Library / /6 of 84

, 2014

Geometry Difficulty 4.6 AIME Prove it United States

Problem:

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

Solution

Solution:

Answer: $1 - \frac{\sqrt{2}}{2}$

Let $M$ be the midpoint of $BC$ , and $D$ the foot of the perpendicular from $I$ to $BC$ . 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 $\cos C = 1 - \frac{\sqrt{2}}{2}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.