Library / Harvard-MIT November Tournament /41 of 74

Harvard-MIT November Tournament

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle, and let $D$ , $E$ , and $F$ be the midpoints of sides $BC$ , $CA$ , and $AB$ , respectively. Let the angle bisectors of $\angle FDE$ and $\angle FBD$ meet at $P$ . Given that $\angle BAC = 37^{\circ}$ and $\angle CBA = 85^{\circ}$ , determine the degree measure of $\angle BPD$ .

Solution

Solution:

Answer: $61^{\circ}$

Because $D$ , $E$ , $F$ are midpoints, we have $ABC \sim DEF$ . Furthermore, we know that $FD \parallel AC$ and $DE \parallel AB$ , so we have
$\angle BDF = \angle BCA = 180^{\circ} - 37^{\circ} - 85^{\circ} = 58^{\circ}$
Also, $\angle FDE = \angle BAC = 37^{\circ}$ . Hence, we have
$\angle BPD = 180^{\circ} - \angle PBD - \angle PDB = 180^{\circ} - \frac{85^{\circ}}{2} - \left(\frac{37^{\circ}}{2} + 58^{\circ}\right) = 61^{\circ}$

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