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.