2001 AMC 10 Problem 20

Below is the professionally curated solution for Problem 20 of the 2001 AMC 10, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 10 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:regular polygonspecial right trianglerationalizing denominator

Difficulty rating: 1580

20.

A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length 2000.2000. What is the length of each side of the octagon?

13(2000)\dfrac13(2000)

2000(21)2000(\sqrt2-1)

2000(22)2000(2-\sqrt2)

10001000

100021000\sqrt2

Solution:

Let each octagon side be x.x. It is the hypotenuse of each cut isosceles right triangle, whose legs are x2.\dfrac{x}{\sqrt2}.

Along one side of the square, two legs and one octagon side give 2x2+x=2000,2\cdot\dfrac{x}{\sqrt2}+x=2000, so x(2+1)=2000x(\sqrt2+1)=2000 and x=20002+1=2000(21).x=\dfrac{2000}{\sqrt2+1}=2000(\sqrt2-1).

Thus, the correct answer is B.

Problem 20 in Other Years