China Tst 2010
2
Problem: Let be an acute triangle with , let be the center of the incircle. Let be the midpoint of and respectively. are on and respectively such that and . A line through parallel to intersects in . Let be the projection of on line . Prove that is on the circumcircle of . Solution: Denote are midpoint of arc and arc respectively, the incircle touches at respectively, are the points where the -excircle and the -excircle touches . intersects at . Since are midpoints of and , and we also have passes through midpoint of , therefore and Next, we need to prove an important lemma: is the center of spiral similarity of triangles . Since Because intersects at , we have triangles are similar in the same direction We also have From , we have triangles are similar in the same direction, hence, there exists a transformation center maps are similar in the same direction. But, as we already have is the spiral similarity of triangles , therefore . This means Denote as the point where the line intersects the
description
mathematical Olympiad
Transcript of China Tst 2010
Problem:Let be an acute triangle with, let be the center of the incircle.Let be the midpoint of and respectively.are on and respectively such that and. A line through parallel to intersects in. Let be the projection of on line. Prove that is on the circumcircle of.Solution:Denote are midpoint of arc and arc respectively, the incircle touches at respectively, are the points where thee!circle and thee!circle touches. intersects at.Since are midpoints of and, and we also havepasses through midpoint of, therefore and "e!t, we need to prove an important lemma:is the center of spiral similarity oftriangles.Since #ecause intersects at, we have triangles are similar in the same direction $e also have %rom, we have triangles are similar in the same direction, hence, there e!ists a transformation center maps are similar in the same direction.#ut, as we already have is the spiral similarity of triangles , therefore.&his means Denote as the point where the line intersects the line draw from perpendicular to and. cuts at. &he problem e'uivalent to proving Since is the center of the transformation maps, we have Also, we have are concyclic points.%rom, we have:.&he proof ends here.
