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

Revision as of 22:48, 15 August 2015

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 $5$ , their 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 $\boxed{\text{(D)}}$ .

@@ Line 10: / Line 10: @@
 Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:
-<cmath>x^4 + 4ix^{3}y + 6x^{2}y^{2} - 4ixy^3 + y^4.</cmath>
+<cmath>x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.</cmath>
 We notice that the only terms with <math>i</math> are the second and the fourth terms. Thus for the expression to be a real number, either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> must be <math>0</math>, or the second term and the fourth term cancel each other out (because in the fourth term, you have <math>i^2 = -1</math>).

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2015 AMC 12A Problems/Problem 24"

Revision as of 22:48, 15 August 2015

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