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

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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>
 
<!-- 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>
  
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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})}
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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})}
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\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 1 (video solution) ==
 
== Solution 1 (video solution) ==
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https://youtu.be/B1OXVB5GDjk
 
https://youtu.be/B1OXVB5GDjk
  
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== Solution 2==
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== Solution 2 (Longer 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>
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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)
  
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~Ileytyn
  
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== Solution 3 (Longer solution-credit: Ileytyn)==
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== Solution 3 (Like Solution 2 but shorter)==
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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>. Also, the length of each side of the square is 2000, and 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 2 times the length of a leg of the isosceles right triangle, 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>, 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 </math> 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 B:<math> \boxed{ 2000(\sqrt{2} -1)}</math>!(not a factorial symbol, just an exclamation point)
 
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All side lengths of an octagon are equal, so the hypotenuse cut off by the isosceles triangle are equal to the sides of the octagon. Setting the side length of the octagon as <math>x</math>, we find that the leg of the isosceles triangle is <math>x \sqrt{2}/2</math> (45-45-90 special triangle). The two legs and a side of the octagon sum up to a side of the square, so we can write the expression <math>2(x \sqrt{2}/2 )+x=2000</math>. Simplifying and factoring out x, we obtain <math>x(\sqrt{2}+1)=2000</math>. Dividing both sides by <math>(\sqrt{2}+1)</math> and rationalizing the denominator, we get our answer: <math>\boxed{\textbf{(B) }2000(\sqrt{2} -1)}</math>
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~goldenuni678
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:33, 9 August 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 (Longer solution)

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)

~Ileytyn

Solution 3 (Like Solution 2 but shorter)

All side lengths of an octagon are equal, so the hypotenuse cut off by the isosceles triangle are equal to the sides of the octagon. Setting the side length of the octagon as $x$, we find that the leg of the isosceles triangle is $x \sqrt{2}/2$ (45-45-90 special triangle). The two legs and a side of the octagon sum up to a side of the square, so we can write the expression $2(x \sqrt{2}/2 )+x=2000$. Simplifying and factoring out x, we obtain $x(\sqrt{2}+1)=2000$. Dividing both sides by $(\sqrt{2}+1)$ and rationalizing the denominator, we get our answer: $\boxed{\textbf{(B) }2000(\sqrt{2} -1)}$ ~goldenuni678

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

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