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

Latest revision as of 13:18, 29 October 2024

Problem

On the dart board shown in the figure below, the outer circle has radius $6$ and the inner circle has radius $3$ . Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?

$[asy] draw(circle((0,0),6)); draw(circle((0,0),3)); draw((0,0)--(0,6)); draw((0,0)--(rotate(120)*(0,6))); draw((0,0)--(rotate(-120)*(0,6))); label('1',(rotate(60)*(0,3/2))); label('2',(rotate(-60)*(0,3/2))); label('2',(0,-3/2)); label('2',(rotate(60)*(0,9/2))); label('1',(rotate(-60)*(0,9/2))); label('1',(0,-9/2)); [/asy]$

$\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}$

Solution 1

To get an odd sum, we must add an even number and an odd number (a 1 and a 2). 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).

The area of this entire smaller circle is $9\pi$ . The area of each smaller section then must be $\frac{9\pi}{3}$ or $3\pi$ . The larger sections trisect a "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 even 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 $\frac{35}{144} \cdot \frac{2}{1} = \boxed{\textbf{(B)} = \frac{35}{72}}$ .

Video Solution 1

https://www.youtube.com/watch?v=93RsgRTJYUM

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=8

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/iFSbbM-suKA

@@ Line 17: / Line 17: @@
 <math>\mathrm{(A)} \frac{17}{36} \qquad \mathrm{(B)} \frac{35}{72} \qquad \mathrm{(C)} \frac{1}{2} \qquad \mathrm{(D)} \frac{37}{72} \qquad \mathrm{(E)} \frac{19}{36}</math>
-==Solution==
+==Solution 1==
-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 get an odd sum, we must add an even number and an odd number (a 1 and a 2). 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).
-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 a "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>.
+The area of this entire smaller 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 a "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>.
@@ Line 43: / Line 43: @@
 ==Video Solution 1==
-https://youtu.be/7kLWbuoNvo8
+https://www.youtube.com/watch?v=93RsgRTJYUM
-== Video Solution 2 ==
+== Video Solution by OmegaLearn ==
 https://youtu.be/tKsYSBdeVuw?t=8
 ~ pi_is_3.14
+==Video Solution by WhyMath==
+https://youtu.be/iFSbbM-suKA
 ==See Also==
 {{AMC8 box|year=2007|num-b=24| after=Last problem}}
 {{MAA Notice}}

2007 AMC 8 (Problems • Answer Key • Resources)
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