Difference between revisions of "2013 USAMO"

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==Day 1==
'''2013 [[USAMO]]''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.
===Problem 1===
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>.
[[2013 USAMO Problems/Problem 1|Solution]]
*[[2013 USAMO Problems]]
*[[2013 USAMO Problems/Problem 1]]
*[[2013 USAMO Problems/Problem 2]]
*[[2013 USAMO Problems/Problem 3]]
*[[2013 USAMO Problems/Problem 4]]
*[[2013 USAMO Problems/Problem 5]]
*[[2013 USAMO Problems/Problem 6]]
===Problem 2===
{{USAMO newbox|year= 2013 |before=[[2012 USAMO]]|after=[[2014 USAMO]]}}
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>.
[[2013 USAMO Problems/Problem 2|Solution]]
{{MAA Notice}}
===Problem 3===
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.
[[2013 USAMO Problems/Problem 3|Solution]]
==Day 2==
===Problem 4===
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===
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.
[[2013 USAMO Problems/Problem 5|Solution]]
===Problem 6===
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>
[[2013 USAMO Problems/Problem 6|Solution]]
== See Also ==
{{USAMO newbox|year= 2013|before=[[2012 USAMO]]|after=[[2014 USAMO]]}}

Latest revision as of 19:35, 7 May 2015

2013 USAMO problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution.

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