MathDB

3

Part of 2008 Sharygin Geometry Olympiad

Problems(4)

inscribed circle in quadrilateral

Source: Sharygin contest. The final raund. 2008. Grade 8. First day. Problem 3

8/31/2008
(D.Shnol) Two opposite angles of a convex quadrilateral with perpendicular diagonals are equal. Prove that a circle can be inscribed in this quadrilateral.
geometry unsolvedgeometry
Inequality with angles, semiperimeter and inradius

Source: Sharygin contest. The final raund. 2008. Grade 9. First day. Problem 3

8/31/2008
(R.Pirkuliev) Prove the inequality \frac1{\sqrt {2\sin A}} \plus{} \frac1{\sqrt {2\sin B}} \plus{} \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {p}{r}}, where p p and r r are the semiperimeter and the inradius of triangle ABC ABC.
inequalitiesgeometryinradiustrigonometrygeometry unsolved
Construction with two circles

Source: Sharygin contest. The final raund. 2008. Grade 10. First day. Problem 3

8/31/2008
(V.Yasinsky, Ukraine) Suppose X X and Y Y are the common points of two circles ω1 \omega_1 and ω2 \omega_2. The third circle ω \omega is internally tangent to ω1 \omega_1 and ω2 \omega_2 in P P and Q Q respectively. Segment XY XY intersects ω \omega in points M M and N N. Rays PM PM and PN PN intersect ω1 \omega_1 in points A A and D D; rays QM QM and QN QN intersect ω2 \omega_2 in points B B and C C respectively. Prove that AB \equal{} CD.
geometrytrapezoidgeometric transformationhomothetypower of a pointradical axisgeometry unsolved
A triangle can be dissected into three equal triangles

Source: Sharygin contest 2008. The correspondence round. Problem 3

9/3/2008
(A.Zaslavsky, 8) A triangle can be dissected into three equal triangles. Prove that some its angle is equal to 60∘ 60^{\circ}.
geometrygeometry proposed