Difference between revisions of "2001 AMC 10 Problems/Problem 20"

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(Solution)
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== See Also ==
 
== See Also ==

Revision as of 20:22, 6 August 2019

Problem

A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?

$\textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})} \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}}$

Solution

[asy] draw((0,0)--(0,10)--(10,10)--(10,0)--cycle); draw((0,7)--(3,10)); draw((7,10)--(10,7)); draw((10,3)--(7,0)); draw((3,0)--(0,3)); label("$x$",(0,1),W); label("$x\sqrt{2}$",(1.5,1.5),NE); label("$2000-2x$",(5,0),S);[/asy]

$2000 - 2x = x\sqrt2$

$2000 = x(2 + \sqrt2)$

$x = \frac {2000}{2 + \sqrt2} =x = \frac {2000(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)}= \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2)$

$x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)}$.


~edited by qkddud~

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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