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: | |
− | x^4 + | + | <cmath>x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.</cmath> |
− | 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( | + | 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>. | |
− | |||
− | |||
− | |||
− | + | 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>\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},</cmath> | |
− | + | <cmath>\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},</cmath> | |
− | + | <cmath>\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.</cmath> | |
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− | |||
− | |||
− | Thus our final answer is (20 + 76) | + | 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>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>76</math>. | ||
+ | |||
+ | <math>\text{Case~2}</math>: The two terms cancel. | ||
+ | |||
+ | We then have: | ||
+ | |||
+ | <cmath>\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).</cmath> | ||
+ | |||
+ | So: | ||
+ | |||
+ | <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>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>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>. | ||
+ | |||
+ | ==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 20:32, 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 |