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]
size(200); 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]
Claim: It is impossible to choose two non-adjacent sides and clip a whole part of it off.
Proof: If you clip adjacent sides, you can cut off at most up to the blue lines; Clipping more is impossible due to the degrees getting larger and larger and more and more circular.
Thus, after infinitely many clips, the hexagon bounded by the blue lines is left, so after finitely many clips, the area left is more than that hexagon.
The side length of that hexagon is 
of the large one, because of 30-30-120 triangles. Thus, the area is 
of the larger one, so we are done.
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. 
