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: $\begin{aligned} f (23, x) & = \frac{\sin x + \sin 2 x + \sin 3 x + \dots + \sin 22 x + \sin 23 x}{\cos x + \cos 2 x + \cos 3 x + \dots + \cos 22 x + \cos 23 x} \\ = \frac{(\sin x + \sin 23 x) + (\sin 2 x + \sin 22 x) + \dots + (\sin 11 x + \sin 13 x) + \sin 12 x}{(\cos x + \cos 23 x) + (\cos 2 x + \cos 22 x) + \dots + (\cos 11 x + \cos 13 x) + \cos 12 x} \\ = \frac{2 \sin 12 x \cos 11 x + 2 \sin 12 x \cos 10 x + \dots + 2 \sin 12 x \cos x + \sin 12 x}{2 \cos 12 x \cos 11 x + 2 \cos 12 x \cos 10 x + \dots + 2 \cos 12 x \cos x + \cos 12 x} \\ = \frac{2 \sin 12 x (\cos 11 x + \cos 10 x + \cos 9 x + \dots + \cos x + 1)}{2 \cos 12 x (\cos 11 x + \cos 10 x + \dots + \cos x + 1)} \\ = \frac{\sin 12 x}{\cos 12 x} \\ = \tan 12 x \end{aligned}$

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 $f(n,x)$ : $\begin{aligned} f (n, x) & = \frac{\sin 0 x + \sin x + \sin 2 x + \sin 3 x + \dots + \sin (n - 1) x + \sin n x}{\cos 0 x + \cos x + \cos 2 x + \cos 3 x + \dots + \cos (n - 1) x + \cos n x - 1} \\ = \frac{\sum_{k = 0}^{n} ℑ (e^{i x})}{(\sum_{k = 0}^{n} ℜ (e^{i x})) - 1} \\ = \frac{ℑ (\sum_{k = 0}^{n} e^{i x})}{ℜ (\sum_{k = 0}^{n} e^{i x}) - 1} \\ = \frac{ℑ (\frac{1 - e^{(n + 1) i x}}{1 - e^{i x}})}{ℜ (\frac{1 - e^{(n + 1) i x}}{1 - e^{i x}}) - 1} \\ = \frac{ℑ (\frac{(1 - e^{(n + 1) i x}) (1 - e^{- i x})}{(1 - e^{i x}) (1 - e^{- i x})})}{ℜ (\frac{(1 - e^{(n + 1) i x}) (1 - e^{- i x})}{(1 - e^{i x}) (1 - e^{- i x})}) - 1} \\ = \frac{ℑ (\frac{1 - e^{(n + 1) i x} - e^{- i x} + e^{n i x}}{2 - e^{i x} - e^{- i x}})}{ℜ (\frac{1 - e^{(n + 1) i x} - e^{- i x} + e^{n i x}}{2 - e^{i x} - e^{- i x}}) - 1} \\ Note that, because sine is odd, & 2 - e^{i x} - e^{- i x} = 2 - \cos x - i \sin x - \cos x + i \sin x \in R . \\ ⟹ f (n, x) & = \frac{ℑ (1 - e^{(n + 1) i x} - e^{- i x} + e^{n i x})}{ℜ (1 - e^{(n + 1) i x} - e^{- i x} + e^{n i x}) - 2 + e^{i x} + e^{- i x}} \\ = \frac{\sin (n x) + \sin x - \sin ((n + 1) x)}{1 - \cos ((n + 1) x) - \cos x + \cos (n x) - 2 + 2 \cos x} \\ = \frac{\sin (n x) + \sin x - \sin ((n + 1) x)}{\cos (n x) + \cos x - \cos ((n + 1) x) - 1} \end{aligned}$ Now, using the sum-to-product identities and double-angle identities, we deduce that: $\begin{aligned} f (n, x) & = \frac{2 \sin (\frac{n + 1}{2} x) \cos (\frac{n - 1}{2} x) - 2 \sin (\frac{n + 1}{2} x) \cos (\frac{n + 1}{2} x)}{2 \cos (\frac{n + 1}{2} x) \cos (\frac{n - 1}{2} x) - 2 \cos^{2} (\frac{n + 1}{2} x) + 1 - 1} \\ = \frac{2 \sin (\frac{n + 1}{2} x) (\cos (\frac{n - 1}{2} x) + \cos (\frac{n + 1}{2} x))}{2 \cos (\frac{n + 1}{2} x) (\cos (\frac{n - 1}{2} x) + \cos (\frac{n + 1}{2} x))} \\ = \frac{\sin (\frac{n + 1}{2} x)}{\cos (\frac{n + 1}{2} x)} \\ = \tan (\frac{n + 1}{2} x) \end{aligned}$ 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 $[0,n]$ : $\begin{aligned} f (n, x) & = \frac{\sin x + \sin 2 x + \sin 3 x + \dots + \sin (n - 1) x + \sin n x}{\cos x + \cos 2 x + \cos 3 x + \dots + \cos (n - 1) x + \cos n x} \\ \approx \frac{n (\frac{1}{n} \int_{0}^{n} \sin (k x) d k)}{n (\frac{1}{n} \int_{0}^{n} \cos (k x) d k)} \\ = \frac{- (\frac{1}{x} \cos (k x)) |_{0}^{n}}{(\frac{1}{x} \sin (k x)) |_{0}^{n}} \\ = \frac{1 - \cos (n x)}{\sin (n x)} \end{aligned}$

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: $\begin{aligned} \frac{1 - \cos (23 x)}{\sin (23 x)} & = \frac{1 - \cos (33 x)}{\sin (33 x)} \\ \tan (\frac{23}{2} x) & = \tan (\frac{33}{2} x) = \tan (\frac{23}{2} x + 5 x) . \end{aligned}$

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: $108+144+180=\boxed{432}.$

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Mock AIME I 2012 Problems/Problem 10

Contents

Problem

Solution 1 (Sum-to-Product)

Solution 2 (Euler's Formula)

Solution 3 (Integral Calculus, non-rigorous)

See Also