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 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? | + | 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 a least one point in common? |
==Solution 1== | ==Solution 1== | ||
Revision as of 16:09, 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.
