{2}(c - a)^{2}}\\geq 0\\] \n\nwhich proves the inequality (and it is unsurprising we are in such a situation, given that there is an infinite curve of rationals). \n\nFor equality, it suffices to show there are infinitely many integer solutions to \n\n\\[a^{2}b + b^{2}c + c^{2}a = 3abc\\iff \\frac{a}{c} +\\frac{b}{a} +\\frac{c}{a} = 3\\] \n\nor equivalently that there are infinitely many rational solutions to \n\n\\[u + v + \\frac{1}{uv} = 3.\\] \n\nFor any \\(0\\neq u\\in \\mathbb{Q}\\) the real solution for \\(u\\) is \n\n\\[v = \\frac{-u + (u - 1)\\sqrt{1 - 4 / u} + 3}{2}\\] \n\nand there are certainly infinitely many rational numbers \\(u\\) for which \\(1 - 4 / u\\) is a rational square (say, \\(u = \\frac{- 4}{q^{2} - 1}\\) for \\(q\\neq \\pm 1\\) a rational number).", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2008/2, proposed by Walther Janous (AUT) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Prove that there are infinitely many positive integers \\(n\\) such that \\(n^{2} + 1\\) has a prime factor greater than \\(2n + \\sqrt{2n}\\) .", "solution": "Problem statement \n\nThe idea is to pick the prime \\(p\\) first! \n\nSelect any large prime \\(p \\geq 2013\\) , and let \\(h = \\left\\lceil \\sqrt{p}\\right\\rceil\\) . We will try to find an \\(n\\) such that \n\n\\[n \\leq \\frac{1}{2} (p - h) \\quad \\text{and} \\quad p \\mid n^{2} + 1.\\] \n\nThis implies \\(p \\geq 2n + \\sqrt{p}\\) which is enough to ensure \\(p \\geq 2n + \\sqrt{2n}\\) . \n\nAssume \\(p \\equiv 1\\) (mod 8) henceforth. Then there exists some \\(\\frac{1}{2} p < x < p\\) such that \\(x^{2} \\equiv - 1\\) (mod \\(p\\) ), and we set \n\n\\[x = \\frac{p + 1}{2} +t.\\] \n\nClaim — We have \\(t \\geq \\frac{h - 1}{2}\\) and hence may take \\(n = p - x\\) . \n\nProof. Assume for contradiction this is false; then \n\n\\[0\\equiv 4(x^{2} + 1)\\pmod{p\\] \\[\\quad = (p + 1 + 2t)^{2} + 4\\] \\[\\quad \\equiv (2t + 1)^{2} + 4\\pmod{p\\] \\[\\quad < h^{2} + 4\\] \n\nSo we have that \\((2t + 1)^{2} + 4\\) is positive and divisible by \\(p\\) , yet at most \\(\\left\\lceil \\sqrt{p}\\right\\rceil^{2} + 4 < 2p\\) . So it must be the case that \\((2t + 1)^{2} + 4 = p\\) , but this has no solutions modulo 8. \\(\\square\\)\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2008/3, proposed by Kestutis Česnavičius (LTU) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\) from the positive reals to the positive reals such that \n\n\\[\\frac{f(w)^{2} + f(x)^{2}}{f(y^{2}) + f(z^{2})} = \\frac{w^{2} + x^{2}}{y^{2} + z^{2}}\\] \n\nfor all positive real numbers \\(w\\) , \\(x\\) , \\(y\\) , \\(z\\) satisfying \\(wx = yz\\) .", "solution": "The answers are \\(f(x) \\equiv x\\) and \\(f(x) \\equiv 1 / x\\) . These work, so we show they are the only ones. \n\nFirst, setting \\((t, t, t, t)\\) gives \\(f(t^{2}) = f(t)^{2}\\) . In particular, \\(f(1) = 1\\) . Next, setting \\((t, 1, \\sqrt{t}, \\sqrt{t})\\) gives \n\n\\[\\frac{f(t)^{2} + 1}{2f(t)} = \\frac{t^{2} + 1}{2t}\\] \n\nwhich as a quadratic implies \\(f(t) \\in \\{t, 1 / t\\}\\) . \n\nNow assume \\(f(a) = a\\) and \\(f(b) = 1 / b\\) . Setting \\((\\sqrt{a}, \\sqrt{b}, 1, \\sqrt{ab})\\) gives \n\n\\[\\frac{a + 1 / b}{f(ab) + 1} = \\frac{a + b}{ab + 1}.\\] \n\nOne can check the two cases on \\(f(ab)\\) each imply \\(a = 1\\) and \\(b = 1\\) respectively. Hence the only answers are those claimed.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2008/4, proposed by Hojoo Lee (KOR) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Let \\(n\\) and \\(k\\) be positive integers with \\(k \\geq n\\) and \\(k - n\\) an even number. There are \\(2n\\) lamps labelled 1, 2, ..., \\(2n\\) each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on). Let \\(N\\) be the number of such sequences consisting of \\(k\\) steps and resulting in the state where lamps 1 through \\(n\\) are all on, and lamps \\(n + 1\\) through \\(2n\\) are all off. Let \\(M\\) be number of such sequences consisting of \\(k\\) steps, resulting in the state where lamps 1 through \\(n\\) are all on, and lamps \\(n + 1\\) through \\(2n\\) are all off, but where none of the lamps \\(n + 1\\) through \\(2n\\) is ever switched on. Determine \\(\\frac{N}{M}\\) .", "solution": "The answer is \\(2^{k - n}\\) . \n\nConsider the following map \\(\\Psi\\) from \\(N\\) - sequences