Difference between revisions of "2012 AMC 12B Problems/Problem 13"
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===Solution 4=== | ===Solution 4=== | ||
− | <math>x^2+ax+b=(x+\frac{a}{2})^2+b-\{a^2}{4}</math> | + | <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-\{c^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. | 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} \ | + | <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 | + | <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> | <math>\therefore 1- \frac{1}{6}\times \frac{5}{6} = 1- \frac{5}{36} = \frac{31}{36}</math> | ||
+ | ~ Ji Yang | ||
== See Also == | == See Also == |
Latest revision as of 23:00, 29 October 2024
Contents
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 |
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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