Difference between revisions of "1991 AIME Problems/Problem 11"
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== Problem == | == Problem == | ||
− | 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>. | + | 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 | ||
− | < | + | <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]\, . | ||
− | </ | + | </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>. | ||
− | In summary, <math>a_{}^{}+b+c= | + | 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 21:32, 11 April 2008
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 one of the
small circles with its two other neighbours, and
its center. 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
In the present problem, and
. It follows that
.
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 |