Problem:
For each integer n>1, let F(n) be the number of solutions of the equation sinx=sinnx on the interval [0,π]. What is ∑n=22007F(n)?
Answer Choices:
A. 2,014,524
B. 2,015,028
C. 2,015,033
D. 2,016,532
E. 2,017,033
Solution:
Note that F(n) is the number of points at which the graphs of y=sinx and y=sinnx intersect on [0,π]. For each n,sinnx≥0 on each interval [(2k−2)π/n,(2k−1)π/n] where k is a positive integer and 2k−1≤n. The number of such intervals is n/2 if n is even and (n+1)/2 if n is odd. The graphs intersect twice on each interval unless sinx=1=sinnx at some point in the interval, in which case the graphs intersect once. This last equation is satisfied if and only if n≡1(mod4) and the interval contains π/2. If n is even, this count does not include the point of intersection at (π,0). Therefore F(n)=2(n/2)+1=n+1 if n is even, F(n)=2(n+1)/2=n+1 if n≡3 (mod4), and F(n)=n if n≡1(mod4). Hence
n=2∑2007F(n)=(n=2∑2007(n+1))−⌊42007−1⌋=2(2006)(3+2008)−501=2,016,532.
The problems on this page are the property of the MAA's American Mathematics Competitions