Difference between revisions of "2005 AIME II Problems/Problem 9"
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== Solution == | == Solution == | ||
| + | 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. Recall the [[trigonometric identities]] <math>\cos \frac{\pi}2 - u = \sin u</math> and <math>\sin \frac{\pi}2 - u = \cos u</math> hold for all real <math>u</math>. If our original equation holds for all <math>t</math>, it must certainly hold for <math>t = \frac{\pi}2 - u</math>. Thus, the question is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</math> we have that <math>(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = \sin n(\frac\pi2 -u) + i\cos n(\frac\pi2 - u)</math> holds for all real <math>u</math>. | ||
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| + | <math>(\sin(\frac\pi2 - u) + i \cos(\frac\pi 2 - u))^n = (\cos u + i \sin u)^n = \cos nu + i\sin nu</math>. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all <math>n</math> such that <math>\cos n u = \sin n(\frac\pi2 - u)</math> and <math>\sin nu = \cos n(\frac\pi2 - u)</math> hold for all real <math>u</math>. | ||
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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(\frac\pi2 - u) = \frac\pi 2 + 2\pi \cdot k</math>, in which case <math>n = 1 + 4k</math>, or we need <math>-nu + n(\frac\pi2 - u) = \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 250 of them in the given range. | ||
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== See Also == | == See Also == | ||
*[[2005 AIME II Problems]] | *[[2005 AIME II Problems]] | ||
Revision as of 09:39, 21 July 2006
Problem
For how many positive integers 
less than or equal to 1000 is 
true for all real 
?
Solution
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 250 of them in the given range.
