# 1999 JBMO Problems/Problem 4

## Problem 4

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$).

## Solution

Its easy to see that $B'$, $C'$, $D$ are collinear (since $\angle B'DC = \angle C'DC$ = 90^\circ $). Applying the sine rule in triangle$ABC $, we get$\frac{\sin BAD }{ \sin CAD} = \frac{BD }{ DC}. $Since$BAB'D $and$CC'AD $are cyclic quadrilaterals,$\angle BAD \angle BB'D $and$\angle CAD = \angle CC'D. $So,$\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC} $and$\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}. $Thus,$BB' = CC' $(the circumcircles$\mathcal{C}_1,\mathcal{C}_2$are congruent). From right triangles$ (Error compiling LaTeX. ! Missing $inserted.)BB'A $and$CC'A $, we have AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2} So$AC' = AB'. $Since$M $is the midpoint of$B'C' $,$AM $is perpendicular to$B'C' $and hence$AM $is parallel to$BC $. So area of$[MBC] = [ABC] $and hence is independent of position of$D $on$BC$.