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14867869 No.14867869 [Reply] [Original] [archived.moe]

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?

>> No.14867913


>> No.14867991

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

>> No.14868012

>a circle is drawn such that it passes through the three points
Do you see the problem with this?

>> No.14868126


>> No.14868134
File: 69 KB, 452x363, 3524344.jpg [View same] [iqdb] [saucenao] [google]

Well, I do, and I am very intelligent, so you better think about it harder.

>> No.14868143

What he said.

>> No.14868144

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.

>> No.14868148

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.

>> No.14868158

pi/4 or sonething like this.

>> No.14868166

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.

>> No.14868175

never used a CAD program?
lol idiot

>> No.14868177
File: 80 KB, 810x1187, image_2022-09-23_010834045.png [View same] [iqdb] [saucenao] [google]

Here's a sample of some Monte Carlo trials. All plot ranges are the unit square.

>> No.14868187

Three points on the same line is also probability zero

>> No.14868190

Any three points are probability zero.

>> No.14868193

Yes, but thats irrelevant.

>> No.14868224

If it's irrelevant, why did you bring it up?

>> No.14868285

I didnt you did.

>> No.14868287

I didn't. you and that other guy both did.

>> No.14868306

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.

>> No.14868315

Extended reals aren't real.

>> No.14868414

This is too difficult for being a homework

>> No.14869085

If you can define three new random variables (radius, centerx centery) the problem becomes a simpler. You could try finding a suitable transformation

>> No.14869113

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.

>> No.14869129
File: 826 KB, 2404x1260, eternalseptember.jpg [View same] [iqdb] [saucenao] [google]

>homework thread
>9000 boring an stupid replies from the unwanted, uninvited interlopers who steadfastly refuse to adopt board culture

>> No.14869265

> le heckin gatekeeping
> not on my board!!!!
I advise you to go back. You contribute nothing to this board.

>> No.14869276

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

>> No.14869914

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

>> No.14870071

Seems pretty close based on the Monte Carlo simulation

>> No.14870108

What is board culture, anon?

>> No.14870260

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.

>> No.14870489

This is neatly done and makes the problem easier to understand.

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.

>> No.14870603

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.

>> No.14872930

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.

>> No.14873239

Thanks for explaining. I understand much better now.

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