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Print59th Ukrainian National Mathematical Olympiad
Ukraine geometry
Problem
Line is perpendicular to the side of acute triangle , and intersects at . intersects circumscribed circle of at and (point in the same half-plane w.r.t. as vertex ). By and are denoted projections of points and to line . Furthermore, vertices belong to the segment . Prove that the center of circumscribed circle of lies on the line, which contains midsegment of , parallel to the side . (Anton Tryhub)
Fig. 31
Solution
We denote by a projection of vertex to line , then quadrilaterals and are inscribed, with diameters and . Then, we have the following equalities of angles:
Therefore, quadrilateral is inscribed, hence, center of circumscribed circle of lies on the perpendicular bisector to the segment . And, clearly, midsegment of parallel to lies on this exact perpendicular bisector. Q.E.D.
Therefore, quadrilateral is inscribed, hence, center of circumscribed circle of lies on the perpendicular bisector to the segment . And, clearly, midsegment of parallel to lies on this exact perpendicular bisector. Q.E.D.
Techniques
Cyclic quadrilateralsAngle chasingTriangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle