2012 AMC 12B Problems/Problem 13
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 's on one side: . Now factor : . If 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.
Solution 3
Clearly, . Imagine the two sides as lines - they will have no solutions when the two lines are parallel (eg. have the same gradient) which is when is not equal to . Also, if and , they're the same line so we must add one case. There are combinations of and , of which they are equal in - but we must subtract 1 as if but they still intersect and have solutions. So we subtract this to obtain .
~ youtube.com/indianmathguy
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |
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