# 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 $a + \tfrac{b}{c} \sqrt{d},$ 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 $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^o$ clockwise rotation about vertex $A$ maps $X$ 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 $XB = 3$ and $X'B = 4$ which tells us that angle $XBX'$ is $90$ because there is a $3$- $4$- $5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120,$ $\angle XCA + \angle XBA = 90,$ $\angle XCB+\angle XBC = 30,$ and $\angle BXC = 150.$ Applying the law of cosines on triangle $BXC$ yields $$BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 3^2+4^2-24 \cdot \cos(150) = 25+12\sqrt{3}$$

and thus the area of $ABC$ equals $$BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.$$

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