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Slovenian Mathematical Olympiad, 2008

Geometry Difficulty 6.2 National Olympiad Prove it Slovenia

Let $ABC$ be an acute triangle. Denote its orthocentre by $H$ and let $A'$ , $B'$ and $C'$ be the feet of the altitudes from $A$ , $B$ and $C$ . Let $P$ be the midpoint of $AH$ , let $Q$ be the intersection of lines $B'P$ and $AB$ and denote the intersection of segments $A'C'$ and $BB'$ by $R$ . Prove that the line $QR$ is perpendicular to the side $BC$ .

Solution

The midpoint of the hypotenuse is also the circumcentre, so $P$ is the circumcentre of the triangle $AHB'$ and $|AP| = |PH| = |PB'|$ . The triangle $HPB'$ is isosceles with the apex at $P$ . Let $QB'B = \alpha$ . Then $\alpha = \angle PB'H = \angle B'HP = \angle BHA'$ . Since $\angle HA'B + \angle BC'H = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ , points $A'$ , $B$ , $C'$ and $H$ are concyclic and $\angle BC'A' = \angle BHA' = \alpha$ . In the quadrilateral $B'QC'R$ we have

$\angle RC'Q + \angle QB'R = \pi - \angle BC'A + \angle QB'B = \pi - \alpha + \alpha = \pi,$
so $B'$ , $Q$ , $C'$ and $R$ are concyclic.

Since $AC'B'H$ is a cyclic quadrilateral we have $\alpha = \angle B'HA = \angle B'C'A$ . The connection between the inscribed angles of a cyclic quadrilateral $B'QC'R$ gives us the equalities $\angle B'RQ = \angle B'C'Q = \angle B'C'A = \alpha$ . So, $\angle B'RQ = \alpha = \angle B'HA$ and $AH$ and $QR$ are parallel. Since $AH$ is perpendicular to $BC$ it is also perpendicular to $QR$ .

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