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

Revision as of 19:34, 8 November 2022

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.

2012 AMC 12B (Problems • Answer Key • Resources)
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.

@@ Line 9: / Line 9: @@
 ===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 <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>.
+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===

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