From Obtuse Triangles to DNA Models and Synthetic Polymers

The Geometry of Random Polygons

Clayton Shonkwiler

Colorado State University


Lewis Carroll's Pillow Problem #58

Random Points on an Infinite Plane

Probability \(p\)





Uh oh!

In Fact...

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.

Gaussians to the Rescue?

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


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”?

Carroll’s Answer

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, 

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


Remember from Geometry that three angles \((\theta_1,\theta_2,\theta_3)\) determine a triangle up to similarity (AAA).




What are the restrictions on the \(\theta_i\)?


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


The Triangle of Triangles


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






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 \(a+b+c=P\).


Not all points in the simplex correspond to triangles




Yet Another Pillow Problem Answer

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




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

Random Triangles...?

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 \(s=1\).

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



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 More Complicated Rotation

A Measure on Triangle Space

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


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

408 random triangles

(collab. w/ A. Harding)

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

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


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

Our Answer

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.

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

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




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

Some Answers


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

Theorem [Blaschke]


Our Answer

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

A Question of Distances

What’s the shortest path from



Shortest Paths Between Planes are Easy!



What’s the shortest path from

An Invariant Representation

Key Feature: This is a representation of \(n\)-gons which is invariant under translations and rotations, so we get the same distance and the same path whether our starting point is


Applications to shape recognition, medical imaging, etc.

Polygons in Space

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.

Wassermann et al., Science 229, 171--174

Lyubchenko, Micron 42, 196--206

Complex Grassmannians and Space Polygons

Polygons in space correspond to planes in complex \(n\)-dimensional space \(\mathbb{C}^n\), where again we understand distances and shortest paths very well.



Again, Invariance

This representation of polygons is orientation-independent... it’s easy to recognize these are the same shape and register them.

Application to Protein Clustering

Needham, 2016

The Grassmannian perspective gives a way to

  • Register 3-D loops
  • Compute distances between 3-D loops

The challenge for the future is to compute distances and find shortest paths while avoiding self-intersection.

Many antibiotics (e.g., Cipro) and many anti-cancer drugs (especially for treating pediatric cancers; e.g. Etoposide) are topoisomerase inhibitors, which seem to work by preventing DNA from unknotting.

Challenge I

Why Does This Matter?

Challenge II

The Grassmannian theory deals with random polygonal loops; find the analogous theory for more complicated polygonal graphs.

Why Does This Matter?

A synthetic \(K_{3,3}\) (Tezuka Lab, Tokyo)

Novel polymer topologies have novel material properties.

More geodesics in polygon space

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.