MathDB
2017 Geometry Tiebreaker #3

Source:

November 19, 2022
2017Geometry Tiebreaker

Problem Statement

Triangle ABCABC has AB=4,BC=6,CA=5AB=4,BC=6,CA=5. Let MM be the midpoint of BC\overline{BC} and PP the point on the circumcircle of ABC\triangle ABC such that MPA=90\angle MPA=90^\circ. Let points DD and EE lie on AC\overline{AC} and AB\overline{AB} respectively such that BDAC\overline{BD}\perp\overline{AC} and CEAB\overline{CE}\perp\overline{AB}. Find PDPE\tfrac{PD}{PE}.