[Guest]

Consider a circle of any size. Then randomly mark 4 points on it. What is the probability of the quadrangle that is formed by these 4 points, to be a "regular quadrangle". Excuse me for this term, as I don't know how to word it. I mean this: A quadrangle that each interior angles of it is smaller than 180⁰ , or with another word: each exterior angles of it is greater than 180⁰

[bonanova]

Since a circle is a convex figure, the interior angle of any three points on a circle is less than 180 degrees.

If the four points are connected "out of order" - forming a "bow-tie" quadrilateral, interior and exterior angles might become confused, but the confusion can be removed by "coloring" the interior, and then it will be seen that the interior angles are all less than 180 degrees.

So, the way the problem seems to be stated, the probability is unity that the quadrangle is regular.

To state another way, the quadrilateral formed by joining each point to two other points is always convex.

Maybe it was meant to ask, what is the probability that the quadrilateral is not a bow tie?

There are six ways to connect four points on a circle. Two of them avoid the bow tie configuration.

That probability is thus 1/3.

[Guest]

Consider a circle of any size. Then randomly mark 4 points on it. What is the probability of the quadrangle that is formed by these 4 points, to be a "regular quadrangle". Excuse me for this term, as I don't know how to word it. I mean this: A quadrangle that each interior angles of it is smaller than 180⁰ , or with another word: each exterior angles of it is greater than 180⁰

For any quadrangle any two consecutive angle being less than 180.. other two will less than 180. and when any two angle being more than 180 it cant be inscribed in a circle...

so probablity is 1 ... <_<

[Guest]

Since a circle is a convex figure, the interior angle of any three points on a circle is less than 180 degrees.

If the four points are connected "out of order" - forming a "bow-tie" quadrilateral, interior and exterior angles might become confused, but the confusion can be removed by "coloring" the interior, and then it will be seen that the interior angles are all less than 180 degrees.

So, the way the problem seems to be stated, the probability is unity that the quadrangle is regular.

To state another way, the quadrilateral formed by joining each point to two other points is always convex.

Maybe it was meant to ask, what is the probability that the quadrilateral is not a bow tie?

There are six ways to connect four points on a circle. Two of them avoid the bow tie configuration.

That probability is thus 1/3.

Excuse me for confusing your brains. Surely any 3 points on a convex surface forms always a triangle, and all interior angles of a triangle are less then 180 degrees. What I mean by "regular quadrangle" is a convex quadrangle. I should use that word, sorry. Let me describe it with another way: If you form a triangle by 3 points, in a circle, and mark a fourth point in the circle, now: if the 4th point is out of the triangle, you will obtain a convex quadrangle; if the 4th point is inside the triangle, the quadrangle won't be convex (I call it irregular:) ) My question is the proportion between them. If the triangle's corners had been placed on the outher border of the circle (as a ring), I could try to calculate it. But if the question is: Corners of triangle may be in any place in the triangle. I'm afraid this is hard enough, even for Bonanova and Prime.

If I haven't succeed to explain my question, please inform me and let me try again.

[bonanova]

Excuse me for confusing your brains. Surely any 3 points on a convex surface forms always a triangle, and all interior angles of a triangle are less then 180 degrees. What I mean by "regular quadrangle" is a convex quadrangle. I should use that word, sorry. Let me describe it with another way: If you form a triangle by 3 points, in a circle, and mark a fourth point in the circle, now: if the 4th point is out of the triangle, you will obtain a convex quadrangle; if the 4th point is inside the triangle, the quadrangle won't be convex (I call it irregular:) ) My question is the proportion between them. If the triangle's corners had been placed on the outher border of the circle (as a ring), I could try to calculate it. But if the question is: Corners of triangle may be in any place in the triangle.

Do you mean any place in the circle?

