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# /sci/ - Science & Math

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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?

 >> Anonymous Fri Sep 23 04:13:09 2022 No.14867913 Two.
 >> Anonymous Fri Sep 23 05:03:45 2022 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
 >> Anonymous Fri Sep 23 05:13:38 2022 No.14868012 >>14867869>a circle is drawn such that it passes through the three pointsDo you see the problem with this?
 >> Anonymous Fri Sep 23 06:21:02 2022 No.14868126
 >> Anonymous Fri Sep 23 06:27:02 2022 No.14868134 File: 69 KB, 452x363, 3524344.jpg [View same] [iqdb] [saucenao] [google] >>14868126Well, I do, and I am very intelligent, so you better think about it harder.
 >> Anonymous Fri Sep 23 06:33:03 2022 No.14868143 >>14868126>>14868134What he said.
 >> Anonymous Fri Sep 23 06:34:01 2022 No.14868144 >>14868134you'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.
 >> Anonymous Fri Sep 23 06:37:53 2022 No.14868148 >>14868144I 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.
 >> Anonymous Fri Sep 23 06:46:58 2022 No.14868158 pi/4 or sonething like this.
 >> Anonymous Fri Sep 23 06:56:42 2022 No.14868166 >>14867869I 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.
 >> Anonymous Fri Sep 23 07:05:50 2022 No.14868175 >>14868012never used a CAD program?lol idiot
 >> Anonymous Fri Sep 23 07:08:35 2022 No.14868177 File: 80 KB, 810x1187, image_2022-09-23_010834045.png [View same] [iqdb] [saucenao] [google] >>14868166Here's a sample of some Monte Carlo trials. All plot ranges are the unit square.
 >> Anonymous Fri Sep 23 07:20:03 2022 No.14868187 >>14868144Three points on the same line is also probability zero
 >> Anonymous Fri Sep 23 07:22:06 2022 No.14868190 >>14868187Any three points are probability zero.
 >> Anonymous Fri Sep 23 07:24:18 2022 No.14868193 >>14868190Yes, but thats irrelevant.
 >> Anonymous Fri Sep 23 07:43:34 2022 No.14868224 >>14868193If it's irrelevant, why did you bring it up?
 >> Anonymous Fri Sep 23 08:36:06 2022 No.14868285 >>14868224I didnt you did.
 >> Anonymous Fri Sep 23 08:37:07 2022 No.14868287 >>14868285I didn't. you and that other guy both did.
 >> Anonymous Fri Sep 23 08:55:30 2022 No.14868306 >>14868190Any three points are probability ~ (dA)^3 ~(dx)^6Integrating 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)^2The case where three points are the same will have total probability ~(dx)^4So as >>14868144 implies, degenerate cases can be ignored.
 >> Anonymous Fri Sep 23 08:59:58 2022 No.14868315 >>14868306Extended reals aren't real.
 >> Anonymous Fri Sep 23 10:35:17 2022 No.14868414 >>14867991This is too difficult for being a homework
 >> Anonymous Fri Sep 23 17:43:32 2022 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
 >> Anonymous Fri Sep 23 17:52:17 2022 No.14869113 >>14868287Noone 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.
 >> Anonymous Fri Sep 23 17:59:52 2022 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
 >> Anonymous Fri Sep 23 19:13:13 2022 No.14869265 >>14869129> le heckin gatekeeping> not on my board!!!!I advise you to go back. You contribute nothing to this board.
 >> Anonymous Fri Sep 23 19:24:25 2022 No.14869276 >>14868166Reiterating 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 readssqrt{R^2 - (x1-x0^2)} + y0 = y1In this way you can eliminate the three times t1,t2,t3 and obtain 3 polynomial equations that determine x0,y0, and Rthen you have to compute a lot of Jacobians and integrate over dx1,dx2,dx3,dy1,dy2,dy3 over the unit square
 >> Anonymous Sat Sep 24 00:01:01 2022 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
 >> Anonymous Sat Sep 24 01:36:22 2022 No.14870071 >>14869914Seems pretty close based on the Monte Carlo simulation
 >> Anonymous Sat Sep 24 02:09:13 2022 No.14870108 >>14869129What is board culture, anon?
 >> Anonymous Sat Sep 24 03:25:18 2022 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.
 >> Anonymous Sat Sep 24 06:00:11 2022 No.14870489 >>14868166>>14868177This is neatly done and makes the problem easier to understand.>>14869914Can 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.
 >> Anonymous Sat Sep 24 07:45:44 2022 No.14870603 >>14869276It might be easier to just use complex numbers.|z-z1|^2 = |z-z2|^2 = |z-z3|^2Take 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 eliminatedTake the right linear combination of these 2 to eliminate z*.Solve for z.
 >> Anonymous Sun Sep 25 03:40:31 2022 No.14872930 >>14870489The 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 squareWe 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.
 >> Anonymous Sun Sep 25 08:07:20 2022 No.14873239 >>14872930Thanks for explaining. I understand much better now.
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