2007 AMC 12A Problems/Problem 24
By looking at various graphs, we obtain that, for most of the graphs
Notice that the solutions are basically reflections across . However, when , 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 .
So if and only if or .
The first occurs whenever , or for some nonnegative integer . Since , . So there are solutions in this case.
The second occurs whenever , or for some nonnegative integer . Here so that there are solutions here.
However, we overcount intersections. These occur whenever
which is equivalent to dividing . If is even, then is odd, so this never happens. If , 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 . In this case, the divisibility becomes dividing . Since and are relatively prime (subtracting twice the second number from the first gives 1), must divide . Since , . Then there is only one intersection, namely when .
Therefore we find is equal to , unless , in which case it is one less, or . 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 is that so we must have since Therefore, we suspect that has an extra point of intersection. Testing, we see this is true.
|2007 AMC 12A (Problems • Answer Key • Resources)|
|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|