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Difference between revisions of "2015 AMC 12A Problems/Problem 24"
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<math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <math>0</math>.  
 
<math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <math>0</math>.  
  
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The two <math>\text{a's}</math> satisfying this are <math>\tfrac{1}{2}</math> and <math>\tfrac{3}{2}</math>, and the two <math>\text{b's}</math> satisfying this are <math>0</math> and <math>1</math>. Because <math>a</math> and <math>b</math> can both be expressed as fractions with a denominator less than or equal to <math>5</math>, their are a total of <math>20</math> possible values for <math>a</math> and <math>b</math>:
+
The two <math>\text{a's}</math> satisfying this are <math>\tfrac{1}{2}</math> and <math>\tfrac{3}{2}</math>, and the two <math>\text{b's}</math> satisfying this are <math>0</math> and <math>1</math>. Because <math>a</math> and <math>b</math> can both be expressed as fractions with a denominator less than or equal to <math>5</math>, there are a total of <math>20</math> possible values for <math>a</math> and <math>b</math>:
  
 
<cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath>  
 
<cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath>  

Revision as of 22:42, 1 January 2016

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?

$\textbf{(A)}\ \frac{3}{5} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}$

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 $\boxed{\text{(D)}}$.

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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