HMMT February

Solution:

Let $G$ be the intersection of the altitude to $\overline{A B}$ at point $D$ with $\overline{A C} \cup \overline{B C}$. We first note that the maximal expected value is obtained when $D G=\frac{[A D G C]}{A D}$, where $[P]$ denotes the area of polygon $P$. Note that if $D G$ were not equal to this value, we could move $D$ either closer or further from $A$ and increase the value of the fraction, which is the expected value of $E F$. We note that this equality can only occur if $D$ is on the side of the altitude to $\overline{A B}$ nearest point $B$. Multiplying both sides of this equation by $A D$ yields $A D \cdot D G=[A D G C]$, which can be interpreted as meaning that the area of the rectangle with base $\overline{A D}$ and height $\overline{D G}$ must have area equal to that of quadrilateral $A D G C$. We can now solve this problem with algebra.

Let $x=B D$. We first compute the area of the rectangle with base $\overline{A D}$ and height $\overline{D G}$. We have that $A D=A B-B D=14-x$. By decomposing the $13$-$14$-$15$ triangle into a $5$-$12$-$13$ triangle and a $9$-$12$-$15$ triangle, and using a similarity argument, we find that $D G=\frac{4}{3} x$. Thus, the area of this rectangle is $\frac{4}{3} x(14-x)=\frac{56}{3} x-\frac{4}{3} x^{2}$.

We next compute the area of quadrilateral $A D G C$. We note that $[A D G C]=[A B C]-[B D G]$. We have that $[A B C]=\frac{1}{2}(12)(14)=84$. We have $B D=x$ and $D G=\frac{4}{3} x$, so $[B D G]=\frac{1}{2}(x)\left(\frac{4}{3} x\right)=\frac{2}{3} x^{2}$. Therefore, we have $[A D G C]=[A B C]-[B D G]=84-\frac{2}{3} x^{2}$.

Equating these two areas, we have
$$\frac{56}{3} x-\frac{4}{3} x^{2}=84-\frac{2}{3} x^{2}$$

or, simplifying,
$$x^{2}-28 x+126=0$$

Solving yields $x=14 \pm \sqrt{70}$, but $14+\sqrt{70}$ exceeds $A B$, so we discard it as an extraneous root. Thus, $B D=14-\sqrt{70}$ and
$$A D=A B-B D=14-(14-\sqrt{70})=\sqrt{70}$$