Difference between revisions of "2012 AMC 12B Problems/Problem 13"
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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 <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=== | ===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. | ||
+ | |||
+ | ===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 | ||
+ | |||
+ | ===Solution 4=== | ||
+ | |||
+ | <math>x^2+ax+b=(x+ \frac{a}{2})^2+b- \frac{a^2}{4}</math> | ||
+ | |||
+ | <math>x^2+cx+d=(x+ \frac{c}{2})^2+d- \frac{c^2}{4}</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | <math>\therefore 1- \frac{1}{6}\times \frac{5}{6} = 1- \frac{5}{36} = \frac{31}{36}</math> | ||
+ | |||
+ | ~ Ji Yang | ||
== See Also == | == See Also == |
Revision as of 23:00, 29 October 2024
Contents
[hide]Problem
Two parabolas have equations and , where and 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?
Solutions
Solution 1
Set the two equations equal to each other: . Now remove the x squared and get 's on one side: . Now factor : . If cannot equal , then there is always a solution, but if , a in chance, leaving a out , always having at least one point in common. And if , then the only way for that to work, is if , a in chance, however, this can occur ways, so a in chance of this happening. So adding one thirty sixth to , we get the simplified fraction of ; answer .
Solution 2
Proceed as above to obtain . The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation has no solution if and only if and . The probability that is while the probability that is . Thus we have for the probability that the parabolas intersect.
Solution 3
Clearly, . 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 is not equal to . Also, if and , they're the same line so we must add one case. There are combinations of and , of which they are equal in - but we must subtract 1 as if but they still intersect and have solutions. So we subtract this to obtain .
~ youtube.com/indianmathguy
Solution 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.
~ Ji Yang
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |
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