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

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==Problem==
 
==Problem==
  
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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 a least one point in common?
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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?
  
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<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>
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<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>
  
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==Solution==
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==Solutions==
  
 
===Solution 1===
 
===Solution 1===
  
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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  thirty sixth to <math>\frac{1080}{1296}</math>, we get the simplified fraction of <math>\frac{31}{36}</math>; answer <math>(D)</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 <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===
 
===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.
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===Solution 3===
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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>.
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~ youtube.com/indianmathguy
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===Solution 4===
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<math>x^2+ax+b=(x+ \frac{a}{2})^2+b- \frac{a^2}{4}</math>
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<math>x^2+cx+d=(x+ \frac{c}{2})^2+d- \frac{c^2}{4}</math>
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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.
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<math>\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}</math>
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<math>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}</math>
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<math>\therefore 1- \frac{1}{6}\times \frac{5}{6} = 1- \frac{5}{36} = \frac{31}{36}</math>
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~ Ji Yang
  
 
== See Also ==
 
== See Also ==

Latest revision as of 23:00, 29 October 2024

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.

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

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

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

~ Ji Yang

See Also

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

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