Problem (Geometry):
In $\triangle A B C, D$ is on $\overline{A C}$ so that $\overline{B D}$ is the angle bisector of $\angle B$. Point $E$ is on $\overline{A B}$ and $\overline{C E}$ intersects $\overline{B D}$ at $P$. Quadrilateral $B C D E$ is cyclic, $B P=12$ and $P E=4$. Compute the ratio $\frac{A C}{A E}$.

Solution:
Let $\omega$ denote the circle that circumscribes quadrilateral $B C D E$. Draw in line segment $\overline{D E}$. Note that $\angle D P E$ and $\angle C P B$ are congruent, and $\angle D E C$ and $\angle D B C$ are congruent, since they cut off the same arc of $\omega$. Therefore, $\triangle B C P$ and $\triangle E D P$ are similar. Thus $\frac{B C}{D E}=\frac{B P}{E P}=$ $\frac{12}{4}=3$.

Because $\angle B C E$ and $\angle B D E$ cut off the same arc of $\omega$, these angles are congruent. Let $\alpha$ be the measure of these angles. Similarly, $\angle D C E$ and $\angle D B E$ cut off the same arc of $\omega$. Let $\beta$ be the measure of these angles. Since $B D$ is an angle bisector, $\mathrm{m} \angle C B D=\beta$.

Note that $\mathrm{m} \angle A D E=180^{\circ}-\mathrm{m} \angle B D E-\mathrm{m} \angle B D C$. It follows that

$$
\begin{aligned}
\mathrm{m} \angle A D E & =180^{\circ}-\mathrm{m} \angle B D E-\left(180^{\circ}-\mathrm{m} \angle C B D-\mathrm{m} \angle B C D\right) \\
\Rightarrow \mathrm{m} \angle A D E & =180^{\circ}-\mathrm{m} \angle B D E-\left(180^{\circ}-\mathrm{m} \angle C B D-\mathrm{m} \angle B C E-\mathrm{m} \angle D C E\right) \\
\Rightarrow \mathrm{m} \angle A D E & =180^{\circ}-\alpha-\left(180^{\circ}-\beta-\alpha-\beta\right) \\
\Rightarrow \mathrm{m} \angle A D E & =2 \beta=\mathrm{m} \angle C B D .
\end{aligned}
$$

Thus $\angle A D E$ is congruent to $\angle C B D$, and it follows that $\triangle A D E$ is similar to $\triangle A B C$. Hence $\frac{B C}{D E}=\frac{A C}{A E}$, and by substituting in given values, we have $\frac{A C}{A E}=\mathbf{3}$.

Final answer: 3