to \\(M\\) - sequences: \n\n- change every instance of \\(n + 1\\) to 1; \n\n- change every instance of \\(n + 2\\) to 2; \n\n: \n\n- change every instance of \\(2n\\) to \\(n\\) . \n\n(For example, suppose \\(k = 9\\) , \\(n = 3\\) ; then \\(144225253 \\mapsto 111222223\\) .) \n\nClearly this is map is well- defined and surjective. So all that remains is: \n\nClaim — Every \\(M\\) - sequence has exactly \\(2^{k - n}\\) pre- images under \\(\\Psi\\) . \n\nProof. Indeed, suppose that there are \\(c_{1}\\) instances of lamp 1. Then we want to pick an odd subset of the 1's to change to \\(n + 1\\) 's, so \\(2^{c_{1} - 1}\\) ways to do this. And so on. Hence the number of pre- images is \n\n\\[\\prod_{i}2^{c_{i} - 1} = 2^{k - n}.\\]", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2008/5, proposed by Bruno Le Floch and Ilia Smilga (FRA) \n"}}
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{"year": "2008", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(ABCD\\) be a convex quadrilateral with \\(BA \\neq BC\\) . Denote the incircles of triangles \\(ABC\\) and \\(ADC\\) by \\(\\omega_{1}\\) and \\(\\omega_{2}\\) respectively. Suppose that there exists a circle \\(\\omega\\) tangent to ray \\(BA\\) beyond \\(A\\) and to the ray \\(BC\\) beyond \\(C\\) , which is also tangent to the lines \\(AD\\) and \\(CD\\) . Prove that the common external tangents to \\(\\omega_{1}\\) and \\(\\omega_{2}\\) intersect on \\(\\omega\\) .", "solution": "Let \\(A B C D \n\nBy the external version of Pitot theorem, the existence of \\(\\omega\\) implies that \n\n\\[B A + A D = C B + C D.\\] \n\nLet \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) be diameters of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) with \\(P,T\\in \\overline{{A C}}\\) . Then the length relation on \\(A B C D\\) implies that \\(P\\) and \\(T\\) are reflections about the midpoint of \\(\\overline{{A C}}\\) \n\nNow orient \\(A C\\) horizontally and let \\(K\\) be the \"uppermost\" point of \\(\\omega\\) , as shown. \n\n\n \n\nConsequently, a homothety at \\(B\\) maps \\(Q\\) , \\(T\\) , \\(K\\) to each other (since \\(T\\) is the uppermost of the excircle, \\(Q\\) of the incircle). Similarly, a homothety at \\(D\\) maps \\(P\\) , \\(S\\) , \\(K\\) to each other. As \\(\\overline{{P Q}}\\) and \\(\\overline{{S T}}\\) are parallel diameters it then follows \\(K\\) is the exsimilicenter of \\(\\omega_{1}\\) and \\(\\omega_{2}\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2008-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2008/6, proposed by Vladimir Shmarov (RUS) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Let \\(n,k\\geq 2\\) be positive integers and let \\(a_{1}\\) , \\(a_{2}\\) , \\(a_{3}\\) , ..., \\(a_{k}\\) be distinct integers in the set \\(\\{1,2,\\ldots ,n\\}\\) such that \\(n\\) divides \\(a_{i}(a_{i + 1} - 1)\\) for \\(i = 1,2,\\ldots ,k - 1\\) . Prove that \\(n\\) does not divide \\(a_{k}(a_{1} - 1)\\) .", "solution": "We proceed indirectly and assume that \n\n\\[a_{i}(a_{i + 1} - 1)\\equiv 0\\pmod {n}\\] \n\nfor \\(i = 1,\\ldots ,k\\) (indices taken modulo \\(k\\) ). We claim that this implies all the \\(a_{i}\\) are equal modulo \\(n\\) . \n\nLet \\(q = p^{e}\\) be any prime power dividing \\(n\\) . Then, \\(a_{1}(a_{2} - 1)\\equiv 0\\) (mod \\(q\\) ), so \\(p\\) divides either \\(a_{1}\\) or \\(a_{2} - 1\\) . \n\n- If \\(p \\mid a_{1}\\) , then \\(p \\nmid a_{1} - 1\\) . Then \n\n\\[a_{k}(a_{1} - 1)\\equiv 0\\pmod {q}\\implies a_{k}\\equiv 0\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{k}\\) . So repeating this argument, we get \\(a_{k - 1} \\equiv 0\\) (mod \\(q\\) ), \\(a_{k - 2} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\n- Similarly, if \\(p \\mid a_{2} - 1\\) then \\(p \\nmid a_{2}\\) , and \n\n\\[a_{2}(a_{3} - 1)\\equiv 0\\pmod {q}\\implies a_{3}\\equiv 1\\pmod {q}.\\] \n\nIn particular, \\(p \\mid a_{3} - 1\\) . So repeating this argument, we get \\(a_{4} \\equiv 0\\) (mod \\(q\\) ), \\(a_{5} \\equiv 0\\) (mod \\(q\\) ), and so on. \n\nEither way, we find \\(a_{i}\\) (mod \\(q\\) ) is constant (and either 0 or 1). \n\nSince \\(q\\) was an arbitrary prime power dividing \\(n\\) , by Chinese remainder theorem we conclude that \\(a_{i}\\) (mod \\(n\\) ) is constant as well. But this contradicts the assumption of distinctness.