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

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== Solution ==
 
== Solution ==
  
Let <math>x</math> represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length <math>x\sqrt{2}/2</math>, so <cmath>2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(\sqrt{2}-1)}.</cmath> boxed{(B)}.<math></math>
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Let <math>x</math> represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length <math>x\sqrt{2}/2</math>, so <cmath>2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(\sqrt{2}-1)}.</cmath> =\boxed{(B)}.<math></math>
  
 
== See Also ==
 
== See Also ==

Revision as of 19:27, 9 January 2020

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

Let $x$ represent the length of each side of the octagon, which is also the length of the hypotenuse of each of the right triangles. Each leg of the right triangles has length $x\sqrt{2}/2$, so \[2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(\sqrt{2}-1)}.\] =\boxed{(B)}.$$ (Error compiling LaTeX. Unknown error_msg)

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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