2013/14 AHSMC Part 1 Problem 15

Problem 15 of 16

A triangle $ABC$ with $AB = 7$ , $BC = 8$ and $CA = 10$ has an interior point $P$ such that $\angle APB = \angle BPC = \angle CPA = 120^\circ$ . Let $r_1$ , $r_2$ and $r_3$ be the radii of the circles passing through the vertices of triangles $PAB$ , $PBC$ and $PCA$ respectively. The value of $r_1^2 + r_2^2 + r_3^2$ is

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Solution

Let $O$ be the centre of the circle passing through the vertices of triangle $PAB$ . Note that since $\angle APB > 90^\circ$ , $O$ lies on the perpendicular bisector of $AB$ outside of triangle $PAB$ . Since $OA = OP = OB$ , $\angle AOB = \angle AOP + \angle POB = 180^\circ - 2\angle APO + 180^\circ - 2\angle BPO = 360^\circ - 2\angle APB = 120^\circ.$ Hence $AB = \sqrt{3}\,OA$ so that $OA = \dfrac{AB}{\sqrt{3}}$ . It follows that $r_1^2 = \dfrac{AB^2}{3}$ . Similarly, $r_2^2 = \dfrac{BC^2}{3}$ and $r_3^2 = \dfrac{CA^2}{3}$ . Hence $r_1^2 + r_2^2 + r_3^2 = \frac{7^2 + 8^2 + 10^2}{3} = 71.$ The answer is (A).

University of Alberta

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