2012 AIME I Problems/Problem 13
Problem 13
Three concentric circles have radii 
and 
An equilateral triangle with one vertex on each circle has side length 
The largest possible area of the triangle can be written as 
where 
and 
are positive integers, 
and 
are relatively prime, and is not divisible by the square of any prime. Find 
Solution
Reinterpret the problem in the following manner. Equilateral triangle 
has a point 
on the interior such that 
and 
A 
clockwise rotation about vertex 
maps to 
and 
to 
Note that angle 
is 
and 
which tells us that triangle is equilateral and that 
We now notice that 
and 
which tells us that angle 
is 
because there is a 
-
-
Pythagorean triple. Now note that 
and 
so 
and 
Applying the law of cosines on triangle 
yields
![\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]](http://latex.artofproblemsolving.com/d/0/8/d08a21af86bf32611c4d590163cff136b6311881.png)
and thus the area of equals ![\[BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.\]](http://latex.artofproblemsolving.com/3/a/5/3a50625e04ad7fd466ccb6d3e6fd52aebbb5bb13.png)
so our final answer is 
See also
| 2012 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
