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

(Problem)
(hmm.. I thought 1 mod 4 something weird happens)
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== Problem ==
 
== 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>?
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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>?
 
 
 
<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>
 
<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>F(2)=3</math>
 
<math>F(2)=3</math>
  
By looking at various graphs, we obtain that
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By looking at various graphs, we obtain that, for most of the graphs
  
<math>F(n+1)=F(n)+1</math>
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<math>F(n) = n + 1</math>
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 +
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), so we get here <math>F(n) = n</math>.
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<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>
  
<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}}
  
[[Category:Trigonometry Problems]]
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[[Category:Introductory Trigonometry Problems]]

Revision as of 20:45, 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 $F(n) = n$.

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

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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