### Clayton Shonkwiler PRO

Mathematician and artist

Clayton Shonkwiler

Colorado State University

09.15.16

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,

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

Text

**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 \(\mathbb{R}^2\).

Three vertices uniformly in the disk:

\(\mathbb{P}(\text{obtuse})=\frac{9}{8}-\frac{4}{\pi^2}\approx 0.72\)

[Edelman–Strang]: Normalize sides so \(a^2+b^2+c^2=1\):

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

— 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\((M)=1\). 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: **\(T\simeq\mathbb{RP}^2=G_2\mathbb{R}^3\)

- The sidelengths \((a,b,c)\) uniquely determine a triangle (remember SSS from high school)
- Obtuseness is scale-invariant

Rescale so that \(a+b+c=2\); then a triangle shape is uniquely specified by a point in the simplex.

Not all points in the simplex correspond to triangles

\(b+c<a\)

\(a+b<c\)

\(a+c<b\)

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

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

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

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

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

**Note:** This is why we chose the normalization \(a+b+c=2\)

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

Then

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

and the triangle inequalities become

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

Text

\((1-s_a)^2+(1-s_b)^2=(1-s_c)^2\)

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

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

\(SO(3)\) obviously acts transitively on the sphere, and the corresponding action on triangles is natural.

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

\(z\) fixed

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

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

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

The pushforward of the uniform measure on the sphere to the simplex \(s_a+s_b+s_c=1\) has density

\(\frac{1}{2\pi \sqrt{s s_a s_b s_c}}\)

This is the Dirichlet(1/2,1/2,1/2) distribution.

By Heron's formula, the density is proportional to \(\frac{1}{\text{Area}}\).

**Corollary [w/ Cantarella, Needham, Stewart]**

The expected area of a random triangle is \(\frac{1}{4\pi}\).

**Corollary [w/ Cantarella, Needham, Stewart]**

The expected curvature of the circumscribed circle to a random triangle is \(\frac{2}{\pi}\).

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.

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

\(\vec{p}=\vec{a} \times \vec{b} = \begin{pmatrix} \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \\ \begin{vmatrix} a_3 & b_3 \\ a_1 & b_1 \end{vmatrix} \\ \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}\end{pmatrix}\)

In general, \(G_k(\mathbb{R}^n)\) is the *Grassmannian* of \(k\)-dimensional linear subspace of \(\mathbb{R}^n\).

If \(P \in G_k(\mathbb{R}^n)\) and \(\{a_1,\ldots , a_n\}\) is a basis for \(P\), then the map \(G_k(\mathbb{R}^n) \to \mathbb{RP}^{\binom{n}{k}-1}\) recording all \(k \times k\) minors of the matrix \([a_1 \cdots a_k]\) is called the *Plücker embedding* of \(G_k(\mathbb{R}^n)\).

Equivalently, \(P \mapsto a_1 \wedge \ldots \wedge a_k \in \mathbb{P}(\bigwedge^k \mathbb{R}^n) \simeq \mathbb{RP}^{\binom{n}{k}-1}\).

The Plücker coordinates satisfy certain relations; in the case of \(G_2(\mathbb{R}^4)\), these reduce to

\(\Delta_{12}\Delta_{34}-\Delta_{13}\Delta_{24}+\Delta_{14}\Delta_{23}=0\)

Suppose \(T\) is a triangle with edge vectors \(e_1, e_2, e_3\). Let \(p \in S^2\) be any lift of \(T\), and let \((a,b)\) be any orthonormal basis of \(p^\bot\).

\((a, b) = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{pmatrix}\)

Let \(z_k = a_k + i b_k\) for each \(k\).

Up to rotation,

\(e_k = z_k^2\) as complex numbers

Let \(e_1, \ldots , e_n\) be the edges of a planar \(n\)-gon with total perimeter 2. Choose \(z_1, \ldots , z_n\) so that \(z_k^2 = e_k\). Let \(z_k = a_k + i b_k\).

The polygon is closed \(\Leftrightarrow e_1 + \ldots e_n = 0\)

\(\sum e_k =\sum z_k^2 = \left(\sum a_k^2 - \sum b_k^2\right) + 2i \sum a_k b_k\)

The polygon is closed \(\Leftrightarrow |a|=|b|\) and \(a \bot b\)

Since \(\sum |e_k| = \sum a_k^2 + \sum b_k^2 = |a|^2 + |b|^2\), we see that \((a,b) \in V_2(\mathbb{R}^n)\), the Stiefel manifold of 2-frames in \(\mathbb{R}^n\).

**Proposition: **Rotating \((a,b)\) in the plane it spans rotates the corresponding \(n\)-gon twice as fast.

**Corollary [Hausmann–Knutson]**

The Grassmannian \(G_2(\mathbb{R}^n)\) is (almost) a \(2^n\)-fold covering of the space of planar \(n\)-gons of perimeter 2.

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

Therefore, \(SO(n)\) acts transitively on \(n\)-gons and preserves the symmetric measure.

In particular, notice that the symmetric measure is invariant under permutations of the edges.

We can package the Plücker coordinates \(\Delta_{ij} = \begin{vmatrix} a_i & b_i \\ a_j & b_j \end{vmatrix}\) of a point in \(G_2(\mathbb{R}^n)\) as a skew-symmetric matrix

\(\Delta = (\Delta_{ij})_{i,j}\)

In other words, this is the matrix of cross products of the \((a_k,b_k)\).

**Semi-Miraculous Fact: **\(-\Delta^2\) is the Gram matrix (i.e., matrix of dot products) of the \((a_k,b_k)\).

**Definition: **Given \(P \in G_2(\mathbb{R}^n)\), the *Plücker sign matrix* of \(P\) is \(\operatorname{sgn} \Delta(P)\) and the *projection sign matrix* of \(P\) is \(\operatorname{sgn}( -\Delta(P)^2)\).

Since \((a_k + i b_k)^2=e_k\), the Plücker sign matrix and projection sign matrix notice when two edges in a polygon become parallel or anti-parallel.

The sign matrices notice, among other things, when a polygon becomes non-convex or non-embedded.

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!

By Clayton Shonkwiler

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

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