Difference between revisions of "1991 AIME Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | {{ | + | 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 any small circle with its two other neighbour circles, and <math>C_{}^{}</math> is the center of any of these small circles. The angle subtended by <math>P_{}^{}OQ</math> is <math>2\pi/n_{}^{}</math>. The segments <math>O_{}^{}P</math> and <math>P_{}^{}C</math> are perpendicular. Therefore, triangle <math>P_{}^{}OC</math> is rectangular and the angle subtended by <math>P_{}^{}OC</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 |
| + | |||
| + | <math> | ||
| + | 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]\, . | ||
| + | </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>. | ||
| + | |||
| + | In summary, <math>a_{}^{}+b+c=12</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}} | ||
Revision as of 19:09, 22 April 2007
Problem
Twelve congruent disks are placed on a circle 
of radius 1 in such a way that the twelve disks cover , 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 
, where 
are positive integers and 
is not divisible by the square of any prime. Find 
.
Solution
Let 
and 
denote the radii of the large and small circles (
), respectively. Suppose that there are 
circles of radius centered on the circumference of the circle having radius . Let 
, 
, and 
label the vertices of the triangle with being at the center of the large circle, whereas and are the tangential points of any small circle with its two other neighbour circles, and 
is the center of any of these small circles. The angle subtended by 
is 
. The segments 
and 
are perpendicular. Therefore, triangle 
is rectangular and the angle subtended by equals 
. Hence, the radius 
. The total area 
of the circles is thus given by
![$A_{n}^{}=n\pi r^{2}=n\pi R^{2}\sin^{2}\left(\frac{\pi}{n}\right)=\frac{1}{2}n\pi R^{2}\left[1-\cos\left(\frac{2\pi}{n}\right)\right]\, .$](http://latex.artofproblemsolving.com/4/4/4/444303944cfb9d1ffc9739294e41cf8164f6554d.png)
In the present problem, 
and 
. It follows that ![$A_{12}^{}=6\pi\left[1-\cos(\pi/6)\right]=\pi\left(6-3\sqrt{3}\right)\equiv \pi\left(a-b\sqrt{c}\right)$](http://latex.artofproblemsolving.com/4/1/0/410e40bd893f180fc771de29e3c3d4bdb1dcfbb1.png)
.
In summary, 
.
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 | ||
