# 2008 AIME I Problems/Problem 2

## Problem

Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$.

## Solution

Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$. This implies that vertex G must be located outside of square $AIME$.

$[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("$$G$$",G,N); label("$$A$$",A,NW); label("$$I$$",I,NE); label("$$M$$",M,NE); label("$$E$$",E,NW); label("$$10$$",(M+E)/2,S); [/asy]$

Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\triangle GXY \sim \triangle GEM$.

Let the height of $GXY$ be $h$. By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = \boxed{025}$.