[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]

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]

just curious if theretically the circle was infentessimally small, would the chance still be 1/4 or would it be 1?

[plainglazed]

...I did not get the requirement that the points of the triangle need be on the circle. With this assumption it looks like you're right, DeeGee. Otherwise you would once again have infinite triangles both containing and not containing the origin in both 1. and 2. which brings us back to the OP's argument in DeeGee's last topic that the probablility cannot be determined for certain.

[Guest]

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

Anything in the OP that states that you can make the assumption that the vertices of the triangle lie on the circle...doesn't this mean that we're back to the argument in your other thread????

[Guest]

Ah! Thanx for pointing this out plainglazed and tpaxatb. I got a bit carried away!

As for your question sparx1, I think it would still be 1/4 as it is independent of the radius of the circle (considering points on the circumference).

Well, if at sub-atomics levels, electrons can exists at different orbits around a nucleus, you can surely distinguish infinite points within an infinitesimally small circle too!

[Guest]

Thinking further about the comments from tpaxatb and plainglazed, the circle situation is slightly easier to analyse even when points are not on the circumference of the circle.

Refer the figure in my previous post:

Mathematically: Consider the first two random points (A' & B') this time anywhere inside the circle. Now, C chould lie in the arc area O-C1-C2 in order for the triangle to contain the origin. The probability is then a ratio of area of O-C1-C2 and the area of the circle. This is again ratio of angle subtended at the arc to 2pi (da/2pi) as in the previous analysis.

Logically: you could consider any random line drawn inside a cricle (A'B') to project at its circumference (AB) too! The key point though is that the random chosen line and its projection on the circumference would subtend the same angle at the centre. So the result of probability would be the same!

So, even in this case, where a random traingle is drawn inside or inscribed in a cricle, the probability that the triangle contains origin would be 1/4.

[plainglazed]

Thinking further about the comments from tpaxatb and plainglazed, the circle situation is slightly easier to analyse even when points are not on the circumference of the circle.

Refer the figure in my previous post:

Mathematically: Consider the first two random points (A' & B') this time anywhere inside the circle. Now, C chould lie in the arc area O-C1-C2 in order for the triangle to contain the origin. The probability is then a ratio of area of O-C1-C2 and the area of the circle. This is again ratio of angle subtended at the arc to 2pi (da/2pi) as in the previous analysis.

Logically: you could consider any random line drawn inside a cricle (A'B') to project at its circumference (AB) too! The key point though is that the random chosen line and its projection on the circumference would subtend the same angle at the centre. So the result of probability would be the same!

So, even in this case, where a random traingle is drawn inside or inscribed in a cricle, the probability that the triangle contains origin would be 1/4.

OK, then what about question 2? Cant visualize the ratios you mentioned in this case. But logically one would think the probability of 1. and 2. would be (or would approach being) the same. Am probably missing something. Will have to keep thinking on this. And when I say "have to" it's because my mind will not give me a choice! Think my boss will understand?

[Guest]

OK, then what about question 2? Cant visualize the ratios you mentioned in this case. But logically one would think the probability of 1. and 2. would be (or would approach being) the same. Am probably missing something. Will have to keep thinking on this. And when I say "have to" it's because my mind will not give me a choice! Think my boss will understand?

:chuckle: i'm in the same boat as you. in fact.....

I started wondering if we wouldn't start getting into the same issues with respect to how you define a random triangle (as in the chord/circle thread)...and it's still rattling around in the back of my mind....

[Guest]

First off, I concur with DeeGee conclusion on question 1 about the 1/4 odds of the triangle containing the origin, regardless of whether A and B are on the circumference of the circle. But to compare question 1 with question 2, my conclusion is that the further the origin is from the center of the circle, the less likely that the third vertex would fall into the desired area to contain the origin.

