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

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== Problem ==
 
For each [[integer]] <math>n>1</math>, let <math>F(n)</math> be the number of solutions to the [[equation]] <math>\sin{x}=\sin{(nx)}</math> on the interval <math>[0,\pi]</math>. What is <math>\sum_{n=2}^{2007} F(n)</math>?
 
For each [[integer]] <math>n>1</math>, let <math>F(n)</math> be the number of solutions to the [[equation]] <math>\sin{x}=\sin{(nx)}</math> on the interval <math>[0,\pi]</math>. What is <math>\sum_{n=2}^{2007} F(n)</math>?
 
 
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<math>\mathrm{(A)}\ 2014524</math>  <math>\mathrm{(B)}\ 2015028</math>  <math>\mathrm{(C)}\ 2015033</math>  <math>\mathrm{(D)}\ 2016532</math>  <math>\mathrm{(E)}\ 2017033</math>
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<math>\mathrm{(A)}\ 2014524\qquad\mathrm{(B)}\ 2015028\qquad\mathrm{(C)}\ 2015033\qquad\mathrm{(D)}\ 2016532\qquad\mathrm{(E)}\ 2017033</math>
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== Solution ==
 
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===Solution 1===
 
  
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== Solution 1 ==
 
<math>F(2)=3</math>
 
<math>F(2)=3</math>
  
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<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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===Solution 2===
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== Solution 2 ==
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<cmath>
 
<cmath>
 
\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)
 
\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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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>. The problem may then be finished as in Solution 1.
 
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>. The problem may then be finished as in Solution 1.
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Note from Williamgolly:
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An easier way to see that there is an extra point of intersection at <math>n \equiv 1 \pmod{4}</math> is that <math>v_2(2j+1)=0,</math> so we must have <math>v_2(n-1)>v_2(n+1)</math> since <math>v_2(2k) \geq 1 >v_2(2j+1).</math>  Therefore, we suspect that <math>n \equiv 1 \pmod{4}</math> has an extra point of intersection.  Testing, we see this is true.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:40, 23 August 2023

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 $\sum_{n=2}^{2007} F(n)$?

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

Solution 1

$F(2)=3$

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

$F(n) = n + 1$

Notice that the solutions are basically reflections across $x = \frac{\pi}{2}$. 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 reflected points), so we get here $F(n) = n$.

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

Solution 2

\[\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)\] 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 $n$. The problem may then be finished as in Solution 1.

Note from Williamgolly: An easier way to see that there is an extra point of intersection at $n \equiv 1 \pmod{4}$ is that $v_2(2j+1)=0,$ so we must have $v_2(n-1)>v_2(n+1)$ since $v_2(2k) \geq 1 >v_2(2j+1).$ Therefore, we suspect that $n \equiv 1 \pmod{4}$ has an extra point of intersection. Testing, we see this is true.

See also

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 12 Problems and Solutions

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