MathDB

4

Part of 2012 Sharygin Geometry Olympiad

Problems(4)

Projection to bisector

Source: Sharygin Geometry Olympiad 2012 - Problem 4

4/28/2012
Given triangle ABCABC. Point MM is the midpoint of side BCBC, and point PP is the projection of BB to the perpendicular bisector of segment ACAC. Line PMPM meets ABAB in point QQ. Prove that triangle QPBQPB is isosceles.
analytic geometrygeometryparallelogramratioperpendicular bisectorcongruent trianglesgeometry unsolved
angle chasing in a triangle 120-30-30

Source: 2012 Sharygin Geometry Olympiad Final Round 8.4

8/3/2018
Let ABCABC be an isosceles triangle with ∠B=120o\angle B = 120^o . Points PP and QQ are chosen on the prolongations of segments ABAB and CBCB beyond point BB so that the rays AQAQ and CPCP intersect and are perpendicular to each other. Prove that ∠PQB=2∠PCQ\angle PQB = 2\angle PCQ.
(A.Akopyan, D.Shvetsov)
Angle Chasingisoscelesgeometry
divide a regular n-gon into equal triangles by several diagonals

Source: 2012 Sharygin Geometry Olympiad Final Round 9.4

8/3/2018
Determine all integer n>3n > 3 for which a regular nn-gon can be divided into equal triangles by several (possibly intersecting) diagonals.
(B.Frenkin)
geometryregular polygondiagonals
locus revisited

Source: 2012 Sharygin Geometry Olympiad Final Round 10.4

8/3/2018
Consider a square. Find the locus of midpoints of the hypothenuses of rightangled triangles with the vertices lying on three different sides of the square and not coinciding with its vertices.
(B.Frenkin)
Locusright trianglesquaregeometry