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}{5} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D) | + | <math> \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}</math> |
==Solution== | ==Solution== | ||
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<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> | + | 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>). |
− | <math>\text{Case~1:}</math> Either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> is <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> | + | 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>: |
<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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<cmath>\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.</cmath> | <cmath>\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.</cmath> | ||
− | Calculating the total number of sets of <math> | + | Calculating the total number of sets of <math>(a,b)</math> results in <math>20 \cdot 20 = 400</math> sets. |
− | Calculating the total number of invalid sets(sets where <math> | + | Calculating the total number of invalid sets (sets where <math>a</math> doesn't equal <math>\tfrac{1}{2}</math> or <math>\tfrac{3}{2}</math> and <math>b</math> doesn't equal <math>0</math> or <math>1</math>), resulting in <math>(20-2) \cdot (20-2) = 324</math>. |
− | Thus the number of valid sets is <math> | + | Thus the number of valid sets is <math>76</math>. |
<math>\text{Case~2}</math>: The two terms cancel. | <math>\text{Case~2}</math>: The two terms cancel. | ||
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<cmath>\cos^2(a\pi) = \sin^2(b\pi),</cmath> | <cmath>\cos^2(a\pi) = \sin^2(b\pi),</cmath> | ||
− | which means for a given value of <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math>, there are <math> | + | which means for a given value of <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math>, there are <math>4</math> valid values(one in each quadrant). |
− | When either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> are equal to <math> | + | When either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> are equal to <math>1</math>, however, there are only two corresponding values. We don't count the sets where either <math>\cos(a\pi)</math> or <math>\sin(b\pi)</math> equals <math>0</math>, 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 <math>a</math> is <math>\tfrac{1}{5}</math>, then <math>b</math> must be <math>\tfrac{3}{10}</math>, which we don't have). Thus the total number of sets for this case is <math>4 \cdot 4 + 2 \cdot 2 = 20</math>. |
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>. |
Revision as of 17:16, 9 March 2015
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 , their 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 .
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 |