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2007 AMC 12A Problems/Problem 24

Revision as of 15:36, 16 September 2017 by Expilncalc (talk | contribs) (→‎Solution 1: Clarified the 1 (mod 4) situation.)

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

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.

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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