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

(Solution 2)
 
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==Solution 2==
 
==Solution 2==
Let <math><BAC = x</math> and <math><BDC = y</math>.
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Let <math><BAC = x</math> and <math><BDC = y</math>. Then by the isosceles triangles manifest in the figure we have <math><DBC = y</math> and <math><ACB = x</math>, so <math><BEA = x+y</math> and <math><EAD = <EDA = \frac{x+y}{2}</math>. Furthermore <math><AEB = 180^\circ - 2x - y</math> and <math><DCE = 180^\circ - x - 2y</math>.
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If <math>AE = DE</math>, then <math>BE \neq CE</math>. But also <math>AB = CD</math>, so by SSA "Incongruence" (aka. the Law of Sines: <math>\frac{AE}{\sin <ABE} = \frac{AB}{\sin <BEA} = \frac{CD}{\sin <CED} = \frac{DE}{\sin <ECD}</math>) we have <math><ABE + <DCE = 180^\circ</math>. This translates into <math>180^\circ = 3x + 3y</math>, or <math>120^\circ = 2x + 2y</math>, which incidentally equals <math><BAD + <ADC</math>, as desired.
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If <math><BAD + <ADC = 120^\circ</math>, then also <math>x + y + <EAD + <EDA = x + y + (x + y) = 120^\circ</math> by the Exterior Angle Theorem, so <math>3x + 3y = 180^\circ</math> and hence <math><ABE</math> and <math><DCE</math> are supplementary. A simple Law of Sines calculation then gives <math>AE = DE</math>, as desired. This completes both directions of the proof.
  
  

Latest revision as of 23:59, 14 September 2014

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 $<BDC = y$. Then by the isosceles triangles manifest in the figure we have $<DBC = y$ and $<ACB = x$, so $<BEA = x+y$ and $<EAD = <EDA = \frac{x+y}{2}$. Furthermore $<AEB = 180^\circ - 2x - y$ and $<DCE = 180^\circ - x - 2y$.

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

If $<BAD + <ADC = 120^\circ$, then also $x + y + <EAD + <EDA = x + y + (x + y) = 120^\circ$ by the Exterior Angle Theorem, so $3x + 3y = 180^\circ$ and hence $<ABE$ and $<DCE$ are supplementary. A simple Law of Sines calculation then gives $AE = DE$, as desired. This completes both directions of the proof.


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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