Difference between revisions of "1997 USAMO Problems/Problem 4"
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| + | == Problem == | ||
| + | To ''clip'' a convex <math>n</math>-gon means to choose a pair of consecutive sides <math>AB, BC</math> and to replace them by three segments <math>AM, MN,</math> and <math>NC,</math> where <math>M</math> is the midpoint of <math>AB</math> and <math>N</math> is the midpoint of <math>BC</math>. In other words, one cuts off the triangle <math>MBN</math> to obtain a convex <math>(n+1)</math>-gon. A regular hexagon <math>P_6</math> of area <math>1</math> is clipped to obtain a heptagon <math>P_7</math>. Then <math>P_7</math> is clipped (in one of the seven possible ways) to obtain an octagon <math>P_8</math>, and so on. Prove that no matter how the clippings are done, the area of <math>P_n</math> is greater than <math>\frac{1}{3}</math>, for all <math>n\ge6</math>. | ||
| + | == Solution == | ||
Revision as of 14:10, 5 July 2011
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 
.
