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What is the maximum probability that three independent random points, chosen from some set of points in 3D space, are the vertices of an acute triangle?

The random points can have any distribution, uniform or otherwise. (Originally, I required that the points are uniformly distributed, but after reading @Karl's comment, I realized that this requirement essentially has no effect, so I removed it.)

Here are some known probabilities (some of them are in 2D, which is a subset of 3D). For these examples, the random points are uniformly distributed.

  • Circle: $\space\frac14$
  • Square lamina: $\space \frac{53}{150}-\frac{\pi}{40}\approx0.275$
  • Disk: $\space\frac{4}{\pi^2}-\frac18\approx0.280$
  • Ball: $\space\frac{33}{70}\approx0.471$
  • Sphere: $\space\frac12$

I guess the maximum probability is $\frac12$, which happens with the sphere with a uniform distribution. It seems like this should be a known result, but I couldn't find any sources.

Context: I am interested in various probabilities that a random triangle is acute. Previous questions of mine are here and here.

Dan
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    Is something similar known in 2D? – Mathieu Rundström Dec 18 '24 at 06:10
  • @MathieuRundström I don't know. Good question. – Dan Dec 18 '24 at 06:12
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    I think the "uniform" condition here (however we might define it) has essentially no effect, since uniform distributions can approximate any distribution (in a sense relevant to this question). We essentially must consider all distributions on $\Bbb R^3$. – Karl Dec 18 '24 at 20:43
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    Not entirely relevant, but as you increase the dimension, the probability for the $n$-sphere approaches $1$. – Varun Vejalla Dec 18 '24 at 22:13
  • @Karl I'm not sure what you mean by the uniform condition having essentially no effect. For example, if points in a unit disk are $(a\cos(2\pi\theta),a\sin(2\pi\theta))$ where $a$ and $\theta$ are uniformly random in $[0,1]$, the distribution is not uniform (they are denser near the centre). This distribution would yield a different probability than a uniform distribution. – Dan Dec 18 '24 at 22:23
  • @VarunVejalla Do you know the probabilities for the $n$-sphere for say $n=4,5,6$? (Mathworld has a page called "Sphere triangle picking") but it is empty. – Dan Dec 18 '24 at 22:24
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    Yes, but we can approximate that non-uniform distribution by a uniform distribution on a (non-uniformly) "dotted" or "squiggly" set. – Karl Dec 18 '24 at 22:24
  • @Karl Oh, I see what you mean. – Dan Dec 18 '24 at 22:26
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    No idea, but as you increase the dimension, the distance between two points gets "more centered" (i.e. less deviation) around $\sqrt{2}$, and the probability of getting two points that are at a distance much less or greater than $\sqrt{2}$ goes to $0$. Not sure how fast the overall probability approaches $1$ though. For $n=3$, it's about $0.65$ (based on 1,000,000 random trials), for $n=4$, it's about $0.75$, and for $n=5$, it's about $0.83$. Maybe $2/3, 3/4$, and $5/6$? – Varun Vejalla Dec 18 '24 at 22:57
  • @VarunVejalla For the $3$-sphere, isn't the probability $1/2$, according to the link in the question? – Dan Dec 18 '24 at 23:02
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    Sorry, I meant the dimension of the sphere. The $2$-sphere would be the normal sphere. – Varun Vejalla Dec 19 '24 at 04:29
  • Related question: Paradox involving probability that a triangle is acute in circles and spheres – Dan Dec 21 '24 at 09:26
  • I updated my answer to address the concerns of a comment and I think I have now shown that the maximum chance of $,3,$ random points forming an acute triangle is $,25%$ – poetasis Dec 29 '24 at 00:38

2 Answers2

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I believe this is an open question. There is a conjecture of Glen Hall that among convex regions, the probability in $\mathbb{R}^n$ is maximized by the $n$-ball [1], which remains open. A few years ago, I proved that the two-dimensional disk is a local maximum in the Gromov-Hausdorff topology and the three-dimensional ball is a local maximum in the $C^1$ topology [2]. However, this only considers convex regions and even less is known in the general case.

[1] Hall, Glen Richard, Acute triangles in the n-ball, J. Appl. Probab. 19, 712-715 (1982). ZBL0492.60014.

[2] Khan, Gabriel, Hall’s conjecture on extremal sets for random triangles, J. Geom. Anal. 30, No. 4, 3413-3457 (2020). ZBL1460.52009.

Gabe K
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We begin by choosing two points and aligning the $\,x$-axis with them.

Below is a $\,2D\,$ diagram representing points in quadrants $\,1\,$ and $\,2\,$ of a Cartesian plane with $\,B\,$ at the origin. All points $\,E\,$ in quadrant $\,1\,$ can only be used to form obtuse triangles with $\,A\,$ and $\,B.\,$ All points $\,D\,$ along the $\,Y$-axis can only be used to form right triangles. All points $\,C\,$ in quadrant $\,2\,$ will form acute triangles unless $\,C\,$ falls within the semicircle defined by diameter $\,AB\,$ or falls to the left of point $\,A.$

If the plane is finite, and $\,C\,$ is to the right of $\,A,\,$ then the probability is $\,50\%\,$ times $\,\dfrac{q-s}{q}\,$ where $\,q\,$ is the area of the finite quadrant and $\,s\,$ is the area of the semicircle.

If the plane is infinite, and $\,C\,$ is to the right of $\,A,\,$ we can see that $\,\dfrac{q-s}{q}=1\,$ so the max would be $\,50\%.\,$ However, if $\,C\,$ is to the left of $\,A,\,$ as pointed out by @Matthew Spam in a comment below, then $\,C\,$ can only form obtuse triangles and that is a problem but it is not unsolvable.

There are infinite points to the left and right of $\,A\,$ and, while it is arguable that these infinities are equal, we can agree that there are not "more" points to the right of $\,A\,$ because $\,A\,$ is a finite distance from the origin while there is infinite distance to the left of $\,A.\quad$ This means that the maximum change of $\,C\,$ being part of an acute triangle is $\,50\%\,$ within the $\,50\%\,$ of "all" points that are to the left of the $\,y$-axis. $\,\therefore \,$The maximum chances of any $\,3\,$ random points forming an acute triangle is
$\,50\%\cdot50\%=25\%.\,$

The results are the same in $\,3D\,$ as can be visualized by rotating this plane about the $\,x$-axis.

enter image description here

poetasis
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