Difference between revisions of "2016 USAMO Problems"
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[[2016 USAMO Problems/Problem 1|Solution]] | [[2016 USAMO Problems/Problem 1|Solution]] | ||
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===Problem 2=== | ===Problem 2=== | ||
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Prove that for any positive integer <math>k,</math> | Prove that for any positive integer <math>k,</math> | ||
<cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath> | <cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath> | ||
is an integer. | is an integer. | ||
+ | [[2016 USAMO Problems/Problem 2|Solution]] | ||
===Problem 3=== | ===Problem 3=== | ||
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Let <math>\triangle ABC</math> be an acute triangle, and let <math>I_B, I_C,</math> and <math>O</math> denote its <math>B</math>-excenter, <math>C</math>-excenter, and circumcenter, respectively. Points <math>E</math> and <math>Y</math> are selected on <math>\overline{AC}</math> such that <math>\angle ABY = \angle CBY</math> and <math>\overline{BE}\perp\overline{AC}.</math> Similarly, points <math>F</math> and <math>Z</math> are selected on <math>\overline{AB}</math> such that <math>\angle ACZ = \angle BCZ</math> and <math>\overline{CF}\perp\overline{AB}.</math> | Let <math>\triangle ABC</math> be an acute triangle, and let <math>I_B, I_C,</math> and <math>O</math> denote its <math>B</math>-excenter, <math>C</math>-excenter, and circumcenter, respectively. Points <math>E</math> and <math>Y</math> are selected on <math>\overline{AC}</math> such that <math>\angle ABY = \angle CBY</math> and <math>\overline{BE}\perp\overline{AC}.</math> Similarly, points <math>F</math> and <math>Z</math> are selected on <math>\overline{AB}</math> such that <math>\angle ACZ = \angle BCZ</math> and <math>\overline{CF}\perp\overline{AB}.</math> | ||
− | Lines <math>I_B F</math> and <math>I_C E</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular. | + | Lines <math>\overleftrightarrow{I_B F}</math> and <math>\overleftrightarrow{I_C E}</math> meet at <math>P.</math> Prove that <math>\overline{PO}</math> and <math>\overline{YZ}</math> are perpendicular. |
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+ | [[2016 USAMO Problems/Problem 3|Solution]] | ||
==Day 2== | ==Day 2== | ||
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[[2016 USAMO Problems/Problem 4|Solution]] | [[2016 USAMO Problems/Problem 4|Solution]] | ||
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===Problem 5=== | ===Problem 5=== | ||
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+ | An equilateral pentagon <math>AMNPQ</math> is inscribed in triangle <math>ABC</math> such that <math>M\in\overline{AB},</math> <math>Q\in\overline{AC},</math> and <math>N, P\in\overline{BC}.</math> Let <math>S</math> be the intersection of <math>\overleftrightarrow{MN}</math> and <math>\overleftrightarrow{PQ}.</math> Denote by <math>\ell</math> the angle bisector of <math>\angle MSQ.</math> | ||
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+ | Prove that <math>\overline{OI}</math> is parallel to <math>\ell,</math> where <math>O</math> is the circumcenter of triangle <math>ABC,</math> and <math>I</math> is the incenter of triangle <math>ABC.</math> | ||
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+ | [[2016 USAMO Problems/Problem 5|Solution]] | ||
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===Problem 6=== | ===Problem 6=== | ||
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+ | Integers <math>n</math> and <math>k</math> are given, with <math>n\ge k\ge 2.</math> You play the following game against an evil wizard. | ||
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+ | The wizard has <math>2n</math> cards; for each <math>i = 1, ..., n,</math> there are two cards labeled <math>i.</math> Initially, the wizard places all cards face down in a row, in unknown order. | ||
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+ | You may repeatedly make moves of the following form: you point to any <math>k</math> of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the <math>k</math> chosen cards and turns them back face-down. Then, it is your turn again. | ||
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+ | We say this game is <i>winnable</i> if there exist some positive integer <math>m</math> and some strategy that is guaranteed to win in at most <math>m</math> moves, no matter how the wizard responds. | ||
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+ | For which values of <math>n</math> and <math>k</math> is the game winnable? | ||
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+ | [[2016 USAMO Problems/Problem 6|Solution]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− | {{USAMO newbox|year= 2016 |before=[[2015 USAMO]]|after=[[2017 USAMO]]}} | + | {{USAMO newbox|year=2016|before=[[2015 USAMO Problems]]|after=[[2017 USAMO Problems]]}} |
Latest revision as of 12:49, 22 November 2023
Contents
[hide]Day 1
Problem 1
Let be a sequence of mutually distinct nonempty subsets of a set . Any two sets and are disjoint and their union is not the whole set , that is, and , for all . Find the smallest possible number of elements in .
Problem 2
Prove that for any positive integer is an integer.
Problem 3
Let be an acute triangle, and let and denote its -excenter, -excenter, and circumcenter, respectively. Points and are selected on such that and Similarly, points and are selected on such that and
Lines and meet at Prove that and are perpendicular.
Day 2
Problem 4
Find all functions such that for all real numbers and ,
Problem 5
An equilateral pentagon is inscribed in triangle such that and Let be the intersection of and Denote by the angle bisector of
Prove that is parallel to where is the circumcenter of triangle and is the incenter of triangle
Problem 6
Integers and are given, with You play the following game against an evil wizard.
The wizard has cards; for each there are two cards labeled Initially, the wizard places all cards face down in a row, in unknown order.
You may repeatedly make moves of the following form: you point to any of the cards. The wizard then turns those cards face up. If any two of the cards match, the game is over and you win. Otherwise, you must look away, while the wizard arbitrarily permutes the chosen cards and turns them back face-down. Then, it is your turn again.
We say this game is winnable if there exist some positive integer and some strategy that is guaranteed to win in at most moves, no matter how the wizard responds.
For which values of and is the game winnable?
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
2016 USAMO (Problems • Resources) | ||
Preceded by 2015 USAMO Problems |
Followed by 2017 USAMO Problems | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |