[a / b / c / d / e / f / g / gif / h / hr / k / m / o / p / r / s / t / u / v / vg / vm / vmg / vr / vrpg / vst / w / wg] [i / ic] [r9k / s4s / vip / qa] [cm / hm / lgbt / y] [3 / aco / adv / an / bant / biz / cgl / ck / co / diy / fa / fit / gd / hc / his / int / jp / lit / mlp / mu / n / news / out / po / pol / pw / qst / sci / soc / sp / tg / toy / trv / tv / vp / vt / wsg / wsr / x / xs] [Settings] [Search] [Mobile] [Home]
Board
Settings Mobile Home
/sci/ - Science & Math


Thread archived.
You cannot reply anymore.




File: problem.png (31 KB, 769x765)
31 KB
31 KB PNG
Three points are chosen randomly inside of a unit square. Then, a circle is drawn such that it passes through the three points.

What is the probability that the circle will be fully contained inside the square?
>>
Two.
>>
it is the first week of the semester and OP is already in over his head and unable to complete his homework.
try passing the prerequisite courses first next time
>>
>>14867869
>a circle is drawn such that it passes through the three points
Do you see the problem with this?
>>
>>14868012
no
>>
File: 3524344.jpg (70 KB, 452x363)
70 KB
70 KB JPG
>>14868126
Well, I do, and I am very intelligent, so you better think about it harder.
>>
>>14868126
>>14868134
What he said.
>>
>>14868134
you're not as smart as you think you are. degenerate cases where the points form a straight line reduce to infinite radius. sampling the same point twice is improbable.
>>
>>14868144
I am extremely smart and not all people are capable of my intellectual prowess, so I will allow you to think about it some more. The problem should become obvious to you eventually.
>>
pi/4 or sonething like this.
>>
>>14867869
I ran a Monte Carlo simulation with 10,000 trials, selecting the three points each time uniformly from the unit square.
A unique solution was found in all 10,000 cases. I used the parametric form {R cos(t) + x0, R sin(t) + y0} with t from 0 to 2pi, to setup equations for the three (x,y) points.
2534 out of the 10,000 trials led to a circle that lay within the unit square. So the answer is 1/4.
>>
>>14868012
never used a CAD program?
lol idiot
>>
>>14868166
Here's a sample of some Monte Carlo trials. All plot ranges are the unit square.
>>
>>14868144
Three points on the same line is also probability zero
>>
>>14868187
Any three points are probability zero.
>>
>>14868190
Yes, but thats irrelevant.
>>
>>14868193
If it's irrelevant, why did you bring it up?
>>
>>14868224
I didnt you did.
>>
>>14868285
I didn't. you and that other guy both did.
>>
>>14868190
Any three points are probability ~ (dA)^3 ~(dx)^6
Integrating vertically and horizontally 3 times each will give a total probability that is not infinitesimal.

The probability of a given circle of radius r is ~ (dx)^3 since each point will have 1 degree of freedom restricted.
Integrating twice (shifting the circle vertically and horizontally) will give the total probability of any of the circles of radius r which is ~ dx which is infinitesimal.
You can pick any subset of R+ of measure zero and ignore circles having those radii.

The case where two points are the same will have total probability ~(dx)^2
The case where three points are the same will have total probability ~(dx)^4

So as >>14868144 implies, degenerate cases can be ignored.
>>
>>14868306
Extended reals aren't real.
>>
>>14867991
This is too difficult for being a homework
>>
If you can define three new random variables (radius, centerx centery) the problem becomes a simpler. You could try finding a suitable transformation
>>
>>14868287
Noone brought that up. He replied to me by saying the event of three sampled points lying on the same line has a probability of zero. That statement is not equal to your statement 'the event of sampling a specific point is zero'. Both are related by measure zero sets with vanishing probability.
>>
File: eternalseptember.jpg (827 KB, 2404x1260)
827 KB
827 KB JPG
>homework thread
>9000 boring an stupid replies from the unwanted, uninvited interlopers who steadfastly refuse to adopt board culture
>>
>>14869129
> le heckin gatekeeping
> not on my board!!!!
I advise you to go back. You contribute nothing to this board.
>>
>>14868166
Reiterating that the answer is 1/4. I may give the analytic proof soon if no one else does it.
Starting from my parametric form {R cos(t) + x0, R sin(t) + y0} you get
{R cos(t1) + x0, R sin(t1) + y0} = {x1,y1}
{R cos(t2) + x0, R sin(t2) + y0} = {x2,y2}
{R cos(t3) + x0, R sin(t3) + y0} = {x3,y3}
This is 6 equations (3 vector equations with 2 components each) for 6 unknowns (R, x0,y0,t1,t2,t3)
Use the x-component of the first equation to solve for t1, then plug that t1 into the y component of the first equation e.g.
t1 = arccos((x1 -x0)/R)
sin(t1) = sqrt{1 - (x1-x0)^2/R^2}
So the y component of the first equation reads
sqrt{R^2 - (x1-x0^2)} + y0 = y1
In this way you can eliminate the three times t1,t2,t3 and obtain 3 polynomial equations that determine x0,y0, and R
then you have to compute a lot of Jacobians and integrate over dx1,dx2,dx3,dy1,dy2,dy3 over the unit square
>>
I'm getting (pi)^3 / 120.
Maybe I'm dumb but I just integrated over all circles of radius < 1/2 that fit in the box.
r in [0,1/2], x in [0,1-2r], y in [0,1-2r] (2*pi*r)^3 dx dy dr
>>
>>14869914
Seems pretty close based on the Monte Carlo simulation
>>
>>14869129
What is board culture, anon?
>>
1/2 , because it is equally likely.

Let 'a' be the side of square,
If you choose any circle(say A) of radius r<=a, there cannot be any circle(B) that satisfies all the three points chosen arbitrarily, but inside circle A and always have finite areas.

No matter how small circle A is, because smallest circle is a point, ie all three points(inside A) are coincident, and the circle B passing through these points is not unique.
>>
>>14868166
>>14868177
This is neatly done and makes the problem easier to understand.

>>14869914
Can you explain the (2*pi*r)^3? I see why you've chosen those ranges, but I don't understand how integrating over possible contained circles gives you a ratio of contained circles vs those that escape the square. As >>14870071 says, your result looks good, I just can't see how you've gotten there.
>>
>>14869276
It might be easier to just use complex numbers.
|z-z1|^2 = |z-z2|^2 = |z-z3|^2
Take the difference between the first 2 to get an equation with |z|^2 eliminated.
Take the difference between the last 2 to get an equation with |z|^2 eliminated
Take the right linear combination of these 2 to eliminate z*.
Solve for z.
>>
>>14870489
The probability that a point will land in some region is A(region)/A(unit square) = A(region).

>I don't understand how integrating over possible contained circles gives you a ratio of contained circles vs those that escape the square
We are trying to find the probability that 3 random points land on some circle that is completely within the unit square.
This can be computed as Sum{A(circle)^3:circle in [0,1]x[0,1]}.
You don't need to worry about double counting since any 3 points on 1 circle will not all be on another circle.
The part I am not confident about is A(circle) = 0 since the circle is 1 dimensional.
I abused notation and used the circumference of the circle for A(circle).

This is very likely incorrect.
>>
>>14872930
Thanks for explaining. I understand much better now.



Delete Post: [File Only] Style:
[Disable Mobile View / Use Desktop Site]

[Enable Mobile View / Use Mobile Site]

All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.