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Difference between revisions of "2005 AIME II Problems/Problem 9"

 
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== Problem ==
 
== Problem ==
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Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The <math>27</math> cubes are randomly arranged to form a <math>3\times 3\times 3</math> cube. Given the probability of the entire surface area of the larger cube is orange is <math>\frac{p^a}{q^b r^c}</math> where <math>p,q,</math> and <math>r</math> are distinct primes and <math>a,b,</math> and <math>c</math> are positive integers, find <math>a+b+c+p+q+r</math>.
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For how many positive integers <math> n </math> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>?
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== Solution ==
 
== Solution ==
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== See Also ==
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=== Solution 1 ===
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*[[2005 AIME II Problems]]
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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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Recall the [[trigonometric identities]] <math>\cos \left(\frac{\pi}2 - u\right) = \sin u</math> and <math>\sin \left(\frac{\pi}2 - u\right) = \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>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)</math> holds for all real <math>u</math>.
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<math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^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\left(\frac\pi2 - u\right)</math> and <math>\sin nu = \cos n\left(\frac\pi2 - u\right)</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\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 ===
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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>.
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=== Solution 3 ===
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We can rewrite <math>\sin(t)</math> as <math>\cos\left(\frac{\pi}{2}-t\right)</math> and <math>\cos(t)</math> as <math>\sin\left(\frac{\pi}{2}-t\right)</math>. This means that <math>\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}</math>. This theorem also tells us that <math>e^{\frac{\pi i}{2}}=i</math>, so <math>\sin t + i\cos t = \frac{i}{e^{it}}</math>. By the same line of reasoning, we have <math>\sin nt + i\cos nt = \frac{i}{e^{int}}</math>.
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For the statement in the question to be true, we must have <math>\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}</math>. The left hand side simplifies to <math>\frac{i^n}{e^{int}}</math>. We cancel the denominators and find that the only thing that needs to be true is that <math>i^n=i</math>. This is true if <math>n\equiv1\pmod{4}</math>, and there are <math>\boxed{250}</math> such numbers between <math>1</math> and <math>1000</math>. Solution by Zeroman
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=== Solution 4 ===
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We are using degrees in this solution instead of radians. I just process stuff better that way.
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We can see that the LHS is <math>cis(n(90^{\circ}-t))</math>, and the RHS is <math>cis(90^{\circ}-nt)</math> So, <math>n(90-t) \equiv 90-nt \mod 360</math> Expanding and canceling the nt terms, we will get <math>90n \equiv 90 \mod 360</math>. Canceling gets <math>n \equiv 1 \mod 4</math>, and thus there are <math>\boxed{250}</math> values of n.
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-AlexLikeMath
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=== Solution 5(CHEAP) ===
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Let <math>t=0</math>. Then, we have <math>i^n=i</math> which means <math>n\equiv 1\pmod{4}</math>. Thus, the answer is <math>\boxed{250}</math>.
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=== Solution 6 ===
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We factor out <math>i^n</math> from <math>(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).</math> We know the final expression must be the same as <math>\sin nt + i \cos nt</math> so we must have <math>i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt</math> in which testing yields <math>n \equiv 1 \pmod{4}</math> is the only mod that works, so we have a total of <math>1000 \cdot\frac{1}{4} = \boxed{250}</math> integers <math>n</math> that work.
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==Solution 7==
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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.
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~coolmath_2018
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==Solution 8==
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<math>(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n</math>
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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>
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[[Category:Intermediate Algebra Problems]]
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==See Also==
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{{AIME box|year=2005|n=II|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 02:30, 11 January 2024

Problem

For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$?

Solution

Solution 1

We know by De Moivre's Theorem that $(\cos t + i \sin t)^n = \cos nt + i \sin nt$ for all real numbers $t$ and all integers $n$. 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 $\cos \left(\frac{\pi}2 - u\right) = \sin u$ and $\sin \left(\frac{\pi}2 - u\right) = \cos u$ hold for all real $u$. If our original equation holds for all $t$, it must certainly hold for $t = \frac{\pi}2 - u$. Thus, the question is equivalent to asking for how many positive integers $n \leq 1000$ we have that $\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)$ holds for all real $u$.