Imagine a circle of radius 1, with its center at (0,0.99). This circle does contain the origin, but in order for the random triangle to contain the origin, at least one of the points of the triangle would have to fall in the tiny area below the x-axis. The odds of this happening are far from the 1/4 odds with the origin at the center of the circle. It appears to me that the origin being the center of the circle would actually be at the highest possible probability and the probability would decrease proportionally toward 0 as the origin approached any edge of the circle.

[Guest]

zero. An infinite set of random triangles can be placed in a circle containing the origin no matter where within the circle the origin lies. A subset of these triangles, also infinite, contain the origin. As a percentage of the original set, however, the subset is infinitely small, or 0%. Here then, is another example of the probability of an event being zero even though the possibility of the event is not.

Edited July 31, 2009 by Brighterthan1000rocks

[Guest]

zero. An infinite set of random triangles can be placed in a circle containing the origin no matter where within the circle the origin lies. A subset of these triangles, also infinite, contain the origin. As a percentage of the original set, however, the subset is infinitely small, or 0%. Here then, is another example of the probability of an event being zero even though the possibility of the event is not.

thats like saying because there are infinitely many more integers than even integers, the probability of an integer being even if you pick it at random is 0%.

[Guest]

thats like saying because there are infinitely many more integers than even integers, the probability of an integer being even if you pick it at random is 0%.

Not quite the same. The subset of even integers is exactly half the set of integers. We can define the percentage and thus the probability, in this case 50%. The same is true for picking a number divisable by ten out of an infinite number of integers. But now, what is the probability of picking a number between 10 and 20 out out the same infinite set.

Nonetheless, I am rethinking my answer to this OP.

[Guest]

There are an infinite number of random triangles that can be drawn inside a circle which will not contain a given point (the origin, perhaps.) There are also an infinite number of triangles that do contain this point. Either it does or it doesn't. Fifty percent.

[bonanova]

DeeGee [with a very nice proof] and others have it for part 1.

Pieater is correct that parts 1 and 2 differ.

Basically, part 2 boils down to finding the average size of a triangle that fits in a unit circle.

Call that area <A>.

The probability of a randomly chosen point [e.g. not necessarily the center] being covered by

a random triangle is then just the ratio of areas:

p_cover = <A>/pi.

BrightRocks, sparx1 has it right.

I started wondering if we wouldn't start getting into the same issues with respect to how you define a random triangle (as in the chord/circle thr

Happily there's no ambiguity in this case.

Select the triangle's vertices from a uniform grid of points within the circle.

If you were going to program this, you could select x and y at random

from the interval [-1, 1] and discard any selection for which x² + y² > 1.

The seemingly equivalent method of selecting r, theta uniformly

from the intervals [0, 1] and [0, 360) won't work, for reasons

that are evident after a little thought.

[Guest]

I'm thinking first about one instance of the first problem. Say we have already chosen x and y randomly inside the unit circle. The probability the choice of a third point would define a triangle that contains the origin does not depend on how far x or y are from the origin. The probability is equal to the smaller of the angles swept between the two points by a ray from the origin divided by 2 pi.

The smallest the angle can be is 0. The largest is pi. Since all the choices are random, independent, and made from a uniform distribution, the expected value of the angle between the first two points is pi/2. Thus, the probability of a random triangle including the origin is pi/2 / 2 pi = 1/4.

[bonanova]

Current consensus says that triangles that cover the center of a circle comprise 25% of any large set of random interior triangles.

Pieater argues persuasively that triangles that cover a point on the circumference comprise 0% of that set.

Intuitively, large triangles tend to contain a mix of central points, while smaller ones can be more local;

their points follow the general point distribution which is weighted toward larger radius values.

So we have big triangles mainly accounting for the large 25% central coverage, and smaller ones

having a higher likelihood of covering points near the perimeter.

The question is still open: what is the average size of triangles within a unit circle?

Dividing by pi will then answer Part 2.

[Guest]

After an hour or more of working I may have the answer to the average triangle...

For a unit circle

3/2pi

which means that the probability is 3/2pi^2.

if any of you are interested I could type up my method/working.

