What's the chance of an acute triangle?

[nuzzle]

I've posted in this thread,

http://www.codeguru.com/forum/showth...ht=acute+angle

but it was a while ago and the OP never came back and it's more of an algorithm question so I continue in a new thread here.

Say you generate the three corners of a triangle in a plane at random. What's the chance the triangle is acute?

I've presented a solution already but I think I have a better one this time.

[nuzzle]

Some consider the problem bad-posed.

A random triangle is generated by the way of three random points and one cannot really generate random points in an infinite plane, or can one? I think so. Maybe not in practice but conceptually.

Say you have an urn filled with N black and white balls. You start drawing balls from the urn. You notice the color and put it back. After a while you detect that one in four is white. According to the frequentist definition of probability this means the chance of a white ball is 1/4. Now someone tells you there aren't N balls in the urn but an infinite number. Does this change anything? Not in my view. The chance of drawing a white ball still is 1/4.

In my view it's not a problem that there are an infinite number of triangles in the plane. From a frequency probabilistic viewpoint the important thing is the number of acute triangles in relation to the total number of triangles. This relation must have a limit but it doesn't matter whether the number of triangles are finite or infinite, just as the urn example shows.

So in my view the problem is not bad-posed. You can conceptually draw triangles at random from an infinite plane and the frequentist probability measure still will exist as the number of acute triangles in relation to total number of triangles.

[nuzzle]

My first approach to a solution was to limit the infinite plane to a region. I tried two versions, a unit square and a unit circle. This allowed me to perform computer simulations. I generated triangels at random within the region and then counted the acute ones in relation to the total number.

The basis for this approach is the assumption that the acute triangle fraction is the same wherever you look in the plane. This would mean that the triangles were "fractal" in nature. Regardless of where you looked and how large a region you looked at the acute triangle fraction would be constant. Unfortunately this is not true. As soon as a region is introduced certain triangles are exclude, namely those that don't fit within the region. Specifically those that have one or two corner points inside but the rest outside.

[nuzzle]

Now to my new approach.

Instead of limiting the plane to a region, the set of all triangles in the infinite plane are transformed while keeping the acute triangle fraction intact. The transformations are those which don't change the inner angles of a triangle, namely rotation, translation and scaling. In other words those transformations don't change whether a triangle is acute or not.

The important thing here is that the acute triangle fraction doesn't change during the transformation. If it does the approach fails. After the transformations the acute triangle fraction must be the same it were before transformations started

The first transformation is that of scale. We have the set of all triangles in the infinite plane. Now all triangles are scaled so the longest side becomes 1.

The second transformation is that of rotation and translation. Now all triangles are rotated and translated so that their length 1 sides becomes aligned, and with the opposite corner in the same direction.

After this a coordinate system is introduced. The left corner of the length 1 side is put in [0, 0] (origo) and the right corner in [1,0].

Remember it was the longest side of all triangles that become length 1 and that means the other two sides are shorter. This allow us to plot the area within which the opposite corner (from the length 1 side) of all triangles will be. The best way to picture this area is to think of an American football. Cut it in half where it's at the narrowest and look at it from the side.

You get this by drawing two circles with radius 1. One circle has its centre in [0,0] and the other in [1,0]. The upper half of their intersection area looks like the half American football I pictured.

[nuzzle]

To recapitulate.

We have transformed the set of all triangles in the infinite plane to the set of all alligned 1 length sided triangles with the opposite corner in the same direction. And most importantly we have done that while keeping the acute triangle fraction intact.

We've fixed the transformed triangle set in a coordinate system so all the triangles have their 1 length sides at [0,0] and [1,0] and their opposite corner within a "half American football" limiting area.

To simulate a solution to the problem we now only have to generate opposite corner points within the limiting area and count how many of these triangles are acute in relation to the total number. That's the probability of getting an acute triangle.

I'll write that program but not today.

My specific question is whether you think my reasoning holds.

[TheGreatCthulhu]

Nice. This should include all the possible types of triangles. Here's something interesting: if you draw a circle with the radius of 0.5 and with the center at (0, 0.5) - when you pick any point on this line and connect it with the ends of [0, 1] segment, you'll get a right triangle. All possible right triangles have the third point on that line. All the triangles with the third point under, or inside this circle are obtuse.

I realized this while I used Paint to visualize what you said, but I just saw that the idea is not new - in fact, there's a text on this topic on the page linked in the thread where all this started (Wolfram Math World: http://mathworld.wolfram.com/ObtuseTriangle.html).

