# 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 1

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 $60^o$ counter-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 $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120$ and $\angle XCA + \angle XBA = 90,$ so $\angle XCB+\angle XBC = 30$ and $\angle BXC = 150.$ Applying the law of cosines on triangle $BXC$ yields

$$BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}$$

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

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

## Note/Solution 2:

Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties:

We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let $\theta$ be the measure of angle $XAC$ so that angle $BAX$ has measure $60-\theta$. Let $AB=BC=AC=x$. The law of cosines on triangles $BAX$ and $XAC$ yields $\cos(60-\theta)=\frac{x^2+9}{10x}$ and $\cos\theta=\frac{x^2+16}{10x}$. Solving this system will yield the value of $x$. Since $\cos\theta=\frac{x^2+16}{10x}$ we have that $\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}$. Substituting these into the equation $\frac{x^2+9}{10x}=\cos(60-\theta)=\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta$ we obtain $\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}$. After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain $x^4-50x^2+193=0$ so that by the quadratic formula $x^2=25\pm12\sqrt{3}$. Under the hypothesis that $O$ lies in the interior of triangle $ABC$, $x^2$ must be $25+12\sqrt{3}$. To see this, note that the other value for $x^2$ is roughly $4.2$ so that $x\approx 2.05$, but since $AX=5$ and $AX\leq x$ we have a contradiction. We then obtain the area as in Solution 1.

Now, suppose $O$ does not lie in the interior of triangle $ABC$. We then obtain convex quadrilateral $OBAC$ with diagonals $CB$ and $OA$ intersecting at $X$. Here $AX=AB=AC=x$. We may let $\alpha$ denote the measure of angle $CAX$ so that angle $XAB$ measures $60-\alpha$. Note that the law of cosines on triangles $CXA$ and $BXA$ yield the same equations as in the first case with $\theta$ replaced with $\alpha$. Thus we obtain again $x^2=25\pm12\sqrt{3}$. If $x^2=25+12\sqrt{3}$ then $x\approx 6.8$, but this is impossible since $AX\leq 5$ but the shortest possible distance from $A$ to $X$ is the height of equilateral triangle $ABC$ which is $\approx6.8\sqrt{3}\approx5.8$; a contradiction. Hence in this case $x^2=25-12\sqrt{3}$. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).