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Difference between revisions of "2007 BMO Problems/Problem 1"

m (wording)
m
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(''Albania'')
 
(''Albania'')
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Let <math> \displaystyle ABCD </math> be a convex quadrilateral with <math> \displaystyle AB=BC=CD </math>, <math> \displaystyle AC \neq \displaystyle BD </math>, and let <math> \displaystyle E </math> be the intersection point of its diagonals. Prove that <math> \displaystyle AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>.
+
Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neqBD </math>, and let <math>E </math> be the intersection point of its diagonals. Prove that <math>AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>.
  
 
== Solution ==
 
== Solution ==
  
−
Since <math> \displaystyle AB = BC </math>, <math> \angle BAC = \angle ACB </math>, and similarly, <math> \angle CBD = \angle BDC </math>.  Since <math> \angle CEB = \angle AED </math>, by consdering triangles <math> \displaystyle CEB, AED </math> we have <math> \angle BAC + \angle BDC = \angle ECB + \angle CBE = \angle EAD + \angle EDA </math>.  It follows that <math> 2 ( \angle BAC + \angle BDC ) = \angle BAD + \angle ADC </math>.
+
Since <math>AB = BC </math>, <math> \angle BAC = \angle ACB </math>, and similarly, <math> \angle CBD = \angle BDC </math>.  Since <math> \angle CEB = \angle AED </math>, by considering triangles <math>CEB, AED </math> we have <math> \angle BAC + \angle BDC = \angle ECB + \angle CBE = \angle EAD + \angle EDA </math>.  It follows that <math> 2 ( \angle BAC + \angle BDC ) = \angle BAD + \angle ADC </math>.
  
 
Now, by the [[Law of Sines]],
 
Now, by the [[Law of Sines]],
Line 14: Line 14:
 
</math>.
 
</math>.
 
</center>
 
</center>
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It follows that <math> \displaystyle AE = DE </math> if and only if
+
It follows that <math>AE = DE </math> if and only if
 
<center>
 
<center>
−
<math> \displaystyle \sin (\pi - 2\alpha - \beta) = \sin (\pi - 2\beta - \alpha) </math>.
+
<math>\sin (\pi - 2\alpha - \beta) = \sin (\pi - 2\beta - \alpha) </math>.
 
</center>
 
</center>
−
Since <math> \displaystyle 0 < \alpha, \beta < \pi/2 </math>,
+
Since <math>0 < \alpha, \beta < \pi/2 </math>,
−
<center><math> \displaystyle 0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi </math>,
+
<center><math>0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi </math>,
 
</center>
 
</center>
 
and
 
and
−
<center><math> \displaystyle -\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi </math>.
+
<center><math>-\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi </math>.
 
</center>
 
</center>
  
−
From these inequalities, we see that <math> \displaystyle \sin (\pi - 2\alpha - \beta ) = \sin (\pi - 2\beta - \alpha) </math> if and only if <math> \displaystyle (\pi - 2\alpha - \beta) = (\pi - 2\beta - \alpha) </math> (i.e., <math> \displaystyle \alpha = \beta </math>) or <math> \displaystyle (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) = \pi </math> (i.e., <math> \displaystyle 3(\alpha + \beta) = \pi </math>).  But if <math> \displaystyle \alpha = \beta </math>, then triangles <math> \displaystyle ABC, BCD </math> are congruent and <math> \displaystyle AC = BD </math>, a contradiction.  Thus we conclude that <math> \displaystyle AE = DE </math> if and only if <math> \displaystyle \alpha + \beta = \pi/3 </math>, Q.E.D.
+
From these inequalities, we see that <math>\sin (\pi - 2\alpha - \beta ) = \sin (\pi - 2\beta - \alpha) </math> if and only if <math>(\pi - 2\alpha - \beta) = (\pi - 2\beta - \alpha) </math> (i.e., <math>\alpha = \beta </math>) or <math>(\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) = \pi </math> (i.e., <math>3(\alpha + \beta) = \pi </math>).  But if <math>\alpha = \beta </math>, then triangles <math>ABC, BCD </math> are congruent and <math>AC = BD </math>, a contradiction.  Thus we conclude that <math>AE = DE </math> if and only if <math>\alpha + \beta = \pi/3 </math>, Q.E.D.
  
  

Revision as of 00:41, 24 October 2012

Problem

(Albania) Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$, $AC \neqBD$ (Error compiling LaTeX. Unknown error_msg), and let $E$ be the intersection point of its diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120^{\circ}$.

Solution

Since $AB = BC$, $\angle BAC = \angle ACB$, and similarly, $\angle CBD = \angle BDC$. Since $\angle CEB = \angle AED$, by considering triangles $CEB, AED$ we have $\angle BAC + \angle BDC = \angle ECB + \angle CBE = \angle EAD + \angle EDA$. It follows that $2 ( \angle BAC + \angle BDC ) = \angle BAD + \angle ADC$.

Now, by the Law of Sines,

$\frac{AE}{DE} = \frac{AE}{EB} \cdot \frac{EB}{EC} \cdot \frac{EC}{DE} = \frac{\sin (\pi - 2\alpha - \beta)}{\sin \alpha} \cdot \frac{\sin \alpha}{\sin \beta} \cdot \frac{\sin \beta}{\sin (\pi - 2\beta - \alpha)} = \frac{\sin (\pi - 2\alpha - \beta)}{\sin (\pi - 2\beta - \alpha)}$.

It follows that $AE = DE$ if and only if

$\sin (\pi - 2\alpha - \beta) = \sin (\pi - 2\beta - \alpha)$.

Since $0 < \alpha, \beta < \pi/2$,

$0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi$,

and

$-\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi$.

From these inequalities, we see that $\sin (\pi - 2\alpha - \beta ) = \sin (\pi - 2\beta - \alpha)$ if and only if $(\pi - 2\alpha - \beta) = (\pi - 2\beta - \alpha)$ (i.e., $\alpha = \beta$) or $(\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) = \pi$ (i.e., $3(\alpha + \beta) = \pi$). But if $\alpha = \beta$, then triangles $ABC, BCD$ are congruent and $AC = BD$, a contradiction. Thus we conclude that $AE = DE$ if and only if $\alpha + \beta = \pi/3$, Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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