Mock Geometry AIME 2011 Problems/Problem 12

Problem

A triangle has the property that its sides form an arithmetic progression, and that the angle opposite the longest side is three times the angle opposite the shortest side. The ratio of the longest side to the shortest side can be expressed as $\frac{a+\sqrt{b}} {c}$ , where $a,b,c$ are positive integers, $b$ is not divisible by the square of any prime, and $a$ and $c$ are relatively prime. Find $a+b+c$ .

Solution

Let the triangle be $ABC$ , such that $AB=a, BC=a+d, CA=a+2d$ , and $\angle ACB=\theta$ , so that $\angle ABC=3\theta$ . Construct two points $X$ and $Y$ on $AC$ such that $\angle ABX=\angle XBY=\angle YBC=\theta=\angle BCA$ .

Since $\angle ABX=\angle ACB$ , $\triangle BAC\sim\triangle XAB$ . Therefore, $\frac{AB}{AX}=\frac{AC}{AB}$ , and $AX=\frac{AB^2}{AC}=\frac{a^2}{a+2d}$ . Now $CX=a+2d-\frac{a^2}{a+2d}=\frac{4ad+4d^2}{a+2d}$ .

Let $BY=CY=x$ . From Angle Bisector Theorem on $\triangle ABY$ with angle bisector $BX$ , $\frac{AX}{AB}=\frac{YX}{YB}$ , so $XY=\frac{AX\cdot YB}{AB}=\frac{\left(\frac{a^2}{a+2d}\right)(x)}{a}=\frac{a^2x}{a^2+2ad}$ .

Therefore, $\frac{4ad+4d^2}{a+2d}=CX=CY+YX=x+\frac{a^2x}{a^2+2ad}=(x)\left(\frac{2a^2+2ad}{a^2+2ad}\right)$ , and so $x= 2d$ . Now we have $XY=\frac{2ad}{a+2d}$ , and $BY=CY=x=2d$ .

Now we use Angle Bisector Theorem on $\triangle CBX$ with angle bisector $BY$ . We have $\frac{XB}{XY}=\frac{CB}{CY}$ , so $XB=\frac{CB\cdot XY}{CY}=\frac{(a+d)\left(\frac{2ad}{a+2d}\right)}{2d}=\frac{a^2+ad}{a+2d}$ .

Finally, notice that $\triangle BXY\sim\triangle CXB$ , so $\frac{BX}{BY}=\frac{CX}{CB}$ .

$\begin{align*} \frac{\frac{a^2+ad}{a+2d}}{2d}&=\frac{\frac{4ad+4d^2}{a+2d}}{a+d} \\ \frac{a^2+ad}{2d}&=4d \\ a^2+ad&=8d^2 \end{align*}$

Now, $a^2+(d)a+(-8d^2)=0$ , so from the Quadratic Formula, $a=\frac{-d+d\sqrt{33}}{2}$ (we neglect the negative root), and so $\frac{a}{d}=\frac{-1+\sqrt{33}}{2}$ .

Therefore, $\frac{AC}{AB}=\frac{a+2d}{a}=1+2\left(\frac{d}{a}\right)=1+2\left(\frac{1+\sqrt{33}}{16}\right)=\frac{9+\sqrt{33}}{8}$ , and the answer is $9+33+8=\boxed{050}$ .

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Mock Geometry AIME 2011 Problems/Problem 12

Problem

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