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Difference between revisions of "2013 USAMO"
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===Problem 3===
 
===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 [i]operation[/i] is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line.  A configuration is called [i]admissible [/i] 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.
+
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]]
 
[[2013 USAMO Problems/Problem 3|Solution]]

Revision as of 18:34, 11 May 2013

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 postive 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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