Difference between revisions of "1997 USAMO Problems/Problem 4"
(added solution tag, USAMO box, and category) |
m |
||
Line 6: | Line 6: | ||
== See Also == | == See Also == | ||
− | {{USAMO | + | {{USAMO newbox|year=1997|num-b=3|num-a=5}} |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 16:13, 12 April 2012
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 |