1999 JBMO Problems/Problem 4

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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 $BB'A$ 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 $BC$.

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