1997 USAMO Problems/Problem 4

Problem

To clip a convex $n$ -gon means to choose a pair of consecutive sides $AB, BC$ and to replace them by three segments $AM, MN,$ and $NC,$ where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$ . In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$ -gon. A regular hexagon $P_6$ of area $1$ is clipped to obtain a heptagon $P_7$ . Then $P_7$ is clipped (in one of the seven possible ways) to obtain an octagon $P_8$ , and so on. Prove that no matter how the clippings are done, the area of $P_n$ is greater than $\frac{1}{3}$ , for all $n\ge6$ .

Solution

$[asy] size(200); draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0)); draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue); draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue); [/asy]$

$\textbf{Claim:}$ It is impossible to choose two non-adjacent sides and clip a whole part of it off.

$\textbf{Proof:}$ 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 $\frac{\sqrt{3}}{3}$ of the large one, because of 30-30-120 triangles. Thus, the area is $\frac13$ of the larger one, so we are done.

1997 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
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1997 USAMO Problems/Problem 4

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