2012 AMC 12B Problems/Problem 13

Revision as of 17:10, 2 March 2012 by Klu2014 (talk | contribs) (Solution 1)


Two parabolas have equations $y= x^2 + ax +b$ and $y= x^2 + cx +d$, where $a, b, c,$ and $d$ 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: $x^2 + ax + b = x^2$ + 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 $x(a-c)=d-b$. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation $x(a-c)=d-b$ has no solution if and only if $a=c$ and $d\neq b$. The probability that $a=c$ is $\frac{1}{6}$ while the probability that $d\neq b$ is $\frac{5}{6}$. Thus we have $1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}$ for the probability that the parabolas intersect.

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