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3

Part of 2015 Sharygin Geometry Olympiad

Problems(3)

isosceles 20-80-80 , M in AC, AM/MC = 1/2, H projection of C on BM, <AHB?

Source: Sharygin Geometry Olympiad 2015 Final 8.3

8/1/2018
In triangle ABCABC we have AB=BC,∠B=20oAB = BC, \angle B = 20^o. Point MM on ACAC is such that AM:MC=1:2AM : MC = 1 : 2, point HH is the projection of CC to BMBM. Find angle AHBAHB.
(M. Yevdokimov)
geometryisoscelesangle
100 discs on the plane so that each two of them have a common point ...

Source: Sharygin Geometry Olympiad 2015 Final 9.3

8/1/2018
Let 100100 discs lie on the plane in such a way that each two of them have a common point. Prove that there exists a point lying inside at least 1515 of these discs.
(M. Kharitonov, A. Polyansky)
geometrycombinatorial geometrycircles
Midpoint geometry

Source: Sharygin geometry olympiad 2015, grade 10, Final Round, Problem 3

7/17/2018
Let A1A_1, B1B_1 and C1C_1 be the midpoints of sides BCBC, CACA and ABAB of triangle ABCABC, respectively. Points B2B_2 and C2C_2 are the midpoints of segments BA1BA_1 and CA1CA_1 respectively. Point B3B_3 is symmetric to C1C_1 wrt BB, and C3C_3 is symmetric to B1B_1 wrt CC. Prove that one of common points of circles BB2B3BB_2B_3 and CC2C3CC_2C_3 lies on the circumcircle of triangle ABCABC.
geometry