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

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== Solution 3 (Longer solution-credit: Ileytyn)==
 
== Solution 3 (Longer solution-credit: Ileytyn)==
  
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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>. 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 2 times the length of a leg of the isosceles right triangle ( the total length of the side is <math>2s+ o</math>, o 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 </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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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 2 times the length of a leg of the isosceles right triangle ( the total length of the side is <math>2s+ o</math>, o 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>, 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 <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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== See Also ==
 
== See Also ==

Revision as of 08:50, 13 August 2021

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

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 \[2 \cdot \frac{x\sqrt{2}}{2} +x=2000, \text{ and } x = \frac{2000}{\sqrt{2}+1}=\boxed{2000(\sqrt{2}-1)}.\]


Solution 3 (Longer solution-credit: Ileytyn)

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 B:$\boxed{ 2000(\sqrt{2} -1)}$!(not a factorial symbol, just an exclamation point)

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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