Difference between revisions of "2005 AIME II Problems/Problem 9"
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==Solution 7== | ==Solution 7== | ||
− | Note that this looks like | + | Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that <math>\sin t = \cos(90 - t)</math> and <math>\cos t = \sin(90 - t)</math>. So we rewrite the expression and setting it equal to the given expression in the problem, we get <math>\cos(90 - nt) + i\sin(90 - nt) = \cos(90n - nt) + i\sin(90n - nt)</math>. Now we can just look at the imaginary parts. Doing so and simplifying, we see that <math>1 + 4k= n</math>. From this we see that <math>n \equiv 1\pmod{4}</math>. So there are <math>\boxed{250}</math> solutions. |
~coolmath_2018 | ~coolmath_2018 | ||
+ | ==Solution 8== | ||
+ | <math>(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n</math> | ||
+ | |||
+ | Hence the required condition is just <math>i^{n} = i</math> which is true for exactly 1 in 4 consecutive numbers. Thus <math>\boxed{250}</math> | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2005|n=II|num-b=8|num-a=10}} | {{AIME box|year=2005|n=II|num-b=8|num-a=10}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 03:30, 11 January 2024
Contents
Problem
For how many positive integers less than or equal to
is
true for all real
?
Solution
Solution 1
We know by De Moivre's Theorem that for all real numbers
and all integers
. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities and
hold for all real
. If our original equation holds for all
, it must certainly hold for
. Thus, the question is equivalent to asking for how many positive integers
we have that
holds for all real
.
. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all
such that
and
hold for all real
.
if and only if either
or
for some integer
. So from the equality of the real parts we need either
, in which case
, or we need
, in which case
will depend on
and so the equation will not hold for all real values of
. Checking
in the equation for the imaginary parts, we see that it works there as well, so exactly those values of
congruent to
work. There are
of them in the given range.
Solution 2
This problem begs us to use the familiar identity . Notice,
since
. Using this,
is recast as
. Hence we must have
. Thus since
is a multiple of
exactly one quarter of the residues are congruent to
hence we have
.
Solution 3
We can rewrite as
and
as
. This means that
. This theorem also tells us that
, so
. By the same line of reasoning, we have
.
For the statement in the question to be true, we must have . The left hand side simplifies to
. We cancel the denominators and find that the only thing that needs to be true is that
. This is true if
, and there are
such numbers between
and
. Solution by Zeroman
Solution 4
We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is , and the RHS is
So,
Expanding and canceling the nt terms, we will get
. Canceling gets
, and thus there are
values of n.
-AlexLikeMath
Solution 5(CHEAP)
Let . Then, we have
which means
. Thus, the answer is
.
Solution 6
We factor out from
We know the final expression must be the same as
so we must have
in which testing yields
is the only mod that works, so we have a total of
integers
that work.
Solution 7
Note that this looks like de Moivre's except switched around. Using de Moivre's as motivation we try to convert the given expression into de Moivre's. Note that and
. So we rewrite the expression and setting it equal to the given expression in the problem, we get
. Now we can just look at the imaginary parts. Doing so and simplifying, we see that
. From this we see that
. So there are
solutions.
~coolmath_2018
Solution 8
Hence the required condition is just which is true for exactly 1 in 4 consecutive numbers. Thus
See Also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.