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2009/1, proposed by Ross Atkins (AUS) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with circumcenter \\(o\\) . The points \\(P\\) and \\(Q\\) are interior points of the sides \\(C A\\) and \\(A B\\) respectively. Let \\(K\\) , \\(L\\) , \\(M\\) be the midpoints of \\(\\overline{{B P}}\\) \\(\\overline{{C Q}}\\) , \\(\\overline{{P Q}}\\) . Suppose that \\(\\overline{{P Q}}\\) is tangent to the circumcircle of \\(\\triangle K L M\\) . Prove that \\(O P = O Q\\) .", "solution": "a point, we have \\(- AQ \\cdot QB = OQ^2 - R^2\\) and \\(- AP \\cdot PC = OP^2 - R^2\\) . Therefore, it suffices to show \\(AQ \\cdot QB = AP \\cdot PC\\) . \n\n\n \n\nAs \\(\\overline{ML} \\parallel \\overline{AC}\\) and \\(\\overline{MK} \\parallel \\overline{AB}\\) we have that \n\n\\[\\angle APQ = \\angle LMP = \\angle LKM\\] \\[\\angle PQA = \\angle KMQ = \\angle MLK\\] \n\nand consequently we have the (opposite orientation) similarity \n\n\\[\\triangle APQ \\sim \\triangle MKL.\\] \n\nTherefore \n\n\\[\\frac{AQ}{AP} = \\frac{ML}{MK} = \\frac{2ML}{2MK} = \\frac{PC}{QB}\\] \n\nid est \\(AQ \\cdot QB = AP \\cdot PC\\) , which is what we wanted to prove.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "2. ", "solution_match": "## \\(\\S 1.2\\) IMO 2009/2, proposed by Sergei Berlov (RUS) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Suppose that \\(s_{1},s_{2},s_{3},\\ldots\\) is a strictly increasing sequence of positive integers such that the sub-sequences \\(s_{s_{1}}\\) , \\(s_{s_{2}}\\) , \\(s_{s_{3}}\\) , ...and \\(s_{s_{1} + 1}\\) , \\(s_{s_{2} + 1}\\) , \\(s_{s_{3} + 1}\\) , ...are both arithmetic progressions. Prove that the sequence \\(s_{1}\\) , \\(s_{2}\\) , \\(s_{3}\\) , ...is itself an arithmetic progression.", "solution": "We present two solutions. \n\n\\(\\P\\) First solution (Alex Zhai). Let \\(s(n):= s_{n}\\) and write \n\n\\[s(s(n)) = Dn + A\\] \\[s(s(n) + 1) = D^{\\prime}n + B.\\] \n\nIn light of the bounds \\(s(s(n))\\leq s(s(n) + 1)\\leq s(s(n + 1))\\) we right away recover \\(D = D^{\\prime}\\) and \\(A\\leq B\\) \n\nLet \\(d_{n} = s(n + 1) - s(n)\\) . Note that \\(\\sup d_{n}< \\infty\\) since \\(d_{n}\\) is bounded above by \\(A\\) \n\nThen we let \n\n\\[m:= \\min d_{n},\\qquad M:= \\max d_{n}.\\] \n\nNow suppose \\(a\\) achieves the maximum, meaning \\(s(a + 1) - s(a) = M\\) . Then \n\n\\[\\underbrace{d_{s(s(a))} + \\cdot\\cdot\\cdot + d_{s(s(a + 1)) - 1}}_{D\\mathrm{~terms}} = \\underbrace{\\left[s(s(s(a + 1))) - s(s(s(a)))\\right]}_{D\\mathrm{~terms}}\\] \\[\\qquad = (D\\cdot s(a + 1) + A) - (D\\cdot s(a) + A) = DM.\\] \n\nNow \\(M\\) was maximal hence \\(M = d_{s(s(a))} = \\dots = d_{s(s(a + 1)) - 1}\\) . But \\(d_{s(s(a))} = B - A\\) is a constant. Hence \\(M = B - A\\) . In the same way \\(m = B - A\\) as desired. \n\n\\(\\P\\) Second solution. We retain the notation \\(D\\) , \\(A\\) , \\(B\\) above, as well as \\(m = \\min_{n}s(n+\\) \\(1) - s(n)\\geq B - A\\) . We do the involution trick first as: \n\n\\[D = \\left[s(s(s(n) + 1)) - s(s(s(n)))\\right] = s(Dn + B) - s(Dn + A)\\] \n\nand hence we recover \\(D\\geq m(B - A)\\) \n\nThe edge case \\(D = B - A\\) is easy since then \\(m = 1\\) and \\(D = s(Dn + B) - s(Dn + A)\\) forces \\(s\\) to be a constant shift. So henceforth assume \\(D > B - A\\) \n\nThe idea is that right now the \\(B\\) terms are \"too big\", so we want to use the involution trick in a way that makes as many \" \\(A\\) minus \\(B\\) \" shape expressions as possible. This motivates considering \\(s(s(s(n + 1))) - s(s(s(n) + 1) + 1) > 0\\) , since the first expression will have all \\(A\\) 's and the second expression will have all \\(B\\) 's. Calculation gives: \n\n\\[s(D(n + 1) + A) - s(Dn + B + 1) = \\left[s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\right]\\] \\[\\qquad = (D s(n + 1) + A) - (D(s(n) + 1) + B)\\] \\[\\qquad = D(s(n + 1) - s(n)) + A - B - D.\\]\n\n\n\nThen by picking \\(n\\) achieving the minimum \\(m\\) , \n\n\\[\\underbrace{m(D + A - B - 1)}_{>0}\\leq s(s(s(n + 1))) - s(s(s(n) + 1) + 1)\\leq Dm + A - B - D\\] \n\nwhich becomes \n\n\\[(D - m(B - A)) + ((B - A) - m)\\leq 0.\\] \n\nSince both of these quantities were supposed to be nonnegative, we conclude \\(m = B - A\\) and \\(D = m^{2}\\) . Now the estimate \\(D = s(Dn + B) - s(Dn + A)\\geq m(B - A)\\) is actually sharp, so it follows that \\(s(n)\\) is arithmetic.