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NMO Selection Tests for BMO and IMO

Geometry Difficulty 8.2 Shortlist Prove it Romania

Let $ABC$ be a triangle, let $A'$ , $B'$ , $C'$ be the orthogonal projections of the vertices $A$ , $B$ , $C$ on the lines $BC$ , $CA$ and $AB$ , respectively, and let $X$ be a point on the line $AA'$ . Let $\gamma_B$ be the circle through $B$ and $X$ , centered on the line $BC$ , and let $\gamma_C$ be the circle through $C$ and $X$ , centered on the line $BC$ . The circle $\gamma_B$ meets the lines $AB$ and $BB'$ again at $M$ and $M'$ , respectively, and the circle $\gamma_C$ meets the lines $AC$ and $CC'$ again at $N$ and $N'$ , respectively. Show that the points $M$ , $M'$ , $N$ and $N'$ are collinear.

Solution

Let $H$ be the orthocenter of the triangle $ABC$ . The line $AH$ is the radical axis of the circles $\gamma_B$ and $\gamma_C$ , hence $HM' \cdot HB = HN' \cdot HC$ and $AM \cdot AB = AN \cdot AC$ , so the lines $M'N'$ and $MN$ are both antiparallel to $BC$ .

The circle $\gamma_B$ meets the line $BC$ again at $B_1$ . Then the lines $MM'$ and $BC$ are antiparallel, since $BMM'B_1$ is a cyclic quadrangle. The conclusion follows.

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