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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...")
 
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==Problem 4==
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Let <math>ABC</math> be a triangle with <math>AB=AC</math>. Also, let <math>D\in[BC]</math> be a point such that <math>BC>BD>DC>0</math>, and let <math>\mathcal{C}_1,\mathcal{C}_2</math> be the circumcircles of the triangles <math>ABD</math> and <math>ADC</math> respectively. Let <math>BB'</math> and <math>CC'</math> be diameters in the two circles, and let <math>M</math> be the midpoint of <math>B'C'</math>. Prove that the area of the triangle <math>MBC</math> is constant (i.e. it does not depend on the choice of the point <math>D</math>).
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== Solution ==
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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>).
 
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>).
  

Revision as of 23:20, 3 December 2018

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 $B'DB$ = $C'DC$ = $90^o$).

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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