Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
(→‎Solution 2)
Line 20: Line 20:
 
==Solution 2==
 
==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:
+
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 isosceles triangle, to give:
  
 
<math>a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta</math>  
 
<math>a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta</math>  

Revision as of 16:13, 23 July 2011

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 1

1990 AIME-12.png

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 is $2 \cdot 12 = 24$.

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 $144 \cdot 5 = \bosed{720}$ (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 $12$, the Law of Cosines can be applied to this isosceles triangle, to give:

$a^2 = 12^2 + 12^2 - 2\cdot 12\cdot 12\cdot \cos \theta$

$a^2 = 2\cdot 12^2 - 2\cdot 12^2 \cos \theta$

$a^2 = 2\cdot 12^2 (1 - \cos \theta)$

$a = 12\sqrt{2} \cdot \sqrt{1 - \cos \theta}$

There are six lengths of sides/diagonals, corresponding to $\theta = {30^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 150^{\circ}, 180^{\circ}}$

Call these lengths $a_1, a_2, a_3, a_4, a_5, a_6$ from shortest to longest. The total length $l$ that is asked for is

$l = 12(a_1 + a_2 + a_3 + a_4 + a_5) + 6a_6$, noting that $a_6$ as written gives the diameter of the circle, which is the longest diagonal.


$l = 12[12\sqrt{2} (\sqrt{1 - \cos 30^{\circ}}+\sqrt{1-\cos 60^{\circ}}+\sqrt{1-\cos 90^{\circ}}+\sqrt{1-\cos 120^{\circ}}+\sqrt{1 - \cos 150^{\circ}})] + 6 \cdot 24$


$l = 144\sqrt{2} \left(\sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 - \frac{1}{2}} + \sqrt{1-0} + \sqrt{1 + \frac{1}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}\right) + 144$ (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 $x$:

$x = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}$

Squaring both sides, the F and L part of FOIL causes the radicals to cancel, leaving $2$:

$x^2 = 2 + 2\sqrt{\left(1 - \frac{\sqrt{3}}{2}}\right)\left(1 + \frac{\sqrt{3}}{2}}\right)$ (Error compiling LaTeX. Unknown error_msg)

$x^2 = 2 + 2\sqrt{1 - \frac{3}{4}}$

$x^2 = 2 + 2\sqrt{\frac{1}{4}}$

$x^2 = 2 + 2\cdot\frac{1}{2}$

$x = \sqrt{3}$


Plugging that sum $x$ back into the equation for $l$:

$l = 144\sqrt{2} \left(\frac{\sqrt{2}}{2}} + 1 + \frac{\sqrt{6}}{2} + \sqrt{3}\right) + 144$ (Error compiling LaTeX. Unknown error_msg)

$l = 144 + 144\sqrt{2} + 144\sqrt{3} + 144\sqrt{6} + 144$


Thus, the quantity asked for is $144\cdot 5 = \boxed{720}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_12&oldid=40614"