Two isosceles triangles with sidelengths $x,x,a$ and $x,x,b$ ($a \neq b$) have equal areas. Find $x$.
Problem
Source: Cono Sur Olympiad, Uruguay 1989, Problem#1
Tags:
t0rajir0u
14.05.2006 08:27
For convenience, let $c = \frac{a}{2}, d = \frac{b}{2}$.
$K = \sqrt{x^2 - c^2} c = \sqrt{x^2 - d^2} d$
$c^2 (x^2 - c^2) = d^2 (x^2 - d^2)$
But $c \neq d$. Therefore
$x^2 - c^2 = d^2$
$x = \boxed{ \frac{\sqrt{a^2 + b^2} }{2} }$.
M4RI0
14.05.2006 08:34
I guess the problems are easy because it was the first year the olympiad was organized.
maraujo68
04.01.2022 00:02
I found the answer: x = sqrt(a^2 + b^2)/2...
Jwenslawski
04.01.2022 00:19
maraujo68 wrote: I found the answer: x = sqrt(a^2 + b^2)/2... Congrats on your first post!