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2009/3, proposed by Gabriel Carroll (USA) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(A B C\\) be a triangle with \\(A B = A C\\) . The angle bisectors of \\(\\angle C A B\\) and \\(\\angle A B C\\) meet the sides \\(B C\\) and \\(C A\\) at \\(D\\) and \\(E\\) , respectively. Let \\(K\\) be the incenter of triangle \\(A D C\\) . Suppose that \\(\\angle B E K = 45^{\\circ}\\) . Find all possible values of \\(\\angle C A B\\) .", "solution": "Here is the solution presented in my book EGM0. \n\nLet \\(I\\) be the incenter of \\(A B C\\) , and set \\(\\angle D A C = 2x\\) (so that \\(0^{\\circ}< x< 45^{\\circ}\\) ). From \\(\\angle A I E = \\angle D I C\\) , it is easy to compute \n\n\\[\\angle K I E = 90^{\\circ} - 2x, \\angle E C I = 45^{\\circ} - x, \\angle I E K = 45^{\\circ}, \\angle K E C = 3x.\\] \n\nHaving chased all the angles we want, we need a relationship. We can find it by considering the side ratio \\(\\frac{I K}{K C}\\) . Using the angle bisector theorem, we can express this in terms of triangle \\(I D C\\) ; however we can also express it in terms of triangle \\(I E C\\) . \n\n\n \n\nBy the law of sines, we obtain \n\n\\[\\frac{I K}{K C} = \\frac{\\sin 45^{\\circ}\\cdot\\frac{E K}{\\sin(90^{\\circ} - 2x)}}{\\sin(3x)\\cdot\\frac{E K}{\\sin(45^{\\circ} - x)}} = \\frac{\\sin 45^{\\circ}\\sin(45^{\\circ} - x)}{\\sin(3x)\\sin(90^{\\circ} - 2x)}.\\] \n\nAlso, by the angle bisector theorem on \\(\\triangle I D C\\) , we have \n\n\\[\\frac{I K}{K C} = \\frac{I D}{D C} = \\frac{\\sin(45^{\\circ} - x)}{\\sin(45^{\\circ} + x)}.\\]\n\n\n\nEquating these and cancelling \\(\\sin \\left(45^{\\circ} - x\\right) \\neq 0\\) gives \n\n\\[\\sin 45^{\\circ}\\sin \\left(45^{\\circ} + x\\right) = \\sin 3x\\sin \\left(90^{\\circ} - 2x\\right).\\] \n\nApplying the product- sum formula (again, we are just trying to break down things as much as possible), this just becomes \n\n\\[\\cos \\left(x\\right) - \\cos \\left(90^{\\circ} + x\\right) = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos \\left(90^{\\circ} + x\\right)\\] \n\nor \\(\\cos x = \\cos \\left(5x - 90^{\\circ}\\right)\\) \n\nAt this point we are basically done; the rest is making sure we do not miss any solutions and write up the completion nicely. One nice way to do this is by using product- sum in reverse as \n\n\\[0 = \\cos \\left(5x - 90^{\\circ}\\right) - \\cos x = 2\\sin \\left(3x - 45^{\\circ}\\right)\\sin \\left(2x - 45^{\\circ}\\right).\\] \n\nThis way we merely consider the two cases \n\n\\[\\sin \\left(3x - 45^{\\circ}\\right) = 0 \\mathrm{and} \\sin \\left(2x - 45^{\\circ}\\right) = 0.\\] \n\nNotice that \\(\\sin \\theta = 0\\) if and only \\(\\theta\\) is an integer multiple of \\(180^{\\circ}\\) . Using the bound \\(0^{\\circ} < x < 45^{\\circ}\\) , it is easy to see that the permissible values of \\(x\\) are \\(x = 15^{\\circ}\\) and \\(x = \\frac{45^{\\circ}}{2}\\) . As \\(\\angle A = 4x\\) , this corresponds to \\(\\angle A = 60^{\\circ}\\) and \\(\\angle A = 90^{\\circ}\\) , which can be seen to work.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2009/4, proposed by Hojoo Lee, Peter Vandendriessche, Jan Vonk (BEL) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that for positive integers \\(a\\) and \\(b\\) , the numbers \n\n\\[a, f(b), f(b + f(a) - 1)\\] \n\nare the sides of a non-degenerate triangle.", "solution": ". \n\nThe only function is the identity function (which works). We prove it is the only one. Let \\(P(a,b)\\) denote the given statement. \n\nClaim — We have \\(f(1) = 1\\) , and \\(f(f(n)) = n\\) . (In particular \\(f\\) is a bijection.) \n\nProof. Note that \n\n\\[P(1,b)\\Rightarrow f(b) = f(b + f(1) - 1).\\] \n\nOtherwise, the function \\(f\\) is periodic modulo \\(N = f(1) - 1\\geq 1\\) . This is impossible since we can fix \\(b\\) and let \\(a\\) be arbitrarily large in some residue class modulo \\(N\\) . \n\nHence \\(f(1) = 1\\) , so taking \\(P(n,1)\\) gives \\(f(f(n)) = n\\) . \n\nClaim — Let \\(\\delta = f(2) - 1 > 0\\) . Then for every \\(n\\) , \n\n\\[f(n + 1) = f(n) + \\delta \\quad \\mathrm{or}\\quad f(n - 1) = f(n) + \\delta\\] \n\nProof. Use \n\n\\[P(2,f(n))\\Rightarrow n - 2< f(f(n) + \\delta)< n + 2.