20. [10] Triangle $\triangle A B C$ has $A B=21, B C=55$, and $C A=56$. There are two points $P$ in the plane of $\triangle A B C$ for which $\angle B A P=\angle C A P$ and $\angle B P C=90^{\circ}$. Find the distance between them.
Answer: $\quad \frac{5}{2} \sqrt{409}$
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
$$
15 \cdot 40 \cdot 55+55 \cdot A D^{2}=21^{2} \cdot 40+56^{2} \cdot 15 \Longrightarrow A D=24 \text {. }
$$

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
$$
P_{1} P_{2}=2 \sqrt{\left(\frac{55}{2}\right)^{2}-\left(\frac{25 \sqrt{3}}{4}\right)^{2}}=\frac{5 \sqrt{409}}{2} .
$$