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>? | 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> | ||
− | == Solution == | + | == Solution 1 == |
<math>F(2)=3</math> | <math>F(2)=3</math> | ||
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<math>F(n) = n + 1</math> | <math>F(n) = n + 1</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), so we get here <math>F(n) = n</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> | <math>3+4+5+5+7+8+9+9+\cdots+2008</math> | ||
<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> | ||
+ | |||
+ | == Solution 2 == | ||
+ | <cmath> | ||
+ | \sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right) | ||
+ | </cmath> | ||
+ | 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 | ||
+ | <cmath> | ||
+ | \frac {2j + 1}{n + 1} = \frac {2k}{n - 1} | ||
+ | </cmath> | ||
+ | |||
+ | <cmath> | ||
+ | k = \frac {(2j + 1)(n - 1)}{2(n + 1)} | ||
+ | </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>. | ||
+ | |||
+ | 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. | ||
== See also == | == See also == | ||
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[[Category:Introductory Trigonometry Problems]] | [[Category:Introductory Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 02:12, 16 February 2021
Contents
Problem
For each integer , let be the number of solutions to the equation on the interval . What is ?
Solution 1
By looking at various graphs, we obtain that, for most of the graphs
Notice that the solutions are basically reflections across . However, when , 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 .
Solution 2
So if and only if or .
The first occurs whenever , or for some nonnegative integer . Since , . So there are solutions in this case.
The second occurs whenever , or for some nonnegative integer . Here so that there are solutions here.
However, we overcount intersections. These occur whenever
which is equivalent to dividing . If is even, then is odd, so this never happens. If , 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 . In this case, the divisibility becomes dividing . Since and are relatively prime (subtracting twice the second number from the first gives 1), must divide . Since , . Then there is only one intersection, namely when .
Therefore we find is equal to , unless , in which case it is one less, or . The problem may then be finished as in Solution 1.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
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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