On that page there is given the chance of obtaining an obtuse triangle, and it's proportional to wholeSurface/obtuseAreaSurface. Since the chance of obtaining a right triangle is proportional to the surface of the line defining the circle centered at (0,0.5), which is 0, the chance for a acute triangle would be 1 - obtuseChance.

1 - 0.63938... = 0.36062... (according to Wolfram Math World)

But, what does this mean. That it's impossible to have true right triangles? That there's an inherent (but generally insignificant) flaw in our number system? (Remember how irrational numbers have no finite decimal representation?) Those triangles do exist, I believe - but the chance of obtaining one is close to 0, but not 0. Or I'm wrong?

However, some might argue that this view of the original problem somewhat alters the nature of the problem at hand: this is more like - what percentage of all existing triangles is acute? The original problem was: what is the chance of getting an acute triangle if you randomly pick 3 points in a plane.
Now, this might or may not be one and the same question depending of what the reader considers to be the core of the problem - the triangles themselves, or the process of picking the points?

In the original thread, when you replied to superbonzo you said: "Math is nothing but a tool you know. Maybe you're applying the wrong yardstick. A problem doesn't become nonsensical just because your favorite yardstick cannot measure it. I wouldn't be too categorical I were you."
Naturally, you have a point, but some problems are not solvable in nature - i.e. the specific question (although it may not seem like that) is absurd because it is in conflict with the very nature of the problem. It cannot be answered.
There are examples in math and science where someone proved that something cannot be proved.

It's like Heisenberg's uncertainty principle in physics - stating that pairs of physical properties, like position and momentum, cannot both be known to arbitrary precision. This has nothing to do with the imperfection of the tool used for the measurement - it's how things are, it's in the nature of a quantum particle.

Now, I'm not familiar with the works related to the problem of picking the random points for the triangle, and I think I can assume that you are not either, so it is our right to have doubts about this conclusion. Wolfram Math World page says that "it is impossible to pick random variables which are uniformly distributed in the plane", and provides a reference to the paper where this is discussed: Eisenberg, B. and Sullivan, R. "Random Triangles n Dimensions." Amer. Math. Monthly 103, 308-318, 1996.
If one perceives the core of the problem to be the operation of picking the points, one could argue that the result might depend on how the points are picked.

It's really all about how you see it.

Another thing - computer simulations can help, but note that there are limitations other than not being able to represent an infinite plane; since the precision of the variables is limited, within the finite surface you would use for the simulation, there exist triangles that the computer is unable to generate. This may or may not affect the final result, but it's certainly something to consider.

[nuzzle]

It's really all about how you see it.

This time I've tried to cut beyond that.

I've argued that this problem is well-posed. It is sensical to talk about "the chance of an acute triangle" although it may be non-sensical to talk about "random points in an infinite plane".

1. From what I can see the basic axioms of probability don't exclude infinite sets.

2. From what I can see it's possible to transform the set of all triangles, to the set I suggested, with the acute triangle fraction intact.

As long as no one disputes the above I'm home. It's how math works.

Thank you for the references and your effort. I'll consider everything in detail.

[superbonzo]

The problem of bad posedness has nothing to do with the use of infinite sets or infinite dimensional geometries of any cardinality; two examples: the interval [0,1] is an infinite non numerable set with a natural uniform probability distribution; the set of right continuous left limited paths in R->R^n is an inifinite dimensional topological space that (equipped with a proper filtration) has a natural uniform probability dstribution (the Wienier process aka Brownian motion) and has a cardinality even greater then the set of "triangles in an infinite plane" as you have defined it ( that, BTW, has the same cardinality of an interval ) !

Here we are not speaking of introducing a probability in the set of all triangles. Of course, this is possible. The problem comes when you try to define a natural notion of "probability uniformity" with infinite sets.

Actually, the same issue happens with finite sets; but finite sets come equipped with a natural measure (the counting measure) that allows you speaking about "chance". This measure is the invariant measure coming from its permutation group.

Conversely, when the set is infinite every point that is contained in an infinite strictly decreasing chain of measureable sets ( like, say, any point in the interval [0,1] ) will have measure 0. This means that if this happens for every point of the set then there can be no probability measure invariant with respect to its permutation group. There's no counting measure for such sets. There's no natural "uniformity" nor "chance".

Of course, you can arbitrarely introduce symmetries and/or probability measure on such sets. But there are inifinitely possible inequivalent choices.

