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BT = tangent from B to omega

Source: All-Russian Olympiad 2006 finals, problem 9.4

May 7, 2006
geometrycircumcircleRussia

Problem Statement

Given a triangle ABCABC. Let a circle ω\omega touch the circumcircle of triangle ABCABC at the point AA, intersect the side ABAB at a point KK, and intersect the side BCBC. Let CLCL be a tangent to the circle ω\omega, where the point LL lies on ω\omega and the segment KLKL intersects the side BCBC at a point TT. Show that the segment BTBT has the same length as the tangent from the point BB to the circle ω\omega.
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