
Re: Random Triangle Problem (LONG summary)
Posted:
Aug 7, 1997 10:06 PM


This is rather long because I have summarised some of the high points of the discussion as well as filling in the proof for the normal case. If you are only interested in this case, you can skip to the end, pausing to look at uniform angles on the way.
There are two kinds of mathematician. Those with insight sometimes end up producing proofs that Paul Erdos would have said are "from the book", but they can easily make mistakes because insight is based on intuition and experience, and one will often have the former before one acquires the latter. Then there are the plodders who sometimes get the right answer in the end from long winded calculations, but have no insight. (That seems to fit one correspondent in this discussion, but I'm not saying who. It is sufficient to say that he hasn't got the right answer yet). You'll see a bit of plodding and a bit of insight below, I think.
Bill Taylor quite rightly pointed out that the probability that a random triangle has an obtuse angle depends upon what is meant by "a random triangle". If you ask someone to draw a random triangle, I have a hunch that you will usually get an acute angled triangle. On the other hand, people are notoriously bad at judging randomness, so this is not a good way to decide what a random triangle is.
While I agree that randomly and independently choosing the angles of the sides to a fixed direction gives a reasonable definition, I want to explore other definitions. Robert Hill has done 1 million simulations for several possible definitions, and I have repeated the experiment with 10 million trials. Based on the binomial distribution, this gives 3 digit accuracy with almost certainty, but the 4th digit is unreliable. To get the 4th digit requires 1000 million trials.
Keith Ramsay has drawn our attention to an American Mathematical Monthly article which has discussed this question and looked at several distributions for the vertices. That's a pity as I was thinking this topic would make a nice Monthly article (jointly with those who have made substantial contributions and acknowledging others, of course). One case they cosidered is when the vertices are independently uniformly distributed over an nsphere. It looks to me as though this should still lead to the probability 3/4 for an obtuse angled triangle. The three vertices determine a plane and that plane intersects the sphere in a circle. Given the plane, the vertices are uniform over the circle and Ilias Kastanas has shown (see below) that this leads to the 3/4 answer. As this is the same for all planes, that is the answer unconditionally too. Similarly, a uniform distribution over the inside of the sphere should lead to the same answer as for a uniform distribution over the inside of a circle. This seems to contradict the article, so what is wrong with this reasoning? I would appreciate any explanation of this.
I will summarise what has been done by other posters as well as adding some results of my own.
First, a random triangle should have a random orientation, but, as the orientation has no effect on the shape, I will also try definitions in which this requirement is dropped. However, I must insist on exchangeability of edges and vertices, that is, the distributions of edge lengths are identical; and so are the joint distributions of lengths of pairs of edges. Similarly for angles. Without exchangeability the angles would be distinguishable in some way other than by their size. This could occur if the angles or vertices were chosen sequentially from different distributions, but this would not lead to a random triangle by most people's standards. Tony Richards is the only exception I know of. I looked at three classes of distribution: defining a triangle (up to similarity) by its angles, defining it by its edge lengths, and defining it by the coordinates of the vertices. For each class I considered several distributions.
As angles are bounded, it would make sense to use a uniform distribution, except that the sum of the angles is pi. This constraint makes independence impossible because, if one is greater than pi/2, the others must be less than pi/2. But we can make the angles of the three sides to a given fixed line uniformly and independently distributed. This was essentially Bill Taylor's solution (he chose the sides perpendicular to three lines drawn at random angles).
For angles, the uniform distribution is attractive as, among all distributions on a bounded range, the uniform distribution has maximum entropy, where entropy is defined to be Shannon's information, the expected value of the log of the density. When I consider lengths of sides or the coordinates of the vertices, bounded distributions are no longer appropriate. Instead I shall fix the standard deviation. This chooses a scale factor but does not affect the shape. For the sides, the distribution is unbounded above, and the maximum entropy distribution is exponential. For the coordinates of vertices, the distribution is unbounded above and below and the maximum entropy distribution is normal.
However, not too much should be made of this as the maximum entropy property is not invariant to transformations. But then, nor is any definition of randomness and this is precisely why this question is illposed.
Ilias Kastanas has given an interesting proof that the probability of an obtuse angle is 3/4 if the vertices are distributed independently and uniformly on a circle. Let the vertices be A, B and C. The triangle is obtuse angled if all the vertices are on the same semicircle. In that case there are 3 mutually exclusive events depending whether A, B or C is in the middle. Given the point A, the 3 are on the same semicircle with B in the middle if B and C are on one of the two semicircles with A at the end (probability = 1/2) and with B between A and C (probability = 1/2 by exchangeability). As these events are independent, the probability that B is between A and C, given A, is 1/4. This is independent of A so the result follows.
