MathDB

6

Part of 2014 Sharygin Geometry Olympiad

Problems(4)

Locus of Intersection of Angle and Perpendicular Bisectors

Source: Sharygin Geometry Olympiad 2014 - Problem 6

11/15/2014
Given a circle with center OO and a point PP not lying on it, let XX be an arbitrary point on this circle and YY be a common point of the bisector of angle POXPOX and the perpendicular bisector to segment PXPX. Find the locus of points YY.
geometryperpendicular bisectorgeometry unsolved
4 tangents to 2 ext. tangent circles (2 each) are also tangent to a third circle

Source: 2014 Sharygin Geometry Olympiad Final Round 8.6

8/3/2018
Two circles k1k_1 and k2k_2 with centers O1O_1 and O2O_2 are tangent to each other externally at point OO. Points XX and YY on k1k_1 and k2k_2 respectively are such that rays O1XO_1X and O2YO_2Y are parallel and codirectional. Prove that two tangents from XX to k2k_2 and two tangents from YY to k1k_1 touch the same circle passing through OO.
(V. Yasinsky)
geometrytangent circlesTangents
incenter and midpoints of arcs of <B,<C collinear iff AC + BC = 3AB

Source: 2014 Sharygin Geometry Olympiad Final Round 9.6

8/3/2018
Let II be the incenter of triangle ABCABC, and M,NM, N be the midpoints of arcs ABCABC and BACBAC of its circumcircle. Prove that points M,I,NM, I, N are collinear if and only ifAC+BC=3AB AC + BC = 3AB.
(A. Polyansky)
geometryincenter
incircles, semicircles, and lines concurrent

Source: 2014 Sharygin Geometry Olympiad Final Round 10.6

8/3/2018
The incircle of a non-isosceles triangle ABCABC touches ABAB at point C′C'. The circle with diameter BC′BC' meets the incircle and the bisector of angle BB again at points A1A_1 and A2A_2 respectively. The circle with diameter AC′AC' meets the incircle and the bisector of angle AA again at points B1B_1 and B2B_2 respectively. Prove that lines AB,A1B1,A2B2AB, A_1B_1, A_2B_2 concur.
(E. H. Garsia)
geometryconcurrencyconcurrent