Otherwise,

1. If all 4 lie on the circle's perimeter, then 100% certainty quadrilateral is convex.
   
2. If 0 or 1, the presence of an arbitrarily large circle doesn't seem all that relevant.
   
3. If 2, symmetry gives you 50% immediately.
   
4. Part of your post suggests 3. But the part in red might be saying 0.
   

[Guest]

Do you mean any place in the circle?

Otherwise,

1. If all 4 lie on the circle's perimeter, then 100% certainty quadrilateral is convex.
   
2. If 0 or 1, the presence of an arbitrarily large circle doesn't seem all that relevant.
   
3. If 2, symmetry gives you 50% immediately.
   
4. Part of your post suggests 3. But the part in red might be saying 0.
   

I meant circle, sorry.

The easy part of my question: 3 points are on the perimeter.

Hard part: None of the points are on the perimeter, but inside the circle.

If 3 p. are on the perimeter, the area of triangle/area of circle might give us the probability. Area of the circle is constant. Area of triangle can be zero or an area of an equilateral triangle. If I simply get average of them I get a proportion of 3*sqrt(3)/8*pi. But this simply average doesn't seem to be the true way.

I couldn't get what Bonanova thought about the hard one.

BTW, I must leave now.

[bonanova]

Calculated probability is 3/4 for 4 arbitrary points within a circle.

Calculated probability is 7/10 for 4 arbitrary points within a square.

Edit: these numbers are wrong.

Corrected results are here.

I'll show my method in a bit; for now, looking at 3 points on perimeter.

Edited November 29, 2008 by bonanova
Incorrect results shown in spoiler

[Prime]

If I understood this problem right…

Take any 3 arbitrary points inside the circle and form a triangle. Probability of the 4th random point inside the circle also to land inside the triangle is the ratio of the area of the triangle to the area of the entire circle.

Any set of 4 points is as likely as another. Thus, the average of the probabilities for all possible sets of points is the probability we seek.

If we express the area of the triangle as a function, then the average probability could be calculated by taking the integral of that function in the interval defined by the circle size and dividing that by the interval length.

Note though, that when we break down a set of 4 points into 3 and 1, we count each 4-poit set 4 times. That’s not a big deal, since we calculate the average, anyway.

Coming up with a manageable function for the triangle area, which could be integrated is another matter. I do not feel like attempting that for now.

Here is one possible route, which does not appear as an easy one. Put the center of the circle into the center of coordinate lines. The sides/area of any triangle could be calculated from the coordinates of the vertices of the triangle.

So, all that seems too complicated. Perhaps, Bonanova could run his simulation software to make an estimate of the probability.

Where did this problem come from? I’m curious.

[bonanova]

If we create a triangle from 3 randomly placed points,

the probability of getting a convex quadrilateral by adding a 4th random point is

the fraction of available area for the 4th point that is outside of the triangle.

.

1. Placing three points at random inside a unit circle - radius=1, area=pi.
   The average area of the triangle formed by the points is .232 [100,000 cases]
   Fraction of circle's area outside the triangle is .926.
   .
2. Placing 3 points at random on the perimeter of a unit circle - radius=1, area=pi.
   The average area of the triangle formed by the points is .478 [100,000 cases]
   Fraction of circle's area outside the triangle is .848.

.

Net:

It's highly likely the quadrilateral is convex: 85% or greater.

The largest possible triangle inside a unit circle is equilateral with points on the perimeter:

area[max] = .75 3^1/2 = 1.299; probability[min] = .586

The smallest triangle has area 0 and probability of 1.

The average triangle area in case 2 is approximately twice that of case 1,

but both are small compared with pi.

I haven't found a closed form for either case.

[Prime]

If we create a triangle from 3 randomly placed points,

the probability of getting a convex quadrilateral by adding a 4th random point is

the fraction of available area for the 4th point that is outside of the triangle.

1. Placing three points at random inside a unit circle - radius=1, area=pi.
   The average area of the triangle formed by the points is .232 [100,000 cases]
   Fraction of circle's area outside the triangle is .926.
   
