Difference between revisions of "2007 BMO Problems/Problem 1"
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(''Albania'') | (''Albania'') | ||
− | Let <math> | + | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neqBD </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. |
Revision as of 00:41, 24 October 2012
Problem
(Albania)
Let be a convex quadrilateral with
, $AC \neqBD$ (Error compiling LaTeX. Unknown error_msg), 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 ,
![$0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi$](http://latex.artofproblemsolving.com/1/3/1/1316ab62513fdc1d1efe0447b00bb009c900e03c.png)
and
![$-\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi$](http://latex.artofproblemsolving.com/9/3/d/93da4b1ff91cea97dfd30ea7ef9472f62f689bd4.png)
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.