Difference between revisions of "2007 BMO Problems/Problem 1"
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Let <math><BAC = x</math> and <math><BDC = y</math>.
Let <math><BAC = x</math> and <math><BDC = y</math>
Latest revision as of 23:59, 14 September 2014
(Albania) Let be a convex quadrilateral with , , and let be the intersection point of its diagonals. Prove that if and only if .
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
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.
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.