2022/23 AHSMC Part 1 Problem 15

Problem 15 of 16

Let $ABC$ be an isosceles right triangle with $AB = AC = 1$ , and let $O$ be the centre of a circle drawn inside $ABC$ and tangent to all three of its sides. The line $BO$ intersects $AC$ at $D$ . The length of the segment $DC$ is

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Solution

Let $E$ , $F$ be the points where the circle is tangent to $AB$ , $AC$ respectively.

If $r$ denotes the radius of the inscribed circle, then $EB = FC = 1 - r$ . The tangents from point $B$ to the circle are equal and so are the tangents from point $C$ , hence $BC = 1 - r + 1 - r = 2 - 2r$ . On the other hand $BC = \sqrt{2}$ , hence $2 - 2r = \sqrt{2}$ thus $r = (2 - \sqrt{2})/2$ . Meanwhile, since $r/AD = (1-r)/1$ (from similar triangles $BAD$ and $BEO$ ), one obtains $AD = \frac{r}{1-r} = \frac{2-\sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1$ and hence $DC = 1 - AD = 2 - \sqrt{2}$ . The answer is (C).

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