Difference between revisions of "2001 AMC 10 Problems/Problem 20"
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<math> 2000 = x(2 + \sqrt2) </math> | <math> 2000 = x(2 + \sqrt2) </math> | ||
| − | <math> x = \frac {2000}{2 + \sqrt2} = \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2) </math> | + | <math> x = \frac {2000}{2 + \sqrt2} =x = \frac {2000(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)}= \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2) </math> |
<math> x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)} </math>. | <math> x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)} </math>. | ||
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 
. What is the length of each side of the octagon?
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]](http://latex.artofproblemsolving.com/4/2/e/42e5c38ee228834593060f611f943b66f1ba396b.png)
.
~helped by qkddud
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
