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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME, harder Prove it United States

Problem:
Triangle $ABC$ has $AB=1$ , $BC=\sqrt{7}$ , and $CA=\sqrt{3}$ . Let $\ell_1$ be the line through $A$ perpendicular to $AB$ , $\ell_2$ the line through $B$ perpendicular to $AC$ , and $P$ the point of intersection of $\ell_1$ and $\ell_2$ . Find $PC$ .

Solution

Solution:
By the Law of Cosines, $\angle BAC = \cos^{-1} \left(\frac{3+1-7}{2 \sqrt{3}}\right) = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = 150^\circ$ .

If we let $Q$ be the intersection of $\ell_2$ and $AC$ , we notice that $\angle QBA = 90^\circ - \angle QAB = 90^\circ - 30^\circ = 60^\circ$ .

It follows that triangle $ABP$ is a $30$ - $60$ - $90$ triangle and thus $PB = 2$ and $PA = \sqrt{3}$ .

Finally, we have $\angle PAC = 360^\circ - (90^\circ + 150^\circ) = 120^\circ$ , and
$PC = \left(PA^2 + AC^2 - 2 PA \cdot AC \cos 120^\circ\right)^{1/2} = (3 + 3 + 3)^{1/2} = 3.$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.