If d+ 1 points are randomly distributed on the (d-1)-sphere in R^d (d=1,2,…), then the convex hull of the d+1 points encloses the center of the (d-1)-sphere with probability 1/(2^d).

The proof is trivial for d=1 and d=2. In 1975, I ‘verified’ the conjecture for d=3 with 10^7 Monte Carlo runs. I cited this result on my website schoengeometry.com (in a footnote in the paper on the Isoperimetric Problem for Polyhedra).

I’m still clueless about how to prove this conjecture.

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I assume that every one of the eight associated triangles is equally likely. This is not at all obvious.

As each particular triangle has a probability of zero it isn’t clear how to compare the probabilities of particular triangles.

Thus Stephen Morris. Can’t we justify the equal probability of the eight triangles based on the equal probability of the points? Those are zero as well, of course, but at least the original problem statement tells us they’re chosen “at random”, which in context I think we can safely take to mean with equal probability.

The zero probability leads me, circularly, back to the “infinite plane” issue, though, and the practical (at least) impossibility of choosing a random real within an interval, let alone uniformly from the set of all reals.

I suppose I needn’t obsess about reals; Carroll’s problem makes sense with a discrete non-dense coordinate grid, after all (though then we can’t necessarily draw our circle, can we?). But picking a random point from a discrete but infinite number line is problematical as well.

What strikes me as particularly elegant about the eight-associated-triangles argument is that it neatly avoids the infinity and zero-probability snarls. Yes, the points are each of zero probability, but unless we toss out the problem statement altogether we have to at least assume that they’re equally probable, and that “equal probability” has a meaning here. (I’m a little doubtful, but I don’t see how any approach to the problem makes sense without that premise.)

Equal probability points ought to give us equal-probability triangles, and if we exclude from consideration right triangles and concern ourselves only with the ratio of acute to obtuse triangles, 1:3 seems fairly convincing, with minimal need to arm-wave away the infinity problems.

The fact that we’re in the same solution ballpark as the Monte Carlo and other finite-area results is of some comfort.

Jeff presented us with one of Caroll’s problems and his neat solution. We have picked apart the problem. Since then we have been searching for a way of reforming the problem so that it is true to the spirit of Caroll’s problem. Ideally we would like Caroll’s solution to be valid. That’s the ultimate goal.

This is rather like the Hitchhiker’s Guide to the Galaxy, we know the answer is 42 but what, exactly, is the question?

How about this for an approach. It is flawed, of course, but teasing out why is getting harder.

Every triangle is associated with a circle where the points of the triangle lie on the circumference of the circle.

Take a circular subset, S, of the infinite plane with radius R centred on some point. Choose three points within S at random to make a triangle, where the chance of a particular point lying within any subset of S is proportional to the area of the subset. If the associated circle does not lie entirely within S then throw this triangle away and start again.

By the argument in my previous comment the triangle is part of a group of 8 triangles of which 1/4 are acute and 3/4 obtuse. All are valid because they share the same associated circle.

For any triangle there is some minimal value of R for which it is valid. It will be valid for all larger values of R.

The proportion of acute and obtuse valid triangles are unchanged as we increase R, it is always 1:3.

Let’s consider some objections.

1. Throwing away triangles where the associated circle crosses the boundary of S might seem to distort the results.

We could combat this by adding another condition: “The radius of the associated circle must be less than sqrt(R).” This means that the likelihood of the associated circle crossing the boundary of S reduces to zero as R tends to infinity.

It is still true that any triangle will be possible for all values of R greater than some value. I believe that fact makes this argument valid.

However, perversely, the chance of having to throw away a randomly chosen triangle because the associated circle is too large (radius>sqrt(R)) increases to 1. I’m not sure what to make of that! I hope I’ve got it wrong. I don’t think it breaks the argument but it is very inconvenient!

2. I assume that every one of the eight associated triangles is equally likely. This is not at all obvious.

As each particular triangle has a probability of zero it isn’t clear how to compare the probabilities of particular triangles.

Again let’s change the method slightly. I take my inspiration from the Monte Carlo method discussed by Mike. Thanks Mike!

We said before that the probability of choosing a point in some subset of S is proportional to the area of the subset. Choose 3 subsets of S; H1, H2 and H3; so that choosing one point from each subset always gives a valid triangle (the associated circle is within S and the radius is less than sqrt(R)). The probability of a chosen triangle having one point in each subset is proportional to the product of the areas of H1, H2 and H3.

Taking our inspiration from Mike and Monte Carlo we could approach it this way. Divide S into a rough grid (it will have to be rough to ensure each section has an equal area, maybe I should have made S a square). Choose three sections from the grid at random to be containers for the three points of the triangle. If these three sections ensure that we have a valid triangle and we can determine if it is obtuse or acute then stop and count 1 to obtuse or acute. If it is not possible for any triangle chosen from these sections to be valid then start again. Otherwise subdivide each section into smaller sections, again with equal area. Repeat until we have an answer.