\\] \n\nLet \\(y = f(f(n) + \\delta)\\) , hence \\(n - 2< y< n + 2\\) and \\(f(y) = f(n) + \\delta\\) . But, remark that if \\(y = n\\) , we get \\(\\delta = 0\\) , contradiction. So \\(y\\in \\{n + 1,n - 1\\}\\) and that is all. \\(\\square\\) \n\nWe now show \\(f\\) is an arithmetic progression with common difference \\(+\\delta\\) . Indeed we already know \\(f(1) = 1\\) and \\(f(2) = 1 + \\delta\\) . Now suppose \\(f(1) = 1,\\ldots ,f(n) = 1 + (n - 1)\\delta\\) . Then by induction for any \\(n\\geq 2\\) , the second case can't hold, so we have \\(f(n + 1) = f(n) + \\delta\\) , as desired. \n\nCombined with \\(f(f(n)) = n\\) , we recover that \\(f\\) is the identity.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2009/5, proposed by Bruno Le Floch (FRA) \n"}}
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{"year": "2009", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) be distinct positive integers and let \\(M\\) be a set of \\(n - 1\\) positive integers not containing \\(s = a_{1} + \\dots +a_{n}\\) . A grasshopper is to jump along the real axis, starting at the point 0 and making \\(n\\) jumps to the right with lengths \\(a_{1}\\) , \\(a_{2}\\) , ..., \\(a_{n}\\) in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in \\(M\\) .", "solution": "The proof is by induction on \\(n\\) . Assume \\(a_{1} < \\dots < a_{n}\\) and call each element of \\(M\\) a mine. Let \\(x = s - a_{n}\\) . We consider four cases, based on whether \\(x\\) has a mine and whether there is a mine past \\(x\\) . \n\n- If \\(x\\) has no mine, and there is a mine past \\(x\\) , then at most \\(n - 2\\) mines in \\([0, x]\\) and so we use induction to reach \\(x\\) , then leap from \\(x\\) to \\(s\\) and win. \n\n- If \\(x\\) has no mine but there is also no mine to the right of \\(x\\) , then let \\(m\\) be the maximal mine. By induction hypothesis on \\(M \\setminus \\{m\\}\\) , there is a path to \\(x\\) using \\(\\{a_{1}, \\ldots , a_{n - 1}\\}\\) which avoids mines except possibly \\(m\\) . If the path hits the mine \\(m\\) on the hop of length \\(a_{k}\\) , we then swap that hop with \\(a_{n}\\) , and finish. \n\n- If \\(x\\) has a mine, but there are no mines to the right of \\(x\\) , we can repeat the previous case with \\(m = x\\) . \n\n- Now suppose \\(x\\) has a mine, and there is a mine past \\(x\\) . There should exist an integer \\(1 \\leq i \\leq n - 1\\) such that \\(s - a_{i}\\) and \\(y = s - a_{i} - a_{n}\\) both have no mine. By induction hypothesis, we can then reach \\(y\\) in \\(n - 2\\) steps (as there are two mines to the right of \\(y\\) ); then \\(y \\to s - a_{i} \\to s\\) finishes. \n\nRemark. It seems much of the difficulty of the problem is realizing induction will actually work. Attempts at induction are, indeed, a total minefield (ha!), and given the position P6 of the problem, it is expected that many contestants will abandon induction after some cursory attempts fail.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2009-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2009/6, proposed by Dmitry Khramtsov (RUS) \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "1", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(f\\colon \\mathbb{R}\\to \\mathbb{R}\\) such that for all \\(x,y\\in \\mathbb{R}\\) \n\n\\[f(\\lfloor x\\rfloor y) = f(x)\\lfloor f(y)\\rfloor .\\]", "solution": "The only solutions are \\(f(x)\\equiv c\\) , where \\(c = 0\\) or \\(1\\leq c< 2\\) . It's easy to see these work. \n\nPlug in \\(x = 0\\) to get \\(f(0) = f(0)\\lfloor f(y)\\rfloor\\) , so either \n\n\\[1\\leq f(y)< 2\\quad \\forall y\\qquad \\mathrm{or}\\qquad f(0) = 0\\] \n\nIn the first situation, plug in \\(y = 0\\) to get \\(f(x)\\lfloor f(0)\\rfloor = f(0)\\) , thus \\(f\\) is constant. Thus assume henceforth \\(f(0) = 0\\) \n\nNow set \\(x = y = 1\\) to get \n\n\\[f(1) = f(1)\\lfloor f(1)\\rfloor\\] \n\nso either \\(f(1) = 0\\) or \\(1\\leq f(1)< 2\\) . We split into cases: \n\n- If \\(f(1) = 0\\) , pick \\(x = 1\\) to get \\(f(y)\\equiv 0\\) . \n\n- If \\(1\\leq f(1)< 2\\) , then \\(y = 1\\) gives \n\n\\[f(\\lfloor x\\rfloor) = f(x)\\] \n\nfrom \\(y = 1\\) , in particular \\(f(x) = 0\\) for \\(0\\leq x< 1\\) . Choose \\((x,y) = \\left(2,\\frac{1}{2}\\right)\\) to get \\(f(1) = f(2)\\left\\lfloor f\\left(\\frac{1}{2}\\right)\\right\\rfloor = 0\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "1. ", "solution_match": "## \\(\\S 1.1\\) IMO 2010/1, proposed by Pierre Bornsztein (FRA) \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "2", "problem_type": null, "exam": "IMO", "problem": "Let \\(I\\) be the incenter of a triangle \\(A B C\\) and let \\(\\Gamma\\) be its circumcircle. Let line \\(A I\\) intersect \\(\\Gamma\\) again at \\(D\\) . Let \\(E\\) be a point on arc \\(\\overrightarrow{B D C}\\) and \\(F\\) a point on side \\(B C\\) such that \n\n\\[\\angle B A F = \\angle C A E< \\frac{1}{2}\\angle B A C.