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

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== Problem ==
 
== Problem ==
  
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length <math> 2000 </math>. What is the length of each side of the octagon?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length <math> 2000 </math>. What is the length of each side of the octagon?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
<math> \textbf{(A)}\ \frac{1}{3}(2000) \qquad \textbf{(B)}\ 2000(\sqrt2-1) \qquad \textbf{(C)}\ 2000(2-\sqrt2)
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<math> \textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})}
\textbf{(D)}\ 1000 \qquad \textbf{(E)}\ 1000\sqrt2 </math>
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\qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}} </math>
  
 
== Solution ==
 
== Solution ==
  
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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>
  
<asy>
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== See Also ==
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>
 
  
<math> 2000 - 2x = x\sqrt2 </math>
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{{AMC10 box|year=2001|num-b=19|num-a=21}}
 
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{{MAA Notice}}
<math> 2000 = x(2 + \sqrt2) </math>
 
 
 
<math> x = \frac {2000}{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>.
 

Latest revision as of 18:49, 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 \[2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(\sqrt{2}-1)}.\]

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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