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Difference between revisions of "2007 AMC 10B Problems/Problem 18"

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<math>\textbf{(A) } \sqrt{2} \qquad\textbf{(B) } 1+\sqrt{2} \qquad\textbf{(C) } \sqrt{6} \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 2+\sqrt{2}</math>
 
<math>\textbf{(A) } \sqrt{2} \qquad\textbf{(B) } 1+\sqrt{2} \qquad\textbf{(C) } \sqrt{6} \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 2+\sqrt{2}</math>
  
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==Solution==
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==Solutions==
  
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'''Solution 1'''
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==Solution 1==
  
 
You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get <math>r+1.</math> You can also add the radius of two outer circles and use a <math>45-45-90</math> triangle to get <math>\frac{2r}{\sqrt{2}} = r\sqrt{2}.</math> Since both representations are for the same thing, you can set them equal to each other.
 
You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get <math>r+1.</math> You can also add the radius of two outer circles and use a <math>45-45-90</math> triangle to get <math>\frac{2r}{\sqrt{2}} = r\sqrt{2}.</math> Since both representations are for the same thing, you can set them equal to each other.
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<cmath>r = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}</cmath>
 
<cmath>r = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}</cmath>
  
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'''Solution 2'''
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==Solution 2==
  
 
You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in <math>2r+2</math>. The two legs are each the length between two large, adjacent circles, thus <math>2r</math>.  
 
You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in <math>2r+2</math>. The two legs are each the length between two large, adjacent circles, thus <math>2r</math>.  

Revision as of 12:46, 3 January 2016

Problem

A circle of radius $1$ is surrounded by $4$ circles of radius $r$ as shown. What is $r$?

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(7pt)); dotfactor=4; real r1=1, r2=1+sqrt(2); pair A=(0,0), B=(1+sqrt(2),1+sqrt(2)), C=(-1-sqrt(2),1+sqrt(2)), D=(-1-sqrt(2),-1-sqrt(2)), E=(1+sqrt(2),-1-sqrt(2)); pair A1=(1,0), B1=(2+2sqrt(2),1+sqrt(2)), C1=(0,1+sqrt(2)), D1=(0,-1-sqrt(2)), E1=(2+2sqrt(2),-1-sqrt(2)); path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r2); path circleD=Circle(D,r2); path circleE=Circle(E,r2); draw(circleA); draw(circleB); draw(circleC); draw(circleD); draw(circleE); draw(A--A1); draw(B--B1); draw(C--C1); draw(D--D1); draw(E--E1); label("$1$",midpoint(A--A1),N); label("$r$",midpoint(B--B1),N); label("$r$",midpoint(C--C1),N); label("$r$",midpoint(D--D1),N); label("$r$",midpoint(E--E1),N); [/asy]

$\textbf{(A) } \sqrt{2} \qquad\textbf{(B) } 1+\sqrt{2} \qquad\textbf{(C) } \sqrt{6} \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 2+\sqrt{2}$

Solutions

Solution 1

You can express the line connecting the centers of an outer circle and the inner circle in two different ways. You can add the radius of both circles to get $r+1.$ You can also add the radius of two outer circles and use a $45-45-90$ triangle to get $\frac{2r}{\sqrt{2}} = r\sqrt{2}.$ Since both representations are for the same thing, you can set them equal to each other. \begin{align*} r+1&=r\sqrt{2}\\ 1&=r(\sqrt{2}-1)\end{align*} \[r = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}\]

Solution 2

You can solve this problem by setting up a simple equation with the Pythagorean Theorem. The hypotenuse would be a segment that includes the radius of two circles on opposite corners and the diameter of the middle circle. This results in $2r+2$. The two legs are each the length between two large, adjacent circles, thus $2r$. Using the Pythagorean Theorem: \begin{align*} (2r+2)^2 = 2(2r)^2\\ 4r^2+8r+4=8r^2\\ r^2+2r+1=2r^2\\ r^2-2r-1=0\\ r=\frac{2+\sqrt{4-(-4)}}{2}=\frac{2+\sqrt{2}}{2}=\boxed{\mathrm{(B) \ } 1 + \sqrt{2}} \end{align*}

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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