   
   
2. Placing 3 points at random on the perimeter of a unit circle - radius=1, area=pi.
   The average area of the triangle formed by the points is .478 [100,000 cases]
   Fraction of circle's area outside the triangle is .848.
   

Net:

It's highly likely the quadrilateral is convex: 85% or greater.

The largest possible triangle inside a unit circle is equilateral with points on the perimeter:

area[max] = .75 3^1/2 = 1.299; probability[min] = .586

The smallest triangle has area 0 and probability of 1.

The average triangle area in case 2 is approximately twice that of case 1,

but both are small compared with pi.

I haven't found a closed form for either case.

I don't think that the case where some points are on the perimeter of the circle need to be considered at all. The OP stated circle, not circumference. The probability for 3 random points to land exactly on the perimeter is zero (since the area of the perimeter is zero, mathematically speaking). Also, was there any special check when placing the points inside the circle to ensure they didn't land on the perimeter?

So, I'll go with your first experiment of .926. And I'm still not up to finding the answer analytically. So I'll settle for this numeric solution.

[bonanova]

I don't think that the case where some points are on the perimeter of the circle need to be considered at all.

The OP stated circle, not circumference.

The probability for 3 random points to land exactly on the perimeter is zero (since the area of the perimeter is zero, mathematically speaking).

Perhaps not, but perimeter points were not considered because you requested it.

Check out post #6 on that matter.

I wonder why you draw a distinction between circumference and circle and seem critical of those who don't.

A circle divides the plane into interior and exterior regions; but a circle is defined as its circumference.

A circle is the locus of points equidistant from its center.

Specifically, a circle is not the locus of points whose distance from a central point is less than or equal to a certain value [radius].

That's the definition of a disk.

As to the probability of random perimeter points, two procedures for generating random points were clearly described

1. At random on the circle. r=1; theta=random. Such probability is not zero - it's unity.

2. At random interior to the circle. Taking random points within a double unit square and rejecting them if distance from origin > 1.

In this case it's possible, with a probability of zero, for points to land on the perimeter.

I find that BrainDen'ers generally do not mind discussions of their posts when comments are offered in a positive and collegial manner.

Let's try to encourage that by example. That's what makes this forum as popular as it is.

[Prime]

Thanks for clarifying the terminology.

...

I find that BrainDen'ers generally do not mind discussions of their posts when comments are offered in a positive and collegial manner.

Let's try to encourage that by example. That's what makes this forum as popular as it is.

I included your post into my previous post as a reference to the only solution provided thus far and not as a criticism. Also, my intention was to disambiguate the statement of the problem.

If my comments and testing of the solutions and statements of problems are perceived as offensive and some kind of personal attacks -- that could discourage me from questioning/analyzing other people’s puzzles in future, rather than change my manner of expression.

I still think that 3 points on the perimeter, 1 inside and 4 points anywhere inside the circle -- are two different problems. Both seem difficult. So let me state another problem yet, which seems related, yet simpler:

Corollary Problem:

What is the average distance between two random points inside a circle of a given radius?

[Guest]

Where did this problem come from? I’m curious.

Thanks for clarifying the terminology.

Corollary Problem:

What is the average distance between two random points inside a circle of a given radius?

I'm also curious, are all problems come from a definite source?

I want to tell that I'm astonished about that terminology. In my native language we use two distinct words for a circle which has an area, and a circle which has only a perimeter (as the shape of ring)

I thought to work on Prime's last problem. But first I tried a simple version of it, and derived this one: " If you get two random points on a 100cm. ruler, what is the average distance between these?"

I found a solution, but Prime's one seems still hard to me. Since nobody worked on his problem for 2 days, maybe you want to work my simple ruler problem. Please don't give an only number result, as a simplest software can get it.

[bonanova]

If you get two random points on a 100cm. ruler, what is the average distance between these?"

What is the average difference of the values of two fair dice?

