What’s the probability that a random triangle is obtuse?


What the heck is a random triangle, anyway?

Clayton Shonkwiler

Colorado State University



Jason Cantarella

U. of Georgia

Thomas Needham

Ohio State

Gavin Stewart


Lewis Carroll's Pillow Problem #58

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, 

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


A Probabilist’s Answer

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 \(a^2+b^2+c^2=1\):


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


  1. The sidelengths \((a,b,c)\) uniquely determine a triangle (remember SSS from high school)
  2. 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




Yet Another Pillow Problem Answer

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




A Change of Coordinates

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

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

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



and the triangle inequalities become

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


The Same Answer Again


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

But there’s no transitive group...

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 \(SO(3)\)...


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

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


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

Our Answer

Theorem [w/ Cantarella, Needham, Stewart]

With respect to the symmetric measure on triangles, the probability that a random triangle is obtuse is

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


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 \(\frac{\pi}{2}\).


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


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




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

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!