MathDB

1

Part of 2012 Sharygin Geometry Olympiad

Problems(4)

Double angles and lengths

Source: Sharygin Geometry Olympiad 2012 - Problem 1

4/28/2012
In triangle ABCABC point MM is the midpoint of side ABAB, and point DD is the foot of altitude CDCD. Prove that ∠A=2∠B\angle A = 2\angle B if and only if AC=2MDAC = 2 MD.
geometrygeometric transformationhomothetycircumcircleangle bisectorgeometry unsolved
Russian angle chasing

Source: 2012 Sharygin Geometry Olympiad Final Round 8.1

8/3/2018
Let MM be the midpoint of the base ACAC of an acute-angled isosceles triangle ABCABC. Let NN be the reflection of MM in BCBC. The line parallel to ACAC and passing through NN meets ABAB at point KK. Determine the value of ∠AKC\angle AKC.
(A.Blinkov)
Angle Chasinggeometry
projections of projections on altitudes result to a segment parallel to 3rd side

Source: 2012 Sharygin Geometry Olympiad Final Round 9.1

8/3/2018
The altitudes AA1AA_1 and BB1BB_1 of an acute-angled triangle ABC meet at point OO. Let A1A2A_1A_2 and B1B2B_1B_2 be the altitudes of triangles OBA1OBA_1 and OAB1OAB_1 respectively. Prove that A2B2A_2B_2 is parallel to ABAB.
(L.Steingarts)
geometryaltitudes
a surface of an n x n x n grid cube, covered by 1 x 2 rectangles

Source: 2012 Sharygin Geometry Olympiad Final Round 10.1

8/3/2018
Determine all integer nn such that a surface of an n×n×nn \times n \times n grid cube can be pasted in one layer by paper 1×21 \times 2 rectangles so that each rectangle has exactly five neighbors (by a line segment).
(A.Shapovalov)
geometry3D geometrygridsquare gridcuberectangle