\\] \n\nFinally, let \\(G\\) be the midpoint of \\(\\overline{{I F}}\\) . Prove that \\(\\overline{{D G}}\\) and \\(\\overline{{E I}}\\) intersect on \\(\\Gamma\\)", "solution": "Let \\(\\overline{{E I}}\\) meet \\(\\Gamma\\) again at \\(K\\) . Then it suffices to show that \\(\\overline{{K D}}\\) bisects \\(\\overline{{I F}}\\) . Let \\(\\overline{{A F}}\\) meet \\(\\Gamma\\) again at \\(H\\) , so \\(\\overline{{H E}}\\parallel \\overline{{B C}}\\) . By Pascal theorem on \n\n\\[A H E K D D\\] \n\nwe then obtain that \\(P = \\overline{{A H}}\\cap \\overline{{K D}}\\) lies on a line through \\(I\\) parallel to \\(\\overline{{B C}}\\) . \n\nLet \\(I_{A}\\) be the \\(A\\) - excenter, and set \\(Q = \\overline{{I_{A}F}}\\cap \\overline{{I P}}\\) , and \\(T = \\overline{{A I D I_{A}}}\\cap \\overline{{B F C}}\\) . Then \n\n\\[-1 = (A I;T I_{A})\\stackrel {E}{=}(I Q;\\infty P)\\] \n\nwhere \\(\\infty\\) is the point at infinity along \\(\\overline{{I P Q}}\\) . Thus \\(P\\) is the midpoint of \\(\\overline{{I Q}}\\) . Since \\(D\\) is the midpoint of \\(\\overline{{I I_{A}}}\\) by \"Fact 5\", it follows that \\(\\overline{{D P}}\\) bisects \\(\\overline{{I F}}\\) . \n\n", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "2. ", "solution_match": "## §1.2 IMO 2010/2, proposed by Tai Wai Ming and Wang Chongli (HKG) \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "3", "problem_type": null, "exam": "IMO", "problem": "Find all functions \\(g\\colon \\mathbb{Z}_{>0}\\to \\mathbb{Z}_{>0}\\) such that \n\n\\[(g(m) + n)(g(n) + m)\\] \n\nis always a perfect square.", "solution": "For \\(c\\geq 0\\) , the function \\(g(n) = n + c\\) works; we prove this is the only possibility. \n\nFirst, the main point of the problem is that: \n\nClaim — We have \\(g(n)\\equiv g(n^{\\prime})\\) (mod \\(p\\) ) \\(\\Longrightarrow n\\equiv n^{\\prime}\\) (mod \\(p\\) ). \n\nProof. Pick a large integer \\(M\\) such that \n\n\\[\\nu_{p}(M + g(n)),\\quad \\nu_{p}(M + g(n^{\\prime}))\\quad \\mathrm{are~both~odd}.\\] \n\n(It's not hard to see this is always possible.) Now, since each of \n\n\\[(M + g(n))(n + g(M))\\] \\[(M + g(n^{\\prime}))(n^{\\prime} + g(M))\\] \n\nis a square, we get \\(n\\equiv n^{\\prime}\\equiv - g(M)\\) (mod \\(p\\) ). \n\nThis claim implies that \n\n- The numbers \\(g(n)\\) and \\(g(n + 1)\\) differ by \\(\\pm 1\\) for any \\(n\\) , and \n\n- The function \\(g\\) is injective. \n\nIt follows \\(g\\) is a linear function with slope \\(\\pm 1\\) , hence done.\n\n\n\n## \\(\\S 2\\) Solutions to Day 2", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "3. ", "solution_match": "## \\(\\S 1.3\\) IMO 2010/3, proposed by Gabriel Carroll (USA) \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "4", "problem_type": null, "exam": "IMO", "problem": "Let \\(P\\) be a point interior to triangle \\(A B C\\) (with \\(C A\\neq C B\\) ). The lines \\(A P\\) , \\(B P\\) and \\(C P\\) meet again its circumcircle \\(\\Gamma\\) at \\(K\\) , \\(L\\) , \\(M\\) , respectively. The tangent line at \\(C\\) to \\(\\Gamma\\) meets the line \\(A B\\) at \\(S\\) . Show that from \\(S C = S P\\) follows \\(M K = M L\\) .", "solution": "## Problem sta \n\nWe present two solutions using harmonic bundles. \n\n\\(\\P\\) First solution (Evan Chen). Let \\(N\\) be the antipode of \\(M\\) , and let \\(NP\\) meet \\(\\Gamma\\) again at \\(D\\) . Focus only on \\(CDMN\\) for now (ignoring the condition). Then \\(C\\) and \\(D\\) are feet of altitudes in \\(\\triangle MNP\\) ; it is well- known that the circumcircle of \\(\\triangle CDP\\) is orthogonal to \\(\\Gamma\\) (passing through the orthocenter of \\(\\triangle MPN\\) ). \n\n\n \n\nNow, we are given that point \\(S\\) is such that \\(\\overline{SC}\\) is