Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

Let $ABC$ be a triangle with circumcenter $O$ , incenter $I, \angle B=45^{\circ}$ , and $OI \parallel BC$ . Find $\cos \angle C$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let $M$ be the midpoint of $BC$ , and $D$ the foot of the perpendicular of $I$ with $BC$ . Because $OI \parallel BC$ , we have $OM=ID$ . Since $\angle BOC=2 \angle A$ , the length of $OM$ is $OA \cos \angle BOM=OA \cos A=R \cos A$ , and the length of $ID$ is $r$ , where $R$ and $r$ are the circumradius and inradius of $\triangle ABC$ , respectively. Thus, $r=R \cos A$ , so $1+\cos A=(R+r) / R$ . By Carnot's theorem, $(R+r) / R=\cos A+\cos B+\cos C$ , so we have $\cos B+\cos C=1$ . Since $\cos B=\frac{\sqrt{2}}{2}$ , we have $\cos C=1-\frac{\sqrt{2}}{2}$ .

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