[Guest]

The above answer was for if the triangle has to be on the circle and not inside it. If the triangle can be placed anywhere inside the circle then the answer for part 2 would be a third of what I originally said.

Just thought I would clear that up .

[bushindo]

After an hour or more of working I may have the answer to the average triangle...

For a unit circle

3/2pi

which means that the probability is 3/2pi^2.

if any of you are interested I could type up my method/working.

I'm interested. Please post your method.

[Guest]

Just put in some thoughts to the average size of triangles and would like to know what the gang thinks!

For any two randomly chosen points inside the circle, the max distance between those points is "2r" and the min is pretty much zero. So, the average distance or rather, the expected distance (if I may use the term considering that all lengths are randomly and evenly distributed) is r.

Thinking on similar lines, choosing any 3 random points should then yield an expected equilateral triangle of side "r".

Then the area of this triangle = root(3).r² / 4

Then the probability is this area divided by area of circle = pi.r²

= root(3) / (4.pi) = 0.1376

[Guest]

I have assumed that a random triangle is drawn by choosing 3 random points in the circumference (A, B, C). Since the location of the first point is irrelevant there are only really 2 variables. The two variables I used were b and c.

Let angle b = AOB (green)

Let angle c = AOC (blue)

0 < a, b < 2pi

Both a and b are taken clockwise from AO.

For now, let’s assume c is constant so we want to find the average triangle, given angle c.

In the diagram:

OA, OB, OC = r

Area OAC = 1/2 * r^2 * sin©

Area OABC = 1/2 * r^2 * sin(b) + 1/2 * r^2 * sin(c-b)

Area ABC = 1/2 * r^2 * (sin(b) + sin(c-b) – sin©)

= 1/2 * r^2 * (sin(b) + sin©cos(b) + cos©sin(b) – sin©) (call this area X)

Where c < b, the angles will have to be swapped round, which will in fact give -X.

Now in order to find the average area of a triangle where c is constant I will integrate between 0 and 2pi and then divide by 2pi. Lets call this new function f(b).

f(b) = 1/(4pi) * r^2 * ( (int(0, b) X db) - (int(b, 2pi) X db)

f(b) = 1/(2pi) * r^2 * (2 – 2cos(b) – b*cos(b) + pi*sin(b ))

Now we can make c a variable. So we do the same again: simply integrate and then divide by 2pi.

Average triangle = 1/(4pi^2) * r^2 * int(0, 2pi) f dc

= 1/(4pi^2) * r^2 * 6pi

= 3 * r^2 / 2 * pi

For the probability, divide this by pi*r^2 as suggested by bonanova, giving

3/2pi^2

---------------------------------------------

If, however, A B & C can be anywhere inside the circle then I got round this by assuming that the radius was a variable too (call this variable R), giving triangles of any size within the circle.

Average triangle = 1/r * int(0,r) 3*R^2/2*pi dR

=1/2pi^2

[Guest]

Just put in some thoughts to the average size of triangles and would like to know what the gang thinks!

For any two randomly chosen points inside the circle, the max distance between those points is "2r" and the min is pretty much zero. So, the average distance or rather, the expected distance (if I may use the term considering that all lengths are randomly and evenly distributed) is r.

Thinking on similar lines, choosing any 3 random points should then yield an expected equilateral triangle of side "r".

Then the area of this triangle = root(3).r² / 4

Then the probability is this area divided by area of circle = pi.r²

= root(3) / (4.pi) = 0.1376

As a tangential member of 'the gang', I hesitate to comment on your accuracy, but I heartily applaud your clarity.

[bushindo]

I have assumed that a random triangle is drawn by choosing 3 random points in the circumference (A, B, C). Since the location of the first point is irrelevant there are only really 2 variables. The two variables I used were b and c.

Let angle b = AOB (green)

Let angle c = AOC (blue)

0 < a, b < 2pi

Both a and b are taken clockwise from AO.

For now, let’s assume c is constant so we want to find the average triangle, given angle c.

