2007 BMO Problems/Problem 1

Revision as of 23:50, 14 September 2014 by Suli (talk | contribs)

Problem

(Albania) Let $ABCD$ be a convex quadrilateral with $AB=BC=CD$, $AC \neq BD$, 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.

Solution 2

Let $<BAC = x$ and $<BDC = y$.


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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