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(\sqrt{2}-1)} \qquad \textbf{(C)} 2000(2-\sqrt{2})
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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})}
\qquad \textbf{(D)} 1000 \qquad \textbf{(E)} 1000\sqrt{2} </math>
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\qquad \textbf{(D) } {1000} \qquad \textbf{(E) } {1000\sqrt{2}} </math>
  
== Solution ==
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== Solution 1 (video solution) ==
  
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https://youtu.be/B1OXVB5GDjk
  
<asy>
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== Solution 2=
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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== Solution 3 (Longer solution-credit: Ileytyn)==
  
<math> 2000 = x(2 + \sqrt2) </math>
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First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let <math>s</math> be the length of a leg of the isosceles right triangle. In terms of <math>s</math>, the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is <math>s \sqrt{2}</math>. Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square (<math>2000</math>) subtracted by <math>2</math> times the length of a leg of the isosceles right triangle ( the total length of the side is <math>2s+ o</math>, <math>o</math> being the length of a side of the regular octagon), which is the same as <math> 2s </math>. As an expression, this is <math>2000-2s</math>, which we can equate to <math>s \sqrt{2}</math>, ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:<math>2000-2s = s \sqrt{2}</math>. By isolating the variable and simplifying the right side, we get the following: <math>2000 = s(2 + \sqrt{2})</math>. Dividing both sides by <math>(2 + \sqrt{2})</math>, we arrive with <math>\frac{2000}{2 + \sqrt{2}} = s</math>, now, to find the length of the side of the octagon, we can plug in <math>s</math> and use the equation <math>2000-2s = o </math>, <math>o</math> being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive <math>2000-2(\frac{2000}{2 + \sqrt{2}})</math>, which is the same as <math>2000-(\frac{4000}{2 + \sqrt{2}})</math>, factoring out a <math> 2000 </math>, we derive the following: <math> 2000(1-(\frac{2}{2 + \sqrt{2}}))</math>, by rationalizing the denominator of <math> \frac{2}{2 + \sqrt{2}} </math>, we get <math> 2000(1-(2 - \sqrt{2})) </math>, after expanding, finally, we get <math>\boxed{\textbf{(B) }2000(\sqrt{2} -1)}</math> !(not a factorial symbol, just an exclamation point)
 
 
<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>.
 
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2001|num-b=19|num-a=21}}
 
{{AMC10 box|year=2001|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 09:59, 1 March 2024

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 1 (video solution)

https://youtu.be/B1OXVB5GDjk

= Solution 2

Solution 3 (Longer solution-credit: Ileytyn)

First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let $s$ be the length of a leg of the isosceles right triangle. In terms of $s$, the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is $s \sqrt{2}$. Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square ($2000$) subtracted by $2$ times the length of a leg of the isosceles right triangle ( the total length of the side is $2s+ o$, $o$ being the length of a side of the regular octagon), which is the same as $2s$. As an expression, this is $2000-2s$, which we can equate to $s \sqrt{2}$, ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:$2000-2s = s \sqrt{2}$. By isolating the variable and simplifying the right side, we get the following: $2000 = s(2 + \sqrt{2})$. Dividing both sides by $(2 + \sqrt{2})$, we arrive with $\frac{2000}{2 + \sqrt{2}} = s$, now, to find the length of the side of the octagon, we can plug in $s$ and use the equation $2000-2s = o$, $o$ being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive $2000-2(\frac{2000}{2 + \sqrt{2}})$, which is the same as $2000-(\frac{4000}{2 + \sqrt{2}})$, factoring out a $2000$, we derive the following: $2000(1-(\frac{2}{2 + \sqrt{2}}))$, by rationalizing the denominator of $\frac{2}{2 + \sqrt{2}}$, we get $2000(1-(2 - \sqrt{2}))$, after expanding, finally, we get $\boxed{\textbf{(B) }2000(\sqrt{2} -1)}$ !(not a factorial symbol, just an exclamation point)

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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