Competition problem — Geometry — Maths Olympiad Prep

Library / /2 of 20

Geometry Difficulty 5.0 AIME, harder Prove it Greece

Let $ABC$ be a triangle with $AB < AC$ , inscribed in a circle $c$ of center $O$ . Let $G$ be its barycenter and $D, E, F$ the foot of the altitudes from $A, B, C$ , respectively. If the rays $AG, GD$ intersect $c$ at $M, N$ respectively, prove that the points $F, E, M, N$ are cocyclic.

Solution

Let $K$ be the midpoint of $BC$ and $P$ the second intersection of $GD$ with $c$ . Then $c$ and the Euler's circle are homothetic with center $G$ and ratio $-2$ , so
$\frac{GP}{GD} = \frac{GA}{GK} = 2 \Rightarrow GP = 2GD.$
From the power of a point theorem we have
$GM \cdot GA = GN \cdot GP \Rightarrow GM \cdot (2GK) = GN \cdot (2GD) \Rightarrow GM \cdot GK = GN \cdot GD, \text{ so the quadrilateral } DKMN \text{ is inscribed in a circle, let it be } c_1.$
Moreover $F, D, K, E$ belong to a circle (the Euler circle), let it be $c_2$ .
Therefore the lines $FE, DK, MN$ are concurrent to the radical center, let it be $T$ , of the circles $c, c_1, c_2$ . To this end, $TF \cdot TE = TD \cdot TK = TN \cdot TM$ , so the points $F, E, M, N$ are cocyclic.

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.