2008 OIM Problems/Problem 5

Problem

Let $ABC$ be a triangle, and $X, Y, Z$ be points on the sides $BC$ , $AC$ , $AB$ respectively. Let $A'$ , $B'$ , and $C'$ be the circumcenters corresponding to the triangles $AZY, BXZ,$ and $CYX$ . Show that

$(A'B'C') \ge \frac{(ABC)}{4}$

and that the equality is fulfilled if and only if the lines $AA',BB',CC'$ have a point in common.

Observation: for any triangle $RST$ , we denote its area by $(RST)$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Page

Toolbox

Search

2008 OIM Problems/Problem 5

Problem

Solution

See also