Difference between revisions of "2015 AMC 12A Problems/Problem 24"
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is a real number? | is a real number? | ||
− | <math> \textbf{(A)}\ \frac{3}{ | + | <math> \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}</math> |
==Solution== | ==Solution== | ||
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: | ||
− | <cmath>x^4 + 4ix^{3}y | + | <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>). | 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>). | ||
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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>. | ||
− | 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>, | + | 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> | ||
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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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>.) | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~zeric | ||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc12a/401 | ||
+ | |||
+ | ~ 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 and are chosen at random among all rational numbers in the interval that can be written as fractions where and are integers with . What is the probability that is a real number?
Solution
Let and . Consider the binomial expansion of the expression:
We notice that the only terms with are the second and the fourth terms. Thus for the expression to be a real number, either or must be , or the second term and the fourth term cancel each other out (because in the fourth term, you have ).
Either or is .
The two satisfying this are and , and the two satisfying this are and . Because and can both be expressed as fractions with a denominator less than or equal to , there are a total of possible values for and :
Calculating the total number of sets of results in sets. Calculating the total number of invalid sets (sets where doesn't equal or and doesn't equal or ), resulting in .
Thus the number of valid sets is .
: The two terms cancel.
We then have:
So:
which means for a given value of or , there are valid values(one in each quadrant).
When either or are equal to , however, there are only two corresponding values. We don't count the sets where either or equals , 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 is , then must be , which we don't have). Thus the total number of sets for this case is .
Thus, our final answer is , which is .
Solution 2
Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for to be a real number, then the angle of must be a multiple of , so satisfies , , , or .
There are possible values of and . at and at . The probability of or is (We overcounted the case where .)
We also consider the case where . This only happens when or . The probability is The total probability is
~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 |