Those differences approximate the distances between two random points along a line of length 6.

Of the 36 possible outcomes, there are two cases where the difference is 5: 1-6 and 6-1.

Similarly, there are four 4's, six 3's, eight 2's, ten 1's and six 0's.

These sum to 70 and average to 1.944 ... and that looks a lot like 2, or 1/3 of our number space.

In fact, if the number space is expanded to 100, [100 dice of 100 sides each] 1.944... becomes 33.33 - a very good approximation to 1/3 of our number space.

That strongly suggests convergence to exactly 1/3 in the limit of zero sampling interval [random real numbers.]

Can we derive this result from first principles?

Sure. Refer to this crude picture of the first randomly chosen point:

0-----------------------------x₁---------------------------------------------------------------------100

Our fist point landed at x₁, and let's define f as the fractional distance from the left end of the stick.

x₁ = 100f where 0<=f<=1.

And now, x₂ lands randomly along the stick.

We can distinguish 2 cases: x₂ lands to the LEFT or x₁ or to the RIGHT of x₁.

Clearly the average point to point distances for these cases occur when the 2nd point lands at the midpoints of these two parts of the stick.

<d_L> = x₁/2 = 100f/2

<d_R> = [100-x₁]/2 = 100[1-f]/2.

By inspection the probability p_L of the 2nd point landing to the left is f; and p_R is [1-f].

The average distance is thus

<d> = p_L <d_L> + p_R <d_R> = [f] [100f/2] + [1-f] 100[1-f]/2 = 100[2f² - 2f + 1]/2

The last step is to find the average of [2f² - 2f + 1]/2 over all values of f from 0 to 1;

that is, over all locations of the first point.

The equation is a quadratic centered at f/2 where its value is 1/4, taking the value 1/2 at f=0 and f=1.

The definite integral of this function from 0 to 1 is, at it turns out, exactly 1/3.

<d> = [1/3] 100, or one third of the length of the stick.

[bonanova]

What is the average distance between two random points inside a circle of a given radius?

Consider a unit circle and its interior 0 <= r <= 1 centered at the origin of a polar coordinate {r, t} system, where r is the distance from the origin, and t is the angle measured ccw from the x-axis.

Without loss of generality let t₁=0 so the first point is at {r₁, 0} [i.e., on the x-axis somewhere between x=0 and x=1.]

Then place the 2nd point arbitrarily at {r₂, t₂} 0<= r₂ <= 1.

The distance between two points in polar coordinates {r₁, t₁} and {r₂, t₂} [setting t₁=0 and letting t₂ = t] is

d = [r₁² + r₂² - 2r₁r₂ cos(t)]^1/2

Now proceed as follows

Average d with respect to t by integrating the exp​ression from 0 to 2 pi radians and dividing by 2 pi. That gives the average distance between two random points at distances r₁ and r₂ from the center of the circle.

Average the angle-averaged d with respect to radius values by noting that the probability of a random point being in an annular region [ring] bounded by r and [r + delta r] increases linearly with r. That simply takes into account that there are more points within a small distance from the circumference than from the center. Multiply the angle-averaged exp​ression by these probabilities and integrate from r=0 to r=1: first with respect to r₁, and then with respect to r₂.

That does it.

If I had the patience, [I don't] I'd attempt that. Since I do have the interest, however, I simulated it:

For a million random pairs of points inside a unit-radius circle,

<d> = 0.905 +/- .001

And just for fun, for a million pairs of points inside the circumscribed square [2 units on a side],

<d> = 1.043 +/- .003 understandably a little larger, given a slightly larger area to roam in.

[Guest]

What is the average difference of the values of two fair dice?

Those differences approximate the distances between two random points along a line of length 6.

Of the 36 possible outcomes, there are two cases where the difference is 5: 1-6 and 6-1.

Similarly, there are four 4's, six 3's, eight 2's, ten 1's and six 0's.

