MathDB

4

Part of 2006 All-Russian Olympiad

Problems(3)

BT = tangent from B to omega

Source: All-Russian Olympiad 2006 finals, problem 9.4

5/7/2006
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.
geometrycircumcircleRussia
Isosceles triangle and tangent circles

Source: All-Russian Olympiad 2006 finals, problem 10.4

5/7/2006
Consider an isosceles triangle ABCABC with AB=ACAB=AC, and a circle ω\omega which is tangent to the sides ABAB and ACAC of this triangle and intersects the side BCBC at the points KK and LL. The segment AKAK intersects the circle ω\omega at a point MM (apart from KK). Let PP and QQ be the reflections of the point KK in the points BB and CC, respectively. Show that the circumcircle of triangle PMQPMQ is tangent to the circle ω\omega.
geometrygeometric transformationreflectioncircumcirclehomothetysymmetryratio
A well-known circle hides in the dark

Source: All-Russian Olympiad 2006 finals, problem 11.4 (problem 4 for grade 11)

5/6/2006
Given a triangle ABC ABC. The angle bisectors of the angles ABC ABC and BCA BCA intersect the sides CA CA and AB AB at the points B1 B_1 and C1 C_1, and intersect each other at the point I I. The line B1C1 B_1C_1 intersects the circumcircle of triangle ABC ABC at the points M M and N N. Prove that the circumradius of triangle MIN MIN is twice as long as the circumradius of triangle ABC ABC.
geometrycircumcirclegeometric transformationhomothetyratiopower of a pointradical axis