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Mock AIME I 2012 Problems/Problem 10

Problem

Consider the function $f(n,x) = \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}}$. Find the sum of all $x$ for which $f(23,x)=f(33,x)$, where $x$ is measured in degrees and $100<x<200$.

Solution 1 (Sum-to-Product)

Recalling the trigonometric sum-to-product identities, we can rearrange terms and evaluate $f(23,x)$ as follows: \begin{align*} f(23, x) &= \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{22x} + \sin{23x}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{22x} + \cos{23x}} \\ &= \dfrac{(\sin{x} + \sin{23x}) + (\sin{2x} + \sin{22x}) + \cdots + (\sin{11x} + \sin{13x}) + \sin{12x}}{(\cos{x} + \cos{23x}) + (\cos{2x} + \cos{22x}) + \cdots + (\cos{11x} + \cos{13x}) + \cos{12x}} \\ &= \dfrac{2\sin{12x}\cos{11x}+2\sin{12x}\cos{10x}+\cdots+2\sin{12x}\cos{x}+\sin{12x}}{2\cos{12x}\cos{11x}+2\cos{12x}\cos{10x}+\cdots+2\cos{12x}\cos{x}+\cos{12x}} \\ &= \dfrac{2\sin{12x}(\cos{11x}+\cos{10x}+\cos{9x}+\cdots+\cos{x}+1)}{2\cos{12x}(\cos{11x}+\cos{10x}+\cdots+\cos{x}+1)} \\ &= \dfrac{\sin{12x}}{\cos{12x}} \\ &= \tan{12x} \end{align*}

Likewise, we can show that \[f(33,x)=\dfrac{\sin{17x}}{\cos{17x}}=\tan{17x}.\]

Now, we desire to find all $x\in(100^{\circ},200^{\circ})$ that satisfy the following equation: \[f(22,x)=f(33,x)\iff\tan{12x}=\tan{17x}.\] Because $\tan x$ has a period of $180^{\circ}$ and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that $5x$ must then be some integer multiple of tangent's period, $180^{\circ}$. Thus, $x$ must be a multiple of $\tfrac{180^{\circ}}5=36^{\circ}$, and so the possible values of $x$ between $100^{\circ}$ and $200^{\circ}$ are $108^{\circ}$, $144^{\circ}$, and $180^{\circ}$.

Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer: \[108+144+180=\boxed{432}.\]

Solution 2 (Euler's Formula)

Recall that, by Euler's Formula, \[e^{ix}=\cos x+i\sin x.\] Therefore, $\cos x= \Re(e^{ix})$, and $\sin x=\Im(e^{ix})$, where $\Re$ and $\Im$ denote the real and imaginary parts, respectively, of a given complex number.

