# 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 ). Applying the sine rule in triangle , we get Since and are cyclic quadrilaterals, and So, and 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 .