Difference between revisions of "2011 USAMO Problems"
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− | =Day 1= | + | ==Day 1== |
− | ==Problem 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 | 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>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath> | <cmath>\frac{ab + 1}{(a + b)^2} + \frac{bc + 1}{(b + c)^2} + \frac{ca + 1}{(c + a)^2} \ge 3.</cmath> | ||
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[[2011 USAMO Problems/Problem 1|Solution]] | [[2011 USAMO Problems/Problem 1|Solution]] | ||
− | ==Problem 2== | + | ===Problem 2=== |
− | An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is 2011. | + | 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]] | ||
− | ==Problem 3== | + | ===Problem 3=== |
In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3 \angle D</math>, <math>\angle C = 3 \angle F</math>, and <math>\angle E = 3 \angle B</math>. Furthermore, <math>AB = DE</math>, <math>BC = EF</math>, and <math>CD = FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent. | In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3 \angle D</math>, <math>\angle C = 3 \angle F</math>, and <math>\angle E = 3 \angle B</math>. Furthermore, <math>AB = DE</math>, <math>BC = EF</math>, and <math>CD = FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent. | ||
[[2011 USAMO Problems/Problem 3|Solution]] | [[2011 USAMO Problems/Problem 3|Solution]] | ||
− | =Day 2= | + | ==Day 2== |
− | ==Problem 4== | + | ===Problem 4=== |
Consider the assertion that for each positive integer <math>n \ge 2</math>, the remainder upon dividing <math>2^{2^n}</math> by <math>2^n - 1</math> is a power of 4. Either prove the assertion or find (with proof) a counterexample. | Consider the assertion that for each positive integer <math>n \ge 2</math>, the remainder upon dividing <math>2^{2^n}</math> by <math>2^n - 1</math> is a power of 4. Either prove the assertion or find (with proof) a counterexample. | ||
[[2011 USAMO Problems/Problem 4|Solution]] | [[2011 USAMO Problems/Problem 4|Solution]] | ||
− | ==Problem 5== | + | ===Problem 5=== |
− | Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>. Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>. Prove that <math>\overline{Q_1 Q_2} \parallel \overline{ | + | Let <math>P</math> be a given point inside quadrilateral <math>ABCD</math>. Points <math>Q_1</math> and <math>Q_2</math> are located within <math>ABCD</math> such that <math>\angle Q_1 BC = \angle ABP</math>, <math>\angle Q_1 CB = \angle DCP</math>, <math>\angle Q_2 AD = \angle BAP</math>, <math>\angle Q_2 DA = \angle CDP</math>. Prove that <math>\overline{Q_1 Q_2} \parallel \overline{BA}</math> if and only if <math>\overline{Q_1 Q_2} \parallel \overline{CD}</math>. |
[[2011 USAMO Problems/Problem 5|Solution]] | [[2011 USAMO Problems/Problem 5|Solution]] | ||
− | ==Problem 6== | + | ===Problem 6=== |
Let <math>A</math> be a set with <math>|A| = 225</math>, meaning that <math>A</math> has 225 elements. Suppose further that there are eleven subsets <math>A_1</math>, <math>\dots</math>, <math>A_{11}</math> of <math>A</math> such that <math>|A_i | = 45</math> for <math>1 \le i \le 11</math> and <math>|A_i \cap A_j| = 9</math> for <math>1 \le i < j \le 11</math>. Prove that <math>|A_1 \cup A_2 \cup \dots \cup A_{11}| \ge 165</math>, and give an example for which equality holds. | Let <math>A</math> be a set with <math>|A| = 225</math>, meaning that <math>A</math> has 225 elements. Suppose further that there are eleven subsets <math>A_1</math>, <math>\dots</math>, <math>A_{11}</math> of <math>A</math> such that <math>|A_i | = 45</math> for <math>1 \le i \le 11</math> and <math>|A_i \cap A_j| = 9</math> for <math>1 \le i < j \le 11</math>. Prove that <math>|A_1 \cup A_2 \cup \dots \cup A_{11}| \ge 165</math>, and give an example for which equality holds. | ||
[[2011 USAMO Problems/Problem 6|Solution]] | [[2011 USAMO Problems/Problem 6|Solution]] | ||
− | = See | + | == See Also == |
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{{USAMO newbox|year= 2011|before=[[2010 USAMO]]|after=[[2012 USAMO]]}} | {{USAMO newbox|year= 2011|before=[[2010 USAMO]]|after=[[2012 USAMO]]}} |
Latest revision as of 12:26, 17 April 2016
Contents
Day 1
Problem 1
Let , , be positive real numbers such that . Prove that
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 from each of the integers at two neighboring vertices and adding to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount 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.
Problem 3
In hexagon , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy , , and . Furthermore, , , and . Prove that diagonals , , and are concurrent.
Day 2
Problem 4
Consider the assertion that for each positive integer , the remainder upon dividing by is a power of 4. Either prove the assertion or find (with proof) a counterexample.
Problem 5
Let be a given point inside quadrilateral . Points and are located within such that , , , . Prove that if and only if .
Problem 6
Let be a set with , meaning that has 225 elements. Suppose further that there are eleven subsets , , of such that for and for . Prove that , and give an example for which equality holds.
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 |