Difference between revisions of "1997 USAMO Problems/Problem 4"
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== Solution == | == 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 == | == See Also == | ||
Revision as of 17:17, 12 October 2019
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. 
