Library / NMO Selection Tests for JBMO /1 of 58

NMO Selection Tests for JBMO

Geometry Difficulty 4.5 AIME Prove it Romania

Let $ABC$ be an acute triangle and $D \in (BC)$ , $E \in (AD)$ be mobile points. The circumcircle of triangle $CDE$ meets the median from $C$ of the triangle $ABC$ at $F$ . Prove that the circumcenter of triangle $AEF$ lies on a fixed line.

Solution

The quadrilateral $EFDC$ is cyclic, hence $\angle FED \equiv \angle FCD$ . Let $P$ be the reflection of point $C$ in the midpoint of the segment line $[AB]$ ; clearly $AP \parallel BC$ and $\angle FCD \equiv \angle FPA$ .

It follows that $\angle FPA \equiv \angle FED$ , which means that the quadrilateral $FEAP$ is cyclic. This shows that the circumcenter $AEF$ lies on the perpendicular bisector of the segment line $[AP]$ (which is a fixed line).

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.