Difference between revisions of "2007 AMC 12A Problems/Problem 24"
Einstein00 (talk | contribs) (→Solution) |
(→See also) |
||
Line 51: | Line 51: | ||
[[Category:Introductory Trigonometry Problems]] | [[Category:Introductory Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 21:32, 3 July 2013
Problem
For each integer , let
be the number of solutions to the equation
on the interval
. What is
?
Solution
Solution 1
By looking at various graphs, we obtain that, for most of the graphs
However, when , the middle apex of the sine curve touches the sine curve at the top only one time (instead of two), so we get here
.
Solution 2
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.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.