HMMT November

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

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

Solution

Solution:

Answer: $86^{\circ}$

Extend $DM$ to hit the circumcircle at $E$ . Then, note that since $ADEB$ is a cyclic quadrilateral, $\angle ABE = 180^{\circ} - \angle ADE = 180^{\circ} - \angle ADM = 180^{\circ} - 68^{\circ} = 112^{\circ}$ .

We also have that $\angle MEC = \angle DEC = \angle DAC = 64^{\circ}$ . But now, since $M$ is the midpoint of $BC$ and since $EM \perp BC$ , triangle $BEC$ is isosceles. This implies that $\angle BEM = \angle MEC = 64^{\circ}$ , and $\angle MBE = 90^{\circ} - \angle MEB = 26^{\circ}$ . It follows that $\angle B = \angle ABE - \angle MBE = 112^{\circ} - 26^{\circ} = 86^{\circ}$ .

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