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Geometry Difficulty 5.4 AIME, harder Prove it Japan

Suppose that three points $A$ , $B$ , $C$ lie on the circumference of a circle $\Gamma$ . Let $P$ be the point of intersection of the lines tangent to $\Gamma$ at $B$ and $C$ . Suppose that the lines $AB$ and $CP$ are parallel, and that $AB = 3$ and $BP = 4$ . Find the length of the line segment $BC$ . Here we represent the length of the line segment $XY$ also by $XY$ .

Solution

$2\sqrt{3}$

Since $AB$ and $CP$ are parallel, $\angle ABC = \angle BCP$ . By a well-known theorem we also see that $\angle CAB = \angle PBC$ must hold. Therefore the triangles $ABC$ and $BCP$ are similar, which implies that $AB : BC = BC : CP$ . From this it follows that $BC = \sqrt{AB \cdot CP}$ . Since $P$ is the point of intersection of the tangent lines to the circle $\Gamma$ , we have $CP = BP = 4$ , from which it follows that we have $BC = \sqrt{3} \cdot 4 = 2\sqrt{3}$ .

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