MathDB

6

Part of 2011 Sharygin Geometry Olympiad

Problems(4)

4 unit circles

Source: 2011 Sharygin Geometry Olympiad #6

5/22/2014
Two unit circles ω1\omega_1 and ω2\omega_2 intersect at points AA and BB. MM is an arbitrary point of ω1\omega_1, NN is an arbitrary point of ω2\omega_2. Two unit circles ω3\omega_3 and ω4\omega_4 pass through both points MM and NN. Let CC be the second common point of ω1\omega_1 and ω3\omega_3, and DD be the second common point of ω2\omega_2 and ω4\omega_4. Prove that ACBDACBD is a parallelogram.
geometry unsolvedgeometry
incenter of a triangle lies on the altitude of another

Source: Sharygin 2011 Final 8.6

12/15/2018
Let BB1BB_1 and CC1CC_1 be the altitudes of acute-angled triangle ABCABC, and A0A_0 is the midpoint of BCBC. Lines A0B1A_0B_1 and A0C1A_0C_1 meet the line passing through AA and parallel to BCBC in points PP and QQ. Prove that the incenter of triangle PA0QPA_0Q lies on the altitude of triangle ABCABC.
geometryincenteraltitudegeometry solvedorthocentercontact triangleincircle
3 parallel lines + collinear related to intersections of circumcircles

Source: Sharygin 2011 Final 9.6

12/20/2018
In triangle ABCABC AA0AA_0 and BB0BB_0 are medians, AA1AA_1 and BB1BB_1 are altitudes. The circumcircles of triangles CA0B0CA_0B_0 and CA1B1CA_1B_1 meet again in point McM_c. Points Ma,MbM_a, M_b are defined similarly. Prove that points Ma,Mb,McM_a, M_b, M_c are collinear and lines AMa,BMb,CMcAM_a, BM_b, CM_c are parallel.
geometrycircumcirclealtitudecollinearcollinearityParallel Lines
l_1^2>\sqrt3 S>l_2^2 , triangle area inequality related to angle bisectors

Source: Sharygin 2011 Final 10.6

3/31/2019
Prove that for any nonisosceles triangle l12>3S>l22l_1^2>\sqrt3 S>l_2^2, where l1,l2l_1, l_2 are the greatest and the smallest bisectors of the triangle and SS is its area.
geometrygeometric inequalityangle bisectorarea of a triangle