MathDB

2

Part of 2002 Tuymaada Olympiad

Problems(3)

Four variables plus a restraint

Source: Tuymaada 2002

4/26/2006
Let a,b,c,da,b,c,d be positive real numbers such that abcd=1abcd=1. Prove that 1+ab1+a+1+bc1+b+1+cd1+c+1+da1+d≥4. \frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+cd}{1+c} + \frac{1+da}{1+d} \geq 4 .
Proposed by A. Khrabrov
inequalitiesinequalities unsolved
Double functional inequation : f(3x-2) <= f(x) <= f(2x-1).

Source: Tuymaada 2002, day 2, problem 2. - Author : A. Golovanov.

5/3/2007
Find all the functions f(x),f(x), continuous on the whole real axis, such that for every real xx f(3x−2)≤f(x)≤f(2x−1).f(3x-2)\leq f(x)\leq f(2x-1).
Proposed by A. Golovanov
functioninductionalgebra proposedalgebra
if the inscribed one is equilateral then the other is also equilateral

Source: Tuymaada Junior 2002 p2

5/11/2019
Points on the sides BC BC , CA CA and AB AB of the triangle ABC ABC are respectively A1 A_1 , B1 B_1 and C1 C_1 such that AC1:C1B=BA1:A1C=CB1:B1A=2:1 AC_1: C_1B = BA_1: A_1C = CB_1: B_1A = 2: 1 . Prove that if triangle A1B1C1 A_1B_1C_1 is equilateral, then triangle ABC ABC is also equilateral.
geometryratioEquilateralEquilateral Triangle