HMMT November

Geometry Difficulty 4.9 AIME Find the answer

Triangle $A B C$ satisfies $\angle B>\angle C$ . Let $M$ be the midpoint of $B C$ , and let the perpendicular bisector of $B C$ meet the circumcircle of $\triangle A B C$ at a point $D$ such that points $A, D, C$ , and $B$ appear on the circle in that order. Given that $\angle A D M=68^{\circ}$ and $\angle D A C=64^{\circ}$ , find $\angle B$ .

Your answer

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

Solution

Extend $D M$ to hit the circumcircle at $E$ . Then, note that since $A D E B$ is a cyclic quadrilateral, $\angle A B E=180^{\circ}-\angle A D E=180^{\circ}-\angle A D M=180^{\circ}-68^{\circ}=112^{\circ}$ . We also have that $\angle M E C=\angle D E C=\angle D A C=64^{\circ}$ . But now, since $M$ is the midpoint of $B C$ and since $E M \perp B C$ , triangle $B E C$ is isosceles. This implies that $\angle B E M=\angle M E C=64^{\circ}$ , and $\angle M B E=90^{\circ}-\angle M E B=26^{\circ}$ . It follows that $\angle B=\angle A B E-\angle M B E=112^{\circ}-26^{\circ}=86^{\circ}$ .

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