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 $\triangle GEM$.
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 $AIME$.


Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of [trapezoid](https://artofproblemsolving.com/wiki/index.php/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 $h + 10 = \boxed{025}$.