Difference between revisions of "2012 AMC 12B Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | Two parabolas have equations <math>y= x^2 + ax +b</math> and <math>y= x^2 + cx +d</math>, where <math>a, b, c,</math> and <math>d</math> are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have | + | Two parabolas have equations <math>y= x^2 + ax +b</math> and <math>y= x^2 + cx +d</math>, where <math>a, b, c,</math> and <math>d</math> are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have at least one point in common? |
− | + | <math>\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1</math> | |
− | + | ==Solutions== | |
− | ==Solution 2== | + | ===Solution 1=== |
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+ | Set the two equations equal to each other: <math>x^2 + ax + b = x^2 + cx + d</math>. Now remove the x squared and get x's on one side: <math>ax-cx=d-b</math>. Now factor <math>x</math>: <math>x(a-c)=d-b</math>. If a cannot equal <math>c</math>, then there is always a solution, but if <math>a=c</math>, a <math>1</math> in <math>6</math> chance, leaving a <math>1080</math> out <math>1296</math>, always having at least one point in common. And if <math>a=c</math>, then the only way for that to work, is if <math>d=b</math>, a <math>1</math> in <math>36</math> chance, however, this can occur <math>6</math> ways, so a <math>1</math> in <math>6</math> chance of this happening. So adding one thirty sixth to <math>\frac{1080}{1296}</math>, we get the simplified fraction of <math>\frac{31}{36}</math>; answer <math>\boxed{(D)}</math>. | ||
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+ | ===Solution 2=== | ||
Proceed as above to obtain <math>x(a-c)=d-b</math>. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation <math>x(a-c)=d-b</math> has no solution if and only if <math>a=c</math> and <math>d\neq b</math>. The probability that <math>a=c</math> is <math>\frac{1}{6}</math> while the probability that <math>d\neq b</math> is <math>\frac{5}{6}</math>. Thus we have <math>1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}</math> for the probability that the parabolas intersect. | Proceed as above to obtain <math>x(a-c)=d-b</math>. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation <math>x(a-c)=d-b</math> has no solution if and only if <math>a=c</math> and <math>d\neq b</math>. The probability that <math>a=c</math> is <math>\frac{1}{6}</math> while the probability that <math>d\neq b</math> is <math>\frac{5}{6}</math>. Thus we have <math>1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}</math> for the probability that the parabolas intersect. |
Latest revision as of 14:24, 7 August 2021
Problem
Two parabolas have equations and , where and are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have at least one point in common?
Solutions
Solution 1
Set the two equations equal to each other: . Now remove the x squared and get x's on one side: . Now factor : . If a cannot equal , then there is always a solution, but if , a in chance, leaving a out , always having at least one point in common. And if , then the only way for that to work, is if , a in chance, however, this can occur ways, so a in chance of this happening. So adding one thirty sixth to , we get the simplified fraction of ; answer .
Solution 2
Proceed as above to obtain . The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation has no solution if and only if and . The probability that is while the probability that is . Thus we have for the probability that the parabolas intersect.
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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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