An earlier post presented an elegant plane geometry theorem discovered by the 19th century school teacher Auguste Miquel. This post presents his pentagon theorem.

Start with a pentagon. It may be irregular, but it needs to be convex.

Extend each of the sides of the pentagon to form a star, then draw give circles, one through each of the triangles formed by a side of the pentagon and a vertex of the star.

The five circles intersect in pairs at ten points: the five vertices of the pentagon and five new points. The five new points lie on a circle.

The converse of this theorem is known as the five circles theorem.

Euclidean geometry dates back at least to Euclid (circa 300 BC), and so you might think it’s been pretty well picked over by now. And yet people still occasionally discover new plane geometry theorems.

Some of these new theorems are complicated, asking question that the ancients would not have asked. But once in a while someone discovers a gem that the ancients could have appreciated but didn’t find.

One example is Miquel’s pivot theorem [1]. The theorem was discovered in 1838, which relative to the timeline of Euclidean geometry makes it a recent discovery.

Choose a point on each side of a triangle. Then for each vertex draw a circle through it and the chosen points on the adjacent sides. Miquel’s theorem says the three circles meet in one point.

Here’s an example. For a trangle ABC, choose points D, E, and F on each side. The three circles described in the theorem intersect at M.

Now the three points D, E, and F don’t have to be limited to the sides of the triangle; they can be on the line segment containing the side. Here’s an example where D is outside the triangle.

And here’s an example where two of the chosen points, D and F, are outside the triangle. The three circles still intersect at one point M.

Related posts

• Equation of a circle through three points
• Van Aubel’s theorem
• Big little hexagon

[1] Miquel, Auguste (1838), “Mémoire de Géométrie”, Journal de Mathématiques Pures et Appliquées, 1: 485–487

There’s a common saying in discussion of fractals that the length of a coastline depends on how small a device you use to measure it. I thought this was a hypothetical, say as applied to the steps in the construction of the Koch snowflake. But the saying has its roots in actually surveying.

Lewis Fry Richardson (1881–1953) noticed that the length of the coast of Scotland depended on the size of segments used to measure it. More specifically, he found that the length followed a power law, i.e. that there’s a linear relation between the log of the coastline length and the log of the ruler length.

Here’s a reproduction of Richardson’s plot, taken from [1].

Mandelbrot built on Richardson’s observation and defined the idea of fractal dimension.

I was under the impression that fractals were invented as mathematical novelties that researchers later found applications for. But as is often the case, the applications came first. Or at least some applications came first.

Ideally there’s always a feedback cycle where applications lead to theory and theory leads to applications. As Donald Knuth put it, “The best theory is inspired by practice. The best practice is inspired by theory.”

Related posts

• Minimalist Mandelbrot set
• The fractal nature of Brownian motion
• Randomly generated dragon

[1] Eoghan Bradley and Mark McCartney. Four hundred years of the fractal coastline of Scotland. The Mathematical Gazette, November 2019, Vol. 103, No. 558 (November 2019), pp. 518-521

The previous post looked at how to interpret cosine similarity, or equivalently angles between word vectors. In a high-dimensional space, randomly chosen vectors are likely nearly perpendicular, and so relatively large angles, such as 50°, indicate very closely related words.

Another way to look at this, as explained in the previous post, is that in high dimensions, a spherical cap of angular radius θ represents a small portion of a sphere, even for moderately large θ.

The proportion of the area inside the spherical cap, given here, involves the “regularized incomplete beta function” and so it’s hard to have an intuition for the value.

For large dimension n, the approximation

n^−1/2 sin^{n − 1}(θ)

gives the proportion of the area inside the cap to within an order of magnitude. It’s easy to see that this function goes to zero quickly as n increases, provided |θ| < π/2.

If you have the cosine similarity c = cos θ rather than θ itself, the approximation becomes

n^−1/2 (1 − c²)^{(n − 1)/2}.

Python script

Let’s try it on the example from the previous post, in which n = 200 and θ = 49°.

import numpy as np
from scipy.special import betainc

# Fraction of S^{n-1} inside a spherical cap of angular radius theta
# theta is measured from the pole
# Assume 0 < theta < pi/2

def cap_fraction(theta, n):
    x = np.sin(theta) ** 2
    return 0.5 * betainc(0.5 * (n - 1), 0.5, x)

def cap_fraction_approx(theta, n):
    return n**(-0.5) * np.sin(theta)**(n-1)

theta = np.deg2rad(49)
print(cap_fraction(theta, 200)) 
print(cap_fraction_approx(theta, 200)) 

This prints 2.03e-26 and 3.37e-26. The order of magnitude is correct as advertised.

A new paper just came out, The Maximum-Area Small Polygon Problem. The paper solves the problem of finding, for each n, the n-gon with diameter 1 and maximum area.

For odd n, the solution is what you might expect: a regular n-gon. I would expect this to be the solution for even n as well, but it’s not.

In 1974 [1] Ron Graham found a solution for n = 6, a hexagon with unit diameter and area larger than a regular hexagon with unit diameter. Polygons with diameter ≤ 1 are called “small”, and he found the “largest” (i.e. maximum area) small hexagon.

The vertices of Graham’s hexagon are given below.

  A = (0.0000000000,  0.0000000000)
  C = (0.4023506913, -0.5000000000)
  F = (0.9390533483, -0.3437714489)
  B = (1.0000000000,  0.0000000000)
  E = (0.9390533483,  0.3437714489)
  D = (0.4023506913,  0.5000000000)

You can verify that the distance between any pair of vertices is no more than 1 and that the area of Graham’s hexagon is 0.674981.

The area of a regular hexagon of diameter 1 is (3/8)√3 = 0.649519, and the area of Graham’s hexagon is about 3.9% larger.

