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 1

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 $9+33+8=\boxed{050}$ .

Solution 2 (Trig)

Let the sides of the triangle be $a$ , $a + k$ , and $a + 2k$ , since they form arithmetic progression. Let the their corresponding opposite angles be $X, 180 - 4X,$ and $3X$ respectively. Using Law of Sines, we find that:

$\frac{\sin{X}}{a} = \frac{\sin{3X}}{a + 2k}$ .

Using the identity $\sin{3X} = \sin{X} (4 \cos^2{X} - 1)$ , we find that $k = a(2 \cos^2{X} - 1)$ . Applying double angle, we find $k = a \cos{2X}$ .

Then we use Law of Sines again, and use the substitution for $k$ , given that $\sin{4X} = \sin{180 - 4X}$ .

$\frac{\sin{3X}}{a + 2a \cos{2X}} = \frac{\sin{4X}}{a + a \cos{2X}}$ .

Then factoring out $a$ and canceling it we get:

$\frac{\sin{3X}}{1 + 2 \cos{2X}} = \frac{\sin{4X}}{1 + \cos{2X}}$ .

Cross multiplying yields:

$\sin{3X} (1 + \cos{2X}) = \sin{4X} (1 + 2 \cos{2X})$

Given that $\cos{2X} = 2 \cos^2{X} - 1$ , and the earlier identity for $\sin {3X}$ we get:

$2 \sin{X} (4 \cos^2{X} - 1) (\cos^2{X}) = \sin{4X} (4 \cos^2{X} - 1)$

We can safely cancel $4 \cos^2{X} - 1$ since the solution, $60^\circ$ would go out of range for the triangle sum of angles.

Expanding $\sin{4X}$ in the ensuing expression,

$2 \sin{X} \cos^2{X} = 4 \sin{X} \cos{X} \cos{2X}$

Cancelling $\sin{X}$ and expanding $\cos{2X}$ yields a cubic equal to $0$ with GCD $\cos{X}$ . Since $90^\circ$ would not work either, we are left with a quadratic. Solving for $\cos{X}$ , and neglecting the negative root since it would go out of bounds too, we find $\cos{X} = \frac{1 + \sqrt{33}}{8}$ .

The problem asks for the ratio of $a + 2k$ to $a$ , using Law of Sines, this is $\frac{\sin{3X}}{\sin{X}}$ . This equals $4 \cos^2{X} - 1$ . Substituting value of $\cos{X}$ , we find the ratio as $\frac{9 + \sqrt{33}}{8}$ , or $\boxed{050}$ .

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Mock Geometry AIME 2011 Problems/Problem 12

Problem

Solution 1

Solution 2 (Trig)