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Difference between revisions of "2011 USAMO Problems"

(Created page with '=Day 1= ==Problem 1== Let <math>a</math>, <math>b</math>, <math>c</math> be positive real numbers such that <math>a^2 + b^2 + c^2 + (a + b + c)^2 \le 4</math>. Prove that <cmath…')
 
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==Problem 2==
 
==Problem 2==
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An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer <math>m</math> from each of the integers at two neighboring vertices and adding <math>2m</math> to the opposite vertex, which is not adajcent to either of the first two vertices. (The amount <math>m</math> and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.
+
An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer <math>m</math> from each of the integers at two neighboring vertices and adding <math>2m</math> to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount <math>m</math> and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.
  
 
[[2011 USAMO Problems/Problem 2|Solution]]
 
[[2011 USAMO Problems/Problem 2|Solution]]

Revision as of 10:11, 29 April 2011

Day 1

Problem 1

Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \le 4$. Prove that \[\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.\]

Solution

Problem 2

An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.) The game is won at a certain vertex if, after some number of turns, that vertex has the number 2011 and the other four vertices have the number 0. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.

Solution

Problem 3

In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A = 3 \angle D$, $\angle C = 3 \angle F$, and $\angle E = 3 \angle B$. Furthermore, $AB = DE$, $BC = EF$, and $CD = FA$. Prove that diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent.

Solution

Day 2

Problem 4

Consider the assertion that for each positive integer $n \ge 2$, the remainder upon dividing $2^{2^n}$ by $2^n - 1$ is a power of 4. Either prove the assertion or find (with proof) a counterexample.

Solution

Problem 5

Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$, $\angle Q_1 CB = \angle DCP$, $\angle Q_2 AD = \angle BAP$, $\angle Q_2 DA = \angle CDP$. Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$.

Solution

Problem 6

Let $A$ be a set with $|A| = 225$, meaning that $A$ has 225 elements. Suppose further that there are eleven subsets $A_1$, $\dots$, $A_{11}$ of $A$ such that $|A_i | = 45$ for $1 \le i \le 11$ and $|A_i \cap A_j| = 9$ for $1 \le i < j \le 11$. Prove that $|A_1 \cup A_2 \cup \dots \cup A_{11}| \ge 165$, and give an example for which equality holds.

Solution

See also

2011 USAMO (Problems • Resources)
Preceded by
2010 USAMO 		Followed by
2012 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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