### Clayton Shonkwiler PRO

Mathematician and artist

Clayton Shonkwiler

Probability \(p\)

\(p\)

\(p\)

\(p\)

\(p\)

Uh oh!

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.

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:

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

Choose three vertices uniformly in the square:

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

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.

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

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

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

and

\(\theta_1=\pi/2\)

\(\theta_2=\pi/2\)

\(\theta_3=\pi/2\)

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

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

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

Not all points in the simplex correspond to triangles

\(b+c<a\)

\(a+b<c\)

\(a+c<b\)

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

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

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

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

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

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!

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.*

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!

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 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

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.

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

But now \(C\) has the parametrization

And the integral reduces to

By Green’s Theorem

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

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

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

**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

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

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.

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

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* \(G_2(\mathbb{R}^n)\).

convex

reflex/reentrant

self-intersecting

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

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

**Theorem [Blaschke]**

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

**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 \(\frac{2}{(n-1)!}\).

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.

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.

By Clayton Shonkwiler

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

- 497

Loading comments...