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SHARYGIN 2010 Problem 15

    Let AA1,BB1AA_1, BB_1 and CC1CC_1 be the altitudes of an acute-angled triangle ABCABC, AA1AA_1 meets B1C1B_1C_1 in a point KK. The circumcircles of the triangles A1KC1A_1KC_1 and A1KB1A_1KB_1 intersect the lines ABAB and ACAC for the second time at points NN and LL respectively. Prove that

    a) the sum of diameters of these two circles is equal to BCBC; b) A1N/BB1+A1L/CC1=1A_1N/BB_1 + A_1L/CC_1 = 1.

    Official text Sharygin Geometry Olympiad 2010 official paper, page 2. Quoted with source.

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