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Difference between revisions of "2010 USAMO Problems"

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=Day 1=
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==Day 1==
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==Problem 1==
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===Problem 1===
 
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter <math>AB</math>. Denote by <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> the feet of the perpendiculars from <math>Y</math> onto lines <math>AX</math>, <math>BX</math>, <math>AZ</math>, <math>BZ</math>, respectively.  Prove that the acute angle formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where <math>O</math> is the midpoint of segment <math>AB</math>.
 
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter <math>AB</math>. Denote by <math>P</math>, <math>Q</math>, <math>R</math>, <math>S</math> the feet of the perpendiculars from <math>Y</math> onto lines <math>AX</math>, <math>BX</math>, <math>AZ</math>, <math>BZ</math>, respectively.  Prove that the acute angle formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where <math>O</math> is the midpoint of segment <math>AB</math>.
  
 
[[2010 USAMO Problems/Problem 1|Solution]]
 
[[2010 USAMO Problems/Problem 1|Solution]]
  
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==Problem 2==
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===Problem 2===
 
There are <math>n</math> students standing in a circle, one behind the other. The students have heights <math>h_1<h_2<\dots <h_n</math>. If a student with height <math>h_k</math> is standing directly behind a student with height <math>h_{k-2}</math> or less, the two students are permitted to switch places. Prove that it is not possible to make more than <math>\binom{n}{3}</math> such switches before reaching a position in which no further switches are possible.
 
There are <math>n</math> students standing in a circle, one behind the other. The students have heights <math>h_1<h_2<\dots <h_n</math>. If a student with height <math>h_k</math> is standing directly behind a student with height <math>h_{k-2}</math> or less, the two students are permitted to switch places. Prove that it is not possible to make more than <math>\binom{n}{3}</math> such switches before reaching a position in which no further switches are possible.
  
 
[[2010 USAMO Problems/Problem 2|Solution]]
 
[[2010 USAMO Problems/Problem 2|Solution]]
  
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==Problem 3==
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===Problem 3===
 
The 2010 positive numbers <math>a_1, a_2, \ldots , a_{2010}</math> satisfy the inequality <math>a_ia_j\leq i+j</math> for all distinct indices <math>i, j</math>. Determine, with proof, the largest possible value of the product <math>a_1a_2\ldots a_{2010}</math>.
 
The 2010 positive numbers <math>a_1, a_2, \ldots , a_{2010}</math> satisfy the inequality <math>a_ia_j\leq i+j</math> for all distinct indices <math>i, j</math>. Determine, with proof, the largest possible value of the product <math>a_1a_2\ldots a_{2010}</math>.
  
 
[[2010 USAMO Problems/Problem 3|Solution]]
 
[[2010 USAMO Problems/Problem 3|Solution]]
  
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=Day 2=
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==Day 2==
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==Problem 4==
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===Problem 4===
 
Let <math>ABC</math> be a triangle with <math>\angle A=90^{\circ}</math>. Points <math>D</math> and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle ABD=\angle DBC</math> and <math>\angle ACE=\angle ECB</math>. Segments <math>BD</math> and <math>CE</math> meet at <math>I</math>. Determine whether or not it is possible for segments <math>AB</math>, <math>AC</math>, <math>BI</math>, <math>ID</math>, <math>CI</math>, <math>IE</math> to all have integer side lengths.
 
Let <math>ABC</math> be a triangle with <math>\angle A=90^{\circ}</math>. Points <math>D</math> and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle ABD=\angle DBC</math> and <math>\angle ACE=\angle ECB</math>. Segments <math>BD</math> and <math>CE</math> meet at <math>I</math>. Determine whether or not it is possible for segments <math>AB</math>, <math>AC</math>, <math>BI</math>, <math>ID</math>, <math>CI</math>, <math>IE</math> to all have integer side lengths.
  
