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

(Created page with "==Problem== A circle of radius <math>1</math> is surrounded by <math>4</math> circles of radius <math>r</math> as shown. What is <math>r</math>? <center><asy> unitsize(3mm); de...")
 
(→‎Solution 2)
 
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<center><asy>
 
<center><asy>
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unitsize(3mm);
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unitsize(5mm);
 
defaultpen(linewidth(.8pt)+fontsize(7pt));
 
defaultpen(linewidth(.8pt)+fontsize(7pt));
 
dotfactor=4;
 
dotfactor=4;
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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==
+
==Solutions==
  
−
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.
+
==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 expressions represent the same length, you can set them equal to each other.
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
r+1&=r\sqrt{2}\\
 
r+1&=r\sqrt{2}\\
 
1&=r(\sqrt{2}-1)\end{align*}</cmath>
 
1&=r(\sqrt{2}-1)\end{align*}</cmath>
−
<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{1(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}</cmath>
 +
 
 +
==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 a segment of length <math>2r+2</math>. The two legs are each the length between the centers of two large, adjacent circles, thus they are each equal to <math>2r</math>.
 +
Using the Pythagorean Theorem:
 +
<cmath>\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+2\sqrt{2}}{2}=\boxed{\mathrm{(B) \ } 1 + \sqrt{2}}
 +
\end{align*}
 +
</cmath>
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2007|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2007|ab=B|num-b=17|num-a=19}}
 +
{{MAA Notice}}

Latest revision as of 13:17, 4 June 2021

Problem

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

[asy] unitsize(5mm); 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 expressions represent the same length, you can set them equal to each other. \begin{align*} r+1&=r\sqrt{2}\\ 1&=r(\sqrt{2}-1)\end{align*} \[r = \frac{1}{\sqrt{2}-1} = \frac{1(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \boxed{\mathrm{(B) \ } 1 + \sqrt{2}}\]

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 a segment of length $2r+2$. The two legs are each the length between the centers of two large, adjacent circles, thus they are each equal to $2r$. 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+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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