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Mock AIME I 2015 Problems/Problem 3

Problem 3

Let $A,B,C$ be points in the plane such that $AB=25$, $AC=29$, and $\angle BAC<90^\circ$. Semicircles with diameters $\overline{AB}$ and $\overline{AC}$ intersect at a point $P$ with $AP=20$. Find the length of line segment $\overline{BC}$.

Solution

First note that $\triangle APB$ is a 3-4-5 scaled right triangle with $\overline{B P}=15$ and $\triangle APC$ is a 20-21-29 right triangle with $\overline{P C}=21$ since $\angle APB$ and $\angle APC$ are inscribed in semicircles, and thus right angles. Also, since $\angle APB$ and $\angle APC$ are right, points B, P and C are collinear and $BC=BP+PC=15+21=\boxed{036}.$

Solution by D. Adrian Tanner

Original Solution

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 \[\cos (\angle BAP+\angle PAC)=\cos\angle BAP\cos\angle PAC - \sin \angle BAP \sin \angle PAC=\dfrac{20}{25}\cdot\dfrac{20}{29}-\dfrac{15}{25}\cdot\dfrac{21}{29}=\dfrac{17}{145}.\] Hence, we have \[BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\dfrac{17}{145}}=\sqrt{1296}=\boxed{036}.\]

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