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Revision as of 17:02, 3 July 2013

Day 1

Problem 1

Are there integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?

Solution

Problem 2

Each cell of an $m\times n$ 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:

(i) The difference between any two adjacent numbers is either $0$ or $1$.

(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to $0$.

Determine the number of distinct gardens in terms of $m$ and $n$.

Solution

Problem 3

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

Day 2

Problem 4

Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$, where we keep track of the order of the summation. For example, $f(4)=6$ because $4$ can be written as $4$, $2+2$, $2+1+1$, $1+2+1$, $1+1+2$, and $1+1+1+1$. Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.

Solution

Problem 5

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

Solution

Problem 6

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

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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