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Difference between revisions of "1999 JBMO Problems/Problem 4"
(Fix LaTeX)
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Applying Sine rule in triangle <math>ABC</math>, we get:
 
Applying Sine rule in triangle <math>ABC</math>, we get:
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<math>Sin(BAD) / Sin(CAD) = BD / DC</math>
+
<math>\sin(BAD) / \sin(CAD) = BD / DC</math>
  
−
Since <math>BAB'D</math> and <math>CC'AD</math> are cyclic quadrilaterals, anlge <math>(BAD)</math> = anlge <math>(BB'D)</math>
+
Since <math>BAB'D</math> and <math>CC'AD</math> are cyclic quadrilaterals, <math>\angle BAD \angle BB'D</math>
−
and <math>angle (CAD) = angle (CC'D)</math>
+
and <math>\angle CAD = \angle CC'D.</math>
  
−
So, <math>Sin(BB'D) / Sin(CC'D) = BD / DC</math>
+
So, <math>\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}.</math>
  
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So <math>BD / Sin(BB'D) = DC / Sin(CC'D)</math>
+
So <math>\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.</math>
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Thus, <math>BB' = CC'</math> (the circumcirlcles <math>\mathcal{C}_1,\mathcal{C}_2</math> are congruent).
+
Thus, <math>BB' = CC'</math> (the circumcircles <math>\mathcal{C}_1,\mathcal{C}_2</math> are congruent).
  
  
−
From right traingles <math>BB'A</math> and <math>CC'A</math>, we have:
+
From right triangles <math>BB'A</math> and <math>CC'A</math>, we have
−
<math>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}</math>
+
<cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}</cmath>
−
So <math>AC' = AB'</math>
+
So <math>AC' = AB'.</math>
  
 
Since <math>M</math> is the midpoint of <math>B'C'</math>, <math>AM</math> is perpendicular to <math>B'C'</math> and hence <math>AM</math> is parallel to <math>BC</math>.
 
Since <math>M</math> is the midpoint of <math>B'C'</math>, <math>AM</math> is perpendicular to <math>B'C'</math> and hence <math>AM</math> is parallel to <math>BC</math>.
  
−
So area of traiangle <math>MBC</math> = area of traingle <math>ABC</math> and hence is independent of position of <math>D</math> on <math>BC</math>.
+
So area of <math>[MBC] = [ABC]</math> and hence is independent of position of <math>D</math> on <math>BC</math>.
−
 
 
−
 
 
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By <math>Kris17</math>
 

Revision as of 16:00, 17 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, $\angle BAD \angle BB'D$ and $\angle CAD = \angle CC'D.$

So, $\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}.$

So $\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.$ 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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