These sum to 70 and average to 1.944 ... and that looks a lot like 2, or 1/3 of our number space.

In fact, if the number space is expanded to 100, [100 dice of 100 sides each] 1.944... becomes 33.33 - a very good approximation to 1/3 of our number space.

That strongly suggests convergence to exactly 1/3 in the limit of zero sampling interval [random real numbers.]

Can we derive this result from first principles?

Sure. Refer to this crude picture of the first randomly chosen point:

0-----------------------------x₁---------------------------------------------------------------------100

Our fist point landed at x₁, and let's define f as the fractional distance from the left end of the stick.

x₁ = 100f where 0<=f<=1.

And now, x₂ lands randomly along the stick.

We can distinguish 2 cases: x₂ lands to the LEFT or x₁ or to the RIGHT of x₁.

Clearly the average point to point distances for these cases occur when the 2nd point lands at the midpoints of these two parts of the stick.

<d_L> = x₁/2 = 100f/2

<d_R> = [100-x₁]/2 = 100[1-f]/2.

By inspection the probability p_L of the 2nd point landing to the left is f; and p_R is [1-f].

The average distance is thus

<d> = p_L <d_L> + p_R <d_R> = [f] [100f/2] + [1-f] 100[1-f]/2 = 100[2f² - 2f + 1]/2

The last step is to find the average of [2f² - 2f + 1]/2 over all values of f from 0 to 1;

that is, over all locations of the first point.

The equation is a quadratic centered at f/2 where its value is 1/4, taking the value 1/2 at f=0 and f=1.

The definite integral of this function from 0 to 1 is, at it turns out, exactly 1/3.

<d> = [1/3] 100, or one third of the length of the stick.

Your dice solution is like using an abacus, and doesn't make sense. Your other solution is certainly correct but I couldn't be sure that taking simply average would be correct, as you did:

The average distance is thus

<d> = p_L <d_L> + p_R <d_R> = [f] [100f/2] + [1-f] 100[1-f]/2 = 100[2f² - 2f + 1]/2

So, I choosed another method, tough it may seem you as dummy :

If we divide the ruler to n points, and take the cases that first point is on near the beginning of the ruler (at 1/n point), according to the place of second point, the average of possible distances will be = 1/n * (0+1+2+..+(n-2)+(n-1)).

If we calculate this for the case that first point is at 2/n---> 1/n * (1+0+1+2+...(n-3)+(n-2)).

3/n ---> 1/n *(2+1+0+1+2+....(n-4)+(n-3))

4/n----> 1/n *(3+2+1+0+1+2+....(n-5)+(n-4))

.........

n/n----> 1/n *((n-1)+(n-2)+...+2+1+0)

if we formulate it:

k/n --> 1/n *((k-1)+(k-2)+....+1+0+1+2+...(n-k+1)+(n-k))

or

k/n---> 1/n * ((k-1)k/2 + ( n(n-k) + n(n-1)/2 ))

If we get sum for all k values: (for all possible positions of first point)

E(sigma)-----> (for k=n) 1/n* E (n² -n +0+ n² -n)/2

E (n² - n) = n(n+1)(2n+1)/6 - n(n+1)/2

For getting average, we must divide it with n, and for calculating its rate in the terms of n unit, we must divide again with n.

Now the upper part of the fraction is n³+..., while the lower part is 3n³. (I don't take lower powers of n into account),

Thus the result is 1/3.

[bonanova]

I couldn't be sure that taking simply average would be correct, as you did:

For every point on one side of the midpoint there is a mirror point on the other side.

The average distance to that pair of points - and thus to all points and their mirrors - is the distance to the midpoint.

Consider:

5 = <5>

5 = <4 5 6>

5 = < 3 4 5 6 7>

5 = < 2 3 4 5 6 7 8>

5 = < 1 2 3 4 5 6 7 8 9>

5 = <1 2 2.537 3 4 5 6 7 7.463 8 9>

etc.

to take into account the freedom of choosing the first point, it's basically what we mean by expectation. The first point could have been placed in the middle, [f=1/2], or at symmetrically placed points about the middle, all with equal probability. It's easy to average f to get 1/2, certainly.