tangent to \\(\\Gamma\\) , and \\(SC = SP\\) . It follows that \\(S\\) is the circumcenter of \\(\\triangle CDP\\) , and hence \\(\\overline{SC}\\) and \\(\\overline{SD}\\) are tangents to \\(\\Gamma\\) . \n\nThen \\(- 1 = (AB;CD) \\stackrel{P}{=} (KL;MN)\\) . Since \\(\\overline{MN}\\) is a diameter, this implies \\(MK = ML\\) . \n\nRemark. I think it's more natural to come up with this solution in reverse. Namely, suppose we define the points the other way: let \\(\\overline{SD}\\) be the other tangent, so \\((AB;CD) = - 1\\) . Then project through \\(P\\) to get \\((KL;MN) = - 1\\) , where \\(N\\) is the second intersection of \\(\\overline{DP}\\) . However, if \\(ML = MK\\) then \\(KMLN\\) must be a kite. Thus one can recover the solution in reverse. \n\n## \\(\\P\\) Second solution (Sebastian Jeon). We have \n\n\\[SP^{2} = SC^{2} = SA\\cdot SB\\Rightarrow \\angle SPA = \\angle PBA = \\angle LBA = \\angle LKA = \\angle LKP\\] \n\n(the latter half is Reim's theorem). Therefore \\(\\overline{SP}\\) and \\(\\overline{LK}\\) are parallel.\n\n\n\nNow, let \\(\\overline{SP}\\) meet \\(\\Gamma\\) again at \\(X\\) and \\(Y\\) , and let \\(Q\\) be the antipode of \\(P\\) on \\((S)\\) . Then \n\n\\[SP^{2} = SQ^{2} = SX\\cdot SY\\Longrightarrow (PQ;XY) = -1\\Longrightarrow \\angle QCP = 90^{\\circ}\\] \n\nthat \\(\\overline{CP}\\) bisects \\(\\angle XCY\\) . Since \\(\\overline{XY} \\parallel \\overline{KL}\\) , it follows \\(\\overline{CP}\\) bisects to \\(\\angle LCK\\) too.", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "4. ", "solution_match": "## \\(\\S 2.1\\) IMO 2010/4, proposed by Marcin Kuczma (POL) \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "5", "problem_type": null, "exam": "IMO", "problem": "Each of the six boxes \\(B_{1}\\) , \\(B_{2}\\) , \\(B_{3}\\) , \\(B_{4}\\) , \\(B_{5}\\) , \\(B_{6}\\) initially contains one coin. The following two types of operations are allowed: \n\na) Choose a non-empty box \\(B_{j}\\) , \\(1\\leq j\\leq 5\\) , remove one coin from \\(B_{j}\\) and add two coins to \\(B_{j + 1}\\) ; \nb) Choose a non-empty box \\(B_{k}\\) , \\(1\\leq k\\leq 4\\) , remove one coin from \\(B_{k}\\) and swap the contents (possibly empty) of the boxes \\(B_{k + 1}\\) and \\(B_{k + 2}\\) . \n\nDetermine if there exists a finite sequence of operations of the allowed types, such that the five boxes \\(B_{1}\\) , \\(B_{2}\\) , \\(B_{3}\\) , \\(B_{4}\\) , \\(B_{5}\\) become empty, while box \\(B_{6}\\) contains exactly \\(2010^{2010^{2010}}\\) coins.", "solution": "Each o \n\nFirst, \n\n\\[(1,1,1,1,1,1) \\to (0,3,1,0,3,1) \\to (0,0,7,0,0,7)\\] \\[\\to (0,0,6,2,0,7) \\to (0,0,6,1,2,7) \\to (0,0,6,1,0,11)\\] \\[\\to (0,0,6,0,11,0) \\to (0,0,5,11,0,0).\\] \n\nand henceforth we ignore boxes \\(B_{1}\\) and \\(B_{2}\\) , looking at just the last four boxes; so we write the current position as \\((5,11,0,0)\\) . \n\nWe prove a lemma: \n\nClaim — Let \\(k \\geq 0\\) and \\(n > 0\\) . From \\((k, n, 0, 0)\\) we may reach \\((k - 1, 2^{n}, 0, 0)\\) . \n\nProof. Working with only the last three boxes for now, \n\n\\[(n,0,0)\\to (n - 1,2,0)\\to (n - 1,0,4)\\] \\[\\to (n - 2,4,0)\\to (n - 2,0,8)\\] \\[\\to (n - 3,8,0)\\to (n - 3,0,16)\\] \\[\\to \\dots \\to (1,2^{n - 1},0)\\to (1,0,2^{n})\\to (0,2^{n},0).\\] \n\nFinally we have \\((k, n, 0, 0) \\to (k, 0, 2^{n}, 0) \\to (k - 1, 2^{n}, 0, 0)\\) . \n\nNow from \\((5,11,0,0)\\) we go as follows: \n\n\\[(5,11,0,0)\\to (4,2^{11},0,0)\\to \\left(3,2^{21},0,0\\right)\\to \\left(2,2^{22^{11}},0,0\\right)\\] \\[\\to \\left(1,2^{22^{11}},0,0\\right)\\to \\left(0,2^{22^{21^{11}}},0,0\\right).\\] \n\nLet \\(A = 2^{22^{22^{11}}} > 2010^{2010^{2010}} = B\\) . Then by using move 2 repeatedly on the fourth box (i.e., throwing away several coins by swapping the empty \\(B_{5}\\) and \\(B_{6}\\) ), we go from \\((0, A, 0, 0)\\) to \\((0, B / 4, 0, 0)\\) . From there we reach \\((0, 0, 0, B)\\) .", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "5. ", "solution_match": "## \\(\\S 2.2\\) IMO 2010/5, proposed by Netherlands \n"}}
