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Difference between revisions of "1991 AIME Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Twelve congruent disks are placed on a circle <math>C^{}_{}</math> of radius 1 in such a way that the twelve disks cover <math>C^{}_{}</math>, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below.  The sum of the areas of the twelve disks can be written in the from <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers and <math>c^{}_{}</math> is not divisible by the square of any prime. Find <math>a+b+c^{}_{}</math>.
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Twelve congruent disks are placed on a [[circle]] <math>C^{}_{}</math> of [[radius]] 1 in such a way that the twelve disks cover <math>C^{}_{}</math>, no two of the disks overlap, and so that each of the twelve disks is [[tangent (geometry)|tangent]] to its two neighbors. The resulting arrangement of disks is shown in the figure below.  The sum of the areas of the twelve disks can be written in the from <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers and <math>c^{}_{}</math> is not divisible by the square of any prime. Find <math>a+b+c^{}_{}</math>.
  
 
[[Image:AIME_1991_Problem_11.gif]]
 
[[Image:AIME_1991_Problem_11.gif]]
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Let <math>R_{}^{}</math> and <math>r_{}^{}</math> denote the radii of the large and small circles (<math>R_{}^{}>r_{}^{}</math>), respectively. Suppose that there are <math>n_{}^{}</math> circles of radius <math>r_{}^{}</math> centered on the circumference of the circle having radius <math>R_{}^{}</math>. Let <math>O_{}^{}</math>, <math>P_{}^{}</math>, and  <math>Q_{}^{}</math> label the vertices of the triangle with <math>O_{}^{}</math> being at the center of the large circle, whereas <math>P_{}^{}</math> and <math>Q_{}^{}</math> are the tangential points of one of the <math>n_{}^{}</math> small circles with its two other neighbours, and <math>S_{}^{}</math> its center. The angle subtended by <math>P_{}^{}OQ</math> is <math>2\pi/n_{}^{}</math>. The segments <math>O_{}^{}P</math> and <math>S_{}^{}P</math> are perpendicular. Therefore, triangle <math>P_{}^{}OS</math> is rectangular and the angle subtended by <math>P_{}^{}OS</math> equals <math>\pi/n_{}^{}</math>. Hence, the radius <math>r_{}^{}=R\sin(\pi/n)</math>. The total area <math>A_{n}^{}</math> of the <math>n_{}^{}</math> circles is thus given by  
 
Let <math>R_{}^{}</math> and <math>r_{}^{}</math> denote the radii of the large and small circles (<math>R_{}^{}>r_{}^{}</math>), respectively. Suppose that there are <math>n_{}^{}</math> circles of radius <math>r_{}^{}</math> centered on the circumference of the circle having radius <math>R_{}^{}</math>. Let <math>O_{}^{}</math>, <math>P_{}^{}</math>, and  <math>Q_{}^{}</math> label the vertices of the triangle with <math>O_{}^{}</math> being at the center of the large circle, whereas <math>P_{}^{}</math> and <math>Q_{}^{}</math> are the tangential points of one of the <math>n_{}^{}</math> small circles with its two other neighbours, and <math>S_{}^{}</math> its center. The angle subtended by <math>P_{}^{}OQ</math> is <math>2\pi/n_{}^{}</math>. The segments <math>O_{}^{}P</math> and <math>S_{}^{}P</math> are perpendicular. Therefore, triangle <math>P_{}^{}OS</math> is rectangular and the angle subtended by <math>P_{}^{}OS</math> equals <math>\pi/n_{}^{}</math>. Hence, the radius <math>r_{}^{}=R\sin(\pi/n)</math>. The total area <math>A_{n}^{}</math> of the <math>n_{}^{}</math> circles is thus given by  
  
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<math>
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<cmath>
 
A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .
 
A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .
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</math>
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</cmath>
  
 
In the present problem, <math>n_{}^{}=12</math> and <math>R_{}^{}=1</math>. It follows that <math>A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)</math>.
 
In the present problem, <math>n_{}^{}=12</math> and <math>R_{}^{}=1</math>. It follows that <math>A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)</math>.
  
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In summary, <math>a_{}^{}+b+c=12</math>.
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In summary, <math>a_{}^{}+b+c=\boxed{012}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=10|num-a=12}}
 
{{AIME box|year=1991|num-b=10|num-a=12}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 20:32, 11 April 2008

Problem

Twelve congruent disks are placed on a circle $C^{}_{}$ of radius 1 in such a way that the twelve disks cover $C^{}_{}$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be written in the from $\pi(a-b\sqrt{c})$, where $a,b,c^{}_{}$ are positive integers and $c^{}_{}$ is not divisible by the square of any prime. Find $a+b+c^{}_{}$.

AIME 1991 Problem 11.gif

Solution

Let $R_{}^{}$ and $r_{}^{}$ denote the radii of the large and small circles ($R_{}^{}>r_{}^{}$), respectively. Suppose that there are $n_{}^{}$ circles of radius $r_{}^{}$ centered on the circumference of the circle having radius $R_{}^{}$. Let $O_{}^{}$, $P_{}^{}$, and $Q_{}^{}$ label the vertices of the triangle with $O_{}^{}$ being at the center of the large circle, whereas $P_{}^{}$ and $Q_{}^{}$ are the tangential points of one of the $n_{}^{}$ small circles with its two other neighbours, and $S_{}^{}$ its center. The angle subtended by $P_{}^{}OQ$ is $2\pi/n_{}^{}$. The segments $O_{}^{}P$ and $S_{}^{}P$ are perpendicular. Therefore, triangle $P_{}^{}OS$ is rectangular and the angle subtended by $P_{}^{}OS$ equals $\pi/n_{}^{}$. Hence, the radius $r_{}^{}=R\sin(\pi/n)$. The total area $A_{n}^{}$ of the $n_{}^{}$ circles is thus given by

\[A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .\]

In the present problem, $n_{}^{}=12$ and $R_{}^{}=1$. It follows that $A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)$.

In summary, $a_{}^{}+b+c=\boxed{012}$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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