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Difference between revisions of "1999 JBMO Problems/Problem 4"

(Created page with "Its easy to see that <math>B'</math>, <math>C'</math>, <math>D</math> are collinear (since angle <math>B'DB</math> = <math>C'DC</math> = <math>90^o</math>). Applying Sine ru...")
(No difference)

Revision as of 17:00, 2 December 2018

Its easy to see that $B'$, $C'$, $D$ are collinear (since angle $B'DB$ = $C'DC$ = $90^o$).

Applying Sine rule in triangle $ABC$, we get: $Sin(BAD) / Sin(CAD) = BD / DC$

Since $BAB'D$ and $CC'AD$ are cyclic quadrilaterals, anlge $(BAD)$ = anlge $(BB'D)$ and $angle (CAD) = angle (CC'D)$

So, $Sin(BB'D) / Sin(CC'D) = BD / DC$

So $BD / Sin(BB'D) = DC / Sin(CC'D)$ Thus, $BB' = CC'$ (the circumcirlcles $\mathcal{C}_1,\mathcal{C}_2$ are congruent).


From right traingles $BB'A$ 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 traiangle $MBC$ = area of traingle $ABC$ and hence is independent of position of $D$ on $BC$.


By $Kris17$

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