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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===
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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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| − | [[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]] |
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| − | ===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>.
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| − | [[2013 USAMO Problems/Problem 2|Solution]]
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| − | ===Problem 3===
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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 3|Solution]]
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| − | ==Day 2==
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| − | ===Problem 4===
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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>
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| − | [[2013 USAMO Problems/Problem 4|Solution]]
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| − | ===Problem 5===
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| − | Given postive 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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| − | [[2013 USAMO Problems/Problem 5|Solution]]
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| − | ===Problem 6===
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| − | 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 6|Solution]]
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| − | == See Also ==
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| − | {{USAMO newbox|year= 2013|before=[[2012 USAMO]]|after=[[2014 USAMO]]}} | |
