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Difference between revisions of "2013 USAMO Problems"

(Created page with "==Day 1== ===Problem 1=== Are there integers <math>a</math> and <math>b</math> such that <math>a^5b+3</math> and <math>ab^5+3</math> are both perfect cubes of integers? [[2013 ...")
 
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==Day 1==
 
==Day 1==
 
===Problem 1===
 
===Problem 1===
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In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively.  Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively.  Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>.
  
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Are there integers <math>a</math> and <math>b</math> such that <math>a^5b+3</math> and <math>ab^5+3</math> are both perfect cubes of integers?
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[[2013 USAMO Problems/Problem 1|Solution]]
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[[2013 USAJMO Problems/Problem 1|Solution]]
 
  
 
===Problem 2===
 
===Problem 2===
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Each cell of an <math>m\times n</math> board is filled with some nonnegative integer.  Two numbers in the filling are said to be ''adjacent'' if their cells share a common side.  (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a ''garden'' if it satisfies the following two conditions:
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For a positive integer <math>n\geq 3</math> plot <math>n</math> equally spaced points around a circle.  Label one of them <math>A</math>, and place a marker at <math>A</math>.  One may move the marker forward in a clockwise direction to either the next point or the point after that.  Hence there are a total of <math>2n</math> distinct moves available; two from each point.  Let <math>a_n</math> count the number of ways to advance around the circle exactly twice, beginning and ending at <math>A</math>, without repeating a move.  Prove that <math>a_{n-1}+a_n=2^n</math> for all <math>n\geq 4</math>.
  
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(i) The difference between any two adjacent numbers is either <math>0</math> or <math>1</math>.
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[[2013 USAMO Problems/Problem 2|Solution]]
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(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to <math>0</math>.
 
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Determine the number of distinct gardens in terms of <math>m</math> and <math>n</math>.
 
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[[2013 USAJMO Problems/Problem 2|Solution]]
 
  
 
===Problem 3===
 
===Problem 3===
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In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively.  Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively.  Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>.
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Let <math>n</math> be a positive integer.  There are <math>\tfrac{n(n+1)}{2}</math> marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing <math>n</math> marks.  Initially, each mark has the black side up.  An ''operation'' is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line.  A configuration is called ''admissible'' if it can be obtained from the initial configuration by performing a finite number of operations.  For each admissible configuration <math>C</math>, let <math>f(C)</math> denote the smallest number of operations required to obtain <math>C</math> from the initial configuration.  Find the maximum value of <math>f(C)</math>, where <math>C</math> varies over all admissible configurations.
  
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[[2013 USAMO Problems/Problem 1|Solution]]
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[[2013 USAMO Problems/Problem 3|Solution]]
  
 
==Day 2==
 
==Day 2==
 
===Problem 4===
 
===Problem 4===
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Let <math>f(n)</math> be the number of ways to write <math>n</math> as a sum of powers of <math>2</math>, where we keep track of the order of the summation.  For example, <math>f(4)=6</math> because <math>4</math> can be written as <math>4</math>, <math>2+2</math>, <math>2+1+1</math>, <math>1+2+1</math>, <math>1+1+2</math>, and <math>1+1+1+1</math>.  Find the smallest <math>n</math> greater than <math>2013</math> for which <math>f(n)</math> is odd.
 
  
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[[2013 USAJMO Problems/Problem 4|Solution]]
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Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
 +
 
 +
[[2013 USAMO Problems/Problem 4|Solution]]
  
 
===Problem 5===
 
===Problem 5===
 +
Given positive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.
  
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Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>.  Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>.  Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>.  Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>.  Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>.  Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath>
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[[2013 USAMO Problems/Problem 5|Solution]]
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[[2013 USAJMO Problems/Problem 5|Solution]]
 
  
 
===Problem 6===
 
===Problem 6===
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Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
+
Let <math>ABC</math> be a triangle.  Find all points <math>P</math> on segment <math>BC</math> satisfying the following property:  If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath>
  
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[[2013 USAMO Problems/Problem 4|Solution]]
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[[2013 USAMO Problems/Problem 6|Solution]]
  
 
== See Also ==
 
== See Also ==
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{{USAJMO newbox|year= 2013|before=[[2012 USAJMO]]|after=[[2014 USAJMO]]}}
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{{USAMO newbox|year= 2013|before=[[2012 USAMO]]|after=[[2014 USAMO]]}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:08, 30 April 2014

Day 1

Problem 1

In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$.

Solution

Problem 2

For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$.

Solution

Problem 3

Let $n$ be a positive integer. There are $\tfrac{n(n+1)}{2}$ marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ marks. Initially, each mark has the black side up. An operation is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration $C$, let $f(C)$ denote the smallest number of operations required to obtain $C$ from the initial configuration. Find the maximum value of $f(C)$, where $C$ varies over all admissible configurations.

Solution

Day 2

Problem 4

Find all real numbers $x,y,z\geq 1$ satisfying \[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]

Solution

Problem 5

Given positive integers $m$ and $n$, prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten.

Solution

Problem 6

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]

Solution

See Also

2013 USAMO (Problems • Resources)
Preceded by
2012 USAMO 		Followed by
2014 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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