2007 AMC 12A Problems/Problem 24

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.

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2007 AMC 12A Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

See also