2011/12 AHSMC Part 1 Problem 15

Problem 15 of 16

The lengths of the sides of triangle $ABC$ are consecutive positive integers. $D$ is the midpoint of $BC$ and $AD$ is perpendicular to the bisector of $\angle C$ . The product of the lengths of the three sides is

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Solution

Let $AD$ intersect the bisector of $\angle C$ at $G$ . Then $\angle CGA = 90^\circ = \angle CGD$ , $\angle GCA = \angle GCD$ and $GC = GC$ . Hence triangles $GCA$ and $GCD$ are congruent, so that $AC = DC$ . It follows that we have $BC = 2DC = 2AC$ . Now among three consecutive positive integers, one is double another. This is only possible if the integers are $1$ , $2$ and $3$ , or $2$ , $3$ and $4$ . The former does not yield a triangle. Hence $AC = 2$ , $AB = 3$ and $BC = 4$ , so that $AB \cdot BC \cdot CA = 24$ . The answer is (A).

University of Alberta

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