Competition problem — Geometry — Maths Olympiad Prep

Library / /20 of 39

Geometry Difficulty 5.4 AIME, harder Prove it Ireland

A triangle *ABC* is inscribed in a circle, centre *O*, and $D$ , $E$ , $F$ are the mid-points of $BC$ , $CA$ , $AB$ respectively. The point $A$ is free to move on the circumference of the circle. Find the locus of $L$ , the midpoint of $FE$ .

Solution

Let $M$ and $N$ be the midpoints of $BD$ and $DC$ respectively. Let $AD$ meet $EF$ at $L'$ . Then $FL' \parallel BD$ and $|FL'| = \frac{1}{2}|BD|$ . Similarly $|L'E| = \frac{1}{2}|DC|$ , hence $|FL'| = |L'E|$ , i.e. $L' = L$ is the midpoint of $FE$ .

Let $P$ be the midpoint of $OD$ . Then $PL \parallel OA$ and $|PL| = \frac{1}{2}|OA|$ . Similarly $|PM| = \frac{1}{2}|OB|$ and $|PN| = \frac{1}{2}|OC|$ . Since $|OA| = |OB| = |OC|$ then $|PL| = |PM| = |PN|$ . So the circle with centre $P$ and radius $|PM|$ passes through $L$ . As $A$ moves on the circumference of the circle, the points $O$ , $P$ , $D$ and $M$ remain fixed. Thus $L$ will always lie on the circumference of the circle centre $P$ and radius equal to half the radius of the circumcircle.

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 and solution reproduced as published; topic and difficulty added by this site.