A Triangle Problem from Caucasus


A Triangle Problem from Caucasus, problem

Solution 1

Let $b+c-a=2x,$ $c+a-b=2y$ and $a+b-c=2z.$ The problem constraint is equivalent to

$\displaystyle \frac{yz}{(x+y)(x+z)}\gt 0$ and $\displaystyle \frac{x(x+y+z)}{(x+y)(x+z)}\gt 0.$

We need to prove that

$\displaystyle \frac{zx}{(y+z)(y+x)}\gt 0$ and $\displaystyle \frac{y(x+y+z)}{(y+z)(y+x)}\gt 0$


$\displaystyle \frac{xy}{(z+x)(z+y)}\gt 0$ and $\displaystyle \frac{z(x+y+z)}{(z+x)(z+y)}\gt 0.$

Now, by weakening the given inequality,

$\displaystyle\begin{align}\left(\frac{yz}{(x+y)(x+z)}\right)\left(\displaystyle \frac{x(x+y+z)}{(x+y)(x+z)}\right)\gt 0\;&\Rightarrow\\ \frac{xyz(x+y+z)}{(x+y)^2(x+z)^2}\gt 0\;&\Rightarrow\\ xyz(x+y+z)\gt 0. \end{align}$

Let $\displaystyle \frac{zx}{(y+z)(y+x)}=A$ and $\displaystyle \frac{y(x+y+z)}{(y+z)(y+x)}=B.$ Then $\displaystyle AB=\frac{xyz(x+y+z)}{(y+z)^2(y+x)^2}\gt 0,$ whereas $\displaystyle A+B=\frac{(y+z)(y+x)}{(y+z)(y+x)}=1,$ which implies that $A,B\gt 0.$ Similarly, $\displaystyle \frac{xy}{(z+x)(z+y)}\gt 0$ and $\displaystyle \frac{z(x+y+z)}{(z+x)(z+y)}\gt 0.$

Solution 2

WLOG, we may assume that $a,b,c$ are positive. We have $\displaystyle \left|\frac{b^2+c^2-a^2}{2bc}\right|\lt 1,$ so that there is an angle $\alpha\in (0^{\circ},180^{\circ})$ such that $\displaystyle \cos\alpha=\frac{b^2+c^2-a^2}{2bc},$ implying


If we form a triangle with sides $b, c$ and angle $\alpha$ in-between then, according to the Law of Cosines, the side opposite $\alpha$ will be exactly $a.$ This means that $a,b,c$ form a triangle and the Law of Cosines applied twice implies the required two inequalities.

Solution 3

A Triangle Problem from Caucasus, solution by NNT


Leo Giugiuc has kindly messaged me this problem from the 2017 Caucasus Mathematics Olympiad (#6), and a fancy solution of his (Solution 1).

Computer assisted Solution 3 is by N. N. Taleb.


|Contact| |Up| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny