Library / /1157 of 1394

, 2020

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle and $\omega$ be its circumcircle. The point $M$ is the midpoint of arc $BC$ not containing $A$ on $\omega$ and $D$ is chosen so that $DM$ is tangent to $\omega$ and is on the same side of $AM$ as $C$ . It is given that $AM = AC$ and $\angle DMC = 38^{\circ}$ . Find the measure of angle $\angle ACB$ .

Solution

Solution:

By inscribed angles, we know that $\angle BAC = 38^{\circ} \cdot 2 = 76^{\circ}$ which means that $\angle C = 104^{\circ} - \angle B$ . Since $AM = AC$ , we have $\angle ACM = \angle AMC = 90^{\circ} - \frac{\angle MAC}{2} = 71^{\circ}$ . Once again by inscribed angles, this means that $\angle B = 71^{\circ}$ which gives $\angle C = 33^{\circ}$ .

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