MathDB

1

Part of 2009 Sharygin Geometry Olympiad

Problems(4)

Equal angles

Source: I.F.Sharygin contest 2009 - Correspondence round - Problem 1

5/31/2009
Points B1 B_1 and B2 B_2 lie on ray AM AM, and points C1 C_1 and C2 C_2 lie on ray AK AK. The circle with center O O is inscribed into triangles AB1C1 AB_1C_1 and AB2C2 AB_2C_2. Prove that the angles B1OB2 B_1OB_2 and C1OC2 C_1OC_2 are equal.
geometry proposedgeometry
trapezoid with bases equal to side and diagonal respectively

Source: 2009 Sharygin Geometry Olympiad Final Round problem 1 grade 8

7/26/2018
Minor base BCBC of trapezoid ABCDABCD is equal to side ABAB, and diagonal ACAC is equal to base ADAD. The line passing through B and parallel to ACAC intersects line DCDC in point MM. Prove that AMAM is the bisector of angle ∠BAC\angle BAC.
A.Blinkov, Y.Blinkov
geometrytrapezoidangle bisector
midpoint and base of altitude symmertic wrt touchpoint with incircle

Source: 2009 Sharygin Geometry Olympiad Final Round problem 1 grade 9

7/26/2018
The midpoint of triangle's side and the base of the altitude to this side are symmetric wrt the touching point of this side with the incircle. Prove that this side equals one third of triangle's perimeter.
(A.Blinkov, Y.Blinkov)
geometryperimetersymmetry
Σ \sqrt{\frac{ab(p- c)}{p}} \ge 6r, geometric inequality

Source: 2009 Sharygin Geometry Olympiad Final Round problem 1 grade 10

7/26/2018
Let a,b,ca, b, c be the lengths of some triangle's sides, p,rp, r be the semiperimeter and the inradius of triangle. Prove an inequality ab(p−c)p+ca(p−b)p+bc(p−a)p≥6r\sqrt{\frac{ab(p- c)}{p}} +\sqrt{\frac{ca(p- b)}{p}} +\sqrt{\frac{bc(p-a)}{p}} \ge 6r
(D.Shvetsov)
geometric inequalitygeometry