But the average point to point distance for all x1 and x2 values is not the average distance to x2 for an initial point at f=1/2; it's given by [2f2 - 2f + 1]/2. So that's what has to be averaged to get the expectation for the point to point distance. Taking the integral and dividing by the full distance does that, and the result of doing that is 1/3.

If you want more about how to calculate expectation, there are Wikipedia articles on the subject.

Edited December 2, 2008 by bonanova
Added the text about [2f2 - 2f + 1]/2

[bonanova]

Your dice solution ... doesn't make sense.

Each die has a random value from 1 to 6.

Rolling two dice is like selecting two random points on a line of length 6.

The difference of the die values is like the distance between the two points.

It's approximate, because it's discrete.

It restricts the places on the stick that can be selected.

If you wanted to select 3.14159265356, for example, you'd have to choose 3 or 4.

So it gives an approximate result: 1.944/6 = 0.324... instead of 0.3333....

It was the first thing I thought of to get a sanity check on the correct result, and it was fairly close.

[Guest]

Each die has a random value from 1 to 6.

Rolling two dice is like selecting two random points on a line of length 6.

The difference of the die values is like the distance between the two points.

It's approximate, because it's discrete.

It restricts the places on the stick that can be selected.

If you wanted to select 3.14159265356, for example, you'd have to choose 3 or 4.

So it gives an approximate result: 1.944/6 = 0.324... instead of 0.3333....

It was the first thing I thought of to get a sanity check on the correct result, and it was fairly close.

Your symmetric solution is nice. To solve my own problem, I used messy chore operations. Whereas you got it simply. Although you used integral (which I'm not well familiar with) at the end, I suppose someone can get the same result with my sigma (sum) method.

Your dice method is surely correct. What I meant was that: If I had been a math. teacher, I wouldnt accept it, because it doesn't prove that ultimate result is 1/3.

[bonanova]

Your symmetric solution is nice. To solve my own problem, I used messy chore operations. Whereas you got it simply. Although you used integral (which I'm not well familiar with) at the end, I suppose someone can get the same result with my sigma (sum) method.

Your dice method is surely correct. What I meant was that: If I had been a math. teacher, I wouldnt accept it, because it doesn't prove that ultimate result is 1/3.

Agreed.

[1] Integrals and sums are closely related; the smaller the intervals in the sum, the closer to an integral it gets.

[2] Dice actually is a good teaching technique.

You'd like students to be able to make comparisons to simpler tasks.

Suppose I had the correct formula and worked through 7 pages of calculations, missed a factor of 2 somewhere and ended up with 2/3 as the result. If I thought the dice [or some other quick and dirty method] would give an approximate result and found .324, it could suggest that 1/3, not 2/3 is the correct answer and give me the encouragement I'd need to look for an extra factor of 2 in my calculations.

That's what I meant by sanity check....

Nice puzzle.

[Prime]

My reasoning for the average distance between two random points is similar to that already posted here. However, I am going to show it anyway, since I think it may help to solve the distance between two points in a circle problem. And, in time, the original problem in the OP.

The case with a die has 6 discrete numbers (steps). Let’s consider a unit segment divided into some large number of equal steps of size dx (delta x).

+----------+----------+------… ------+----------+

0<--dx-->---------------------------------------> 1

Consider a distance between two points segment of the length x, where 0 < x <= 1 and x is a whole number of steps dx. Such segment can be placed inside the unit segment (1 – x) /dx times aligning its edges on the dx boundaries. That is the longer the segment, the less ways to place it inside the unit segment. For example, if the segment is equal to the entire unit, there is only one way to fit it. If the segment is just one step (dx) shorter, there are two ways to fit it: starting at 0 and starting at 1, if segment is 2 steps shorter, there are 3 ways to place it, and so on.

