Difference between revisions of "1990 AIME Problems/Problem 12"
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A [[regular polygon|regular]] 12-gon is inscribed in a [[circle]] of [[radius]] 12. The [[sum]] of the lengths of all sides and [[diagonal]]s of the 12-gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math> where <math>a^{}_{}</math>, <math>b^{}_{}</math>, <math>c^{}_{}</math>, and <math>d^{}_{}</math> are positive integers. Find <math>a + b + c + d^{}_{}</math>. | A [[regular polygon|regular]] 12-gon is inscribed in a [[circle]] of [[radius]] 12. The [[sum]] of the lengths of all sides and [[diagonal]]s of the 12-gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math> where <math>a^{}_{}</math>, <math>b^{}_{}</math>, <math>c^{}_{}</math>, and <math>d^{}_{}</math> are positive integers. Find <math>a + b + c + d^{}_{}</math>. | ||
− | == Solution == | + | == Solution 1== |
[[Image:1990 AIME-12.png]] | [[Image:1990 AIME-12.png]] | ||
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Adding all of these up, we get <math>12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24</math> | Adding all of these up, we get <math>12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24</math> | ||
− | <math>= 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}</math>. Thus, the answer is <math>144 \cdot 5 = 720</math>. | + | <math>= 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}</math>. Thus, the answer is <math>144 \cdot 5 = \bosed{720}</math>. |
+ | ==Solution 2== | ||
+ | |||
+ | A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Since the distance from the center to a vertex of the 12-gon is <math>12</math>, the Law of Cosines can be applied to this isoceles triangle, to give: | ||
+ | |||
+ | <math>a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta</math> | ||
+ | |||
+ | <math>a^2 = 2\cdot 12^2 - 2\cdot 12^2 \cos \theta</math> | ||
+ | |||
+ | <math>a^2 = 2\cdot 12^2 (1 - \cos \theta)</math> | ||
+ | |||
+ | <math>a = 12\sqrt{2} \cdot \sqrt{1 - \cos \theta}</math> | ||
+ | |||
+ | There are six lengths of sides/diagonals, corresponding to <math>\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}</math> | ||
+ | |||
+ | Call these lengths <math>a_1, a_2, a_3, a_4, a_5, a_6</math> from shortest to longest. The total length <math>l</math> that is asked for is | ||
+ | |||
+ | <math>l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6</math>, noting that a_6 as written gives the diameter of the circle, which is the longest diagonal. | ||
+ | |||
+ | |||
+ | |||
+ | <math>l = 12[12\sqrt{2} (\sqrt{1 - \cos 30^{\circ}}+\sqrt{1-\cos 60^{\circ}}+\sqrt{1-\cos 90^{\circ}}+\sqrt{1-\cos 120^{\circ}}+\sqrt{1 - \cos 150^{\circ}})] + 6 \cdot 24</math> | ||
+ | |||
+ | |||
+ | <math>l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}\right) + 144</math> | ||
+ | |||
+ | |||
+ | <math>l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2}} + 1 + \frac{\sqrt{6}}{2} + \sqrt{1 + \frac{\sqrt{3}}{2}\right) + 144</math> | ||
+ | |||
+ | |||
+ | To simplify the two nested radicals, add them, and call the sum <math>x</math>: | ||
+ | |||
+ | <math>x = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}</math> | ||
+ | |||
+ | Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving <math>2</math>: | ||
+ | |||
+ | <math>x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}}\right)\left(1 + \frac{\sqrt{3}}{2}}\right)</math> | ||
+ | |||
+ | <math>x^2 = 2 + 2\sqrt{1 - \frac{3}{4}}</math> | ||
+ | |||
+ | <math>x^2 = 2 + 2\sqrt{\frac{1}{4}}</math> | ||
+ | |||
+ | <math>x^2 = 2 + 2\cdot\frac{1}{2}</math> | ||
+ | |||
+ | <math>x = \sqrt{3}</math> | ||
+ | |||
+ | |||
+ | Plugging that sum <math>x</math> back into the equation for <math>l</math>: | ||
+ | |||
+ | <math>l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2}} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144</math> | ||
+ | |||
+ | <math>l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144</math> | ||
+ | |||
+ | |||
+ | Thus, the quantity asked for is <math>144\cdot 5 = \boxed{720}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=11|num-a=13}} | {{AIME box|year=1990|num-b=11|num-a=13}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 16:39, 7 June 2011
Contents
[hide]Problem
A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form where , , , and are positive integers. Find .
Solution 1
The easiest way to do this seems to be to find the length of each of the sides and diagonals. To do such, draw the radii that meet the endpoints of the sides/diagonals; this will form isosceles triangles. Drawing the altitude of those triangles and then solving will yield the respective lengths.
- The length of each of the 12 sides is . .
- The length of each of the 12 diagonals that span across 2 edges is (or notice that the triangle formed is equilateral).
- The length of each of the 12 diagonals that span across 3 edges is (or notice that the triangle formed is a right triangle).
- The length of each of the 12 diagonals that span across 4 edges is .
- The length of each of the 12 diagonals that span across 5 edges is .
- The length of each of the 6 diameters is .
Adding all of these up, we get
. Thus, the answer is $144 \cdot 5 = \bosed{720}$ (Error compiling LaTeX. Unknown error_msg).
Solution 2
A second method involves drawing a triangle connecting the center of the 12-gon to two vertices of the 12-gon. Since the distance from the center to a vertex of the 12-gon is , the Law of Cosines can be applied to this isoceles triangle, to give:
There are six lengths of sides/diagonals, corresponding to
Call these lengths from shortest to longest. The total length that is asked for is
, noting that a_6 as written gives the diameter of the circle, which is the longest diagonal.
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}\right) + 144$ (Error compiling LaTeX. Unknown error_msg)
$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2}} + 1 + \frac{\sqrt{6}}{2} + \sqrt{1 + \frac{\sqrt{3}}{2}\right) + 144$ (Error compiling LaTeX. Unknown error_msg)
To simplify the two nested radicals, add them, and call the sum :
Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving :
$x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}}\right)\left(1 + \frac{\sqrt{3}}{2}}\right)$ (Error compiling LaTeX. Unknown error_msg)
Plugging that sum back into the equation for :
$l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2}} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144$ (Error compiling LaTeX. Unknown error_msg)
Thus, the quantity asked for is
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |