[bonanova]

To complement DeeGee's thought provoking puzzle of

.

1. A triangle is drawn at random in a circle centered at the origin.
   What is the probability the origin lies inside the triangle?
   .
   
2. A triangle is drawn at random in a circle that contains the origin.
   What is the probability the origin lies inside the triangle?

.

I could comment on whether the answers are the same, but I won't.

Enjoy!

[Guest]

So

i had an interesting idea inspired by psychic mind so blame it on him if im wrong

so Mr. Psych found that the average area is 3r²/2pi for the points on the circumference. So every triangle defines its own circle. So every random triangle inside the unit circle defines a smaller circle. Therefore i arrived at the formula of

integral[0tor]((3X²piR²)/(2pi²X²))

this formula comes from your average triangle defined by circle A with radius a is 3a²/2pi

the amount of triangles in this smaller circle is related to its comparative area so its area is a²/r²

so the integral evaluates to 3r³/2pi

divide by r to get the average triangle size and then pir² to get the percentage of the circle and you get

3/2pi²

so instead of the circles value or percentage being determined by its area it should be determined by its perimeter. because the triangles aren't in the circle they're on it. so the formula changes to

integral[0tor]((3X²/2pi)(r/a)

which evaluates to 3r³/(4pi)

do the dividing for the average size and average percentage and get .075999999 or 3/(4pi²)

I wrote a program to solve this problem and this answer is slightly off I will post my answer if you want by I figured you guys would want to solve this.

[Guest]

just realized some of the circles defined will be only partialy within the unit circle while i only consider circles defined inside the unit circle... oh well back to the drawing board

[bonanova]

For one million random triangles in the unit circle:

Area distribution for one million random triangles in increments of .01.

Area of inscribed equilateral triangle = 1.299.... [max]

Average of 10⁶ areas = 0.2321 sq units or 7.39% of the area of the circle.

On average a random triangle covers 7.4% of the points in a unit circle.

Answer to part 2:

The probability that a random triangle will cover the origin

when drawn inside a circle that contains the origin is 7.39%.

Recall the probability of covering the center is 25%.

How it matters in a finite space where the point to be covered lies with respect to the center.

As opposed to infinite spaces, where there is no center.

The percentage of cases where certain points in the circle were covered.

Points taken along an "x" radius: (0, 0), (0.1, 0), (0.2, 0) ... etc. out to (0.99, 0)

_r__ - _Pct_

0.00 - 24.96 [center of circle: 25% = theoretical value]

0.10 - 24.35

0.02 - 22.58

0.03 - 19.84

0.04 - 16.36

0.05 - 12.50 [half-radius = half percent of center coverage]

0.06 - 08.58

0.07 - 05.05

0.08 - 02.26

0.09 - 00.55

0.95 - 00.13

0.99 - 00.01 [almost on circumference]

Interpolation of these numbers gives 7.39% for roughly

r = .66 or 2/3 the radius of the circle.

[Guest]

lol i must've been sleepy or something when i wrote my last post. for some reason i thought someone had asked the chance of a random circle containing the origin but looking back now, nobody did. So please disregard my last post. lol

[Guest]

lol i must've been sleepy or something when i wrote my last post. for some reason i thought someone had asked the chance of a random circle containing the origin but looking back now, nobody did. So please disregard my last post. lol

I did, but as post 37 in Origin in a triangle.

[Guest]

is that a program you wrote bonanva? just asking because I wrote one and got a different answer? I used polar coordinates but I dont think that should matter in fact i think it is the only way to do it unless you choose more then two random numbers for each point (could be wrong). Anyway I made an organized array of all my points(after 1000 points) and the distribution of points looks right. for clarification i ran the program in the millions of trials but for a testing i ran it for 333.

[bushindo]

is that a program you wrote bonanva? just asking because I wrote one and got a different answer? I used polar coordinates but I dont think that should matter in fact i think it is the only way to do it unless you choose more then two random numbers for each point (could be wrong). Anyway I made an organized array of all my points(after 1000 points) and the distribution of points looks right. for clarification i ran the program in the millions of trials but for a testing i ran it for 333.

Just some comments. If bonanova were sampling uniformly from the cartesian plane, and you were sampling uniformly from polar coordinates (as in uniformly sample radius in (0, 1) and theta in (0, 360) ), then you would arrive at different answers since the two methods are not equivalent. Uniformly sampling on polar coordinates result in a distribution that favor points near the origin. That is, if you pick a random circular region E near the edge of the circle, and compare that to a circular region C of the same area near the center, sampling uniformly on polar coordinates is more likely to produce a point in C than in E. Both methods are equally valid, of course. We just need to consider which one is closer to the intended meaning of `random' in the OP.

[Guest]

yah your right i forgot about that and cartesian I would say is more valid, but how can you do it. Do you just have to label each point thats some uniform distance apart and then randomly choose a number between 1 and total points. This is why i ended up using polar.

[bushindo]

yah your right i forgot about that and cartesian I would say is more valid, but how can you do it. Do you just have to label each point thats some uniform distance apart and then randomly choose a number between 1 and total points. This is why i ended up using polar.

There are many different methods of sampling. My favored method is to uniformly sample x from (-1, 1) and y from (-1, 1). If the coordinate (x,y) is outside the circle (that is, if x^2 + y^2 > 1 ) then I discard the coordinate and sample again. Conceptually, we sample from the square circumscribing the circle, and discard all points outside the circle. It is equivalent to uniformly sampling on the circle.

[Guest]

I wrote a program too and I agree totally with bonanova. Shame no one can use good old maths to work this out.

[Guest]

There are many different methods of sampling. My favored method is to uniformly sample x from (-1, 1) and y from (-1, 1). If the coordinate (x,y) is outside the circle (that is, if x^2 + y^2 > 1 ) then I discard the coordinate and sample again. Conceptually, we sample from the square circumscribing the circle, and discard all points outside the circle. It is equivalent to uniformly sampling on the circle.

I dont know why i didnt think of something so simple. I thought about choosing x axis first then out of the possible y's, but that obviously wouldnt work, so i used polar but then you informed me that isnt uniform either i didnt even think about just randomly choosing and then deleting the bad ones but anyway heres some numbers

as bonanva said cartesian 7.4%

polar coordinates 4.6%

sampling across the whole rectangle but dividing by the circles area 19.5%

[bonanova]

is that a program you wrote bonanva? just asking because I wrote one and got a different answer? I used polar coordinates but I dont think that should matter in fact i think it is the only way to do it unless you choose more then two random numbers for each point (could be wrong). Anyway I made an organized array of all my points(after 1000 points) and the distribution of points looks right. for clarification i ran the program in the millions of trials but for a testing i ran it for 333.

Yes, and the point [vertex] sampling was done as Bushindo describes.

I actually did the calculation in response to a that asks the probability

that four darts tossed at random on a circular dartboard form a convex set of points.

[Guest]

Hi,

Sorry to return on an old thread...

Can someone explain why "the probability that the two randomly chosen points A & B subtend an angle of "da" at the centre is da/pi" ?

I understand the probability that 2 random points chosen such that the angle is < "da" gives da/pi.

Thanks

Lets say the first two randomly chosen points (A & B) subtend an angle "da" at the centre (O).

Then in order for the triangle to contain the origin, the third point must lie between C1 and C2 which also are angle "da" apart

Now, the probability that the two randomly chosen points A & B subtend an angle of "da" at the centre is da/pi

as Pi is the max angle that the two points can have at the centre and all angles from 0 to pi are equally likely

Now, for the point C, the probability that it lies between C1 and C2 is da/2pi

Probability that the triangle will enclose the origin then is:

{dp = {{d2a /(2 pi2)

Integral from 0 to pi to cover all possible values of pi from 0 to pi

Then, P = 1/(2 pi2) . (pi2/2)

Thus, P = 1/4

Going by the same logic as in Part 1, even if the origin is not at the centre, the probability would be the same (1/4).

As a general solution, any point (x,y) inside a circle has a probability of 1/4 to be enclosed in a triangle drawn by three random points on the circumference of the circle

[Guest]

Hello friends,

I was inspired to register to this forum in order to share the solution of the second problem. The solution is clearly 3/8. Here is how I solved it:

http://www.cphpvb.net/probability/6649-position-of-point-in-circle-and-triangle/

I am sorry, but the solution on this page is in Bulgarian language. Briefly I parametrize the area of the triangle by it's angles "A" and "B". In this solution I used circle with radius R=1. The general solution is the same.

P.S. In this solution I used the vertexes of the triangle to be on the edge of the circle. I actually paraphrased the problem more general as following:

"We have a circle "k" with radius R and three random points A, B and C on the circle edge. What is the probability that a random point from the circle will be inside the triangle?"

If the vertexes of the triangle are inside the circle, then it's not that easy...

Edited December 24, 2010 by Philip Petrov

[Guest]

The solution is clearly 3/8

Sorry - it's 3/4

[Guest]

Sorry - it's 3/4

Ok, total shame on me... This, divided by π (Pi), so:

P = 3/4Pi