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

Clayton Shonkwiler

09.15.16

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,

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

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 $$\mathbb{R}^2$$.

### And more...

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

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

### ​Observations

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

### Idea

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

### Problem

Not all points in the simplex correspond to triangles

$$b+c<a$$

$$a+b<c$$

$$a+c<b$$

### Yet Another Pillow Problem Answer

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

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

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

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

### A Change of Coordinates

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

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

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

### 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

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

$$c=1-z^2$$ fixed

$$z$$ fixed

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

### 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

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

### Naturality

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 $$\frac{1}{\text{Area}}$$.

### Expectations

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 $$\frac{2}{\pi}$$.

### 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.

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

### Plücker coordinates

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

### Plücker coordinates

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

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

### Triangles and the Grassmannian

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

### Polygons and Stiefel Manifolds

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 $$\mathbb{R}^n$$.

### Polygons and Grassmannians

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.

### The Symmetric Measure

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.

### Plücker Matrices

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 $$(a_k,b_k)$$.

### Sign Matrices

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.

### 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!

### References

#### 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?

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