In the diagram:

OA, OB, OC = r

Area OAC = 1/2 * r^2 * sin©

Area OABC = 1/2 * r^2 * sin(b) + 1/2 * r^2 * sin(c-b)

Area ABC = 1/2 * r^2 * (sin(b) + sin(c-b) – sin©)

= 1/2 * r^2 * (sin(b) + sin©cos(b) + cos©sin(b) – sin©) (call this area X)

Where c < b, the angles will have to be swapped round, which will in fact give -X.

Now in order to find the average area of a triangle where c is constant I will integrate between 0 and 2pi and then divide by 2pi. Lets call this new function f(b).

f(b) = 1/(4pi) * r^2 * ( (int(0, b) X db) - (int(b, 2pi) X db)

f(b) = 1/(2pi) * r^2 * (2 – 2cos(b) – b*cos(b) + pi*sin(b ))

Now we can make c a variable. So we do the same again: simply integrate and then divide by 2pi.

Average triangle = 1/(4pi^2) * r^2 * int(0, 2pi) f dc

= 1/(4pi^2) * r^2 * 6pi

= 3 * r^2 / 2 * pi

For the probability, divide this by pi*r^2 as suggested by bonanova, giving

3/2pi^2

---------------------------------------------

If, however, A B & C can be anywhere inside the circle then I got round this by assuming that the radius was a variable too (call this variable R), giving triangles of any size within the circle.

Average triangle = 1/r * int(0,r) 3*R^2/2*pi dR

=1/2pi^2

The double integral is a nice way of finding the average size of all triangle constructed from 3 points on a circle's circumfence. Kudos on a job well done. However, the extension to A, B, C anywhere in the last part is the average triangle size of all triangle whose vertices are equidistant from the origin. This method would not count the triangles with vertices of different lengths to the origin, so I think it's safe to say that we determined the lower bound to part two is 1/2pi^2.

[Guest]

Just put in some thoughts to the average size of triangles and would like to know what the gang thinks!

For any two randomly chosen points inside the circle, the max distance between those points is "2r" and the min is pretty much zero. So, the average distance or rather, the expected distance (if I may use the term considering that all lengths are randomly and evenly distributed) is r.

Thinking on similar lines, choosing any 3 random points should then yield an expected equilateral triangle of side "r".

Then the area of this triangle = root(3).r² / 4

Then the probability is this area divided by area of circle = pi.r²

= root(3) / (4.pi) = 0.1376

I agree with this. I had something similarly suggested in the "origin" problem, but i did a little worse at estimation (i got .159). The problem is the difficultly to prove this. That is why i gave up this line of thinking other then the fact that It doesnt apply to the infinite plane.

[bushindo]

Just put in some thoughts to the average size of triangles and would like to know what the gang thinks!

For any two randomly chosen points inside the circle, the max distance between those points is "2r" and the min is pretty much zero. So, the average distance or rather, the expected distance (if I may use the term considering that all lengths are randomly and evenly distributed) is r.

Thinking on similar lines, choosing any 3 random points should then yield an expected equilateral triangle of side "r".

Then the area of this triangle = root(3).r² / 4

Then the probability is this area divided by area of circle = pi.r²

= root(3) / (4.pi) = 0.1376

The generalization from the line to the circle is not necessarily true.

I think so far we have agreed that `random sampling' refers to uniform sampling from the cartesian plane. The claim is that the distribution of segment length from 1D generalizes to 2D. That is, if we draw 2 points randomly from the unit circle, the expected distance between them is 1 . I wrote a program that uniformly sample 2 points from the unit circle, and computing the distance between them. After half a million samples, the distribution is not uniform, nor is it symmetric around 1.

The expected distance is 0.9056518.

Given that the generalization from 2 points on a line to 2 points on circle doesn't work, I would hesitate to generalize the result from 2 points on the circle to 3 points on the circle. Guess we'll have to wait until someone feels courageous enough to tackle a sextuple integral (the name sounds more fun/naughty than it really is), or comes up with an ingenius method to solve this.

Edited August 3, 2009 by bushindo