Using this fact and the geometric series formula, we can now manipulate our expression for $f(n,x)$: \begin{align*} f(n,x) &= \frac{\sin{0x} + \sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{0x} + \cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx} - 1} \\ &= \frac{\sum_{k=0}^n\Im(e^{ix})}{\left(\sum_{k=0}^n\Re(e^{ix})\right)-1} \\ &= \frac{\Im(\sum_{k=0}^ne^{ix})}{\Re(\sum_{k=0}^ne^{ix})-1} \\ &= \frac{\Im\left(\frac{1-e^{(n+1)ix}}{1-e^{ix}}\right)}{\Re\left(\frac{1-e^{(n+1)ix}}{1-e^{ix}}\right)-1} \\ &= \frac{\Im\left(\frac{(1-e^{(n+1)ix})(1-e^{-ix})}{(1-e^{ix})(1-e^{-ix})}\right)}{\Re\left(\frac{(1-e^{(n+1)ix})(1-e^{-ix})}{(1-e^{ix})(1-e^{-ix})}\right)-1} \\ &= \frac{\Im\left(\frac{1-e^{(n+1)ix}-e^{-ix}+e^{nix}}{2-e^{ix}-e^{-ix}}\right)}{\Re\left(\frac{1-e^{(n+1)ix}-e^{-ix}+e^{nix}}{2-e^{ix}-e^{-ix}}\right)-1} \\ \text{Note that, because sine is odd, }&2-e^{ix}-e^{-ix}=2-\cos x -i\sin x - \cos x +i\sin x \in \mathbb{R}.\\ \implies f(n,x) &= \frac{\Im(1-e^{(n+1)ix}-e^{-ix}+e^{nix})}{\Re(1-e^{(n+1)ix}-e^{-ix}+e^{nix})-2+e^{ix}+e^{-ix}} \\ &= \frac{\sin(nx)+\sin x-\sin((n+1)x)}{1-\cos((n+1)x)-\cos x+\cos(nx)-2+2\cos x} \\ &= \frac{\sin(nx)+\sin x-\sin((n+1)x)}{\cos(nx)+\cos x-\cos((n+1)x)-1} \end{align*} Now, using the sum-to-product identities and double-angle identities, we deduce that: \begin{align*} f(n,x) &= \frac{2\sin\left(\frac{n+1}2x\right)\cos\left(\frac{n-1}2x\right)-2\sin\left(\frac{n+1}2x\right)\cos\left(\frac{n+1}2x\right)}{2\cos\left(\frac{n+1}2x\right)\cos\left(\frac{n-1}2x\right)-2\cos^2\left(\frac{n+1}2x\right)+1-1} \\ &= \frac{2\sin\left(\frac{n+1}2x\right)(\cos\left(\frac{n-1}2x\right)+\cos\left(\frac{n+1}2x\right))}{2\cos\left(\frac{n+1}2x\right)(\cos\left(\frac{n-1}2x\right)+\cos\left(\frac{n+1}2x\right))} \\ &= \frac{\sin\left(\frac{n+1}2x\right)}{\cos\left(\frac{n+1}2x\right)} \\ &= \tan\left(\frac{n+1}2x\right) \end{align*} Now, we know that $f(23,x)=\tan{12x}$, and $f(33,x)=\tan{17x}$. From here, we proceed as in Solution 1 to obtain our answer of $\boxed{432}$.

Solution 3 (Integral Calculus, non-rigorous)

Recall that the average value of some function $g(x)$ over the interval $[a,b]$ is given by \[\frac1{b-a}\int_a^b g(x)dx.\] Then, we can approximate (hopefully well enough) the above expression for $f(n,x)$ by multiplying $n$ by the average of the function $\sin(kx)$ over the interval $[0,n]$: \begin{align*} f(n,x) &= \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}} \\ &\approx \dfrac{n(\frac1n\int_0^n\sin(kx)dk)}{n(\frac1n\int_0^n\cos(kx)dk)} \\ &= \dfrac{-(\frac1x\cos(kx))|^n_0}{(\frac1x\sin(kx))|^n_0} \\ &= \dfrac{1-\cos(nx)}{\sin(nx)} \end{align*}

Now, we desire to find $x\in(100^{\circ},200^{\circ})$ that satisfy the equation: \[f(23,x)=f(33,x)\iff\frac{1-\cos(23x)}{\sin(23x)}=\frac{1-\cos(33x)}{\sin(33x)}.\] Recalling the tangent half-angle identities, we can solve as follows: \begin{align*} \frac{1-\cos(23x)}{\sin(23x)} &= \frac{1-\cos(33x)}{\sin(33x)} \\ \tan\left(\frac{23}2x\right) &= \tan\left(\frac{33}2x\right) = \tan\left(\frac{23}2x+5x\right). \end{align*}

Now, because $\tan x$ has a period of $180^{\circ}$ and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that $5x$ must then be some integer multiple of tangent's period, $180^{\circ}$. Thus, $x$ must be a multiple of $\tfrac{180^{\circ}}5=36^{\circ}$, and so the possible values of $x$ between $100^{\circ}$ and $200^{\circ}$ are $108^{\circ}$, $144^{\circ}$, and $180^{\circ}$.

Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer: \[108+144+180=\boxed{432}.\]

See Also

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