+----------+----------+------… ------+----------+

0<------------(1 - 1dx)------------->________1

0________<------------(1 - 1dx)--------------->

Thus the function representing the number of placements for the segment of the length x inside a unit segment is (1 - x)/dx. That divided by the total possible placements of all possible segments would represent probability/ratio for the segment of the length x.

The total number of placements is the sum of all possible values for (1 - x)/dx. When we break down the interval into more and smaller steps, dx tends to zero, such sum is represented by integral. So the Integral of (1 - x)dx where x ranges from 0 to 1 represents total placements of all possible distances.

Intgr (1 - x)dx = x - x²/2.

Plugging in the boundaries, we get 1 - 1/2 - (0 - 0) = 1/2.

That represents total placements of all segments.

The average length is the sum of all possible lengths times their probabilities.

So if each length is represented by x and its probability by (1 - x)/(1/2) dx then such a sum would be the Integral x*(1 - x)/(1/2) dx.

Solving we find: Intgr (2x - 2x²) dx = x² - 2x³/3.

Plugging in the boundaries for x (0 and 1), we find: 1 - 2/3 - 0 = 1/3.

From this we could move to constructing a similar solution for the distance between two points in a circle. And from there to the original problem -- distance between a random segment in a circle and anohter random point. From which we could find triangles' area and thus the probability of the 4th random point falling inside it.

[Guest]

Awesome problem!!!

I won't rehash the proof that the average distance between two points selected at random on a segment is 1/3 the length of the segment. That's been shown.

So, knowing that, we can move on to the average distance between two points selected at random from the unit disc. This must be 1/3 the average distance across the circle. I decided to use all the vertical chords within the unit circle:

Avg chord length = (1/2)[integral from -1 to 1 of 2(sqrt(1-x^2))] = Pi/2

Thus, the average distance between two points at random within the unit disc is (1/3)(Pi/2) = Pi/6.

If you rotate the circle so that the first two random points are horizontal, then the base of the triangle is Pi/6. When we choose the third point, it must be, on average, Pi/6 units from the base. So, the area of the triangle is (1/2)(Pi/6)(Pi/6) = (Pi^2)/72.

The area outside the triangle is: Pi - (Pi^2)/72 = Pi [1 - Pi/72]

Thus, the ratio of the area outside the triangle is: (Pi [1 - Pi/72]) / (Pi) = 1 - Pi/72.

This is approximately 95.64% which seems to match the simulated result fairly well.

:wub: This problem :wub:

[Prime]

Awesome problem!!!

I won't rehash the proof that the average distance between two points selected at random on a segment is 1/3 the length of the segment. That's been shown.

So, knowing that, we can move on to the average distance between two points selected at random from the unit disc. This must be 1/3 the average distance across the circle. I decided to use all the vertical chords within the unit circle:

Avg chord length = (1/2)[integral from -1 to 1 of 2(sqrt(1-x^2))] = Pi/2

Thus, the average distance between two points at random within the unit disc is (1/3)(Pi/2) = Pi/6.

If you rotate the circle so that the first two random points are horizontal, then the base of the triangle is Pi/6. When we choose the third point, it must be, on average, Pi/6 units from the base. So, the area of the triangle is (1/2)(Pi/6)(Pi/6) = (Pi^2)/72.

The area outside the triangle is: Pi - (Pi^2)/72 = Pi [1 - Pi/72]

Thus, the ratio of the area outside the triangle is: (Pi [1 - Pi/72]) / (Pi) = 1 - Pi/72.

This is approximately 95.64% which seems to match the simulated result fairly well.

This problem

This is an interesting solution! Much simpler than I thought. But, I think it needs a bit more work.

