Competition problem — Geometry — Maths Olympiad Prep

Library / /99 of 220

Geometry Difficulty 5.8 AIME, harder Prove it Ukraine

Consider an isosceles triangle $ABC$ , where $AC$ is a base, $P$ is an arbitrary point on $AC$ , $T$ is a projection of $P$ onto $BC$ . Determine in the ratio in which a symmedian drawn from a vertex $C$ of a triangle $\triangle PBC$ divides $AT$ ? Symmedian $CS$ , $S \in BP$ of a $\triangle PBC$ is a reflection of median $CF$ over the angle bisector $CL$ .

Solution

Let $BH$ be an altitude of $\triangle ABC$ , then $H$ is a midpoint of $AC$ , $BH$ and $PT$ are altitudes of $\triangle PBC$ (Fig. 39). Recall that symmedian bisects a line segment whose endpoints are feet of the altitudes. Therefore, a symmedian of $\triangle PBC$ is passing through $M$ which is a midpoint of $HT$ . Let $CM$ intersect $AT$ at $K$ , then applying the Menelaus theorem on the triangle $\triangle ATH$ we get:
$\frac{AK}{KT} \cdot \frac{TM}{MH} \cdot \frac{HC}{CA} = 1 \Rightarrow \frac{AK}{KT} = \frac{AC}{CH} = 2.$

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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