In triangle ABC, AB=125, AC=117, BC=120. The angle bisector of A intersect BC at L and the angle bisector of B intersect AC at K. Let M and N be the feet of perpendiculars from C to BK and AL respectively. Find MN.
I tried to coordinate bash this with the origin C. But after I calculated the coordinate of A and B I gave up. Point to a line with square roots is just crazy. Is there another way to do this (without crazy calculation)?
Any help appreciated.
Coordinate-bashing definitely isn't pretty, but it turns out that the solution is straightforward using good ol' geometry:
Let $M^\prime$ and $N^\prime$ (which are necessarily on $\overline{AB}$) be the reflections of $C$ over bisectors $\overline{BM}$ and $\overline{AN}$, respectively. Then $\triangle ACN^\prime$ and $\triangle BCM^\prime$ are isosceles, so that $|AN^\prime| = b$ and $|BM^\prime| = a$, and we readily find that $|M^\prime N^\prime| = a+b-c$. Since $\overline{MN}$ is a midsegment of $\triangle M^\prime C N^\prime$, it is parallel to $\overline{M^\prime N^\prime}$ (bonus fact!) as well as half its length: $|MN| = \frac12(a+b-c)$. $\square$
