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

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== Problem ==
 
== Problem ==
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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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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 ==
 
<math>F(2)=3</math>
 
<math>F(2)=3</math>
  
−
By looking at various graphs, we obtain that
+
By looking at various graphs, we obtain that, for most of the graphs
 +
 
 +
<math>F(n) = n + 1</math>
 +
 
 +
Notice that the solutions are basically reflections across <math>x = \frac{\pi}{2}</math>.
 +
However, when <math>n \equiv 1 \pmod{4}</math>, 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 <math>F(n) = n</math>.
 +
 
 +
<math>3+4+5+5+7+8+9+9+\cdots+2008</math>
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<math>= (1+2+3+4+5+\cdots+2008) - 3 - 501</math>
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<math>= \frac{(2008)(2009)}{2} - 504 = 2016532</math>  <math>\mathrm{(D)}</math>
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 +
== Solution 2 ==
 +
<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>.
 +
 
 +
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.
 +
 
 +
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.
 +
 
 +
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>
 +
 
 +
<cmath>
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k = \frac {(2j + 1)(n - 1)}{2(n + 1)}
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</cmath>
 +
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.
 +
 
 +
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>. The problem may then be finished as in Solution 1.
  
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<math>F(n+1)=F(n)+1</math>
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Note from Williamgolly:
 +
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.
  
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<math>3+4+5+\cdots+2008=2017033</math>  <math>\mathrm{(E)}</math>
 
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2007|num-b=23|num-a=25|ab=A}}
 
{{AMC12 box|year=2007|num-b=23|num-a=25|ab=A}}
  
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[[Category:Trigonometry Problems]]
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[[Category:Introductory Trigonometry Problems]]
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{{MAA Notice}}

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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