Question

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?

Answer 1

you're basically finding the probability of the two parabolas intersecting (where a does not equal c, and b does not equal d)

Answer 2

a,,b,c,d are all whole numbers between 1 and 6 (including both 1 and 6)

Answer 3

how do i find the chance that a will be equal to c?

Answer 4

waiiit, there a 1/6 chance of a=c and 1/6 chance of b=d right?

Answer 5

yes correct

Answer 6

what do i do from here though?

Answer 7

what happens when you set the two equations y= x^2 + ax +b and y= x^2 + cx +d, equal to one another? what do you notice cancelling?

Answer 8

well, the x^2

Answer 9

so we'll have ax + b = cx + d

solve for x to get ??

Answer 10

x(a-c) = d-b

Answer 11

keep going

Answer 12

well, \[\frac{ d-b }{ a-c }=x \]

Answer 13

ok let me think for a second

Answer 14

ok I just realized that if a = c and b = d, then we'll have 2 identical parabolas which means that every point on either parabola is on the other parabola. Since they say "AT LEAST ONE" point in common, this means that a = c and b = d is possible.

For instance, if a = 2, b = 3 and a = c, b = d, then

y= x^2 + ax +b and y= x^2 + cx +d

turn into

y= x^2 + 2x +3 and y= x^2 + 2x +3

they'll have infinitely many points in common since they are the same parabola

Answer 15

as for generating every possible case, I'm still thinking that one over

Answer 16

this is a tough problem, so I looked it up and found this page

http://www.artofproblemsolving.com/Wiki/index.php/2012_AMC_12B_Problems/Problem_13

apparently it's part of the AMC (American Mathematics Competitions) and they're known to ask really tough questions

Answer 17

could you explain their solution 1 to me

Answer 18

which part of solution 1?

Answer 19

1296 = 6^4. There are 4 dice and 6 faces per die, so 6^4 = 1296 possible ways to roll 4 dice.

they are getting 1080 from computing 5*6^3. They say "assume a is not equal to c", to that means one of the dice (either a or c) has 1 less choice to land on.

Answer 20

the 1080/1296 represents the cases when a does not equal c

the 1/6 represents when a = c (d = b would also have to be true)

Answer 21

wait, why 5 times 6^3 why 5?

Answer 22

because if a doesn't equal c, then either a or c has one less choice. Let's say a = 2

if a doesn't equal c, then c cannot choose 2 to land on. It can only choose from {1, 3, 4, 5, 6} so 5 choices

Answer 23

normally it is 6^4 or 6*6*6*6

but one of the die has been restricted to 5 choices, so 5*6*6*6 = 5*6^3 = 1080

Answer 24

isnt that dependent on b=d?

Answer 25

no if a doesn't equal c, then b = d is not important

b could equal d, or it could not, we don't care really

Answer 26

b = d is only crucial when a = c

Answer 27

because if a = c and \(b \ne d\) leads to a problem of division by zero leading to no solutions at all

Answer 28

what are the chance of b=d AND a=c occuring

Answer 29

1/6 times 1/6

Answer 30

so the answer can be defined as \[\frac{ 5\times6^{3} }{ 6^{4} }+\frac{ 1 }{ 36 }\]

Answer 31

Why does that work?

Answer 32

Ahhh, is it because we are adding the total amount of ways that a is not equal to c, with the total amount of ways that a=c AND b=d?

Answer 33

does that sum up every possibility?

Answer 34

they made a typo: on the solution 1, it *should* say "adding 1/36 to 1080/1296" so you are correct

because: 1080/1296+1/36 = 31/36

while: 1080/1296+1/6 = 1

Answer 35

but yes you have the correct idea

Answer 36

ok, since i need to use this for a brief paper, i need to generalize it.

Answer 37

would it be fair to say, that given two parabolas with 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 n-sided die, the probability that the graphs of the two intersect is equal to
\[1-\frac{ 1 }{ n}\times \frac{ n-1 }{ n } = \frac{ n^{2}-n+1 }{ n^{2} }\]

Answer 38

yes you can use solution 2 to generalize as it might be easier

using solution 1 gives you

\[\Large \frac{(n-1)*n^3}{n^4} + \frac{1}{n^2}\]
\[\Large \frac{(n-1)*n^3}{n^4} + \frac{n^2}{n^4}\]
\[\Large \frac{(n-1)*n^3+n^2}{n^4}\]
\[\Large \frac{n^4-n^3+n^2}{n^4}\]
\[\Large \frac{n^2(n^2-n+1)}{n^4}\]
\[\Large \frac{n^2-n+1}{n^2}\]
so either way works

Answer 39

ok great, thanks. One more thing though, would it be extremely tedious (and maybe impossible) to generalize this for an n-th degree equation?

