Difference between revisions of "2007 AMC 12A Problems/Problem 24"

(hmm.. I thought 1 mod 4 something weird happens)
(Solution: rigorous proof by adalton)
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<math>= (1+2+3+4+5+\cdots+2008) - 3 - 501</math>
 
<math>= (1+2+3+4+5+\cdots+2008) - 3 - 501</math>
 
<math>= \frac{(2008)(2009)}{2} - 504 = 2016532</math>  <math>\mathrm{(D)}</math>
 
<math>= \frac{(2008)(2009)}{2} - 504 = 2016532</math>  <math>\mathrm{(D)}</math>
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 +
----
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<cmath>
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\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)
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</cmath>
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So <math>\sin nx = \sin x</math> if and only if <math>\cos \frac {n + 1}{2}x = 0</math> or <math>\sin \frac {n - 1}{2}x = 0</math>.
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The first occurs whenever <math>\frac {n + 1}{2}x = (j + 1/2)\pi</math>, or <math>x = \frac {(2j + 1)\pi}{n + 1}</math> for some nonnegative integer <math>j</math>. Since <math>x\leq \pi</math>, <math>j\leq n/2</math>. So there are <math>1 + \lfloor n/2 \rfloor</math> solutions in this case.
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The second occurs whenever <math>\frac {n - 1}{2}x = k\pi</math>, or <math>x = \frac {2k\pi}{n - 1}</math> for some nonnegative integer <math>k</math>. Here <math>k\leq \frac {n - 1}{2}</math> so that there are <math>\left\lfloor \frac {n + 1}{2}\right\rfloor</math> solutions here.
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However, we overcount intersections. These occur whenever
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<cmath>
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\frac {2j + 1}{n + 1} = \frac {2k}{n - 1}
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</cmath>
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<cmath>
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k = \frac {(2j + 1)(n - 1)}{2(n + 1)}
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</cmath>
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which is equivalent to <math>2(n + 1)</math> dividing <math>(2j + 1)(n - 1)</math>. If <math>n</math> is even, then <math>(2j + 1)(n - 1)</math> is odd, so this never happens. If <math>n\equiv 3\pmod{4}</math>, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.
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 +
This leaves <math>n\equiv 1\pmod{4}</math>. In this case, the divisibility becomes <math>\frac {n + 1}{2}</math> dividing <math>(2j + 1)\frac {n - 1}{4}</math>. Since <math>\frac {n + 1}{2}</math> and <math>\frac {n - 1}{4}</math> are relatively prime (subtracting twice the second number from the first gives 1), <math>\frac {n + 1}{2}</math> must divide <math>2j + 1</math>. Since <math>j\leq \frac {n - 1}{2}</math>, <math>2j + 1\leq n < 2\cdot \frac {n + 1}{2}</math>. Then there is only one intersection, namely when <math>j = \frac {n - 1}{4}</math>.
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Therefore we find <math>F(n)</math> is equal to <math>1 + \lfloor n/2 \rfloor + \left \lfloor \frac {n + 1}{2}\right\rfloor = n + 1</math>, unless <math>n\equiv 1\pmod{4}</math>, in which case it is one less, or <math>n</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:47, 28 September 2007

Problem

For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\sin{x}=\sin{(nx)}$ on the interval $[0,\pi]$. What is $\sum_{n=2}^{2007} F(n)$?

$\mathrm{(A)}\ 2014524$ $\mathrm{(B)}\ 2015028$ $\mathrm{(C)}\ 2015033$ $\mathrm{(D)}\ 2016532$ $\mathrm{(E)}\ 2017033$

Solution

$F(2)=3$

By looking at various graphs, we obtain that, for most of the graphs

$F(n) = n + 1$

However, when $n \equiv 1 \pmod{4}$, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two), so we get here $F(n) = n$.

$3+4+5+5+7+8+9+9+\cdots+2008$ $= (1+2+3+4+5+\cdots+2008) - 3 - 501$ $= \frac{(2008)(2009)}{2} - 504 = 2016532$ $\mathrm{(D)}$


\[\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)\] So $\sin nx = \sin x$ if and only if $\cos \frac {n + 1}{2}x = 0$ or $\sin \frac {n - 1}{2}x = 0$.

The first occurs whenever $\frac {n + 1}{2}x = (j + 1/2)\pi$, or $x = \frac {(2j + 1)\pi}{n + 1}$ for some nonnegative integer $j$. Since $x\leq \pi$, $j\leq n/2$. So there are $1 + \lfloor n/2 \rfloor$ solutions in this case.

The second occurs whenever $\frac {n - 1}{2}x = k\pi$, or $x = \frac {2k\pi}{n - 1}$ for some nonnegative integer $k$. Here $k\leq \frac {n - 1}{2}$ so that there are $\left\lfloor \frac {n + 1}{2}\right\rfloor$ solutions here.

However, we overcount intersections. These occur whenever \[\frac {2j + 1}{n + 1} = \frac {2k}{n - 1}\]

\[k = \frac {(2j + 1)(n - 1)}{2(n + 1)}\] which is equivalent to $2(n + 1)$ dividing $(2j + 1)(n - 1)$. If $n$ is even, then $(2j + 1)(n - 1)$ is odd, so this never happens. If $n\equiv 3\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4.

This leaves $n\equiv 1\pmod{4}$. In this case, the divisibility becomes $\frac {n + 1}{2}$ dividing $(2j + 1)\frac {n - 1}{4}$. Since $\frac {n + 1}{2}$ and $\frac {n - 1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $\frac {n + 1}{2}$ must divide $2j + 1$. Since $j\leq \frac {n - 1}{2}$, $2j + 1\leq n < 2\cdot \frac {n + 1}{2}$. Then there is only one intersection, namely when $j = \frac {n - 1}{4}$.

Therefore we find $F(n)$ is equal to $1 + \lfloor n/2 \rfloor + \left \lfloor \frac {n + 1}{2}\right\rfloor = n + 1$, unless $n\equiv 1\pmod{4}$, in which case it is one less, or $n$.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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