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

 
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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> and <math> \displaystyle AC </math> not equal to <math> \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>.
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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>.
  
 
== Solution ==
 
== Solution ==

Revision as of 23:45, 4 May 2007

Problem

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

Solution

Since $\displaystyle AB = BC$, $\angle BAC = \angle ACB$, and similarly, $\angle CBD = \angle BDC$. Since $\angle CEB = \angle AED$, by consdering triangles $\displaystyle CEB, AED$ 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 $\displaystyle AE = DE$ if and only if

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

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

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

and

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

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