# Difference between revisions of "2012 AMC 12B Problems/Problem 13"

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==Solution 1== | ==Solution 1== | ||

− | Set the two equations equal to each other: <math>x^2 + ax + b = | + | Set the two equations equal to each other: <math>x^2 + ax + b = x^2</math> + cx + d. Now remove the x squared and get x's on one side: ax-cx=d-b. Now factor x: x(a-c)=d-b. If a cannot equal c, then there is always a solution, but if a=c, a 1 in 6 chance, leaving a 1080 out 1296, always having at least one point in common. And if a=c, then the only way for that to work, is if d=b, a 1 in 36 chance, however, this can occur 6 ways, so a 1 in 6 chance of this happening. So adding one sixth to 1080/1296, we get the simplified fraction of 31/36; answer D. |

==Solution 2== | ==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. |

## Revision as of 17:10, 2 March 2012

## 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 a least one point in common?

## Solution 1

Set the two equations equal to each other: + cx + d. Now remove the x squared and get x's on one side: ax-cx=d-b. Now factor x: x(a-c)=d-b. If a cannot equal c, then there is always a solution, but if a=c, a 1 in 6 chance, leaving a 1080 out 1296, always having at least one point in common. And if a=c, then the only way for that to work, is if d=b, a 1 in 36 chance, however, this can occur 6 ways, so a 1 in 6 chance of this happening. So adding one sixth to 1080/1296, we get the simplified fraction of 31/36; answer D.

## 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.