Difference between revisions of "2007 BMO Problems/Problem 1"
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− | Let <math> | + | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neq BD </math>, and let <math>E </math> be the intersection point of its diagonals. Prove that <math>AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>. |
== Solution == | == Solution == | ||
− | Since <math> | + | Since <math>AB = BC </math>, <math> \angle BAC = \angle ACB </math>, and similarly, <math> \angle CBD = \angle BDC </math>. Since <math> \angle CEB = \angle AED </math>, by considering triangles <math>CEB, AED </math> we have <math> \angle BAC + \angle BDC = \angle ECB + \angle CBE = \angle EAD + \angle EDA </math>. It follows that <math> 2 ( \angle BAC + \angle BDC ) = \angle BAD + \angle ADC </math>. |
Now, by the [[Law of Sines]], | Now, by the [[Law of Sines]], | ||
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</math>. | </math>. | ||
</center> | </center> | ||
− | It follows that <math> | + | It follows that <math>AE = DE </math> if and only if |
<center> | <center> | ||
− | <math> | + | <math>\sin (\pi - 2\alpha - \beta) = \sin (\pi - 2\beta - \alpha) </math>. |
</center> | </center> | ||
− | Since <math> | + | Since <math>0 < \alpha, \beta < \pi/2 </math>, |
− | <center><math> | + | <center><math>0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi </math>, |
</center> | </center> | ||
and | and | ||
− | <center><math> | + | <center><math>-\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi </math>. |
</center> | </center> | ||
− | From these inequalities, we see that <math> | + | From these inequalities, we see that <math>\sin (\pi - 2\alpha - \beta ) = \sin (\pi - 2\beta - \alpha) </math> if and only if <math>(\pi - 2\alpha - \beta) = (\pi - 2\beta - \alpha) </math> (i.e., <math>\alpha = \beta </math>) or <math>(\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) = \pi </math> (i.e., <math>3(\alpha + \beta) = \pi </math>). But if <math>\alpha = \beta </math>, then triangles <math>ABC, BCD </math> are congruent and <math>AC = BD </math>, a contradiction. Thus we conclude that <math>AE = DE </math> if and only if <math>\alpha + \beta = \pi/3 </math>, Q.E.D. |
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
+ | |||
Latest revision as of 22:59, 14 September 2014
Contents
[hide]Problem
(Albania) Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .
Solution
Since , , and similarly, . Since , by considering triangles we have . It follows that .
Now, by the Law of Sines,
.
It follows that if and only if
.
Since ,
and
From these inequalities, we see that if and only if (i.e., ) or (i.e., ). But if , then triangles are congruent and , a contradiction. Thus we conclude that if and only if , Q.E.D.
Solution 2
Let and . Then by the isosceles triangles manifest in the figure we have and , so and . Furthermore and .
If , then . But also , so by SSA "Incongruence" (aka. the Law of Sines: ) we have . This translates into , or , which incidentally equals , as desired.
If , then also by the Exterior Angle Theorem, so and hence and are supplementary. A simple Law of Sines calculation then gives , 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.