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Difference between revisions of "2016 USAMO Problems"

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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>
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<cmath>\left(k^2\right)!\cdot\product_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath>
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<cmath>\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}</cmath>
 
is an integer.
 
is an integer.
  
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[[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>
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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]]
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==Day 2==
 
==Day 2==
 
===Problem 4===
 
===Problem 4===
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[[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 <math>\textit{winnable}</math> 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]]|after=[[2017 USAMO]]}}

Revision as of 08:46, 3 August 2016

Day 1

Problem 1

Let $X_1, X_2, \ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_i\cap X_{i+1}=\emptyset$ and $X_i\cup X_{i+1}\neq S$, for all $i\in\{1, \ldots, 99\}$. Find the smallest possible number of elements in $S$.

Solution

Problem 2

Prove that for any positive integer $k,$ \[\left(k^2\right)!\cdot\prod_{j=0}^{k-1}\frac{j!}{\left(j+k\right)!}\] is an integer.

Solution

Problem 3

Let $\triangle ABC$ be an acute triangle, and let $I_B, I_C,$ and $O$ denote its $B$-excenter, $C$-excenter, and circumcenter, respectively. Points $E$ and $Y$ are selected on $\overline{AC}$ such that $\angle ABY = \angle CBY$ and $\overline{BE}\perp\overline{AC}.$ Similarly, points $F$ and $Z$ are selected on $\overline{AB}$ such that $\angle ACZ = \angle BCZ$ and $\overline{CF}\perp\overline{AB}.$

Lines $\overleftrightarrow{I_B F}$ and $\overleftrightarrow{I_C E}$ meet at $P.$ Prove that $\overline{PO}$ and $\overline{YZ}$ are perpendicular.

Solution

Day 2

Problem 4

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x$ and $y$, \[(f(x)+xy)\cdot f(x-3y)+(f(y)+xy)\cdot f(3x-y)=(f(x+y))^2.\]

Solution

Problem 5

An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB},$ $Q\in\overline{AC},$ and $N, P\in\overline{BC}.$ Let $S$ be the intersection of $\overleftrightarrow{MN}$ and $\overleftrightarrow{PQ}.$ Denote by $\ell$ the angle bisector of $\angle MSQ.$

Prove that $\overline{OI}$ is parallel to $\ell,$ where $O$ is the circumcenter of triangle $ABC,$ and $I$ is the incenter of triangle $ABC.$

Solution

Problem 6

Integers $n$ and $k$ are given, with $n\ge k\ge 2.$ You play the following game against an evil wizard.

The wizard has $2n$ cards; for each $i = 1, ..., n,$ there are two cards labeled $i.$ 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 $k$ 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 $k$ chosen cards and turns them back face-down. Then, it is your turn again.

We say this game is $\textit{winnable}$ if there exist some positive integer $m$ and some strategy that is guaranteed to win in at most $m$ moves, no matter how the wizard responds.

For which values of $n$ and $k$ is the game winnable?

Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

2016 USAMO (Problems • Resources)
Preceded by
2015 USAMO 		Followed by
2017 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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