Difference between revisions of "1997 USAMO Problems/Problem 4"
m |
|||
Line 9: | Line 9: | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:33, 4 July 2013
Problem
To clip a convex -gon means to choose a pair of consecutive sides and to replace them by three segments and where is the midpoint of and is the midpoint of . In other words, one cuts off the triangle to obtain a convex -gon. A regular hexagon of area is clipped to obtain a heptagon . Then is clipped (in one of the seven possible ways) to obtain an octagon , and so on. Prove that no matter how the clippings are done, the area of is greater than , for all .
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
1997 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.