2015 Mock AIME I Problem 3: Intersection of Circles

First note that $\triangle APB$ is a $3-4-5$ right triangle and $\triangle APC$ is a $20-21-29$ right triangle.

Thus, we have $\cos \angle BAP = \dfrac{20}{25}=\dfrac{4}{5}$ and $\cos \angle PAC = \dfrac{20}{29}$ .

Similarly, we have $\sin \angle BAP = \dfrac{15}{25}=\dfrac{3}{5}$ and $\sin \angle PAC = \dfrac{21}{29}$ .

Applying the Law of Cosines on $\angle BAC$ gives us:

$BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\cos (\angle BAP+\angle PAC)}$

We also know that:

$\cos (\angle BAP+\angle PAC)=\cos\angle BAP\cos\angle PAC - \sin \angle BAP \sin \angle PAC$

$\qquad\qquad\qquad\qquad\quad\;\;=\dfrac{20}{25}\cdot\dfrac{20}{29}-\dfrac{15}{25}\cdot\dfrac{21}{29}=\dfrac{17}{145}.$

Hence, we have $BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\dfrac{17}{145}}=\sqrt{1296}=\boxed{36}$ .