Difference between revisions of "2007 BMO Problems/Problem 1"
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− | Let <math> \displaystyle ABCD </math> be a convex quadrilateral with <math> \displaystyle AB=BC=CD </math> | + | Let <math> \displaystyle ABCD </math> be a convex quadrilateral with <math> \displaystyle AB=BC=CD </math>, <math> \displaystyle AC \neq \displaystyle BD </math>, and let <math> \displaystyle E </math> be the intersection point of its diagonals. Prove that <math> \displaystyle AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>. |
== Solution == | == Solution == |
Revision as of 23:45, 4 May 2007
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 consdering triangles
we have
. It follows that
.
Now, by the Law of Sines,
.
It follows that if and only if
.
Since ,
![$\displaystyle 0 < (\pi - 2\alpha - \beta) + (\pi - 2\beta - \alpha) < 2\pi$](http://latex.artofproblemsolving.com/6/a/9/6a9bf77bf7369ce7783ab68a24ea857d15e49030.png)
and
![$\displaystyle -\pi < (\pi - 2\alpha - \beta) - (\pi - 2\beta - \alpha) < \pi$](http://latex.artofproblemsolving.com/b/9/7/b97accab92b52609c6c341d5a2f96ccc4f63bf50.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.