4. Let $A B C$ be a right triangle at $C$ with sides $B C=3$ and $A B=12$. Let $M$ be the midpoint of $A B$, and $D$ the intersection of $A C$ and the circumcircle of $B C M$. Finally, let $P$ be the intersection of $B C$ and $M D$. What is the length of the segment $P A$?
(A) $\frac{28}{5} \sqrt{15}$
(B) $6 \sqrt{15}$
(C) 24
(D) $12 \sqrt{15}$
(E) 48
4. The answer is (C). Since $B C D M$ is cyclic, we have $\angle B M D=\angle B C D=90^{\circ}$. Triangles $A B C$ and $B M P$ are thus similar because they are right-angled and share the angle at $B$. Therefore, $B M: B C=B P: B A$, which means $6: 3=B P: 12$, from which $B P=24$ and $C P=21$. Finally, triangle $A C P$ is also right-angled, and applying the Pythagorean theorem we get $A P=\sqrt{C P^{2}+C A^{2}}=$ $\sqrt{441+\left(A B^{2}-B C^{2}\right)}=\sqrt{576}=24$.

Second solution: As in the first solution, we have $D M \perp A B$, so $P M$ is the perpendicular bisector of segment $A B$. In particular, $A B P$ is isosceles, so $A P=B P=24$, as calculated in the first solution.

![](https://cdn.mathpix.com/cropped/2024_04_17_ffee1d4705848828ad5fg-13.jpg?height=494&width=688&top_left_y=136&top_left_x=724)