# 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 = 90^\circABC\frac{\sin BAD }{ \sin CAD} = \frac{BD }{ DC}.BAB'DCC'AD\angle BAD \angle BB'D\angle CAD = \angle CC'D.\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.BB' = CC'\mathcal{C}_1,\mathcal{C}_2$are congruent).

From right triangles$ (Error compiling LaTeX. ! Missing $ inserted.)BB'ACC'AAC' = AB'.MB'C'AMB'C'AMBC[MBC] = [ABC]DBC$.