Difference between revisions of "2012 AMC 12B Problems/Problem 13"
(→Solution 1) |
(→Solution 1) |
||
| Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
| − | Set the two equations equal to each other: <math>x^2 + ax + b = x^2 + cx + d</math>. Now remove the x squared and get x's on one side: <math>ax-cx=d-b</math>. Now factor <math>x</math>: <math>x(a-c)=d-b</math>. If a cannot equal <math>c</math>, then there is always a solution, but if <math>a=c</math>, a <math>1</math> in <math>6</math> chance, leaving a <math>1080</math> out <math>1296</math>, always having at least one point in common. And if <math>a=c</math>, then the only way for that to work, is if <math>d=b</math>, a <math>1</math> in <math>36</math> chance, however, this can occur <math>6</math> ways, so a <math>1</math> in <math>6</math> chance of this happening. So adding one sixth to <math>\frac{1080}{1296}</math>, we get the simplified fraction of <math>3136</math>; answer <math>D</math>. | + | Set the two equations equal to each other: <math>x^2 + ax + b = x^2 + cx + d</math>. Now remove the x squared and get x's on one side: <math>ax-cx=d-b</math>. Now factor <math>x</math>: <math>x(a-c)=d-b</math>. If a cannot equal <math>c</math>, then there is always a solution, but if <math>a=c</math>, a <math>1</math> in <math>6</math> chance, leaving a <math>1080</math> out <math>1296</math>, always having at least one point in common. And if <math>a=c</math>, then the only way for that to work, is if <math>d=b</math>, a <math>1</math> in <math>36</math> chance, however, this can occur <math>6</math> ways, so a <math>1</math> in <math>6</math> chance of this happening. So adding one sixth to <math>\frac{1080}{1296}</math>, we get the simplified fraction of <math>3136</math>; answer <math>D\box</math>. |
==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 16:34, 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: 
. Now remove the x squared and get x's on one side: 
. Now factor 
: 
. If a 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 sixth to 
, we get the simplified fraction of 
; answer $D\box$ (Error compiling LaTeX. Unknown error_msg).
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.
