Difference between revisions of "2007 BMO Problems/Problem 1"
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− | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \ | + | 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 == | ||
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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. | 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. | ||
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+ | ==Solution 2== | ||
+ | Let <math><BAC = x</math> and <math><BDC = y</math>. | ||
+ | |||
Revision as of 22:50, 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 .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.