1999 JBMO Problems/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 ).
Its easy to see that , , are collinear (since angle = = ).
Applying Sine rule in triangle , we get:
Since and are cyclic quadrilaterals, and
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 .