The average segment length over an average chord seems to represent the average distance between two random points inside a circle. Although not all chords are of equal probability. When you consider a set of all parallel chords, each length occurs twice -- once on either side of the circle with a diameter as symmetry axis. The diameter, however, occurs only once. But since the diameter is just a line with a zero area, it may not affect the overall average length.

The second part -- distance between random point and an average length segment inside the circle may be a little off. I think that the actual average may be a bit bigger. The reason: not every possible distance belongs to a chord inside the circle. Here is an illustration:

[Guest]

The second part -- distance between random point and an average length segment inside the circle may be a little off. I think that the actual average may be a bit bigger. The reason: not every possible distance belongs to a chord inside the circle.

Hrm...I've got to think on that a bit longer...something nags at me saying it's irrelevant...but....meh.

[bonanova]

Determine the probability p(convex) that 4 random points placed on the unit disk form a convex quadrilateral.
Here is an algorithm for averaging the exact result for a million random triangles.

[1] Create a random triangle on the unit disk.

Choose x at random from 20000 equally spaced points in [-1, 1]
Choose y at random from 20000 equally spaced points in [-1, 1]
Reject the pair if x²+y² > 1.
Repeat twice to get 3 random points P₁[x₁, y₁], P₂[x₂, y₂], P₃[x₃, y₃] that define a random triangle T [P₁ P₂ P₃] on the unit disk.

[2] Determine the region where a 4th point creates a convex quadrilateral

Referring to the figure, we see the quadrilateral is convex iff the 4th point falls outside the red area. For the case where T is an inscribed triangle, it is the only red area. That is, when a vertex P_i lies on the perimeter, T_i and S_i vanish. But when P_i is an interior point, they must be taken into account: subtracting only the average value of T [.232, or about 7% of the disk] gives a convex probability about 93%. That is clearly an optimistic estimate.

[3] Find the red area.

Extend the sides of T to intersect the circle at the six points marked CP.
Calculate the areas T, T₁, T₂, T₃ of the four triangles [P₁, P₂, P₃], [P₁ CP₁₂ CP₁₃], [P₂ CP₂₃ CP₂₁], and [P₃ CP₃₁ CP₃₂]


.. . . | x₁ y₁ 1|
T = 1/2| x₂ y₂ 1|
. . .. | x₃ y₃ 1| , etc.

Determine the angles t₁, t₂, t₃ subtended by the pairs [CP₁₂ CP₁₃], [CP₂₃ CP₂₁] and [CP₃₁ CP₃₂] respectively
Calculate the areas S₁, S₂, S₃ of the circular segments.

S = 1/2 [t - sin(t)]

RedArea = S₁ + S₂ + S₃ + T₁ + T₂ + T₃ + T

[4] Find the convex probability

The the convex probability is the fraction of the unit disk area [pi] that is green: GreenArea = pi - RedArea
p(convex) = GreenArea/pi.

[5] Find the expected probability for any T.

Repeat for 100,000 triangles T. Average the results.
Repeat ten times and average the averages - one million total cases.
Result: <p(convex)> = 0.7044 +/- 0.0005

This result looks suspiciously like 0.7071067812 = sqrt(2)/2, which suggests the problem may have a closed form solution.

Finally, the average size of T was noted: <T> = 0.2321 +/- 0.0002 [7.388 % of the disk]

[bonanova]

Awesome problem!!!

I won't rehash the proof that the average distance between two points selected at random on a segment is 1/3 the length of the segment.

That's been shown.

So, knowing that, we can move on to the average distance between two points selected at random from the unit disc.

This must be 1/3 the average distance across the circle.

I decided to use all the vertical chords within the unit circle:

Avg chord length = (1/2)[integral from -1 to 1 of 2(sqrt(1-x^2))] = Pi/2

Thus, the average distance between two points at random within the unit disc is (1/3)(Pi/2) = Pi/6.

The average length of the sides of the 3 million triangles of the previous post [random vertices] was .664....

That is just slightly less than 1/3 of the diameter, not of the average chord length.

If that helps our thinking.