HMMT February

Geometry Difficulty 5.4 AIME, harder Find the answer

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

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

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

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.