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Ukrainian National Mathematical Olympiad - Third Round

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Ukraine

Let BMBM be a median in an acute triangle ABCABC, whose sides ABAB and BCBC have different length. The extension of BMBM intersects the circumcircle of ABCABC at a point NN. Let DD be a point on the circumcircle such that ∠BDH=90∘\angle BDH = 90^\circ, where HH is the orthocenter of ABCABC. Let KK be a point chosen so that ANCKANCK is a parallelogram. Prove that the lines ACAC, KHKH and BDBD are concurrent.

(Ihor Nahel)

Solution

Since BC1HA1BC_1HA_1 and BANCBANC are cyclic quadrilaterals, it follows that ∠A1HC1=180∘−∠B\angle A_1HC_1 = 180^\circ - \angle B and ∠ANC=180∘−∠B\angle ANC = 180^\circ - \angle B. This implies ∠A1HC1=∠ANC\angle A_1HC_1 = \angle ANC. But ∠ANC=∠A1HC1\angle ANC = \angle A_1HC_1 (as vertical angles) and ∠AKC=∠ANC\angle AKC = \angle ANC (as opposite angles in a parallelogram). Hence we obtain
∠AHC=∠AKC\angle AHC = \angle AKC, which means AKHCAKHC is cyclic. This implies ∠KHC1=∠KAC\angle KHC_1 = \angle KAC. Then ∠KAC=∠ACN\angle KAC = \angle ACN, because AK∥NCAK \parallel NC and ∠ACN=∠ABN\angle ACN = \angle ABN, as these angles are inscribed in the circumcircle of ABCABC. Therefore, ∠KHC1=∠KBC1\angle KHC_1 = \angle KBC_1, which implies that the point KK also lies on the circle with diameter BHBH.

Denote by ω1\omega_1 the circumcircle of ABCABC, by ω2\omega_2 the circle with diameter BHBH, and by ω3\omega_3 the circumcircle of AKHCAKHC. Then ACAC, KHKH and BDBD are radical axes of circles ω1\omega_1 and ω3\omega_3, ω2\omega_2 and ω3\omega_3, ω1\omega_1 and ω2\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>BCAB > BC, lines ACAC, KHKH and BDBD are concurrent.

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