Let $A B C D$ be a quadrilateral with an inscribed circle $\omega$ and let $P$ be the intersection of its diagonals $A C$ and $B D$. Let $R_{1}, R_{2}, R_{3}, R_{4}$ be the circumradii of triangles $A P B, B P C, C P D, D P A$ respectively. If $R_{1}=31$ and $R_{2}=24$ and $R_{3}=12$, find $R_{4}$.
Note that $\angle A P B=180^{\circ}-\angle B P C=\angle C P D=180^{\circ}-\angle D P A$ so $\sin A P B=\sin B P C=\sin C P D=$ $\sin D P A$. Now let $\omega$ touch sides $A B, B C, C D, D A$ at $E, F, G, H$ respectively. Then $A B+C D=$ $A E+B F+C G+D H=B C+D A$ so \[ \frac{A B}{\sin A P B}+\frac{C D}{\sin C P D}=\frac{B C}{\sin B P C}+\frac{D A}{\sin D P A} \] and by the Extended Law of Sines this implies \[ 2 R_{1}+2 R_{3}=2 R_{2}+2 R_{4} \] which immediately yields $R_{4}=R_{1}+R_{3}-R_{2}=19$. $\fbox{19}$.