# 2012 AIME I Problems/Problem 13

## Problem 13

Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \frac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

## Solution

Reinterpret the problem in the following manner. Equilateral triangle ABC has a point X on the interior such that AX = 5, BX = 4, CX = 3. A 60 degree clockwise (or counter clockwise) rotation about vertex A maps B to B', X to X' and C to C'. Note that angle XAX' is 60 and XA = X'A = 5 which tells us that triangle XAX' is equilateral and that XX' = 5. We now notice that XB = 3, X'B = 4 which tells us that angle XBX' is 90 because there is a 3,4,5 pythagorean triple. Now note that angle ABC + angle ACB = 120, angle XCA + angle XBA = 90, angle XCB+angle XBC = 30, angle BXC = 150. Applying the law of cosines on triangle BXC yields

$BX^2+CX^2-2*BX*CX*cos(150) = BC^2 = 3^2+4^2-24*cos(150) = 25+12\sqrt{3}$ and since we are looking for the area we have $BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9$

so our final answer is 3+4+25+9 = $\fbox{041}$.