[1] R. L. Graham. The Largest Small Hexagon. Journal of Combinatorial Theory (A) 18, 165–170 (1975). The paper was submitted February 22, 1974 and published in 1975.

Yesterday Levent Alpöge announced that he and his colleagues had discovered a new Hadamard matrix using Claude AI. That motivated a post I wrote this morning on how to construct Hadamard matrices. I mentioned in that post that these matrices arise in applications.

This evening I gave an example, describing how NASA used a Hadamard matrix of order 32 to transmit photos from the Mariner 9 spacecraft in 1971. This post will give another application: sphere packing.

Conway and Sloane [1] give a correspondence between binary codes and sphere packings that they call Construction A. Given an (n, M, d) binary code C, center a sphere on a point x if and only if x is a congruent (mod 2) to codeword in C.

Here (n, M, d) means an error correcting code that encodes M bits of data as strings of n bits, with a minimum Hamming distance between code words of d, i.e. all codewords differ in at least d bits.

The previous post described how to create a (32, 6, 16) code by stacking a Hadamard matrix H of order 32 on top of −H and turning −1’s into 0’s. The analogous construction for a (8, 4, 4) Hadamard code gives E₈, the densest packing in ℝ⁸.

We start with the Hadamard matrix

and obtain the matrix

whose centers form the sphere packing.

This doesn’t look like the E8 sphere packing as it is usually presented, but it’s isomorphic.

[1] J. H. Conway and N. J. A. Sloane. Sphere Packings, Lattices and Groups. Springer. 1999.

The volume of a sphere of radius r is

V = 4πr³ / 3

and the surface area is

A = 4πr²

and so the ratio of volume to area is

V / A = r / 3.

Surprisingly, the same ratio holds for all regular solids if r is the radius of the largest sphere that can be inscribed inside the regular solid.

For example, if the edge of a cube is a, then r = a/2. The volume is 8r³, the area is 24r², and the ratio is r/3.

The relationship between edge length and radius, and between radius and volume, is more complicated for the four other regular solids (tetrahedron, octahedron, dodecahedron, and icosahedron). However, in each case the ratio of volume to area is r/3.

The proof is surprisingly simple. Pick a face and form a pyramid by connecting each face vertex to the center of the inscribed sphere. The pyramid has height r and volume equal to B/3 where B is the area of the base. If the regular solid has f faces, the volume of the solid is fBr / 3 and the area is fB. So the ratio of volume to area is r/3.

The theorem generalizes to n > 3 dimensions. The formula for the volume of a pyramid in n dimensions is Bh/n where B is the (n − 1)-dimensional volume of the base, and so the ratio of n-dimensional volume of a regular solid to (n − 1)-dimensional volume of its boundary is r/n.

I ran across a geometry theorem with the following diagram.

The theorem corresponding to the diagram is interesting, but I found reproducing the diagram more interesting.

The segment AB is a diameter and the line CD is perpendicular to the diameter.

Assume the outer circle is a unit circle. I guessed C = (cos(1), sin(1)) and made the following diagram.

I guessed the value of C by eyeballing it, but in retrospect this would have been a convenient value for the creator of the original diagram to have chosen.

Drawing the blue circle inscribed in the triangle was easy using the equations for the center and radius from this post. Drawing the other two circles, the green and orange circles, was harder. They are also inscribed circles, but not inscribed in a triangle. They’re inscribed in a three-sided figure with two perpendicular sides and a circular arc.

The radius r of the green circle is the distance from the center of the circle to each of its tangent lines. Also, the distance from the origin to the center of the circle must be 1 − r. This is enough information to set up a quadratic equation for r. The same reasoning applies to the orange circle.

The original diagram comes from [1] and the theorem it illustrates says the diameter of the blue circle equals the sum of the radii of the green and orange circles.

Python code

In case you’re interested, here’s the code that created the diagram.

#!/usr/bin/env -S uv run --script

# /// script
# dependencies = ["numpy", "matplotlib"]
# ///

import numpy as np
import matplotlib.pyplot as plt

def connect(A, B, color='gray'):
    plt.plot([A[0], B[0]], [A[1], B[1]], color=color, linewidth=2)

def circle(c, r, color='gray'):
    t = np.linspace(0, 2*np.pi)
    plt.plot(c[0] + r*np.cos(t), c[1] + r*np.sin(t), color=color, linewidth=2)

def quadratic(a, b, c):
    det = b**2 - 4*a*c
    return ((-b - det**0.5)/(2*a), (-b + det**0.5)/(2*a))

A = np.array([-1, 0])
B = np.array([ 1, 0])
C = np.array([np.cos(1), np.sin(1)])
a = np.linalg.norm(B - C)
b = np.linalg.norm(A - C)
c = np.linalg.norm(B - A)
s = (a + b + c)/2

circle([0,0], 1)
connect(A, B,)
connect(A, C)
connect(C, B)
connect(C, C*np.array([1, -1]))

center = (a*A + b*B + c*C)/(2*s)
radius = 0.5*a*b/s
circle(center, radius, 'C0')

Ex = C[0]
roots = quadratic(1, 2 + 2*Ex, Ex**2 - 1)
r = roots[1] # Smaller root is negaive
print(roots)
center = (r + Ex, -r)
circle(center, r, 'C1')

roots = quadratic(1, 2 - 2*Ex, Ex**2 - 1)
r = roots[1] # Smaller root is negaive
center = (Ex - r, -r)
circle(center, r, 'C2')

plt.gca().set_aspect("equal")
plt.axis("off")
plt.show()

[1] Leon Bankoff. A Geometrical Coincidence. Mathematics Magazine, Vol. 37, No. 5 (Nov., 1964), p. 324.