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

(Created page with "==Problem== 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...")
 
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==Solution==
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==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.
 
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.
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==Solution 2==
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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 23:20, 29 February 2012

Problem

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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