1997 USAMO Problems/Problem 4
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
[asy] draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0)); draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue); draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue); [/asy]
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. 
