Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 7.0 National olympiad, round 2 Prove it Ukraine

Let $BM$ be a median in an acute triangle $ABC$ , whose sides $AB$ and $BC$ have different length. The extension of $BM$ intersects the circumcircle of $ABC$ at a point $N$ . Let $D$ be a point on the circumcircle such that $\angle BDH = 90^\circ$ , where $H$ is the orthocenter of $ABC$ . Let $K$ be a point chosen so that $ANCK$ is a parallelogram. Prove that the lines $AC$ , $KH$ and $BD$ are concurrent.

(Ihor Nahel)

Solution

Since $BC_1HA_1$ and $BANC$ are cyclic quadrilaterals, it follows that $\angle A_1HC_1 = 180^\circ - \angle B$ and $\angle ANC = 180^\circ - \angle B$ . This implies $\angle A_1HC_1 = \angle ANC$ . But $\angle ANC = \angle A_1HC_1$ (as vertical angles) and $\angle AKC = \angle ANC$ (as opposite angles in a parallelogram). Hence we obtain
$\angle AHC = \angle AKC$ , which means $AKHC$ is cyclic. This implies $\angle KHC_1 = \angle KAC$ . Then $\angle KAC = \angle ACN$ , because $AK \parallel NC$ and $\angle ACN = \angle ABN$ , as these angles are inscribed in the circumcircle of $ABC$ . Therefore, $\angle KHC_1 = \angle KBC_1$ , which implies that the point $K$ also lies on the circle with diameter $BH$ .

Denote by $\omega_1$ the circumcircle of $ABC$ , by $\omega_2$ the circle with diameter $BH$ , and by $\omega_3$ the circumcircle of $AKHC$ . Then $AC$ , $KH$ and $BD$ are radical axes of circles $\omega_1$ and $\omega_3$ , $\omega_2$ and $\omega_3$ , $\omega_1$ and $\omega_2$ . As it is known, the radical axes of three circles either intersect at the same point, which is their radical center, or are parallel.

Since $AB > BC$ , lines $AC$ , $KH$ and $BD$ are concurrent.

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