# 2007 BMO Problems/Problem 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 $ and $. Then by the isosceles triangles manifest in the figure we have $ and $, so $ and $. Furthermore $ and $.

If $AE = DE$, then $BE \neq CE$. But also $AB = CD$, so by SSA "Incongruence" (aka. the Law of Sines: $\frac{AE}{\sin ) we have $. This translates into $180^\circ = 3x + 3y$, or $120^\circ = 2x + 2y$, which incidentally equals $, as desired.

If $, then also $x + y + by the Exterior Angle Theorem, so $3x + 3y = 180^\circ$ and hence $ and $ are supplementary. A simple Law of Sines calculation then gives $AE = DE$, as desired. This completes both directions of the proof.