2
Part of 2012 Tuymaada Olympiad
Problems(3)
Easy polynomial problem
Source: Tuymaada 2012, Problem 2, Day 1, Seniors
7/20/2012
Let be a real quadratic trinomial, so that for all the inequality holds. Find the sum of the roots of .Proposed by A. Golovanov, M. Ivanov, K. Kokhas
algebrapolynomialquadraticsinequalitiessum of rootsalgebra unsolved
Both cyclic and circumscribed quadrilateral
Source: Tuymaada 2012, Problem 6, Day 2, Seniors
7/21/2012
Quadrilateral is both cyclic and circumscribed. Its incircle touches its sides and at points and , respectively. The perpendiculars to and drawn at and , respectively, meet at point ; those drawn at and meet at point , and finally, those drawn at and meet at point . Prove that points , and are collinear.Proposed by A. Golovanov
geometryratiotrapezoidtrigonometrytrig identitiesLaw of Sinesgeometry proposed
Angle bisector in a rectangle
Source: Tuymaada 2012, Problem 2, Day 1, Juniors
7/21/2012
A rectangle is given. Segment is equal to and lies on the half-line . is the midpoint of . Prove that is the angle bisector of .Proposed by S. Berlov
geometryrectanglesymmetryangle bisectorgeometry proposed