MathDB

8

Part of 2015 Sharygin Geometry Olympiad

Problems(3)

angle chasing practice for Russian 8-graders

Source: Sharygin Geometry Olympiad 2015 Final 8.8

8/1/2018
Points C1,B1C_1, B_1 on sides AB,ACAB, AC respectively of triangle ABCABC are such that BB1⊥CC1BB_1 \perp CC_1. Point XX lying inside the triangle is such that ∠XBC=∠B1BA,∠XCB=∠C1CA\angle XBC = \angle B_1BA, \angle XCB = \angle C_1CA. Prove that ∠B1XC1=90o−∠A\angle B_1XC_1 =90^o- \angle A.
(A. Antropov, A. Yakubov)
geometryangles
Circle joining centers tangent to original circumcircle

Source: Sharygin 2015 Finals Grade 9 Problem 8

7/22/2018
A perpendicular bisector of side BCBC of triangle ABCABC meets lines ABAB and ACAC at points ABA_B and ACA_C respectively. Let OaO_a be the circumcenter of triangle AABACAA_BA_C. Points ObO_b and OcO_c are defined similarly. Prove that the circumcircle of triangle OaObOcO_aO_bO_c touches the circumcircle of the original triangle.
geometrycircumcircle
Dividing rectangle geometry

Source: Sharygin geometry olympiad 2015, grade 10, Final Round, Problem 8

7/17/2018
Does there exist a rectangle which can be divided into a regular hexagon with sidelength 11 and several congruent right-angled triangles with legs 11 and 3\sqrt{3}?
geometryrectangle