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

(Solution 1)
(Solution 1)
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==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\Box</math>.
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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>.
  
 
==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 16:35, 2 March 2012

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 a least one point in common?

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 sixth to $\frac{1080}{1296}$, we get the simplified fraction of $3136$; answer $(D)$.

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.

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