HMMT November

Since a regular pentagon has interior angles $108^{\circ}$, we can compute $\angle P D E=66^{\circ}, \angle P A E=60^{\circ}$, and $\angle A P D=360^{\circ}-\angle A E D-\angle P D E-\angle P A E=126^{\circ}$. Now observe that drawing $P E$ divides quadrilateral $P A E D$ into equilateral triangle $P A E$ and isosceles triangle $P E D$, where $\angle D P E=\angle E D P=66^{\circ}$. That is, we get $P A=P E=s$, where $s$ is the side length of the pentagon. Now triangles $P A B$ and $P E D$ are congruent (with angles $48^{\circ}-66^{\circ}-66^{\circ}$), so $P D=P B$ and $\angle P D C=\angle P B C=42^{\circ}$. This means that triangles $P D C$ and $P B C$ are congruent (side-angle-side), so $\angle B P C=\angle D P C$. Finally, we compute $\angle B P C+\angle D P C=2 \angle B P C=360^{\circ}-\angle A P B-\angle E P A-\angle D P E=168^{\circ}$, meaning $\angle B P C=84^{\circ}$.