Solution to Problem #14
Douglas Grimm at the Australian National University, Phillipe Fondanaiche
of Paris (France), Chris Welty of London (England), Dennis Franciskovich
of London (England), D.K. Trenner of Loveland CO, Matt Hudelson of
Washington State University, Ross Millikan of San Mateo CA, and Brian
Glover of St. Paul MN solved the problem. Dennis Franciskovich notes that
this problem is similar to one on the 1981 AHSME. Here is Dennis
Franciskovich's solution:
The answer is 1275. The solution is as follows:
We rewrite the equation as x/2000 = sin x , and see that we are looking
for the number of times that y = x/2000 and y = sin x intersect. We
first note that x = 0 is a solution. We also note that if x = a is a
solution, then so is x = -a (because sin(-a) = -sin a).
Positive solutions must be less than or equal to 2000, because abs(x) =
2000 abs(sin x) <= 2000. So we determine the number of cycles that sin x
completes from x = 0 to x = 2000, and calculate the number of times the
graphs intersect. 2000/(2*pi) = 318.3..., and between x = 318*2*pi and x
= 2000, the graphs intersect twice (this is the first third or so of the
sine cycle), and for each whole cycle the graphs also intersect twice, so
the number of intersections between 0 and 2000 inclusive is 2*319 = 638.
We double again to count the negative solutions, and then subtract one
because we have counted x = 0 twice. So the number of real solutions is
2*638 - 1 = 1275.
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