Answer 40

that was my ultimate goal, but I'm not so sure if its possible any more

Answer 41

so instead of quadratics, any two polynomials of degree n?

Answer 42

right

Answer 43

well the nice thing about the quadratics one we were given is that the quadratic portion cancelled out leaving us with linear equations. Quadratics are hard enough, cubics are even more impossible (still doable). Same can be said about quartics.

according to this, quintics can't even be solved exactly

http://mathworld.wolfram.com/QuinticEquation.html

and the same is true for any degree larger than 5

Answer 44

sure you can get numerical approximations, so that's at least something. But basically I'm trying to say that it gets even more complicated as n increases

Answer 45

well, would it be fair to say in my paper, that for any n-th degree polynomial that maintains just the four variables at a, b, c, and d, Such as the equations
y=x^5 + x^4 + x^3 + x^2 + ax+ b
y= x^5 + x^4 + x^3 + x^2 + cx+ d
the proability will be (n^2 - n +1)/n^2, dependent on the n-value of the die?

Answer 46

oh you're still sticking to that format, I see

Answer 47

because no matter how high the degree of the polynomial, the x^m powers can be subtracted off to get the original ax +b = cx+d?

Answer 48

is that true?

Answer 49

so all that other extra complicated stuff would cancel

Answer 50

yes that looks good

Answer 51

i suppose that the paper would be better if i tried an equation where one of the variable was behind an x^2, but i dont think i could solve it, so i will stick to this simplified generalization

Answer 52

so yeah, you can say "the \(\large x^m\) terms cancel where \(\large 2\le m \le n\)"

Answer 53

yeah I don't know how you'd solve it that way

Answer 54

wait, why must you include the n in the inequality?

Answer 55

i could have an 8th degree polynomial with a 6 sided die couldnt i?

Answer 56

I put in the n to restrict how high m goes

Answer 57

when you are referring to the nth degree polynomials

Answer 58

but why does it need to be restricted

Answer 59

because if n = 12, then m = 13 isn't possible

Answer 60

if n = 12, all the terms from x^2 to x^12 will cancel

Answer 61

wait, n is not referring to the sides of the die is it?

Answer 62

no, the two sides of the equation

Answer 63

the degrees of the two polynomials

Answer 64

oh wait, you used n to be the sides of the die

Answer 65

ok, i understand, i thought you were referring to n- sides of the die, as in the previous generalization. I understand now

Answer 66

hmm, then you'll have to use another variable for the generalized polynomials

Answer 67

so for a z-th degree polynomial, what would my inequailty be?

Answer 68

\[\Large 2 \le m \le z\]

Answer 69

ok, so the x^m terms cancel, when m is greater than or equal to 2, and less than or equal to the degree of the polynomial?

Answer 70

you'd say "the \(\large x^m\) terms cancel where \(\large 2\le m \le z\)" or something like that

Answer 71

got it, great Thanks so much for your help. By the way, do you know what a mathematical paper abstract should look like given this:
"The abstract is fundamental. First, it identifies the subject and it gives details that didn't fit into the title. Then, it describes the main issues and briefly mentions the discussions included in the paper. It should not include any general or background information. It can be viewed as a table of contents, but written in prose."

Answer 72

like for this problem what should i put in the abstarct?

Answer 73

The abstract is basically what you're trying to do in the entire paper. It's like a summary of the entire paper for those who really don't want to read all of it. It's helpful to save time when you're conducting research. If you're trying to learn more about topic X, and topic X is mentioned in the abstract, then you're probably going to read the paper. Ideally you'd find your answer, but that doesn't happen every time. If you don't get your answer, at least you might learn some cool new trick or property or something.

Answer 74

so basically, it needs to be brief and to the point?

Answer 75

yeah relatively short

Answer 76

So in your case, you're going to describe the problem. In addition, you're going to briefly mention how you found the solutions for the particular and general cases. Of course, the actual full descriptions will be in the body of the paper and not in the abstract.

Answer 77

so no lengthy mathematical equations in the abstract?

Answer 78

No, just the basic idea of them. The abstract is about a paragraph in length I think. Usually it's all words.

Answer 79

ok, got it thank you very much i appreciate your help

Answer 80

you're welcome