Difference between revisions of "1991 AIME Problems/Problem 11"
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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>. | 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>. | ||
− | + | <asy> | |
+ | unitsize(100); | ||
+ | draw(Circle((0,0),1)); | ||
+ | dot((0,0)); | ||
+ | draw((0,0)--(1,0)); | ||
+ | label("$1$", (0.5,0), S); | ||
+ | |||
+ | for (int i=0; i<12; ++i) | ||
+ | { | ||
+ | dot((cos(i*pi/6), sin(i*pi/6))); | ||
+ | } | ||
+ | |||
+ | for (int a=1; a<24; a+=2) | ||
+ | { | ||
+ | dot(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))); | ||
+ | draw(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12))--((1/cos(pi/12))*cos((a+2)*pi/12), (1/cos(pi/12))*sin((a+2)*pi/12))); | ||
+ | draw(Circle(((1/cos(pi/12))*cos(a*pi/12), (1/cos(pi/12))*sin(a*pi/12)), tan(pi/12))); | ||
+ | }</asy> | ||
+ | |||
+ | _Diagram by 1-1 is 3_ | ||
== Solution == | == Solution == | ||
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We thus know that the [[apothem]] of the [[dodecagon]] is equal to <math>1</math>. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote <math>A, M,</math> and <math>O</math> respectively. Notice that <math>OM=1</math>, and that <math>\triangle OMA</math> is a right triangle with [[hypotenuse]] <math>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles. The area of one circle is thus <math>\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})</math>, so the area of all <math>12</math> circles is <math>\pi (84 - 48 \sqrt {3})</math>, giving an answer of <math>84 + 48 + 3 = \boxed{135}</math>. | We thus know that the [[apothem]] of the [[dodecagon]] is equal to <math>1</math>. To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote <math>A, M,</math> and <math>O</math> respectively. Notice that <math>OM=1</math>, and that <math>\triangle OMA</math> is a right triangle with [[hypotenuse]] <math>OA</math> and <math>m \angle MOA = 15^\circ</math>. Thus <math>AM = (1) \tan{15^\circ} = 2 - \sqrt {3}</math>, which is the radius of one of the circles. The area of one circle is thus <math>\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})</math>, so the area of all <math>12</math> circles is <math>\pi (84 - 48 \sqrt {3})</math>, giving an answer of <math>84 + 48 + 3 = \boxed{135}</math>. | ||
+ | |||
+ | ----- | ||
+ | Note that it is very useful to understand the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contests. The side lengths are in the ratio <math>\sqrt {3} - 1</math>, <math>\sqrt {3} + 1</math>, and <math>2\sqrt {2}</math> | ||
<!--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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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:43, 6 December 2019
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 .
_Diagram by 1-1 is 3_
Solution
We wish to find the radius of one circle, so that we can find the total area.
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length that is tangent to the larger circle at the midpoint of the two centers. Thus, we have essentially have a regular dodecagon whose vertices are the centers of the smaller triangles circumscribed about a circle of radius .
We thus know that the apothem of the dodecagon is equal to . To find the side length, we make a triangle consisting of a vertex, the midpoint of a side, and the center of the dodecagon, which we denote and respectively. Notice that , and that is a right triangle with hypotenuse and . Thus , which is the radius of one of the circles. The area of one circle is thus , so the area of all circles is , giving an answer of .
Note that it is very useful to understand the side lengths of a 15-75-90 triangle, as these triangles often appear on higher level math contests. The side lengths are in the ratio , , and
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.