Problem:
In triangle ABC,AB=125,AC=117, and BC=120. The angle bisector of angle A intersects BC at point L, and the angle bisector of angle B intersects AC at point K. Let M and N be the feet of the perpendiculars from C to BK and AL, respectively. Find MN.
Solution:
Extend CM and CN to meet AB at points P and Q, respectively. Triangles BPM and BCM are congruent. Thus M is the midpoint of PC. Analogously, N is the midpoint of QC​. Hence MN is a midline of triangle PQC, and MN=2PQ​. Furthermore,
PQ=AQ+BP−AB=AC+BC−AB=112
Thus MN=56​.
The problems on this page are the property of the MAA's American Mathematics Competitions
