MathDB

2017 Oral Moscow Geometry Olympiad

Part of Oral Moscow Geometry Olympiad

Subcontests

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intersecting squares, black triangle area equals to sum of gray areas

Two squares are arranged as shown. Prove that the area of the black triangle equal to the sum of the gray areas. https://2.bp.blogspot.com/-byhWqNr1ras/XTq-NWusg2I/AAAAAAAAKZA/1sxEZ751v_Evx1ij7K_CGiuZYqCjhm-mQCK4BGAYYCw/s400/Oral%2BSharygin%2B2017%2B8.9%2Bp5.png

orthocenter, incenter of different triangles and midpoint are collinear

The inscribed circle of the non-isosceles triangle ABCABC touches sides AB,BCAB, BC and ACAC at points C1,A1C_1, A_1 and B1B_1, respectively. The circumscribed circle of the triangle A1BC1A_1BC_1 intersects the lines B1A1B_1A_1 and B1C1B_1C_1 at the points A0A_0 and C0C_0, respectively. Prove that the orthocenter of triangle A0BC0A_0BC_0, the center of the inscribed circle of triangle ABCABC and the midpoint of the ACAC lie on one straight line.
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another locus, center of the circumcircle moves on a straight line

Around triangle ABCABC with acute angle C is circumscribed a circle. On the arc ABAB, which does not contain point CC, point DD is chosen. Point D′D' is symmetric on point DD with respect to line ABAB. Straight lines AD′AD' and BD′BD' intersect segments BCBC and ACAC at points EE and FF. Let point CC move along its arc ABAB. Prove that the center of the circumscribed circle of a triangle CEFCEF moves on a straight line.

Geometry

Given acute angled traingle ABCABC and altitudes AA1AA_1, BB1BB_1, CC1CC_1. Let MM midpoint of BCBC. PP point of intersection of circles (AB1C1)(AB_1C_1) and (ABC)(ABC) . TT is point of intersection of tangents to (ABC)(ABC) at BB and CC. SS point of intersection of ATAT and (ABC)(ABC). Prove that P,A1,SP,A_1,S and midpoint of MTMT collinear.