What’s the Probability a Random Triangle is Obtuse?

Lewis Carroll’s Pillow Problem #58

Earlier versions

W. S. B. Woolhouse, Educational Times 18 (1865), p. 189

J. J. Sylvester, Educational Times 18 (1865), p. 68

W. S. B. Woolhouse, The Lady's and Gentleman's Diary 158 (1861), p. 76

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,

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

Random Points on an Infinite Plane

Probability \(p\)

\(p\)

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.

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:

\(\mathbb{P}(\text{obtuse})=\frac{97}{150}-\frac{\pi}{40}\approx 0.7252\)

J.J. Sylvester, Educational Times, April 1864

Sylvester’s four point problem

Sylvester and Cayley’s solution

J.J. Sylvester, Phil. Trans. R. Soc. London 154 (1864), p. 654, footnote 64(b)

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VII (1867), p. 81

A. De Morgan’s solution

A. De Morgan, Trans. Cambridge Phil. Soc. XI (1871), pp. 147–148

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VI (1866), p. 52

C.M. Ingleby, Mathematical Questions with Their Solutions V (1865), p. 82

C.M. Ingleby’s solution

G. C. De Morgan’s solution

G.C. De Morgan, Mathematical Questions with Their Solutions V (1865), p. 109

W.S.B. Woolhouse’s solution

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VI (1866), p. 52

W.S.B. Woolhouse’s solutions

W.S.B. Woolhouse, Mathematical Questions with Their Solutions VIII (1868), p. 105

Quadrilaterals in convex bodies

\(\mathbb{P}(\text{reflex})=\frac{1}{3}\)

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

Theorem [Blaschke, 1917]

\(\frac{35}{12\pi^2}\leq\mathbb{P}(\text{reflex})\leq\frac{1}{3}\)

J.M. Wilson’s solution

J.M. Wilson, Mathematical Questions with Their Solutions V (1866), p. 81

We need measure theory!

W.A. Whitworth, Mathematical Questions with Their Solutions VIII (1868), p. 36

Report on J.J. Sylvester’s presentation of his paper “On a Special Class of Questions on the Theory of Probabilities” to the British Association for the Advancement of Science, 1865

We need geometry!

Random polygons?

Are these questions really about choosing random points, or are they actually about choosing random polygons?

How would you choose a polygon “at random”?

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

Geometric idea

The space of all \(n\)-gons should be a nice space \(P_n\) with a transitive symmetry group. We should use the invariant probability measure on \(P_n\). Then we should compute the measure of the subset of \(n\)-gons satisfying our favorite condition.

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 \(z_k = u_k + i v_k\).

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

\(\sum e_k =\sum z_k^2 = \left(\sum u_k^2 - \sum v_k^2\right) + 2i \sum u_k v_k\)

The polygon is closed \(\Leftrightarrow \|\vec{u}\|=\|\vec{v}\|\) and \(\vec{u} \bot \vec{v}\)

Since \(\sum |e_k| = \sum u_k^2 + \sum v_k^2 = \|\vec{u}\|^2 + \|\vec{v}\|^2\), we see that \((\vec{u},\vec{v})\) is an orthonormal pair of vectors in \(n\)-dimensional space; the collection of such things is called a Stiefel manifold.

Polygons and Grassmannians

Proposition: Rotating \((\vec{u},\vec{v})\) in the plane it spans rotates the corresponding \(n\)-gon twice as fast.

Corollary [Hausmann–Knutson]

The \(n\)-gons of perimeter 2 correspond to the collection of planes in \(n\)-dimensional space. This collection is called a Grassmann manifold.

Triangles

A triangle corresponds to a plane in \(\mathbb{R}^3\) with orthonormal basis \((\vec{u},\vec{v})\)...

or, equivalently, the line through

\(\vec{p}=\vec{u}\times \vec{v}\)

\begin{bmatrix} | & | \\ \vec{u} & \vec{v} \\ | & | \end{bmatrix} = \begin{bmatrix} - z_1 - \\ - z_2 - \\ - z_3 - \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \vec{p} = \vec{u} \times \vec{v} = \begin{bmatrix} -\frac{i}{2}(\overline{z}_2 z_3 - \overline{z}_3 z_2) \\ -\frac{i}{2}(\overline{z}_3 z_1 - \overline{z}_1 z_3) \\ -\frac{i}{2}(\overline{z}_1 z_2 - \overline{z}_2 z_1) \end{bmatrix}

Translating to side lengths

\begin{bmatrix} x \\ y \\ z \end{bmatrix}

z^2 = \left(-\frac{i}{2}(\overline{z}_1 z_2 - \overline{z}_2 z_1)\right)^2 = \dots = 1-\overline{z}_3 z_3 = 1-c

= \begin{bmatrix} \sqrt{1-a} \\ \sqrt{1-b} \\ \sqrt{1-c} \end{bmatrix}

= \begin{bmatrix} -\frac{i}{2}(\overline{z}_2 z_3 - \overline{z}_3 z_2) \\ -\frac{i}{2}(\overline{z}_3 z_1 - \overline{z}_1 z_3) \\ -\frac{i}{2}(\overline{z}_1 z_2 - \overline{z}_2 z_1) \end{bmatrix}