Difference between revisions of "1990 AIME Problems/Problem 12"
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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 = \ | + | <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 = \boxed{720}</math>. |
==Solution 2== | ==Solution 2== | ||
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− | <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}} + \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 | + | <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> |
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Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving <math>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 | + | <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{1 - \frac{3}{4}}</math> | ||
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<math>x^2 = 2 + 2\cdot\frac{1}{2}</math> | <math>x^2 = 2 + 2\cdot\frac{1}{2}</math> | ||
− | <math>x = \sqrt{3}</math> | + | <math>x = \sqrt{3}.</math> |
− | Plugging that sum <math>x</math> back into the equation for <math>l</math> | + | Plugging that sum <math>x</math> back into the equation for <math>l</math>, we find |
− | <math>l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2 | + | <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> | + | <math>l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144</math>. |
− | Thus, the quantity | + | Thus, the desired quantity is <math>144\cdot 5 = \boxed{720}</math>. |
== See also == | == See also == |
Revision as of 16:04, 13 March 2015
Contents
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 .
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 isosceles 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 as written gives the diameter of the circle, which is the longest diagonal.
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 :
Plugging that sum back into the equation for , we find
.
Thus, the desired quantity 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.