Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Russia

Let $BH$ be an altitude in an acute-angled triangle $ABC$ . Let $M$ and $N$ be the midpoints of $AH$ and $CH$ , respectively. Let $BB'$ be a diameter in the circumcircle of $\triangle BMN$ . Prove that $AB' = CB'$ .

Solution

Let $O$ be the center of the circle $\Omega$ ; that is, $O$ is the midpoint of the diameter $BB'$ . Denote by $H'$ and $P$ the projections of the points $B'$ and $O$ , respectively, onto the line $AC$ (see Fig. 9). Since $O$ lies on the perpendicular bisector of $MN$ , we get that $P$ is the midpoint of $MN$ . Since $O$ is the midpoint of $BB'$ , $P$ is the midpoint of $HH'$ . Thus, $H$ and $H'$ are symmetric with respect to the midpoint of $MN$ , from which $HM = H'N$ and $H'M = HN$ . We have $AH' = AM + H'M = HM + H'M = H'N + HN = H'N + CN = CH'$ . Thus, $B'H'$ is the perpendicular bisector of the segment $AC$ , consequently, $AB' = CB'$ , as required.

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