Difference between revisions of "2005 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
We know by [[De Moivre's Theorem]] that <math>(\cos t + i \sin t)^n = \cos nt + i \sin nt</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. | We know by [[De Moivre's Theorem]] that <math>(\cos t + i \sin t)^n = \cos nt + i \sin nt</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. | ||
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<math>\sin x = \cos y</math> if and only if either <math>x + y = \frac \pi 2 + 2\pi \cdot k</math> or <math>x - y = \frac\pi2 + 2\pi\cdot k</math> for some integer <math>k</math>. So from the equality of the real parts we need either <math>nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n = 1 + 4k</math>, or we need <math>-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n</math> will depend on <math>u</math> and so the equation will not hold for all real values of <math>u</math>. Checking <math>n = 1 + 4k</math> in the equation for the imaginary parts, we see that it works there as well, so exactly those values of <math>n</math> congruent to <math>1 \pmod 4</math> work. There are <math>\boxed{250}</math> of them in the given range. | <math>\sin x = \cos y</math> if and only if either <math>x + y = \frac \pi 2 + 2\pi \cdot k</math> or <math>x - y = \frac\pi2 + 2\pi\cdot k</math> for some integer <math>k</math>. So from the equality of the real parts we need either <math>nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n = 1 + 4k</math>, or we need <math>-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n</math> will depend on <math>u</math> and so the equation will not hold for all real values of <math>u</math>. Checking <math>n = 1 + 4k</math> in the equation for the imaginary parts, we see that it works there as well, so exactly those values of <math>n</math> congruent to <math>1 \pmod 4</math> work. There are <math>\boxed{250}</math> of them in the given range. | ||
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+ | === Solution 2 === | ||
+ | This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</math>. Notice, <math>\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}</math> since <math>\sin(-t) = -\sin(t)</math>. Using this, <math>(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)</math> is recast as <math>(i e^{-it})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> is a multiple of <math>4</math> exactly one quarter of the residues are congruent to <math>1</math> hence we have <math>\boxed{250}</math>. | ||
== See also == | == See also == |
Revision as of 01:17, 11 May 2009
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
.
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 |