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{"year": "2010", "tier": "T0", "problem_label": "6", "problem_type": null, "exam": "IMO", "problem": "Let \\(a_{1},a_{2},a_{3},\\ldots\\) be a sequence of positive real numbers, and \\(s\\) be a positive integer, such that \n\n\\[a_{n} = \\max \\{a_{k} + a_{n - k}\\mid 1\\leq k\\leq n - 1\\} \\mathrm{~for~all~}n > s.\\] \n\nProve there exist positive integers \\(\\ell \\leq s\\) and \\(N\\) , such that \n\n\\[a_{n} = a_{\\ell} + a_{n - \\ell}\\mathrm{~for~all~}n\\geq N.\\]", "solution": "Let \n\n\\[w_{1} = \\frac{a_{1}}{1},\\quad w_{2} = \\frac{a_{2}}{2},\\quad \\ldots ,\\quad w_{s} = \\frac{a_{s}}{s}.\\] \n\n(The choice of the letter \\(w\\) is for \"weight\".) We claim the right choice of \\(\\ell\\) is the one maximizing \\(w_{\\ell}\\) . \n\nOur plan is to view each \\(a_{n}\\) as a linear combination of the weights \\(w_{1},\\ldots ,w_{s}\\) and track their coefficients. \n\nTo this end, let's define an \\(n\\) - type to be a vector \\(T = \\langle t_{1},\\ldots ,t_{s}\\rangle\\) of nonnegative integers such that \n\n\\(n = t_{1} + \\dots +t_{s}\\) ; and \n\n\\(t_{i}\\) is divisible by \\(i\\) for every \\(i\\) . \n\nWe then define its valuation as \\(v(T) = \\sum_{i = 1}^{s}w_{i}t_{i}\\) . \n\nNow we define a \\(n\\) - type to be valid according to the following recursive rule. For \\(1\\leq n\\leq s\\) the only valid \\(n\\) - types are \n\n\\[T_{1} = \\langle 1,0,0,\\ldots ,0\\rangle\\] \\[T_{2} = \\langle 0,2,0,\\ldots ,0\\rangle\\] \\[T_{3} = \\langle 0,0,3,\\ldots ,0\\rangle\\] \\[\\vdots\\] \\[T_{s} = \\langle 0,0,0,\\ldots ,s\\rangle\\] \n\nfor \\(n = 1,\\ldots ,s\\) , respectively. Then for any \\(n > s\\) , an \\(n\\) - type is valid if it can be written as the sum of a valid \\(k\\) - type and a valid \\((n - k)\\) - type, componentwise. These represent the linear combinations possible in the recursion; in other words the recursion in the problem is phrased as \n\n\\[a_{n} = \\max_{T\\mathrm{~is~a~valid~}n\\mathrm{-type}}v(T).\\] \n\nIn fact, we have the following description of valid \\(n\\) - types: \n\nClaim — Assume \\(n > s\\) . Then an \\(n\\) - type \\(\\langle t_{1},\\ldots ,t_{s}\\rangle\\) is valid if and only if either \n\nthere exist indices \\(i< j\\) with \\(i + j > s\\) , \\(t_{i}\\geq i\\) and \\(t_{j}\\geq j\\) ; or \n\nthere exists an index \\(i > s / 2\\) with \\(t_{i}\\geq 2i\\) .\n\n\n\nProof. Immediate by forwards induction on \\(n > s\\) that all \\(n\\) - types have this property. \n\nThe reverse direction is by downwards induction on \\(n\\) . Indeed if \\(\\sum_{i} \\frac{t_{i}}{i} > 2\\) , then we may subtract off on of \\(\\{T_{1}, \\ldots , T_{s}\\}\\) while preserving the condition; and the case \\(\\sum_{i} \\frac{t_{i}}{i} = 2\\) is essentially by definition. \\(\\square\\) \n\nRemark. The claim is a bit confusingly stated in its two cases; really the latter case should be thought of as the situation \\(i = j\\) but requiring that \\(t_{i} / i\\) is counted with multiplicity. \n\nNow, for each \\(n > s\\) we pick a valid \\(n\\) - type \\(T_{n}\\) with \\(a_{n} = v(T_{n})\\) ; if there are ties, we pick one for which the \\(\\ell\\) th entry is as large as possible. \n\nClaim — For any \\(n > s\\) and index \\(i \\neq \\ell\\) , the \\(i\\) th entry of \\(T_{n}\\) is at most \\(2s + \\ell i\\) . \n\nProof. If not, we can go back \\(i\\ell\\) steps to get a valid \\((n - i\\ell)\\) - type \\(T\\) achieved by decreasing the \\(i\\) th entry of \\(T_{n}\\) by \\(i\\ell\\) . But then we can add \\(\\ell\\) to the \\(\\ell\\) th entry \\(i\\) times to get another \\(n\\) - type \\(T^{\\prime}\\) which obviously has valuation at least as large, but with larger \\(\\ell\\) th entry. \\(\\square\\) \n\nNow since all other entries in \\(T_{n}\\) are bounded, eventually the sequence \\((T_{n})_{n > s}\\) just consists of repeatedly adding 1 to the \\(\\ell\\) th entry, as required. \n\nRemark. One big step is to consider \\(w_{k} = a_{k} / k\\) . You can get this using wishful thinking or by examining small cases. (In addition this normalization makes it easier to see why the largest \\(w\\) plays an important role, since then in the definition of type, the \\(n\\) - types all have a sum of \\(n\\) . Unfortunately, it makes the characterization of valid \\(n\\) - types somewhat clumsier too.)", "metadata": {"resource_path": "IMO/segmented/en-IMO-2010-notes.jsonl", "problem_match": "6. ", "solution_match": "## \\(\\S 2.3\\) IMO 2010/6, proposed by Morteza Saghafiyan (IRN) \n"}}
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