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

 
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== Solution ==
 
== Solution ==
  
Its easy to see that  <math>B'</math>, <math>C'</math>, <math>D</math> are collinear (since <math>\angle B'DC = \angle C'DC</math> = 90^\circ<math>). Applying the sine rule in triangle </math>ABC<math>, we get </math>\frac{\sin BAD  }{ \sin CAD} = \frac{BD }{ DC}.<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> So, </math>\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}<math> and </math>\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.<math> Thus, </math>BB' = CC'<math> (the circumcircles </math>\mathcal{C}_1,\mathcal{C}_2<math> are congruent).
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Its easy to see that  <math>B'</math>, <math>C'</math>, <math>D</math> are collinear (since <math>\angle B'DC = \angle C'DC = 90^\circ</math>). Applying the sine rule in triangle <math>ABC</math>, we get <cmath>\frac{\sin BAD  }{ \sin CAD} = \frac{BD }{ DC}.</cmath> 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> So, <cmath>\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}</cmath> and <cmath>\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.</cmath> Thus, <math>BB' = CC'</math> (the circumcircles <math>\mathcal{C}_1,\mathcal{C}_2</math> are congruent).
  
From right triangles </math>BB'A<math> and </math>CC'A<math>, we have
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From right triangles <math>BB'A</math> and <math>CC'A</math>, we have
<cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}</cmath> 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>. So area of </math>[MBC] = [ABC]<math> and hence is independent of position of </math>D<math> on </math>BC$.
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<cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}.</cmath> 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>. So area of <math>[MBC] = [ABC]</math> and hence is independent of position of <math>D</math> on <math>BC</math>.

Latest revision as of 15:04, 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 $\angle B'DC = \angle C'DC = 90^\circ$). 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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