HMMT February

Let $P_{1}$ and $P_{2}$ be the two possible points $P$, with $A P_{1}<A P_{2}$. Both lie on the $\angle A$-bisector and the circle $\gamma$ with diameter $B C$. Let $D$ be the point where the $\angle A$-bisector intersects $B C$, let $M$ be the midpoint of $B C$, and let $X$ be the foot of the perpendicular from $M$ onto the $\angle A$-bisector. Since we know the radius of $\gamma$, to compute $P_{1} P_{2}$ it suffices to compute $M X$. By the angle bisector theorem we find $B D=15$ and $D C=40$, so Stewart's theorem gives $15 \cdot 40 \cdot 55+55 \cdot A D^{2}=21^{2} \cdot 40+56^{2} \cdot 15 \Longrightarrow A D=24$. Then $\cos \angle A D B=\frac{-21^{2}+15^{2}+24^{2}}{2 \cdot 15 \cdot 24}=\frac{1}{2}$, so $\angle A D B=\angle M D X=60^{\circ}$. Since $D M=B M-B D=\frac{55}{2}-15=\frac{25}{2}$, we get $M X=D M \sin \angle M D X=\frac{25 \sqrt{3}}{4}$. Hence $P_{1} P_{2}=2 \sqrt{\left(\frac{55}{2}\right)^{2}-\left(\frac{25 \sqrt{3}}{4}\right)^{2}}=\frac{5 \sqrt{409}}{2}$.