2012 AMC 12B Problems/Problem 13

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 at least one point in common?

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

Solutions

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

Solution 4

$x^2+ax+b=(x+ \frac{a}{2})^2+b- \frac{a^2}{4}$

$x^2+cx+d=(x+ \frac{c}{2})^2+d- \frac{c^2}{4}$

The only case where these two functions have no intersections is when the x-values of the turning point are the same but the y-values are not the same.

$\therefore \frac{a}{2}= \frac{c}{2} \implies a=c \quad\quad probability = \frac{1}{6} \times \frac{1}{6} \times 6= \frac{1}{6}$

$b- \frac{a^2}{4} \neq d- \frac{c^2}{4} \implies b \neq d \quad\quad probability = 1- \frac{1}{6}\times \frac{1}{6}\times 6= \frac{5}{6}$

$\therefore 1- \frac{1}{6}\times \frac{5}{6} = 1- \frac{5}{36} = \frac{31}{36}$

~ JiYang

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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