Difference between revisions of "1996 USAMO Problems/Problem 3"
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Let <math>ABC</math> be a triangle. Prove that there is a line <math>l</math> (in the plane of triangle <math>ABC</math>) such that the intersection of the interior of triangle <math>ABC</math> and the interior of its reflection <math>A'B'C'</math> in <math>l</math> has area more than <math>\frac{2}{3}</math> the area of triangle <math>ABC</math>. | Let <math>ABC</math> be a triangle. Prove that there is a line <math>l</math> (in the plane of triangle <math>ABC</math>) such that the intersection of the interior of triangle <math>ABC</math> and the interior of its reflection <math>A'B'C'</math> in <math>l</math> has area more than <math>\frac{2}{3}</math> the area of triangle <math>ABC</math>. | ||
| − | == | + | ==Solution== |
| − | + | {{solution}} | |
| + | |||
| + | == See Also == | ||
| + | {{USAMO box|year=1996|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Olympiad Geometry Problems]] | ||
Revision as of 09:25, 20 July 2016
Problem
Let 
be a triangle. Prove that there is a line 
(in the plane of triangle ) such that the intersection of the interior of triangle and the interior of its reflection 
in has area more than 
the area of triangle .
Solution
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See Also
| 1996 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
