Difference between revisions of "2015 AMC 12A Problems/Problem 24"
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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>). |
Revision as of 23:48, 15 August 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 |