HMMT, 2013

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

In triangle $ABC$ , $\angle A = 45^{\circ}$ and $M$ is the midpoint of $\overline{BC}$ . $\overline{AM}$ intersects the circumcircle of $ABC$ for the second time at $D$ , and $AM = 2MD$ . Find $\cos \angle AOD$ , where $O$ is the circumcenter of $ABC$ .

Solution

Solution:

$\cos \angle AOD = -\frac{1}{8}$ . $\angle BAC = 45^{\circ}$ , so $\angle BOC = 90^{\circ}$ . If the radius of the circumcircle is $r$ , $BC = \sqrt{2} r$ , and $BM = CM = \frac{\sqrt{2}}{2} r$ . By power of a point, $BM \cdot CM = AM \cdot DM$ , so $AM = r$ and $DM = \frac{1}{2} r$ , and $AD = \frac{3}{2} r$ . Using the law of cosines on triangle $AOD$ gives $\cos \angle AOD = -\frac{1}{8}$ .

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