Unless you add them to their geometry structure a priori: examples are compact Lie group's (the circle, the set of rotations of any dimenions,...) which always have a left invariant probability measure (that comes from the set of its diffeomorphisms ) (but note that to each Lie group structure on the same set corresponds a different prob.measure); non-compact Lie groups (the set of translation, the Lorentz group, the Heinsenberg group ,...) that admit "uniform" prob.measure represented in their Banach duals ( for example, plane waves in ordinary quantum mechanics ) and so on.

So, if the problem is "find a family of probability measures over the set of triangles whose limiting measure is translation invariant and calculate the limit of the measure of the subsets of acute triangles" then the problem is well posed and solvable (but there are many solutions dependeing on the way you choose that family); conversely, if the problem is "find the chance of a triangle being acute" then the problem is bad posed (unless you have a very naive notion of a probability theory that reduce the problem to the formulation above...)

[nuzzle]

The problem comes when you try to define a natural notion of "probability uniformity" with infinite sets.

I haven't defined anything really.

I've assumed that the traditional axioms of probability are valid also with infinite sets. This would mean that the notion of probability in the frequentists sense also exists for infinite sets. And this would mean that it's indeed possible to determine the fraction of acute angles in the infinite set of all triangles.

So, if the problem is "find a family of probability measures over the set of triangles whose limiting measure is translation invariant and calculate the limit of the measure of the subsets of acute triangles" then the problem is well posed and solvable (but there are many solutions dependeing on the way you choose that family);

I haven't chosen any "family of probability measures". I've used the standard frequentist notion of probability.

Consider the urn example I gave. It doesn't matter whether the urn has a finite or an infinite number of balls. By drawing balls you can determine what fraction is white and that's the same as the probability of getting a white ball. The same with the triangles. By drawing from a transformed set which is equivalent to the infinite set of all triangles and counting the acute ones you can determine the probability of an acute triangle.

Thank you for your reply but I would appreciate if you were a little more specific in your criticisms.

I depend heavily on the urn example. Do you agree that it doesn't matter whether the urn contains a finite or an infinite number of balls? Do you agree that the notion of a fraction of white balls also exists in the infinite case?

I've assumed it's possible to transform the infinite set of all triangles to another set thereby preserving the original acute angle fraction. Do you agree?

[superbonzo]

>> Thank you for your reply but I would appreciate if you were a little more specific in your criticisms.

I'd appreciate it too , but there are so many foundational issues involved here, ranging from the foundation of mathematics to physics, probability and plain natural philosophy... it would take pages only for introducing the general problem rigorously...

>> I've assumed it's possible to transform the infinite set of all triangles to another set thereby preserving the original acute angle fraction. Do you agree?

So, are you conjecturing that any tranformation f:R^2*3->R^2*3 that preserves angles and whose image is bounded give the same result ?

for example, given a triangle translate and scale it in such a way that the circumscribed circle is centered in the origin and has radius one. Then rotate it in such a way one of its side is vertical. Obviously our assumptions are satisfied; also, the image is representable as a simple pair of angles, thus it have the same "dimensionality" of the image of your transformation. Let's take such pair of angles uniformly on the circle. Is the result the same ?

a quick calculation (non tested, but I suppose is correct) with wolfram mathematica gives me 0.360+-0.001 for your transformation and 0.248+-0.001 for the transformation above. Both angle preseving, different "chances".

>>I depend heavily on the urn example. Do you agree that it doesn't matter whether the urn contains a finite or an infinite number of balls? Do you agree that the notion of a fraction of white balls also exists in the infinite case?

in your urn example the cardinality of the set of balls is inessential a priori; that is you assume from the beginning that the empirical ratio of white to black balls is fixed. The number of balls does not play any role in that specific probabilistic model.

BTW, you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong; I'll try to give you an example ( that does not nearly exhaust the many foundational issues involved here ):

take, say, a linear array of N leds; under each led put a small circuit that blinks its led randomly at a fixed rate R (say, a random number generator with Poisson statistics ); suppose also that R is small compared to N in such a way that the probability of two leds blinking simultaneously is low. Under these assumptions you'll see a uniformly distributed flash over the array length.

