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Team Selection Test for the 54th IMO

Geometry Difficulty 8.4 Shortlist Prove it Bulgaria

Points $M$ , $N$ and $P$ lie on the sides $BC$ , $CA$ , $AB$ of triangle $ABC$ . Triangles $CNM$ , $APN$ and $BMP$ are acute and let $H_C$ , $H_A$ and $H_B$ be their respective orthocenters. Prove that if the three lines $AH_A$ , $BH_B$ and $CH_C$ are concurrent then $MH_A$ , $NH_B$ and $PH_C$ are also concurrent.

Solution

Denote by $A_1$ , $B_1$ and $C_1$ the projections of $A$ , $B$ and $C$ on $NP$ , $PM$ and $MN$ , respectively. It follows from the condition of the problem (Carnot's theorem) that $(NC_1^2 - MC_1^2) + (MB_1^2 - PB_1^2) + (PA_1^2 - NA_1^2) = 0$ . Hence $0 = (CN^2 - CM^2) + (BM^2 - BP^2) + (AP^2 - AN^2) = (CN^2 - AN^2) + (AP^2 - BP^2) + (BM^2 - CM^2)$ and the same theorem implies that the perpendiculars from $M$ , $N$ and $P$ to $BC$ , $AC$ and $AB$ are concurrent at a point $O$ .

Since $OMH_CN$ and $OMH_BP$ are parallelograms we conclude that $PH_BH_CN$ is also a parallelogram. Therefore the segments $PH_C$ and $NH_B$ bisect each other at some point. The segment $MH_A$ passes through the same point.

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