HMMT-Nov 2014 guts, Problem 13

Let $A B C$ be a triangle with $A B=A C=\frac{25}{14} B C$. Let $M$ denote the midpoint of $\overline{B C}$ and let $X$ and $Y$ denote the projections of $M$ onto $\overline{A B}$ and $\overline{A C}$, respectively. If the areas of triangle $A B C$ and quadrilateral $A X M Y$ are both positive integers, find the minimum possible sum of these areas.

Solution: By similar triangles, one can show that $[A X M Y]=2 \cdot[A M X]=\left(\frac{24}{25}\right)^{2} \cdot 2[A B M]=$ $\left(\frac{24}{25}\right)^{2} \cdot[A B C]$. Thus the answer is $25^{2}+24^{2}=1201$. $\fbox{1201}$.

Answer: 1201