2012 AMC 12B Problems/Problem 13


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 at least one point in common?

$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1$


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 thirty sixth to $\frac{1080}{1296}$, we get the simplified fraction of $\frac{31}{36}$; answer $\boxed{(D)}$.

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.

Solution 3

Clearly, $ax + b = cx + d$. Imagine the two sides as lines - they will have no solutions when the two lines are parallel (eg. have the same gradient) which is when $a$ is not equal to $c$. Also, if $b = d$ and $a = c$, they're the same line so we must add one case. There are $36$ combinations of $a$ and $c$, of which they are equal in $6$ - but we must subtract 1 as if $a=c$ but $b=d$ they still intersect and have solutions. So we subtract this to obtain $\frac{36}{36} - \frac{5}{36} = \frac{31}{36}$.

~ youtube.com/indianmathguy

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png