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Difference between revisions of "2007 AMC 8 Problems/Problem 25"

(Created page with 'In our diagram, the three smaller sections trisect a circle with radius <math>3</math>. Thus, the area of the entire section is <math>9\pi</math>. The area of each section then m…')
 
(Extensive editing to the text for clarity of language and solution method)
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In our diagram, the three smaller sections trisect a circle with radius <math>3</math>. Thus, the area of the entire section is <math>9\pi</math>. The area of each section then must be <math>\frac{9\pi}{3}</math> or <math>3\pi</math>. The area of the larger regions trisect an area which is the difference of two circles, one with radius <math>3</math>, the other radius <math>6</math>. So, the area of the region is <math>36\pi - 9\pi</math> or <math>27\pi</math>. The area of each section must be <math>\frac{27\pi}{3}</math> or <math>9\pi</math>. To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. You could either get an odd number then an even number, or an even number then an odd number.
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To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).
  
  
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CASE 1:
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To find the areas of the sections, notice that the three smaller sections trisect a circle with radius <math>3</math>. The area of this entire circle is <math>9\pi</math>. The area of each smaller section then must be <math>\frac{9\pi}{3}</math> or <math>3\pi</math>. The larger sections trisect an "ring" which is the difference of two circles, one with radius <math>3</math>, the other radius <math>6</math>. So, the area of the ring (''annulus'') is <math>36\pi - 9\pi</math> or <math>27\pi</math>. The area of each larger section must be <math>\frac{27\pi}{3}</math> or <math>9\pi</math>. Note that the area of the whole circle is <math>36\pi</math>.
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On the first throw, you will get an odd number (<math>1</math>). Only one <math>3\pi</math> section holds a <math>1</math>, but two <math>9\pi</math> sections hold a <math>1</math>, so, <math>3\pi + (2 \cdot 9\pi) = 21\pi</math>. Since the area of the whole circle is <math>36\pi</math>, the probability of getting an odd is <math>\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}</math>. Since all that is left is even numbers, the probability of getting an even is <math>1 - \text{ pobability of getting an odd }</math> (do you see why). So it is <math>\frac{12}{12} - \frac{7}{12} = \frac{5}{12}</math>. So, we now multiply both probabilities to get <math>\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}</math>.
 
  
  
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CASE 2:
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One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is <math>3\pi + (2 \cdot 9\pi) = 21\pi</math>. Since the area of the whole circle is <math>36\pi</math>, the probability of getting an odd is <math>\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}</math>.
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This time you will get an even then an odd number. But this will be the same thing as an odd then an even, because first we will find the even prob. (<math>e</math>) and multiply that by odd prob. (<math>o</math>), which is exactly what we just did! So this probability is just <math>\frac{35}{144}</math>.  
 
  
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Since the remaining sections contain even numbers (that is, 2), the probability of throwing an odd is the complement, or <math>1 - \frac{7}{12} = \frac{5}{12}</math>.
  
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Instead of adding (which we could do), since they are the same number we can just multiply by <math>2</math> (n + n does equal 2 *n). We will then get <math>\frac{35}{144} \cdot \frac{2}{1} = \text{ ( cross cancel ) } = \frac{35}{144 \div 2} = \boxed{\frac{35}{72} \text{ which is } \bold{(B)}}</math>.
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Now, the two cases: You could either get an odd then an even, or an even then an odd.
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CASE 1: Odd then even
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Multiply the probabilities to get <math>\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}</math>.
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CASE 2: Even then odd
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Multiply the probabilities to get <math>\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}</math>. Notice that this is the same.
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Thus, the total probability of an odd sum is <math>\frac{35}{144} \cdot \frac{2}{1}</math> = <math>\boxed{\frac{35}{72}}</math>.

Revision as of 20:58, 11 October 2010

To get an odd sum, we must add an even number and an odd number. So we have a little casework to do. Before we do that, we also have to figure out some relative areas. You could either find the absolute areas (as is done below for completeness), or apply some proportional reasoning and symmetry to realize the relative areas (which is really all that matters).


To find the areas of the sections, notice that the three smaller sections trisect a circle with radius $3$. The area of this entire circle is $9\pi$. The area of each smaller section then must be $\frac{9\pi}{3}$ or $3\pi$. The larger sections trisect an "ring" which is the difference of two circles, one with radius $3$, the other radius $6$. So, the area of the ring (annulus) is $36\pi - 9\pi$ or $27\pi$. The area of each larger section must be $\frac{27\pi}{3}$ or $9\pi$. Note that the area of the whole circle is $36\pi$.


One smaller section and two larger sections contain an odd number (that is, 1). So the probability of throwing an odd number is $3\pi + (2 \cdot 9\pi) = 21\pi$. Since the area of the whole circle is $36\pi$, the probability of getting an odd is $\frac{21\pi}{36\pi} = \frac{21}{36} = \frac{7}{12}$.

Since the remaining sections contain even numbers (that is, 2), the probability of throwing an odd is the complement, or $1 - \frac{7}{12} = \frac{5}{12}$.


Now, the two cases: You could either get an odd then an even, or an even then an odd.


CASE 1: Odd then even Multiply the probabilities to get $\frac{7}{12} \cdot \frac{5}{12} = \frac{35}{144}$.


CASE 2: Even then odd Multiply the probabilities to get $\frac{5}{12} \cdot \frac{7}{12} = \frac{35}{144}$. Notice that this is the same.


Thus, the total probability of an odd sum is $\frac{35}{144} \cdot \frac{2}{1}$ = $\boxed{\frac{35}{72}}$.

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