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Difference between revisions of "2008 AIME II Problems/Problem 8"

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===Solution 1===
 
===Solution 1===
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By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a [[telescope|telescoping series]]:  
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By the [[trigonometric identity|product-to-sum identities]], we have that <math>2\cos a \sin b = \sin (a+b) - \sin (a-b)</math>. Therefore, this reduces to a telescope series:  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\
 
\sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\
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\end{align*}</cmath>
 
\end{align*}</cmath>
  
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Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. It easily follows that <math>n = \boxed{251}</math> is the smallest such integer.
+
Thus, we need <math>\sin \left(\frac{n(n+1)\pi}{2008}\right)</math> to be an integer; this can be only <math>\{-1,0,1\}</math>, which occur when <math>2 \cdot \frac{n(n+1)}{2008}</math> is an integer. Thus <math>1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1</math>. We know that <math>n</math> cannot be <math>250</math> as <math>250</math> isn't divisible by <math>4</math>, so 1004 doesn't divide <math>n(n+1) = 250 \cdot 251</math>. Therefore, it is clear that <math>n = \boxed{251}</math> is the smallest such integer.
  
 
===Solution 2===
 
===Solution 2===
−
We proceed with complex trigonometry. We know that for all <math>\theta</math>, we have <math>\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)</math> and <math>\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)</math> for some complex number <math>z</math>. Similarly, we have <math>\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)</math> and <math>\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)</math>. Thus, we have <math>\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)</math>
+
We proceed with complex trigonometry. We know that for all <math>\theta</math>, we have <math>\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)</math> and <math>\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)</math> for some complex number <math>z</math> on the unit circle. Similarly, we have <math>\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)</math> and <math>\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)</math>. Thus, we have <math>\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)</math>
  
 
<math>= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)</math>
 
<math>= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)</math>

Latest revision as of 17:26, 16 January 2023

Problem

Let $a = \pi/2008$. Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer.

Solution

Solution 1

By the product-to-sum identities, we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$. Therefore, this reduces to a telescope series: \begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\ &= -\sin(0) + \sin(n(n+1)a) = \sin(n(n+1)a) \end{align*}

Thus, we need $\sin \left(\frac{n(n+1)\pi}{2008}\right)$ to be an integer; this can be only $\{-1,0,1\}$, which occur when $2 \cdot \frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \cdot 251 | n(n+1) \Longrightarrow 251 | n, n+1$. We know that $n$ cannot be $250$ as $250$ isn't divisible by $4$, so 1004 doesn't divide $n(n+1) = 250 \cdot 251$. Therefore, it is clear that $n = \boxed{251}$ is the smallest such integer.

Solution 2

We proceed with complex trigonometry. We know that for all $\theta$, we have $\cos \theta = \dfrac{1}{2} \left( z + \dfrac{1}{z} \right)$ and $\sin \theta = \dfrac{1}{2i} \left( z - \dfrac{1}{z} \right)$ for some complex number $z$ on the unit circle. Similarly, we have $\cos n \theta = \dfrac{1}{2} \left( z^n + \dfrac{1}{z^n} \right)$ and $\sin n \theta = \dfrac{1}{2i} \left(z^n - \dfrac{1}{z^n} \right)$. Thus, we have $\cos n^2 a \sin n a = \dfrac{1}{4i} \left( z^{n^2} + \dfrac{1}{z^{n^2}} \right) \left( z^{n} - \dfrac{1}{z^n} \right)$

$= \dfrac{1}{4i} \left( z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} - z^{n^2 - n} + \dfrac{1}{z^{n^2 - n}} \right)$

$= \dfrac{1}{2} \left( \dfrac{1}{2i} \left(z^{n^2 + n} - \dfrac{1}{z^{n^2 + n}} \right) - \dfrac{1}{2i} \left(z^{n^2 - n} - \dfrac{1}{z^{n^2 - n}} \right) \right)$

$= \dfrac{1}{2} \left( \sin ((n^2 + n)a) - \sin ((n^2 - n)a) \right)$

$= \dfrac{1}{2} \left( \sin(((n+1)^2 - (n+1))a) - \sin((n^2 - n)a) \right)$

which clearly telescopes! Since the $2$ outside the brackets cancels with the $\dfrac{1}{2}$ inside, we see that the sum up to $n$ terms is

$\sin ((2^2 - 2)a) - \sin ((1^2 - 1)a) + \sin ((3^3 - 3)a) - \sin ((2^2 - 2)a) \cdots + \sin (((n+1)^2 - (n+1))a) - \sin ((n^2 - n)a)$

$= \sin (((n+1)^2 - (n+1))a) - \sin(0)$

$= \sin ((n^2 + n)a) - 0$

$= \sin \left( \dfrac{n(n+1) \pi}{2008} \right)$.

This expression takes on an integer value iff $\dfrac{2n(n+1)}{2008} = \dfrac{n(n+1)}{1004}$ is an integer; that is, $1004 \mid n(n+1)$. Clearly, $1004 = 2^2 \cdot 251$, implying that $251 \mid n(n+1)$. Since we want the smallest possible value of $n$, we see that we must have ${n,n+1} = 251$. If $n+1 = 251 \rightarrow n=250$, then we have $n(n+1) = 250(251)$, which is clearly not divisible by $1004$. However, if $n = 251$, then $1004 \mid n(n+1)$, so our answer is $\boxed{251}$.

It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.

See also

2008 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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