This method is consistant choosing points so that the probability of a point occuring in a subset of S is proportional to the area of the subset.

There are two situations in which we will have to go to a very fine level of granularity before we can get an answer. One if where the three points are very close together, and almost form a line. The other is where two points are close together and the third is some distance away but almost equidistant. I think these two situations are equivalent. Take the second case, reflecting the distant point across the centre of the associated circle will bring it pretty much into line with the other two.

Therefore I think that the probability of coming up with any of the eight associated circles is pretty much the same.

I’m not sure if I’ve come up with a genuine advance here or if the same fundamental problem exists and I’m just obscuring it with clever sounding stuff.

]]>We have a strong intuition, after all, that if we choose an integer from the set of all integers, we have a 1/10 chance of picking an integer evenly divisible by 10, even though we there’s a 1:1 correspondence between them and the other 9/10 of the integers, or all the integers. So at least in that sense, 3/4 obtuse seems pretty compelling.

(I notice that “cute” is an unintentional pun, being a shortening of “acute”.) ]]>

The eight triangles ABC, ABC’, AB’C, AB’C’, A’BC, A’BC’, A’B’C and A’B’C’ are all centred on X. The operation of replacing one point with the opposite point on the circle will give one of the other eight triangles. The triangles are paired, for example ABC is the same as A’B’C’, just rotated 180 degrees about X.

Of these four pairs of triangles:

One is acute

One has A/A’ obtuse

One has B/ B’ obtuse

One has C/ C’ obtuse

So we have a way of splitting all triangles into groups of eight, of which one quarter are acute.

This doesn’t get round the problem that the problem is ill-defined. Since there are an infinite number of acute and obtuse triangles we could group them in different ways and get a different answer. For example we could pair every acute triangle with an obtuse triangle and get an answer of 1/2.

]]>Let’s reign in the insanity of this by recalling a few old classic theorems.

AAA theorem states that two triangles that have the same angle measures are similar (not necessarily congruent). Next, (I don’t recall the name of the theorem) states that any three points are cocircular.

Considering this, any circle drawn on a plane will have an equal distribution of triangles of all arrangements of angles. Therefore, any circle will represent the distribution of all other sets of triangles sharing circles on an infinite plane.

The position of a triangle relative to its circumscribing circle has no weight on its angles, we can also fix the position of any triangle in question and still represent the distribution of all triangles in the circle

Now that we have fixed some factors, let’s draw a circle on a cartesian plane and put point A on the positive X axis at (1,0).

For a triangle to be obtuse, assume a line drawn through the center point of our cartesian plane, which splits the circle into two. Let’s fix this line also, at the X axis. To produce an obtuse triangle, both points must be on either the positive plane or the negative plane. Landing a point on the opposite side of the circle (-1, 0) will produce a right triangle. However, with an infitnite number of points on the circle, the odds of landing that point with B or C is lim(n->infinity) 1/n. Let’s set this value equal to P(R) (probability of a right-triangle event)

Point A is fixed. Point B is arbitrarily on the circle, so the odds of point B being on the circle is 1. The odds of point C being on the same side as point B is 1/2 (equal sides of the circle).

Therefore, the odds of an obtuse triangle are:

P(A) * P(B) * P(C)

1 * (1-P(R)) * (1 -P(R)/2)

Extrapolate the limit in P(R) to equal zero, and that equals 1/2, but we can’t forget that you can still get a right triangle. ]]>

I drew a picture. The picture assumes that we’ve done a coordinate transformation to make AB a unit line and put C (unshown) in the first quadrant.

A little googling shows that a lot of ink has been expended on this particular problem. The only universal conclusion seems to be that Carroll’s solution is mistaken, and that the stated conditions are at best problematical–ill-posed, in Mike Jarvis’s words.

One semi-satisfying approach to the problem is to change the conditions, implicitly or explicitly, from “infinite plane” to a large but finite plane with a discrete coordinate system. If we do that, though, it appears that the probability of an obtuse triangle depends on the shape of the plane’s boundary (is it an arbitrarily large circle, or square, or what have you?), as Mike again points out. This very dependency ought to make us suspicious of extrapolating any answer to the “infinite plane”.

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Picture two lines perpendicular to segment AB, one which goes through A and the other which goes through B. These two lines divide the plan into three regions, namely, the region between the parallel lines, the region on the outside adjacent to A and the region to the outside adjacent to B.

Any point C that falls on the parallel lines will form a right triangle. Any point C that falls between the lines will form an acute triangle. Any point C that falls outside the two lines forms an obtuse angle. We ignore the possibility that the three are collinear.

So the probability would be the ratio between the two “outside” regions and the whole space.

I think… ]]>