HMMT November

Geometry Difficulty 4.6 AIME Find the answer

Let $A B C$ be a triangle, and let $D, E$ , and $F$ be the midpoints of sides $B C, C A$ , and $A B$ , respectively. Let the angle bisectors of $\angle F D E$ and $\angle F B D$ meet at $P$ . Given that $\angle B A C=37^{\circ}$ and $\angle C B A=85^{\circ}$ , determine the degree measure of $\angle B P D$ .

Your answer

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

Solution

Because $D, E, F$ are midpoints, we have $A B C \sim D E F$ . Furthermore, we know that $F D \| A C$ and $D E \| A B$ , so we have $\angle B D F=\angle B C A=180-37-85=58^{\circ}$ Also, $\angle F D E=\angle B A C=37^{\circ}$ . Hence, we have $\angle B P D=180^{\circ}-\angle P B D-\angle P D B=180^{\circ}-\frac{85^{\circ}}{2}-\left(\frac{37^{\circ}}{2}+58^{\circ}\right)=61^{\circ}$

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