 
[[2010 USAMO Problems/Problem 4|Solution]]
 
[[2010 USAMO Problems/Problem 4|Solution]]
  
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==Problem 5==
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===Problem 5===
 
Let <math>q = \dfrac{3p-5}{2}</math> where <math>p</math> is an odd prime, and let
 
Let <math>q = \dfrac{3p-5}{2}</math> where <math>p</math> is an odd prime, and let
 
<center>
 
<center>
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</cmath>
 
</cmath>
 
</center>
 
</center>
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Prove that if <math>\dfrac{1}{p}-2S_q = \dfrac{m}{n}</math> for integers
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Prove that if <math>\dfrac{1}{p}-2S_q = \dfrac{m}{n}</math> for relatively prime integers
 
<math>m</math> and <math>n</math>, then <math>m-n</math> is divisible by <math>p</math>.
 
<math>m</math> and <math>n</math>, then <math>m-n</math> is divisible by <math>p</math>.
  
 
[[2010 USAMO Problems/Problem 5|Solution]]
 
[[2010 USAMO Problems/Problem 5|Solution]]
  
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==Problem 6==
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===Problem 6===
 
A blackboard contains 68 pairs of nonzero integers.  Suppose that for each positive integer <math>k</math> at most one of the pairs <math>(k, k)</math> and <math>(-k, -k)</math> is written on the blackboard.  A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0.  The student then scores one point for each of the 68 pairs in which at least one integer is erased.  Determine, with proof, the largest number <math>N</math> of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.
 
A blackboard contains 68 pairs of nonzero integers.  Suppose that for each positive integer <math>k</math> at most one of the pairs <math>(k, k)</math> and <math>(-k, -k)</math> is written on the blackboard.  A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0.  The student then scores one point for each of the 68 pairs in which at least one integer is erased.  Determine, with proof, the largest number <math>N</math> of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.
  
 
[[2010 USAMO Problems/Problem 6|Solution]]
 
[[2010 USAMO Problems/Problem 6|Solution]]
  
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= See also =
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== See Also ==
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*[[USAMO Problems and Solutions]]
 
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{{USAMO newbox|year= 2010|before=[[2009 USAMO]]|after=[[2011 USAMO]]}}
 
{{USAMO newbox|year= 2010|before=[[2009 USAMO]]|after=[[2011 USAMO]]}}
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{{MAA Notice}}

Latest revision as of 12:44, 4 July 2013

Day 1

Problem 1

Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P$, $Q$, $R$, $S$ the feet of the perpendiculars from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where $O$ is the midpoint of segment $AB$.

Solution

Problem 2

There are $n$ students standing in a circle, one behind the other. The students have heights $h_1<h_2<\dots <h_n$. If a student with height $h_k$ is standing directly behind a student with height $h_{k-2}$ or less, the two students are permitted to switch places. Prove that it is not possible to make more than $\binom{n}{3}$ such switches before reaching a position in which no further switches are possible.

Solution

Problem 3

The 2010 positive numbers $a_1, a_2, \ldots , a_{2010}$ satisfy the inequality $a_ia_j\leq i+j$ for all distinct indices $i, j$. Determine, with proof, the largest possible value of the product $a_1a_2\ldots a_{2010}$.

Solution

Day 2

Problem 4

Let $ABC$ be a triangle with $\angle A=90^{\circ}$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD=\angle DBC$ and $\angle ACE=\angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$ to all have integer side lengths.

Solution

Problem 5

Let $q = \dfrac{3p-5}{2}$ where $p$ is an odd prime, and let

\[S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q\cdot (q+1) \cdot (q+2)}.\]

Prove that if $\dfrac{1}{p}-2S_q = \dfrac{m}{n}$ for relatively prime integers $m$ and $n$, then $m-n$ is divisible by $p$.

Solution

Problem 6

A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer $k$ at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point for each of the 68 pairs in which at least one integer is erased. Determine, with proof, the largest number $N$ of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.

Solution

See Also

2010 USAMO (Problems • Resources)
Preceded by
2009 USAMO 		Followed by
2011 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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