$\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu$. 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 $n$ such that $\cos n u = \sin n\left(\frac\pi2 - u\right)$ and $\sin nu = \cos n\left(\frac\pi2 - u\right)$ hold for all real $u$.

$\sin x = \cos y$ if and only if either $x + y = \frac \pi 2 + 2\pi \cdot k$ or $x - y = \frac\pi2 + 2\pi\cdot k$ for some integer $k$. So from the equality of the real parts we need either $nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n = 1 + 4k$, or we need $-nu + n\left(\frac\pi2 - u\right) = \frac\pi 2 + 2\pi \cdot k$, in which case $n$ will depend on $u$ and so the equation will not hold for all real values of $u$. Checking $n = 1 + 4k$ in the equation for the imaginary parts, we see that it works there as well, so exactly those values of $n$ congruent to $1 \pmod 4$ work. There are $\boxed{250}$ of them in the given range.

Solution 2

This problem begs us to use the familiar identity $e^{it} = \cos(t) + i \sin(t)$. Notice, $\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}$ since $\sin(-t) = -\sin(t)$. Using this, $(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)$ is recast as $(i e^{-it})^n = i e^{-itn}$. Hence we must have $i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}$. Thus since $1000$ is a multiple of $4$ exactly one quarter of the residues are congruent to $1$ hence we have $\boxed{250}$.

Solution 3

We can rewrite $\sin(t)$ as $\cos\left(\frac{\pi}{2}-t\right)$ and $\cos(t)$ as $\sin\left(\frac{\pi}{2}-t\right)$. This means that $\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}$. This theorem also tells us that $e^{\frac{\pi i}{2}}=i$, so $\sin t + i\cos t = \frac{i}{e^{it}}$. By the same line of reasoning, we have $\sin nt + i\cos nt = \frac{i}{e^{int}}$.

For the statement in the question to be true, we must have $\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}$. The left hand side simplifies to $\frac{i^n}{e^{int}}$. We cancel the denominators and find that the only thing that needs to be true is that $i^n=i$. This is true if $n\equiv1\pmod{4}$, and there are $\boxed{250}$ such numbers between $1$ and $1000$. 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 $cis(n(90^{\circ}-t))$, and the RHS is $cis(90^{\circ}-nt)$ So, $n(90-t) \equiv 90-nt \mod 360$ Expanding and canceling the nt terms, we will get $90n \equiv 90 \mod 360$. Canceling gets $n \equiv 1 \mod 4$, and thus there are $\boxed{250}$ values of n.

-AlexLikeMath

Solution 5(CHEAP)

Let $t=0$. Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$. Thus, the answer is $\boxed{250}$.

Solution 6

We factor out $i^n$ from $(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).$ We know the final expression must be the same as $\sin nt + i \cos nt$ so we must have $i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt$ in which testing yields $n \equiv 1 \pmod{4}$ is the only mod that works, so we have a total of $1000 \cdot\frac{1}{4} = \boxed{250}$ integers $n$ 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 $\sin t = \cos(90 - t)$ and $\cos t = \sin(90 - t)$. So we rewrite the expression and setting it equal to the given expression in the problem, we get $\cos(90 - nt) + i\sin(90 - nt) = \cos(90n - nt) + i\sin(90n - nt)$. Now we can just look at the imaginary parts. Doing so and simplifying, we see that $1 + 4k= n$. From this we see that $n \equiv 1\pmod{4}$. So there are $\boxed{250}$ solutions.

~coolmath_2018

Solution 8

$(\sin\theta + i\cos\theta)^{n} = i^{n} (\cos\theta - i\sin\theta)^n$

Hence the required condition is just $i^{n} = i$ which is true for exactly 1 in 4 consecutive numbers. Thus $\boxed{250}$

See Also

2005 AIME II (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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