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 1 == | |
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<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> | ||
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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> | ||
− | + | == 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: | ||
+ | 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 20:21, 27 March 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.
Note from Williamgolly: An easier way to see that there is an extra point of intersection at is that so we must have since Therefore, we suspect that 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 |
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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