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Difference between revisions of "2007 AMC 10A Problems/Problem 15"

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Three circles are inscribed in a rectangle of width w and height h as shown. Two
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== Problem ==
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of the circles are congruent and are each tangent to two adjacent sides of the
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Four circles of radius <math>1</math> are each tangent to two sides of a square and externally tangent to a circle of radius <math>2</math>, as shown. What is the area of the square?
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rectangle and to each other. The other circle is larger and is tangent to three sides
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<center>
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of the rectangle and to the two smaller circles. What the ratio of h to w? Express
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<asy>
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your answer as a decimal to the nearest hundredth.
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unitsize(5mm);
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defaultpen(linewidth(.8pt)+fontsize(8pt));
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real h=3*sqrt(2)/2;
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pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h);
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pair X=O0+2*dir(30), Y=O2+dir(45);
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draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle);
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draw(Circle(O0,2));
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draw(Circle(O1,1));
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draw(Circle(O2,1));
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draw(Circle(O3,1));
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draw(Circle(O4,1));
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draw(O0--X);
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draw(O2--Y);
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label("$2$",midpoint(O0--X),NW);
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label("$1$",midpoint(O2--Y),SE);
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</asy>
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</center>
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<math>\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}</math>
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==Solution 1==
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Draw a square connecting the centers of the four small circles of radius <math>1</math>. This square has a diagonal of length <math>6</math>, as it includes the diameter of the big circle of radius <math>2</math> and two radii of the small circles of radius <math>1</math>. Therefore, the side length of this square is <cmath>\frac{6}{\sqrt{2}} = 3\sqrt{2}.</cmath> The side length of the larger square is <math>2</math> units greater than the one found by connecting the midpoints, so its side length is <cmath>2 + 3\sqrt{2}.</cmath> The area of the larger square is <math>(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}</math> <math>(B).</math>
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== Solution 2 ==
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We draw the diagonal of the square. This diagonal yields <math>2\sqrt{2}+1+1+2+2=2\sqrt{2}+6</math>. We know that the side length <math>s</math> in terms of the diagonal <math>d</math> is <math>s=\frac{d}{\sqrt{2}}</math>, so our side length is <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math>. However, we are trying to look for the area, so squaring <math>\frac{2\sqrt{2}+6}{\sqrt{2}}</math> yields <math>\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 09:29, 4 June 2021

Problem

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square?

[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); real h=3*sqrt(2)/2; pair O0=(0,0), O1=(h,h), O2=(-h,h), O3=(-h,-h), O4=(h,-h); pair X=O0+2*dir(30), Y=O2+dir(45); draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); draw(Circle(O0,2)); draw(Circle(O1,1)); draw(Circle(O2,1)); draw(Circle(O3,1)); draw(Circle(O4,1)); draw(O0--X); draw(O2--Y); label("$2$",midpoint(O0--X),NW); label("$1$",midpoint(O2--Y),SE); [/asy]

$\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}$

Solution 1

Draw a square connecting the centers of the four small circles of radius $1$. This square has a diagonal of length $6$, as it includes the diameter of the big circle of radius $2$ and two radii of the small circles of radius $1$. Therefore, the side length of this square is \[\frac{6}{\sqrt{2}} = 3\sqrt{2}.\] The side length of the larger square is $2$ units greater than the one found by connecting the midpoints, so its side length is \[2 + 3\sqrt{2}.\] The area of the larger square is $(2+3\sqrt{2})^2 = 22 + 12\sqrt{2}$ $(B).$

Solution 2

We draw the diagonal of the square. This diagonal yields $2\sqrt{2}+1+1+2+2=2\sqrt{2}+6$. We know that the side length $s$ in terms of the diagonal $d$ is $s=\frac{d}{\sqrt{2}}$, so our side length is $\frac{2\sqrt{2}+6}{\sqrt{2}}$. However, we are trying to look for the area, so squaring $\frac{2\sqrt{2}+6}{\sqrt{2}}$ yields $\frac{44+24\sqrt{2}}{2}=\boxed{\text{(B)}22+12\sqrt{2}}$

See Also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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