### What the heck is a random triangle, anyway?

Clayton Shonkwiler

http://shonkwiler.org

10.30.16

Jason Cantarella

U. of Georgia

Thomas Needham

Ohio State

Gavin Stewart

NYU

### Lewis Carroll's Pillow Problem #58

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,

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

Proposition [Portnoy]: If the distribution of $$(x_1,y_1,x_2,y_2,x_3,y_3)\in\mathbb{R}^6$$ is spherically symmetric (for example, a standard Gaussian), then

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

Consider the vertices $$(x_1,y_1),(x_2,y_2),(x_3,y_3)$$ as determining a single point in $$\mathbb{R}^6$$.

For example, when the vertices of the triangle are chosen from independent, identically-distributed Gaussians on $$\mathbb{R}^2$$.

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”?

### 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

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

$$b^2+c^2=a^2$$

$$a^2+b^2=c^2$$

$$a^2+c^2=b^2$$

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

### Translation for Geometers

The space of all triangles should be a (preferably compact) manifold $$T$$ with a transitive isometry group. We should use the left-invariant metric on $$T$$, scaled so vol$$(T)=1$$. Then the Riemannian volume form induced by this metric is a natural probability measure on $$T$$, and we should compute the volume of the subset of obtuse triangles.

Ideally, this should generalize to $$n$$-gons.

Spoiler: $$T\simeq\mathbb{RP}^2=G_2\mathbb{R}^3$$

### 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 group action!

### Take Square Roots!

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

$$x^2=s_a, \quad y^2 = s_b, \quad z^2 = s_c$$

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 Measure on Triangle Space

Since the uniform measure is the unique (up to scale) measure on $$S^2$$ invariant under the action of $$SO(3)$$...

Definition

The symmetric measure on triangle space is the probability measure proportional to the uniform measure on the sphere.

### Right Triangles

The right triangles are exactly those satisfying

$$a^2+b^2=c^2$$  & permutations

Since $$a=1-s_a=1-x^2$$, etc., the right triangles are determined by the quartic

$$(1-x^2)^2+(1-y^2)^2=(1-z^2)^2$$  & permutations

$$x^2 + x^2y^2 + y^2 = 1$$,  etc.

### Obtuse Triangles

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

But now $$C$$ has the parametrization

And the integral reduces to

### Solution to the Pillow Problem

By Stokes’ Theorem

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

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

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

Theorem [w/ Cantarella, Needham, Stewart]

With respect to the symmetric measure on triangles, 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 simplex approach doesn‘t obviously generalize.

Key Observation: The coordinates $$(x,y,z)$$ of a point on the sphere are the Plücker coordinates of the perpendicular 2-plane.

$$\vec{p}=\vec{a} \times \vec{b}$$

### Planes and Polygons

In general, we can identify the collection of planar $$n$$-gons with $$G_2(\mathbb{R}^n)$$, the Grassmannian of 2-planes through the origin in $$\mathbb{R}^n$$.

Definition [w/ Cantarella & Deguchi]

The symmetric measure on $$n$$-gons of perimeter 2 up to translation and rotation is the pushforward of Haar measure on $$G_2(\mathbb{R}^n)$$.

### Sylvester’s Four Point Problem

convex

reflex/reentrant

self-intersecting

Modern Reformulation: What is the probability that all vertices of a random quadrilateral lie on its convex hull?

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

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

Theorem [Blaschke]

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

Theorem [w/ Cantarella, Needham, Stewart]

With respect to the symmetric measure, 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]

With respect to any permutation-invariant measure on $$n$$-gon space, 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 polygons in $$\mathbb{R}^3$$ as well.

Example Theorem [w/ Cantarella, Grosberg, Kusner]

The expected total curvature of a random space $$n$$-gon is exactly

$$\frac{\pi}{2}n + \frac{\pi}{4} \frac{2n}{2n-3}$$

The polygon space is $$G_2(\mathbb{C}^n)$$ and the analog of the squaring map is the Hopf map.

### Open Problem

What is the manifold of equilateral planar $$n$$-gons up to translation and rotation?

Is there a good parametrization of this manifold?

# Thank you!

### 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.