Problem (Geometry):
Let $A R M L$ be a trapezoid with bases $\overline{A R}$ and $\overline{M L}$, such that $M R=R A=A L$ and $L R=$ $A M=M L$. Point $P$ lies inside the trapezoid such that $\angle R M P=12^{\circ}$ and $\angle R A P=6^{\circ}$. Diagonals $A M$ and $R L$ intersect at $D$. Compute the measure, in degrees, of angle $A P D$.

Solution:
First, determine the angles of $A R M L$. Let $\mathrm{m} \angle M=x$. Then $\mathrm{m} \angle L R M=x$ because $\triangle L R M$ is isosceles, and $\mathrm{m} \angle R L M=180^{\circ}-2 x$. Because $\overline{A R} \| \overline{L M}, \mathrm{~m} \angle A R M=180^{\circ}-x$ and $\mathrm{m} \angle A R L=180^{\circ}-2 x$, as shown in the diagram below.



However, $\triangle A R L$ is also isosceles (because $A R=A L$ ), so $\mathrm{m} \angle A L R=180^{\circ}-2 x$, yielding $\mathrm{m} \angle A L M=360^{\circ}-4 x$. Because $\mathrm{m} \angle R M L=\mathrm{m} \angle A L M$, conclude that $360^{\circ}-4 x=x$, so $x=72^{\circ}$. Therefore the base angles $L$ and $M$ have measure $72^{\circ}$ while the other base angles $A$ and $R$ have measure $108^{\circ}$. Finally, the angle formed by diagonals $\overline{A M}$ and $\overline{L R}$ is as follows: $\mathrm{m} \angle R D M=180^{\circ}-\mathrm{m} \angle L R M-\mathrm{m} \angle A M R=180^{\circ}-72^{\circ}-36^{\circ}=72^{\circ}$.

Now construct equilateral $\triangle R O M$ with $O$ on the exterior of the trapezoid, as shown below.



Because $A R=R M=R O$, triangle $O A R$ is isosceles with base $\overline{A O}$. The measure of $\angle A R O$ is $108^{\circ}+60^{\circ}=168^{\circ}$, so $\mathrm{m} \angle R A O=(180-168)^{\circ} / 2=6^{\circ}$. Thus $P$ lies on $\overline{A O}$. Additionally, $\mathrm{m} \angle P O M=\mathrm{m} \angle A O M=60^{\circ}-6^{\circ}=54^{\circ}$, and $\mathrm{m} \angle P M O=60^{\circ}+12^{\circ}=72^{\circ}$ by construction. Thus $\mathrm{m} \angle M P O=180^{\circ}-72^{\circ}-54^{\circ}=54^{\circ}$, hence $\triangle P M O$ is isosceles with $P M=O M$. But because $O M=R M, \triangle R M P$ is isosceles with $R M=M P$, and $R M=D M$ implies that $\triangle P D M$ is also isosceles. But $\mathrm{m} \angle R M P=12^{\circ}$ implies that $\mathrm{m} \angle P M D=36^{\circ}-12^{\circ}=24^{\circ}$, so $\mathrm{m} \angle D P M=78^{\circ}$. Thus $\mathrm{m} \angle A P D=180^{\circ}-\mathrm{m} \angle O P M-\mathrm{m} \angle D P M=180^{\circ}-54^{\circ}-78^{\circ}=48^{\circ}$.

Final answer: 48