, 2019

Solution:

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 $\angle B P C = 84^{\circ}$.