Problem 14. The answer is (C).

Let $\alpha=\widehat{D C O}=\widehat{O D C}$ and $\beta=\widehat{A P C}=\widehat{O P D}=\widehat{D O P}$. By the exterior angle theorem applied to the angle at $O$ of triangle $C O P$, we have $\widehat{A O C}=\alpha+\beta$; and by the same theorem applied to the angle at $D$ of triangle $O D P$, we have $\alpha=2 \beta$. Therefore, $\widehat{A O C}=3 \beta$ and $\widehat{A P C}=20^{\circ}$.

(Problem proposed by F. Caceffo)
14. Given a circle $\gamma$ with center $O$ and diameter $A B$ of length $12 \, \text{cm}$, let $C$ be a point on $\gamma$ such that $A \widehat{O C}=60^{\circ}$, and let $P$ be a point on the extension of the diameter $A B$, on the side of $B$, such that $O D=D P$ (where $D$ is the intersection point between $P C$ and $\gamma$ between $C$ and $P$). What is the measure of the angle $A \widehat{P} C$?
(A) $15^{\circ}$
(B) $\frac{45^{\circ}}{2}$
(C) $20^{\circ}$
(D) $18^{\circ}$
(E) $24^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_17_82169a1997fb4f3442bcg-2.jpg?height=253&width=351&top_left_y=1093&top_left_x=1059)