Suppose the edges are at random uniform angles. Using Bill's idea, translating an edge has no effect on the shape of the triangle (unless the triangle becomes degenerate), so we could translate them so they touch a circle of radius 1 (on either side). The points of contact would then be uniformly distributed around the circle and could be described by angles between 0 and 2pi. The circle could be either an incircle of the triangle or an escribed circle. In the latter case, translating _one_ of the three sides to the opposite side of the circle makes the circle into an inscribed one, while, in the former case, translating _any_ side to the opposite side of the circle makes the circle an escribed one. By Ilias's argument for the circumcircle, the probability that this gives an incircle is 1/4. (I think Charles Giffen was the first to point out in this thread that the circle is not always an incircle).
Because of complication about whether the circle is an incircle or not, I will work with the angles of the sides to a fixed direction instead. We can describe these as independently uniform between 0 and pi. The joint distribution is uniform on a cube with sides from 0 to pi. That is the density is 1/pi^3. The joint distribution of the order statistics, U, V, W, has density 6/pi^3 with 0 < U < V < W < pi. That is they are uniform over a certain tetrahedron. The joint density of U, A = VU and B = WV, where U defines the orientation and the A and B are two interior angles of the triangle, is 6/pi^3 with 0 < U + A + B < pi, U > 0, A > 0, B > 0. So the marginal density of A and B is 6(piAB)/pi^3 with 0 < A + B < pi, A > 0, B > 0. The marginal density of A is then 3(pi  A)^2 /pi^3, 0 < A < pi, and the probability that A > pi/2 is 1/8. Similarly, the probability that B > pi/2 is 1/8. Now here comes a surprise. The probability that the third angle, C, is greater than pi/2 is the probability that A + B is less than pi/2. This is not 1/8 but 1/2 as you can show by direct integration. This gives the required result. Why the asymmetry? It was implicit in sorting the angles in order of size. Angle C is supplementary to the difference between the greatest and least of the three angles to a fixed line, and this introduces the asymmetry.
This argument is not as neat as Bill's but it's good old fashioned straightforward analysis.
To restore the symmetry we have to think about permuting the vertices. The joint density of A and C is 6C/pi^3, 0 < A + C < pi, A > 0, C > 0 because A and B gives identical information to A and C and the transformation is linear (Jacobian = 1). As all permuations of the vertices are equally likely, we want an equally weighted mixture of the 6 densities obtained by interchanging A, B and C. I.e. (2(piAB) + 2A + 2B)/pi^3 = 2/pi^2 and The joint density of A and B is uniform over the the region 0 < A + B < pi, A > 0, B > 0.
This gives a much easier way to get the result. It might be possible to preserve the symmetry throughout the argument. But, anyway, note that the marginal distribution for any angle is triangular. That is why each angle has less than an equal chance of being obtuse. By integrating the triangular distribution directy, it is easy to see tha the probability os 1/4 for the obtuseness of each vertex.
One can represent a set of variables with constrained sum using homogeneous coordinates. For three variables, represent the point as a point in a cube. This may not satisfy the constraint so we project this point onto the plane containing three vertices that does satisfy the constraint. The part of this plane in the unit cube is an equilateral triangle and the coordinates are proportional to the distances of the point from its edges. This is essentially the coodinate system above, except that the triangle has been distorted. (In fact, the projection of the triangle on each coordinate plane gives the coordinate system above for each pair of vertices). Our constraint is that the sum of the angles is pi, and we see that we have captured the idea of a constrained uniform distribution very well. The marginals are triangular when there is a constraint.
One answer to the meaning of a random triangle is therefore that the angles are uniform when expressed in homogeneous coordinates. In other words we choose random uniform angles from 0 to pi and then scale them so that the sum is pi.
We could also choose the side lengths at random. As a scale factor doesn't affect the shape of the triangle, we can condition on any given perimeter. Whatever the distribution of the perimeter, we will get the same probability of an obtuse angle if, given the perimeter, the event "triangle is obtuse" is independent of the distribution of the perimeter. Then it is sensible to make the conditional distributions of the side lengths uniform subject to the constraint on the sum and subject to the triangle inequality. This suggests using homogenous coordinates (which was first proposed by a correspondent from the Netherlands, I forget who) and then rejecting "triangles" which don't satisfy the triangle inequality. This means choosing a point uniformly in an equilateral triangle of height 1 and returning the homogeneous coordinates if the point is in the middle triangle in the partition of the triangle into four. This is equivalent to choosing uniformly in a triangle 1/2 the size, and returning 1/2  the usual coordinates. A simulation of this gave a probability for obtuse triangles of 0.682.
