Difference between revisions of "2015 AMC 12A Problems/Problem 24"
Mathgeek2006 (talk | contribs) m (→Problem) |
m (→Solution) |
||
Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
+ | |||
+ | ===Solution 1=== | ||
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: | ||
Line 43: | Line 45: | ||
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=== | ||
+ | (WIP) | ||
== 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}} |
Revision as of 11:58, 4 January 2020
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
Solution 1
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
(WIP)
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 |