MathDB

7

Part of 2012 Sharygin Geometry Olympiad

Problems(4)

Bisector and side lengths

Source: Sharygin Geometry Olympiad 2012 - Problem 7

4/28/2012
In a non-isosceles triangle ABCABC the bisectors of angles AA and BB are inversely proportional to the respective sidelengths. Find angle CC.
trigonometrygeometrytrig identitiesLaw of Sinesgeometry unsolved
3 right angles, 2 of them by the altiudes

Source: 2012 Sharygin Geometry Olympiad Final Round 8.7

8/3/2018
The altitudes AA1AA_1 and CC1CC_1 of an acute-angled triangle ABCABC meet at point HH. Point QQ is the reflection of the midpoint of ACAC in line AA1AA_1, point PP is the midpoint of segment A1C1A_1C_1. Prove that ∠QPH=90o\angle QPH = 90^o.
(D.Shvetsov)
geometryaltitudes
dividing a convex 5gon by all its diagonals into 10 triangles and 1 smaller 5gon

Source: 2012 Sharygin Geometry Olympiad Final Round 9.7

8/3/2018
A convex pentagon PP is divided by all its diagonals into ten triangles and one smaller pentagon P′P'. Let NN be the sum of areas of five triangles adjacent to the sides of PP decreased by the area of P′P'. The same operations are performed with the pentagon P′P', let N′N' be the similar difference calculated for this pentagon. Prove that N>N′N > N'.
(A.Belov)
geometrypentagondiagonalsgeometric inequalityconvex
two intersecting circles, prove that a line bisects a segment

Source: 2012 Sharygin Geometry Olympiad Final Round 10.7

8/3/2018
Consider a triangle ABCABC. The tangent line to its circumcircle at point CC meets line ABAB at point DD. The tangent lines to the circumcircle of triangle ACDACD at points AA and CC meet at point KK. Prove that line DKDK bisects segment BCBC.
(F.Ivlev)
geometrycirclesbisection