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Difference between revisions of "1990 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
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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
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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>.
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<center><math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math></center>
 
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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 ==
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{{solution}}
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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 [[radius|radii]] that meet the endpoints of the sides/diagonals; this will form [[isosceles triangle]]s. Drawing the [[altitude]] of those [[triangle]]s and then solving will yield the respective lengths.
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*The length of each of the 12 sides is <math>2 \cdot 12\sin 15</math>. <math>24\sin 15 = 24\sin (45 - 30) = 24\frac{\sqrt{6} - \sqrt{2}}{4} = 6(\sqrt{6} - \sqrt{2})</math>.
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*The length of each of the 12 diagonals that span across 2 edges is <math>2 \cdot 12\sin 30 = 12</math> (or notice that the triangle formed is [[equilateral triangle|equilateral]]).
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*The length of each of the 12 diagonals that span across 3 edges is <math>2 \cdot 12\sin 45 = 12\sqrt{2}</math> (or notice that the triangle formed is a <math>45 - 45 - 90</math> [[right triangle]]).
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*The length of each of the 12 diagonals that span across 4 edges is <math>2 \cdot 12\sin 60 = 12\sqrt{3}</math>.
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*The length of each of the 12 diagonals that span across 5 edges is <math>2 \cdot 12\sin 75 = 24\sin (45 + 30) = 24\frac{\sqrt{6}+\sqrt{2}}{4} = 6(\sqrt{6}+\sqrt{2})</math>.
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*The length of each of the 6 [[diameter]]s are <math>2 \cdot 12 = 24</math>.
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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 = 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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1990|num-b=11|num-a=13}}
 
{{AIME box|year=1990|num-b=11|num-a=13}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 11:55, 3 March 2007

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 $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},$ where $a^{}_{}$, $b^{}_{}$, $c^{}_{}$, and $d^{}_{}$ are positive integers. Find $a + b + c + d^{}_{}$.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $2 \cdot 12\sin 15$. $24\sin 15 = 24\sin (45 - 30) = 24\frac{\sqrt{6} - \sqrt{2}}{4} = 6(\sqrt{6} - \sqrt{2})$.
  • The length of each of the 12 diagonals that span across 2 edges is $2 \cdot 12\sin 30 = 12$ (or notice that the triangle formed is equilateral).
  • The length of each of the 12 diagonals that span across 3 edges is $2 \cdot 12\sin 45 = 12\sqrt{2}$ (or notice that the triangle formed is a $45 - 45 - 90$ right triangle).
  • The length of each of the 12 diagonals that span across 4 edges is $2 \cdot 12\sin 60 = 12\sqrt{3}$.
  • The length of each of the 12 diagonals that span across 5 edges is $2 \cdot 12\sin 75 = 24\sin (45 + 30) = 24\frac{\sqrt{6}+\sqrt{2}}{4} = 6(\sqrt{6}+\sqrt{2})$.
  • The length of each of the 6 diameters are $2 \cdot 12 = 24$.

Adding all of these up, we get $12[6(\sqrt{6} - \sqrt{2}) + 12 + 12\sqrt{2} + 12\sqrt{3} + 6(\sqrt{6}+\sqrt{2})] + 6 \cdot 24 = 12(12 + 12\sqrt{2} + 12\sqrt{3} + 12\sqrt{6}) + 144 = 288 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6}$. Thus, the answer is $144 \cdot 5 = 720$.

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
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