What’s the probability that a random triangle is obtuse?

or:

What the heck is a random triangle, anyway?

Clayton Shonkwiler

Lewis Carroll's Pillow Problem #58

Random Points on an Infinite Plane

Probability \(p\)

\(p\)

\(p\)

\(p\)

\(p\)

Uh oh!

In Fact...

There is no way to choose points randomly in the plane so that the probability that a point lies in a region depends only on the size of the region.

Fancier Version

The uniform measure on the plane is not a probability measure.

Gaussians to the Rescue?

One way to choose points randomly in the plane is according to the Gaussian (normal) distribution.

Proposition: Suppose the vertices \((x_1,y_1),(x_2,y_2),(x_3,y_3)\) of the triangle are chosen independently from the standard 2-variable Gaussian distribution. Then

\(\mathbb{P}(\text{obtuse})=\frac{3}{4}\)

Choose three vertices uniformly in the disk:

\(\mathbb{P}(\text{obtuse})=\frac{9}{8}-\frac{4}{\pi^2}\approx 0.7197\)

Restricted Domain?

Choose three vertices uniformly in the square:

\(\mathbb{P}(\text{obtuse})=\frac{97}{150}-\frac{\pi}{40}\approx 0.7252\)

Random Triangles?

Is Carroll’s question really about choosing random points, or is it actually about choosing random triangles?

How would you choose a triangle “at random”?

Key observation: Obtuseness is scale-invariant.

Angles?!

Remember from Geometry that three angles \((\theta_1,\theta_2,\theta_3)\) determine a triangle up to similarity (AAA).

\(\theta_1\)

\(\theta_2\)

\(\theta_3\)

What are the restrictions on the \(\theta_i\)?

\(\theta_1+\theta_2+\theta_3=\pi\)

\(0<\theta_1, 0 < \theta_2, 0<\theta_3\)

and

The Triangle of Triangles

\(\theta_1+\theta_2+\theta_3=\pi\)

\(0<\theta_1, 0 < \theta_2, 0<\theta_3\)

and

\(\theta_1=\pi/2\)

\(\theta_2=\pi/2\)

\(\theta_3=\pi/2\)

\(\mathbb{P}(\text{obtuse})=\frac{3}{4}\)

Side Lengths?!

Remember the sidelengths \((a,b,c)\) uniquely determine a triangle (SSS).

Obtuseness is scale-invariant, so pick a perimeter \(P\) and we have \(a+b+c=P\).

Problem

Not all points in the simplex correspond to triangles

\(b+c<a\)

\(a+b<c\)

\(a+c<b\)

Yet Another Pillow Problem Answer

\(\mathbb{P}(\text{obtuse})=9-12\ln 2 \approx 0.68\)

\(b^2+c^2=a^2\)

\(a^2+b^2=c^2\)

\(a^2+c^2=b^2\)

Carroll’s Answer

Suppose \(AB\) is the longest side. Then

\(\mathbb{P}(\text{obtuse})=\frac{\pi/8}{\pi/3-\sqrt{3}/4} \approx 0.64\)

But if \(AB\) is the second longest side, 

\(\mathbb{P}(\text{obtuse}) = \frac{\pi/2}{\pi/3+\sqrt{3}/2} \approx 0.82\)

— Stephen Portnoy, Statistical Science 9 (1994), 279–284

Random Triangles...?

Transitive Transformation Group

To choose objects from a collection at random, there must be some sense in which the objects are indistinguishable from each other.

Or: there should be some way of transforming any object in the collection into any other object.

The collection of transformations is called a transformation group. The fact that you can transform any object into any other object means the group acts transitively. Take MATH 366 for more!

Haar Measure

Given a transitive group of transformations acting on a collection of objects, there is a natural way to assign a size to any subset of the objects: just take the average of the sizes of all transformations of the subset.

This way of assigning sizes to subsets is called Haar measure.

We can then choose objects randomly according to the following rule:

The probability that a randomly chosen object will be in any given subset is proportional to the size of the subset.

Back to Triangles

Let \(s=\frac{1}{2}(a+b+c)\) and define

Note: It’s convenient to choose \(s=1\).

\(s_a=s-a, \quad s_b = s-b, \quad s_c = s-c\)

Then

\(s_a+s_b+s_c=3s-(a+b+c)=3s-2s=s\)

and the triangle inequalities become

\(s_a>0, \quad s_b > 0, \quad s_c > 0\)

But there's still no transitive transformation group!

