Difference between revisions of "2007 AMC 12A Problems/Problem 24"
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This leaves <math>n\equiv 1\pmod{4}</math>. In this case, the divisibility becomes <math>\frac {n + 1}{2}</math> dividing <math>(2j + 1)\frac {n - 1}{4}</math>. Since <math>\frac {n + 1}{2}</math> and <math>\frac {n - 1}{4}</math> are relatively prime (subtracting twice the second number from the first gives 1), <math>\frac {n + 1}{2}</math> must divide <math>2j + 1</math>. Since <math>j\leq \frac {n - 1}{2}</math>, <math>2j + 1\leq n < 2\cdot \frac {n + 1}{2}</math>. Then there is only one intersection, namely when <math>j = \frac {n - 1}{4}</math>. | This leaves <math>n\equiv 1\pmod{4}</math>. In this case, the divisibility becomes <math>\frac {n + 1}{2}</math> dividing <math>(2j + 1)\frac {n - 1}{4}</math>. Since <math>\frac {n + 1}{2}</math> and <math>\frac {n - 1}{4}</math> are relatively prime (subtracting twice the second number from the first gives 1), <math>\frac {n + 1}{2}</math> must divide <math>2j + 1</math>. Since <math>j\leq \frac {n - 1}{2}</math>, <math>2j + 1\leq n < 2\cdot \frac {n + 1}{2}</math>. Then there is only one intersection, namely when <math>j = \frac {n - 1}{4}</math>. | ||
| − | Therefore we find <math>F(n)</math> is equal to <math>1 + \lfloor n/2 \rfloor + \left \lfloor \frac {n + 1}{2}\right\rfloor = n + 1</math>, unless <math>n\equiv 1\pmod{4}</math>, in which case it is one less, or <math>n</math>. | + | Therefore we find <math>F(n)</math> is equal to <math>1 + \lfloor n/2 \rfloor + \left \lfloor \frac {n + 1}{2}\right\rfloor = n + 1</math>, unless <math>n\equiv 1\pmod{4}</math>, in which case it is one less, or <math>n</math>. The problem may then be finished as in Solution 1. |
== See also == | == See also == | ||
Revision as of 22:15, 17 February 2012
Problem
For each integer 
, let 
be the number of solutions to the equation 
on the interval ![$[0,\pi]$](http://latex.artofproblemsolving.com/e/c/e/ece27ba559bd93cdf2416aa0f88fc8d7d85e5791.png)
. 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
![\[\sin nx - \sin x = 2\left( \cos \frac {n + 1}{2}x\right) \left( \sin \frac {n - 1}{2}x\right)\]](http://latex.artofproblemsolving.com/8/7/1/871a5070b1c511018316319158347e8bfae88996.png)
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
![\[\frac {2j + 1}{n + 1} = \frac {2k}{n - 1}\]](http://latex.artofproblemsolving.com/e/c/e/eced9caeb4854046455cc26885c31312bfde4ca2.png)
![\[k = \frac {(2j + 1)(n - 1)}{2(n + 1)}\]](http://latex.artofproblemsolving.com/a/f/d/afdc681460be8e130fe26710fb88978f356aa940.png)
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 | |
