Jason Cantarella

U. of Georgia

Thomas Needham

Ohio State

Gavin Stewart

NYU

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:

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

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”?

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

\(b+c<a\)

\(a+b<c\)

\(a+c<b\)

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

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

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

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

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

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: **\(T\simeq\mathbb{RP}^2=G_2\mathbb{R}^3\)

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!

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 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

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.

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.

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

But now \(C\) has the parametrization

And the integral reduces to

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

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

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.

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

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

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

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.

What is the manifold of *equilateral* planar \(n\)-gons up to translation and rotation?

Is there a good parametrization of this manifold?

- J. Cantarella, T. Deguchi, and C. Shonkwiler. Probability theory of random polygons from the quaternionic perspective.
*Communications on Pure and Applied Mathematics***67**(2014), 1658–1699. - J. Cantarella, A. Y. Grosberg, R. Kusner, and C. Shonkwiler. Expected total curvature of random polygons.
*American Journal of Mathematics***137**(2015), 411–438. - J. Cantarella, T. Needham, C. Shonkwiler, and G. Stewart. Random triangles and polygons in the plane. Coming soon!

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.