However, the definition of a random triangle ought to be generalisable to random polygons. While basing the definition on angles or sides defines classes of similar triangles, it doesn't define classes of similar polygons. However, if we are only interested in properties of the angles of polygons, the exterior angles may be sufficient. If, on the other hand, we want to determine the shape, then the lengths of the sides are also needed. We might require that the lengths of the sides should have uniform distributions, but a uniform distribution on an infinite range does not exist unless we weaken the axioms of probability either by only requiring finite additivity, or by allowing probabilities to be greater than 1. Also, the lengths of the sides do determine a triangle, but do not determine a polygon.
Instead let's choose the vertices at random. This corresponds to the original question. We could choose the coordinates of each vertex from a bounded uniform distribution, but this would mean that the vertices are chosen randomly from a square. This would not be very satisfying. We would like a distribution with circular symmetry. If the coordinates are independent then, apart from a degenerate distribution at the origin, this implies they have the same normal distribution. In addition we will assume that all the vertices are chosen independently. We have already remarked that this gives maximum entropy, a property that no longer applies in polar coordinates. However, the polar form of the distribution of each vertex does have maximum entropy if the coordinates are taken to be the angle and the square of the radius.
We would certainly like circular symmetry. If we want to weaken the normality condition we have to allow the coordinates of each vertex to be dependent. In polar coordinates, we want the radius and angle to be independent. One attractive possibility is to choose the vertices randomly in a circle. This will achieve maximum entropy for the bounded range and this was the distribution Ilias considered.
It is difficult to imagine any other vertex distributions apart from the two given above that would correspond to our intuitive concept of random triangles or polygons.
A difficulty with random polygons is that the vertices do not determine a polygon uniquely. If we join the vertices in an arbitrary order there will be a high probability of producing a selfintersecting polygon. Suppose the vertices are joined so that, in polar coordinates with the origin at the centroid of the points, each point is joined to the next in cyclic order. An interesting question might be to find the probability that the polygon is convex? I must confess, that, if we are only interested in the angles, the lengths of the sides are not important, so two polygons that have the same angles, in the same order could be regarded as equivalent for some purposes. As the sum of the exterior angles of any nonselfintersecting polygon is 2pi, we could just try to choose the angles as uniformly as possible subject to this condition. We could do this using homogeneous coordinates.
To return to triangles (I won't consider other polygons), the coordinates of the three vertices form a six variate normal distribution. Without loss of generality I will assume they have independent standard normal distributions. Changing the mean or the variance only changes the location or scale (so long as the variances are all equal).
One might imagine that for any circularly symmetric distribution that the angles of the triangle are uniformly distributed. They are certainly not independent as their sum is pi. In fact this leads to the correlation between angles being 0.5 whatever their distribution so long as they are identically distributed. We have already seen, however, that a triangular distribution occurs under some circumstances, and I shall show that this is true for a normal distribution.
Let the vertices be vectors A, B and C. (This is a change of notation, these are no longer the angles of the triangle). These form a 6 variate normal distribution with all variables independent. The sides of the triangle are BA, CB and AC and each coordinate has correlation 0.5 with the corresponding coordinate of the others (but the x coordinates are all independent of the y coordinates). However, it is easier to condition on A. The conditional distribution of B and C given A is normal and the 4 components are all independent. But we are now regarding A as fixed, so the distribution of U = BA and V = CA can be found by substitution. U and V are conditionally independent even though, unconditionally, their components have correlation 0.5. If we use polar coordinates centred on A it follows that, conditionally, U and V have rotationally symmetric independent distributions. Thus we are in the case of uniformly and independently distributed angles. The angle between these then has the same triangular distribution discussed previously and, as this is independent of A, the probability that angle A is obtuse is 1/4. This proves the result.
Alternatively, we can condition on the mid point of BC and the vector CB. Then A will have a bivariate normal distribution, and using the fact that the squared length of A  (B+C)/2 has an exponential distribution, we can find the probability that A is in the circle with BC as diameter. This should lead to the same answer.
Some might wonder whether the first argument would apply to any distribution with circular symmetry. It doesn't. We used the properties of conditional distributions based on a multivariate normal distribtuion. These do not apply to other distributions. For example, for a uniform distribution over a circle, if A is near the circumference, then B and C are very much constrained and the argument breaks down. The point is that the conditional distributions are no longer symmetric.
I have done the theory above for those distributions that interest me. For these and some other distributions I carried out 10 million random trials with the following results:
square circle triangle angles sides normal Cauchy 0.725 0.720 0.748 0.750 0.682 0.750 0.833
The first three are uniform distributions of vertices over the respective shapes (the triangle being equilateral), "sides" means sides uniform in homogeneous coordinates as discussed above, and "Cauchy" means independent Cauchy distributed coordinates. I didn't look at truncated Cauchy radius and uniform angles.

Terry Moore, Statistics Department, Massey University, New Zealand.
Theorems! I need theorems. Give me the theorems and I shall find the proofs easily enough. Bernard Riemann

