2012/13 AHSMC Part 1 Problem 14

Problem 14 of 16

The incircle of triangle $ABC$ is tangent to $AB$ and $AC$ at $F$ and $E$ respectively. If $BC = 1$ , $\angle A = 90^\circ$ and $\angle B \ne \angle C$ , what is the distance from the midpoint of $BC$ to $EF$ ?

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Solution

Let $M$ be the midpoint of $BC$ and $D$ the point where the circle is tangent to $BC$ . Let $B'$ , $C'$ and $M'$ be the respective projections of $B$ , $C$ and $M$ on $EF$ . Now $AEF$ is a right isosceles triangle. Hence so are $BB'F$ and $CC'E$ . Hence $BB' = \dfrac{BF}{\sqrt{2}}$ and $CC' = \dfrac{CE}{\sqrt{2}}$ , so that $MM' = \frac{1}{2}(BB' + CC') = \frac{1}{2\sqrt{2}}(BF + CE) = \frac{1}{2\sqrt{2}}(BD + CD) = \frac{BC}{2\sqrt{2}} = \frac{\sqrt{2}}{4}.$ The answer is (A).

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