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

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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

@@ Line 1: / Line 1: @@
 ==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 at least one point in common?
-==Solution 1==
+<math>\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</math>
-Set the two equations equal to each other: <math>x^2 + ax + b = X^2</math> + 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.
+==Solutions==
-==Solution 2==
+===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 <math>x</math>'s on one side: <math>ax-cx=d-b</math>. Now factor <math>x</math>: <math>x(a-c)=d-b</math>. If <math>a</math> 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  thirty sixth to <math>\frac{1080}{1296}</math>, we get the simplified fraction of <math>\frac{31}{36}</math>; answer <math>\boxed{(D)}</math>.
+===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.
+===Solution 3===
+Clearly, <math>ax + b = cx + d</math>. 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 <math>a</math> is not equal to <math>c</math>. Also, if <math>b = d</math> and <math>a = c</math>, they're the same line so we must add one case. There are <math>36</math> combinations of <math>a</math> and <math>c</math>, of which they are equal in <math>6</math> - but we must subtract 1 as if <math>a=c</math> but <math>b=d</math> they still intersect and have solutions. So we subtract this to obtain <math>\frac{36}{36} - \frac{5}{36} = \frac{31}{36}</math>.
+~ youtube.com/indianmathguy
+== See Also ==
+{{AMC12 box|year=2012|ab=B|num-b=12|num-a=14}}
+{{MAA Notice}}

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