MathDB

1

Part of 2006 Vietnam Team Selection Test

Problems(2)

Problem 1 vietnamese tst 2006

Source: Vietnamese TST 2006

4/17/2006
Given an acute angles triangle ABCABC, and HH is its orthocentre. The external bisector of the angle ∠BHC\angle BHC meets the sides ABAB and ACAC at the points DD and EE respectively. The internal bisector of the angle ∠BAC\angle BAC meets the circumcircle of the triangle ADEADE again at the point KK. Prove that HKHK is through the midpoint of the side BCBC.
geometrycircumcircletrigonometryparallelogramratiogeometry proposed
Problem 4 vietnamese tst 2006

Source: Vietnam Team Selection Tests 2006, Day 2, Problem 1

4/18/2006
Prove that for all real numbers x,y,z∈[1,2]x,y,z \in [1,2] the following inequality always holds: (x+y+z)(1x+1y+1z)≥6(xy+z+yz+x+zx+y). (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}). When does the equality occur?
inequalitiesfunctionconvex-concave inequalitiesthree variable inequality