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Difference between revisions of "2015 AMC 12A Problems/Problem 24"

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is a real number?
 
is a real number?
  
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<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>
+
<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==
  
−
First we consider consider the binomial expansion of the expression: (let cos(aπ) be x and sin(bπ) be y)
+
Let <math>\cos(a\pi) = x</math> and <math>\sin(b\pi) = y</math>. Consider the binomial expansion of the expression:
−
x^4 + i * 4x^3*y + 6x^2*y^2 - i * 4xy^3 + y^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(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).  
+
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>).  
−
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:
 
  
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0, 1, 1/2, 3/2, 1/3,
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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>.  
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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
+
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>:
−
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:
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<cmath>0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},</cmath>
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For the two terms to cancel, then
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<cmath>\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},</cmath>
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cos(aπ)^3*sin(bπ) = cos(aπ)*sin(bπ)^3
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<cmath>\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},</cmath>
−
So
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<cmath>\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.</cmath>
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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).
+
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 19:32, 26 May 2022

Problem

Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0,2)$ that can be written as fractions $\frac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5$. What is the probability that \[(\text{cos}(a\pi)+i\text{sin}(b\pi))^4\] is a real number?

$\textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}$

Solution

Let $\cos(a\pi) = x$ and $\sin(b\pi) = y$. Consider the binomial expansion of the expression: \[x^4 + 4ix^{3}y - 6x^{2}y^{2} - 4ixy^3 + y^4.\]

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 $\cos(a\pi)$ or $\sin(b\pi)$ must be $0$, or the second term and the fourth term cancel each other out (because in the fourth term, you have $i^2 = -1$).

$\text{Case~1:}$ Either $\cos(a\pi)$ or $\sin(b\pi)$ is $0$.

The two $\text{a's}$ satisfying this are $\tfrac{1}{2}$ and $\tfrac{3}{2}$, and the two $\text{b's}$ satisfying this are $0$ and $1$. Because $a$ and $b$ can both be expressed as fractions with a denominator less than or equal to $5$, there are a total of $20$ possible values for $a$ and $b$:

\[0, 1, \frac{1}{2}, \frac{3}{2}, \frac{1}{3},\] \[\frac{2}{3}, \frac{4}{3}, \frac{5}{3}, \frac{1}{4}, \frac{3}{4},\] \[\frac{5}{4}, \frac{7}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5},\] \[\frac{4}{5}, \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \text{and} \frac{9}{5}.\]

Calculating the total number of sets of $(a,b)$ results in $20 \cdot 20 = 400$ sets. Calculating the total number of invalid sets (sets where $a$ doesn't equal $\tfrac{1}{2}$ or $\tfrac{3}{2}$ and $b$ doesn't equal $0$ or $1$), resulting in $(20-2) \cdot (20-2) = 324$.

Thus the number of valid sets is $76$.

$\text{Case~2}$: The two terms cancel.

We then have:

\[\cos^3(a\pi) \cdot \sin(b\pi) = \cos(a\pi) \cdot \sin^3(b\pi).\]

So:

\[\cos^2(a\pi) = \sin^2(b\pi),\]

which means for a given value of $\cos(a\pi)$ or $\sin(b\pi)$, there are $4$ valid values(one in each quadrant).

When either $\cos(a\pi)$ or $\sin(b\pi)$ are equal to $1$, however, there are only two corresponding values. We don't count the sets where either $\cos(a\pi)$ or $\sin(b\pi)$ 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 $\tfrac{1}{5}$, then $b$ must be $\tfrac{3}{10}$, which we don't have). Thus the total number of sets for this case is $4 \cdot 4 + 2 \cdot 2 = 20$.

Thus, our final answer is $\frac{(20 + 76)}{400} = \frac{6}{25}$, which is $\boxed{\text{(D)}}$.

Solution 2

Multiplying complex numbers is equivalent to multiplying their magnitudes and summing their angles. In order for $(\cos(a\pi)+i\sin(b\pi))^4$ to be a real number, then the angle of $\cos(a\pi)+i\sin(b\pi)$ must be a multiple of $45^{\circ}$, so $\cos(a\pi)+i\sin(b\pi)$ satisfies $\cos(a\pi)=0$, $\sin(b\pi)=0$, $\cos(a\pi)=\sin(b\pi)$, or $\cos(a\pi)=-\sin(b\pi)$.

There are $20$ possible values of $a$ and $b$. $\sin(x\pi) = 0$ at $x = \{0,1\}$ and $\cos(x\pi) = 0$ at $x = \{\frac12, \frac32\}$. The probability of $\sin(b\pi) = 0$ or $\cos(a\pi) = 0$ is $\frac{1}{10} + \frac{1}{10} - \frac{1}{100}$ (We overcounted the case where $\sin = \cos = 0$.)

We also consider the case where $\sin(b\pi) = \pm \cos(a\pi)$. This only happens when $a = \{0,1\}, b = \{\frac12, \frac32\}$ or $a,b = \{ \frac14, \frac34, \frac54, \frac74 \}$. The probability is $\frac{1}{100} + \frac{1}{25}$ The total probability is $\frac15 + \frac{1}{25} = \frac{6}{25} \rightarrow\boxed{\text{(D)}}$

~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
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