2024/25 AHSMC Part 1 Problem 16

Problem 16 of 16

In triangle $ABC$ , $D$ is the midpoint of $BC$ and $AD$ is perpendicular to the bisector of angle $B$ . The side lengths of the triangle are positive integers, and one of these sides has length $4$ . The number of different triangles possible is:

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Solution

Let $BE$ be the angle bisector of $\angle B$ , with $E$ on $AC$ . In $\triangle ABD$ the bisector of $\angle B$ is also an altitude, so $\triangle ABD$ is isosceles with $AB = BD$ , hence $BC = 2\,AB$ . Conversely, $BC = 2\,AB$ forces the median $AD$ to be perpendicular to the bisector of $\angle B$ .

We want integer-sided triangles with $BC = 2\,AB$ and one side equal to $4$ .

If $AB = 4$ then $BC = 8$ and $5 \le AC \le 11$ , giving seven triangles $(4,5,8),\ldots,(4,11,8)$ .

If $AC = 4$ and $AB = n$ then $BC = 2n$ and $1 < n < 4$ , so $n = 2$ or $3$ : triangles $(2,4,4)$ and $(3,4,6)$ .

If $BC = 4$ then $AB = 2$ and $2 < AC < 6$ , giving $(2,3,4)$ , $(2,4,4)$ , $(2,4,5)$ .

The triangle $(2,4,4)$ is counted twice, so there are $7+2+3-1 = 11$ such triangles. The answer is (B).

University of Alberta

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