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2007 AMC 12A Problem 24

Problem 24 of 25HarderAlgebraGeometryCounting & Probability

For each integer n>1,n\gt 1, let F(n)F(n) be the number of solutions of the equation sinx=sinnx\sin x=\sin nx on the interval [0,π].[0,\pi]. What is n=22007F(n)?\displaystyle\sum_{n=2}^{2007}F(n)?

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Solution

On each interval where sinnx0,\sin nx\ge 0, the graphs of sinx\sin x and sinnx\sin nx meet twice, unless they share the value 11 there, in which case they meet once. Counting the humps and the endpoint at (π,0)(\pi,0) gives F(n)=n+1F(n)=n+1 when nn is even or n3(mod4),n\equiv 3\pmod 4, and F(n)=nF(n)=n when n1(mod4).n\equiv 1\pmod 4. Thus n=22007F(n)=n=22007(n+1)#{n1 ⁣ ⁣(mod4)}. \begin{aligned} &\sum_{n=2}^{2007}F(n) \\ &=\sum_{n=2}^{2007}(n+1) \\ &\quad {}-\#\{n\equiv 1\!\!\pmod 4\}. \end{aligned} The first sum is 2,017,033,2{,}017{,}033, and there are 501501 values n1(mod4)n\equiv 1\pmod 4 in the range, giving 2,017,033501=2,016,532.2{,}017{,}033-501=2{,}016{,}532. Thus, the correct answer is D.

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Concepts: trigonometry · counting intersections · summation

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