Now, put yourself in the center of the array and start spinning at angular speed w (I mean you, not the array); given that the blinking rate depends only on the local time of the led+circuit system you'll observe the rates of leds at a distance r to be decreased to ~ R * sqrt( 1 - (r*w/c)^2) ; thus the distribution won't be uniform anymore even if the phenomena generating the random events (the led array) have not been touched. Thus, uniformity is an observer dependent property (and things get even worse in general relativity and/or quantum field theory ). Of course, you can define locally a uniform point-distribution. But you can do that only because the "local" representation of space ( eg a subset of an euclidean/Minkowsky/... space ) has a priori a definition of uniformity (actually, a local symmetry group strictly related to its geometrical structure ).

[D_Drmmr]

Consider the urn example I gave. It doesn't matter whether the urn has a finite or an infinite number of balls.

Actually, it's the other way round. It doesn't matter how many balls there are in the urn, as long as 1/4th of them are white and you draw balls uniform randomly (meaning each ball has the same chance of being draw each time) then the probability that you draw a white ball is independent of the number of balls in the urn. That's why it is reasonable to state that it also works with an infinite number of balls.

The problem would be very different if 1/4th of the top 100 balls are white and everything else black. If you then draw uniform randomly from the top 100 balls, the probability of drawing a white ball is 1/4. But that doesn't mean that 1/4th of all the balls are white.

That's the whole point. These two problems look the same, but they are very different. I believe your reasoning is flawed, because you are generalizing a consequence of one problem into a general rule.

[TheGreatCthulhu]

D_Drmmr:
As for the urn example, I think that it is implied that nuzzle was not talking about an urn with any sort of special arrangement of the balls ("top 100 balls are white"). Also, the urn is just a more "reader-friendly" term for a set, so you shouldn't think of it in terms of it's physical properties, like only being able to pick the topmost balls - imagine that, somehow, you are able to pick any of the balls.

But, the urn-balls form a countable set, while the set of points is uncountable; perhaps this is not the right parallel to draw?

superbonzo:
Would it make any difference if the points could somehow... "pop-up" on their own? If there wasn't any sentient-being-introduced picking involved? Or this is just the same?

All:
What we need to do here is to correctly define the actual problem that nuzzle is trying to solve, and to make a distinction between it and all other discussion that is actually philosophical in nature.

So, essentially: what is the percentage of acute triangles in the set of all possible triangles? Is there an inherent flaw in such a problem proposition, too?

[TheGreatCthulhu]

Disk Triangle Picking: http://mathworld.wolfram.com/DiskTrianglePicking.html

It says:
"The probability P_2 that three random points in a disk form an acute triangle is
P_2=4/(pi^2)-1/8=0.280284... "

But it also uses those nasty words "Pick three points [...] distributed independently and uniformly"...

[nuzzle]

Actually, it's the other way round. It doesn't matter how many balls there are in the urn, as long as 1/4th of them are white and you draw balls uniform randomly (meaning each ball has the same chance of being draw each time) then the probability that you draw a white ball is independent of the number of balls in the urn. That's why it is reasonable to state that it also works with an infinite number of balls.

The problem would be very different if 1/4th of the top 100 balls are white and everything else black. If you then draw uniform randomly from the top 100 balls, the probability of drawing a white ball is 1/4. But that doesn't mean that 1/4th of all the balls are white.

That's the whole point. These two problems look the same, but they are very different. I believe your reasoning is flawed, because you are generalizing a consequence of one problem into a general rule.

Yes there are two ways of seeing it depending on whether you know the white ball fraction or whether you want to determine it by drawing balls.

You claim the triangles are not drawn with equal probability. How is that?

I may be mathematically naive but if you can define the infinite plane as a set of points then you can also pick points from this set at random. Why not? Each point has an equal probability of being picked and the probability of a specific point being picked is zero.

And if you pick three points independently at random in the infinite plane you've picked a triangle at random in the infinite plane. If you pick triangles like that doesn't each triangle have an equal chance of being picked? If not, exactly what triangles are favoured?

[nuzzle]

Disk Triangle Picking: http://mathworld.wolfram.com/DiskTrianglePicking.html

It says:
"The probability P_2 that three random points in a disk form an acute triangle is
P_2=4/(pi^2)-1/8=0.280284... "

But it also uses those nasty words "Pick three points [...] distributed independently and uniformly"...

I mentioned this in the orginal thread post #9.

http://www.codeguru.com/forum/showth...ht=acute+angle

I perform simulations there on a unit square and got the same results (and I also used a unit circle).

But as I mentioned in #3 in this thread, as soon as you limit the infinite plane to a region you skew the fraction of acute triangles so that's a fawlty approach I wanted to avoid this time.