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

Latest revision as of 22:59, 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.

Resources

@@ Line 28: / Line 28: @@
 ==Solution 2==
-Let <math><BAC = x</math> and <math><BDC = y</math>.
+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>.
+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.
+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.

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