# 1999 JBMO Problems/Problem 4

## Problem 4

Let be a triangle with . Also, let be a point such that , and let be the circumcircles of the triangles and respectively. Let and be diameters in the two circles, and let be the midpoint of . Prove that the area of the triangle is constant (i.e. it does not depend on the choice of the point ).

## Solution

Its easy to see that , , are collinear (since angle = = ).

Applying Sine rule in triangle , we get:

Since and are cyclic quadrilaterals, and

So,

So Thus, (the circumcircles are congruent).

From right triangles and , we have
So

Since is the midpoint of , is perpendicular to and hence is parallel to .

So area of and hence is independent of position of on .