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Difference between revisions of "2015 AMC 12A Problems/Problem 24"

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==Solution==
 
==Solution==
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===Solution 1===
 
  
 
Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:
 
Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:
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Thus, our final answer is <math>\frac{(20 + 76)}{400} = \frac{6}{25}</math>, which is <math>\boxed{\text{(D)}}</math>.
 
Thus, our final answer is <math>\frac{(20 + 76)}{400} = \frac{6}{25}</math>, which is <math>\boxed{\text{(D)}}</math>.
  
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===Solution 2===
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==Solution 2==
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(WIP)
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Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for <math>(\cos(a\pi)+i\sin(b\pi))^4</math> to be a real number, then the angle of <math>\cos(a\pi)+i\sin(b\pi)</math> must be a multiple of <math>45^{\circ}</math>, so <math>\cos(a\pi)+i\sin(b\pi)</math> satisfies <math>\cos(a\pi)=0</math>, <math>\sin(b\pi)=0</math>, <math>\cos(a\pi)=\sin(b\pi)</math>, or <math>\cos(a\pi)=-\sin(b\pi)</math>.
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There are <math>20</math> possible values of <math>a</math> and <math>b</math>. <math>\sin(x\pi) = 0</math> at <math>x = \{0,1\}</math> and <math>\cos(x\pi) = 0</math> at <math>x = \{\frac12, \frac32\}</math>. The probability of <math>\sin(b\pi) = 0</math> or <math>\cos(a\pi) = 0</math> is <math>\frac{1}{10} + \frac{1}{10} - \frac{1}{100}</math> (We overcounted the case where <math>\sin = \cos = 0</math>.)
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We also consider the case where <math>\sin(b\pi) = \pm \cos(a\pi)</math>. This only happens when <math>a = \{0,1\}, b = \{\frac12, \frac32\}</math> or <math>a,b = \{ \frac14, \frac34, \frac54, \frac74 \}</math>. The probability is <math>\frac{1}{100} + \frac{1}{25}</math> The total probability is <math>\frac15 + \frac{1}{25} = \frac{6}{25} \rightarrow\boxed{\text{(D)}}</math>
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~zeric
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=== Video Solution by Richard Rusczyk ===
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https://artofproblemsolving.com/videos/amc/2015amc12a/401
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~ dolphin7
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2015|ab=A|num-b=23|num-a=25}}

Latest revision as of 19:32, 26 May 2022

Problem

Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0,2)$ that can be written as fractions $\frac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5$. What is the probability that \[(\text{cos}(a\pi)+i\text{sin}(b\pi))^4\] is a real number?

$\textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}$

Solution

Let $\cos(a\pi) = x$ and $\sin(b\pi) = y$. Consider the binomial expansion of the expression: \[x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.\]

We notice that the only terms with $i$ are the second and the fourth terms. Thus for the expression to be a real number, either $\cos(a\pi)$ or $\sin(b\pi)$ must be $0$, or the second term and the fourth term cancel each other out (because in the fourth term, you have $i^2 = -1$).

$\text{Case~1:}$ Either $\cos(a\pi)$ or $\sin(b\pi)$ is $0$.

The two $\text{a's}$ satisfying this are $\tfrac{1}{2}$ and $\tfrac{3}{2}$, and the two $\text{b's}$ satisfying this are $0$ and $1$. Because $a$ and $b$ can both be expressed as fractions with a denominator less than or equal to $5$, there are a total of $20$ possible values for $a$ and $b$:

\[0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},\] \[\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},\] \[\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},\] \[\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.\]

Calculating the total number of sets of $(a,b)$ results in $20 \cdot 20 = 400$ sets. Calculating the total number of invalid sets (sets where $a$ doesn't equal $\tfrac{1}{2}$ or $\tfrac{3}{2}$ and $b$ doesn't equal $0$ or $1$), resulting in $(20-2) \cdot (20-2) = 324$.

Thus the number of valid sets is $76$.

$\text{Case~2}$: The two terms cancel.

We then have:

\[\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).\]

So:

\[\cos^2(a\pi) = \sin^2(b\pi),\]

which means for a given value of $\cos(a\pi)$ or $\sin(b\pi)$, there are $4$ valid values(one in each quadrant).

When either $\cos(a\pi)$ or $\sin(b\pi)$ are equal to $1$, however, there are only two corresponding values. We don't count the sets where either $\cos(a\pi)$ or $\sin(b\pi)$ equals $0$, for we would get repeated sets. We also exclude values where the denominator is an odd number, for we cannot find any corresponding values(for example, if $a$ is $\tfrac{1}{5}$, then $b$ must be $\tfrac{3}{10}$, which we don't have). Thus the total number of sets for this case is $4 \cdot 4 + 2 \cdot 2 = 20$.

Thus, our final answer is $\frac{(20 + 76)}{400} = \frac{6}{25}$, which is $\boxed{\text{(D)}}$.

Solution 2

Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for $(\cos(a\pi)+i\sin(b\pi))^4$ to be a real number, then the angle of $\cos(a\pi)+i\sin(b\pi)$ must be a multiple of $45^{\circ}$, so $\cos(a\pi)+i\sin(b\pi)$ satisfies $\cos(a\pi)=0$, $\sin(b\pi)=0$, $\cos(a\pi)=\sin(b\pi)$, or $\cos(a\pi)=-\sin(b\pi)$.

There are $20$ possible values of $a$ and $b$. $\sin(x\pi) = 0$ at $x = \{0,1\}$ and $\cos(x\pi) = 0$ at $x = \{\frac12, \frac32\}$. The probability of $\sin(b\pi) = 0$ or $\cos(a\pi) = 0$ is $\frac{1}{10} + \frac{1}{10} - \frac{1}{100}$ (We overcounted the case where $\sin = \cos = 0$.)

We also consider the case where $\sin(b\pi) = \pm \cos(a\pi)$. This only happens when $a = \{0,1\}, b = \{\frac12, \frac32\}$ or $a,b = \{ \frac14, \frac34, \frac54, \frac74 \}$. The probability is $\frac{1}{100} + \frac{1}{25}$ The total probability is $\frac15 + \frac{1}{25} = \frac{6}{25} \rightarrow\boxed{\text{(D)}}$

~zeric

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc12a/401

~ dolphin7

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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