Difference between revisions of "2015 AMC 12A Problems/Problem 24"
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| − | + | We consider the binomial expansion of the expression: (let cos(aπ) be x and sin(bπ) be y) | |
x^4 + i * 4x^3*y + 6x^2*y^2 - i * 4xy^3 + y^4. | x^4 + i * 4x^3*y + 6x^2*y^2 - i * 4xy^3 + y^4. | ||
Revision as of 12:03, 7 February 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
![\[(\text{cos}(a\pi)+i\text{sin}(b\pi))^4\]](http://latex.artofproblemsolving.com/8/8/7/887b74a71c3746ec5fd9fcee5cfd177bf7857b95.png)
is a real number?
$\textbf{(A)}\ \frac{3}{5} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}$ (Error compiling LaTeX. Unknown error_msg)
Solution
We consider the binomial expansion of the expression: (let cos(aπ) be x and sin(bπ) be y) x^4 + i * 4x^3*y + 6x^2*y^2 - i * 4xy^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(aπ) or sin(bπ) 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). Let's calculate the probability of the first case: For either cos(aπ) or sin(bπ) to be 0, the two valid a's that would work are 1/2 and 3/2, and the two valid b's that would work are 0 and 1. Because a and b are can both be expressed as fractions with a denominator less or equal to 5, their are a total of 20 possible values for a and b:
0, 1, 1/2, 3/2, 1/3, 2/3, 4/3, 5/3, 1/4, 3/4, 5/4, 7/4, 1/5, 2/5, 3/5, 4/5, 6/5, 7/5, 8/5, 9/5.
We calculate the total number of sets of (a,b) we can find: 20*20=400. We calculate the total number of invalid sets(as in when a doesn't equal 1/2 or 3/2 and b doesn't equal 0 or 1): (20-2)*(20-2) = 324. Thus our number of valid sets is 76.
Let's calculate the probability for the second case: For the two terms to cancel, then
cos(aπ)^3*sin(bπ) = cos(aπ)*sin(bπ)^3
So
cos(aπ)^2 = sin(bπ)^2
Which means for a given value of cos(aπ) or sin(bπ), there are 4 valid values for the corresponding (one in each quadrant). When either cos(aπ) or sin(bπ) is equal to 1 however, there are only two corresponding values. We don't count the sets where either cos(aπ) or sin(bπ) 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 1/5, then b must be 3/10, which we don't have) Thus our total number of sets for this case is 4*4 + 2*2 = 20.
Thus our final answer is (20 + 76)/400 = 6/25, which is (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 | |
