MathDB

7

Part of 2011 Sharygin Geometry Olympiad

Problems(4)

P,Q on sides AB , AC of ABC with PB = QC, prove PQ < BC

Source: 2011 Sharygin Geometry Olympiad Correspondence Round P7

4/2/2019
Points PP and QQ on sides ABAB and ACAC of triangle ABCABC are such that PB=QCPB = QC. Prove that PQ<BCPQ < BC.
geometrygeometric inequalityequal segments
locus of points symmetric to M wrt PQ so that &lt;PMQ is right, Q in x'x, P in y'y

Source: Sharygin 2011 Final 8.7

12/15/2018
Let a point MM not lying on coordinates axes be given. Points QQ and PP move along YY - and XX-axis respectively so that angle PMQP M Q is always right. Find the locus of points symmetric to MM wrt PQP Q.
geometryLocusLocus problemsright angleaxes
equal segments starting with two circles inscribed into same angle

Source: Sharygin 2011 Final 9.7

12/20/2018
Circles ω\omega and Ω\Omega are inscribed into the same angle. Line ℓ\ell meets the sides of angles, ω\omega and Ω\Omega in points AA and F,BF, B and C,DC, D and EE respectively (the order of points on the line is A,B,C,D,E,FA,B,C,D,E, F). It is known thatBC=DE BC = DE. Prove that AB=EFAB = EF.
geometryequal segmentscirclesanglecommon tangents
concurrent circumcircles related to 3 cyclic quadrilaterals, circumcenter

Source: Sharygin 2011 Final 10.7

3/31/2019
Point OO is the circumcenter of acute-angled triangle ABCABC, points A1,B1,C1A_1,B_1, C_1 are the bases of its altitudes. Points A′,B′,C′A', B', C' lying on lines OA1,OB1,OC1OA_1, OB_1, OC_1 respectively are such that quadrilaterals AOBC′,BOCA′,COAB′AOBC', BOCA', COAB' are cyclic. Prove that the circumcircles of triangles AA1A′,BB1B′,CC1C′AA_1A', BB_1B', CC_1C' have a common point.
geometrycyclic quadrilateralCircumcenterconcurrentcirclesaltitudescircumcircle