Take Square Roots!

Consider \((x,y,z)\) so that

\(x^2=s_a=1-a, \quad y^2=s_b=1-b, \quad z^2=s_c=1-c\)

The unit sphere is a \(2^3\)-fold cover of triangle space

The Transitive Group

The rotations are natural transformations of the sphere, and the corresponding action on triangles is natural.

\(c=1-z^2\) fixed

\(z\) fixed

\(C(\theta) = (\frac{z^2+1}{2}\cos 2\theta, z \sin 2\theta)\)

The equal-area-in-equal-time parametrization of the ellipse

A More Complicated Rotation

Right Triangles

The right triangles are exactly those satisfying

\(a^2+b^2=c^2\) or \(b^2+c^2=a^2\) or \(c^2+a^2=b^2\)

Since \(a=1-x^2\), etc., the right triangles are determined by the quartic

\((1-x^2)^2+(1-y^2)^2=(1-z^2)^2\)  or ...

\(x^2 + x^2y^2 + y^2 = 1\),  etc.

Obtuse Triangles

\(\mathbb{P}(\text{obtuse})=\frac{1}{4\pi}\text{Area} = \frac{24}{4\pi} \iint_R d\theta dz\)

But now \(C\) has the parametrization

And the integral reduces to

Solution to the Pillow Problem

By Green’s Theorem

\(\frac{6}{\pi} \iint_R d\theta dz=\frac{6}{\pi}\oint_{\partial R}z d\theta = \frac{6}{\pi}\left(\int_{z=0} zd\theta + \int_C zd\theta \right)\)

\(\left(\sqrt{\frac{1-y^2}{1+y^2}},y,y\sqrt{\frac{1-y^2}{1+y^2}}\right)\)

\(\frac{6}{\pi} \int_0^1 \left(\frac{2y}{1+y^4}-\frac{y}{1+y^2}\right)dy\)

Our Answer

Theorem [w/ Cantarella, Needham, Stewart]

Thinking of random triangles as points chosen uniformly on the sphere, the probability that a random triangle is obtuse is

\(\frac{3}{2}-\frac{3\ln 2}{\pi}\approx0.838\)

Generalization

For \(n>3\), the sidelengths do not uniquely determine an \(n\)-gon, so the previous approach doesn‘t obviously generalize.

Key Observation: Choosing a point on the sphere is equivalent to choosing the perpendicular plane.

 

\(\vec{p}=\vec{a} \times \vec{b}\)

Planes and Polygons

In general, we can identify the collection of planar \(n\)-gons with the collection of 2-dimensional planes through the origin in \(n\)-dimensional space \(\mathbb{R}^n\).

 

This space is called the Grassmann manifold or

Grassmannian \(G_2(\mathbb{R}^n)\).

Sylvester’s Four Point Problem

convex

reflex/reentrant

self-intersecting

Some Answers

\(\mathbb{P}(\text{reflex})=\frac{1}{3}\)

\(\mathbb{P}(\text{reflex})=\frac{35}{12\pi^2}\approx 0.296\)

Theorem [Blaschke]

\(\frac{35}{12\pi^2}\leq\mathbb{P}(\text{reflex})\leq\frac{1}{3}\)

Our Answer

Theorem [w/ Cantarella, Needham, Stewart]

Under the correspondence of quadrilaterals with planes in \(\mathbb{R}^4\), each of the three classes of quadrilaterals occurs with equal probability. In particular, \(\mathbb{P}(\text{reflex})=\frac{1}{3}\).

More generally...

Theorem [w/ Cantarella, Needham, Stewart]

The probability that a random \(n\)-gon is convex is \(\frac{2}{(n-1)!}\).

Polygons in Space

There is a version of this story for \(n\)-gons in space as well.

Polygons in space provide a foundational theoretical and computational model for ring polymers like bacterial DNA.

Thank you for listening!

Square Roots of Angles

It turns out that if we take square roots of angles instead, we get:

\(\mathbb{P}(\text{obtuse}) = \frac{4-2\sqrt{2}}{\sqrt{\pi}} \approx 0.661\)

But this approach doesn’t seem to generalize nearly as well as taking square roots of edgelengths.

Random Triangles

By Clayton Shonkwiler

Random Triangles

What's the probability that a random triangle is obtuse? Or, what the heck is a random triangle, anyway?

  • 375
Loading comments...

More from Clayton Shonkwiler