Difference between revisions of "2005 AIME II Problems/Problem 9"
(→See also: category) |
(→Solution) |
||
Line 10: | Line 10: | ||
<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. | <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. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | We can rewrite <math>(\sin t + i \cos t)^n</math> as <math>(\cos(90-t) + i\sin(90-t))^n</math> which, by De Moivre's Theorem is equal to <math>\cos{n(90-t)} + i\sin{n(90-t)}</math>, but we know that is is equal to <math>\sin nt + i \cos nt</math>, now if we replace <math>\sin nt + i \cos nt</math> with <math>\cos{90-nt} + i\sin{90-nt}</math>. This gives us the equation: | ||
+ | |||
+ | |||
+ | <math>\cos{90n - nt} + i\sin{90 - nt} = \cos{90-nt} + i\sin{90-nt}</math> | ||
+ | |||
+ | Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get | ||
+ | |||
+ | <math>90n - nt = 90 - nt</math> | ||
+ | |||
+ | but <math>90</math> in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is: | ||
+ | |||
+ | <math>90n = 90 + 360x</math> for <math>x</math> being an integer. | ||
+ | |||
+ | <math>n = 4x + 1</math> | ||
+ | |||
+ | we know that <math>n \le 1000</math> | ||
+ | |||
+ | so we want all numbers that are less than or equal to <math>1000</math> and also <math>\equiv 1 (\mbox{mod} 4)</math> and there are <math>\fbox{250}</math> such numbers | ||
== See also == | == See also == |
Revision as of 20:49, 15 July 2008
Contents
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.
Solution 2
We can rewrite as which, by De Moivre's Theorem is equal to , but we know that is is equal to , now if we replace with . This gives us the equation:
Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get
but in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is:
for being an integer.
we know that
so we want all numbers that are less than or equal to and also and there are such numbers
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 |