2025-11-05 19:45:50
Here’s an interesting theorem that leads to some aesthetically pleasing images. It’s known as the Japanese cyclic polygon theorem.
For all triangulations of a cyclic polygon, the sum of inradii of the triangles is constant. Conversely, if the sum of inradii is independent of the triangulation, then the polygon is cyclic.
The image above shows two triangulations of an irregular hexagon. By the end of the post we’ll see the code that created these images, and we’ll see one more triangulations. And we’ll see that the sum of the radii of the circles is the same in each image.
In case any of the terms in the theorem above are unfamiliar, here’s a quick glossary.
A triangulation of a polygon is a way to divide the polygon into triangles, with the restriction that the triangle vertices come from the polygon vertices.
A polygon is cyclic if there exists a circle that passes through all of its vertices. Incidentally, if a polygon has n + 2 sides, the number of possible triangulations is the nth Catalan number. More on that here.
The inradius of a triangle is the radius of the largest circle that fits inside the triangle. More on inner and outer radii here.
We’d like to illustrate the theorem by drawing some images and by adding up the inradii. This means we need to be able to find the inradius and the incenter.
The inradius of a triangle equals its area divided by its semiperimeter. The area we can find using Heron’s formula. The semiperimeter is half the perimeter.
The incenter is the weighted sum of the vertices, weighted by the lengths of the opposite sides, and normalized. If the vertices are A, B, and C, then the incenter is
where A is the vertex opposite side a, B is the vertex opposite side b, and C is the vertex opposite side c.
The following Python code will draw a triangle with its incircle. Calling this function for each triangle in the triangulation will create the illustrations we’re after.
from numpy import *
import matplotlib.pyplot as plt
def draw_triangle_with_incircle(A, B, C):
a = linalg.norm(B - C)
b = linalg.norm(A - C)
c = linalg.norm(A - B)
s = (a + b + c)/2
r = sqrt(s*(s-a)*(s-b)*(s-c))/s
center = (a*A + b*B + c*C)/(a + b + c)
plt.plot([A[0], B[0]], [A[1], B[1]], color="C0")
plt.plot([B[0], C[0]], [B[1], C[1]], color="C0")
plt.plot([A[0], C[0]], [A[1], C[1]], color="C0")
t = linspace(0, 2*pi)
plt.plot(cos(t) + center[0], sin(t) + center[1], color="C2")
return r
Next, we pick six points not quite evenly spaced around a circle.
angle = [0, 50, 110, 160, 220, 315] # degrees p = [array([cos(deg2rad(angle[i])), sin(deg2rad(angle[i]))]) for i in range(6)]
Now we draw three different triangulations of the hexagon defined by the points above. We also print the sum of the inradii to verify that each sum is the same.
def draw_polygon(triangles):
rsum = 0
for t in triangles:
rsum += draw_triangle_with_incircle(p[t[0]], p[t[1]], p[t[2]])
print(rsum)
plt.axis("off")
plt.gca().set_aspect("equal")
plt.show()
draw_polygon([[0,1,2], [0,2,3], [0,3,4], [0,4,5]])
draw_polygon([[0,1,2], [2,3,4], [4,5,0], [0,2,4]])
draw_polygon([[0,1,2], [0,2,3], [0,3,5], [3,4,5]])
This produces the two images at the top of the post the one below. All the inradii sum are 1.1441361217691244 with a little variation in the last decimal place.
2025-11-04 00:12:02
A tetrahedron has four triangular faces. Suppose three of those faces come together like the corner of a cube, each perpendicular to the other. Let A1, A2, and A3 be the areas of the three triangles that meet at this corner and let A0 be the area of remaining face, the one opposite the right corner.
De Gua’s theorem, published in 1783, says
A0² = A1² + A2² + A3².
We will illustrate De Gua’s theorem using Python.
from numpy import *
def area(p1, p2, p3):
return 0.5*abs(linalg.norm(cross(p1 - p3, p2 - p3)))
p0 = array([ 0, 0, 0])
p1 = array([11, 0, 0])
p2 = array([ 0, 3, 0])
p3 = array([ 0, 0, 25])
A0 = area(p1, p2, p3)
A1 = area(p0, p2, p3)
A2 = area(p0, p1, p3)
A3 = area(p0, p1, p2)
print(A0**2 - A1**2 - A2**2 - A3**2)
This prints 0, as expected.
A natural question is whether De Gua’s theorem generalizes to higher dimensions, and indeed it does.
Suppose you have a 4-simplex where one corner fits in the corner of a hypercube. The 4-simplex has 5 vertices. If you leave out any vertex, the remaining 4 points determine a tetrahedron. Let V0 be the volume of the tetrahedron formed by leaving out the vertex at the corner of the hypercube, and let V1, V2, V3, and V4 be the volumes of the tetrahedra formed by dropping out each of the other vertices. Then
V0² = V1² + V2² + V3² + V4².
You can extend the theorem to even higher dimensions analogously.
Let’s illustrate the theorem for the 4-simplex in Python. The volume of a tetrahedron can be computed as
V = det(G)1/2/6
where G is the Gram matrix computed in the code below.
def volume(p1, p2, p3, p4):
u = p1 - p4
v = p2 - p4
w = p3 - p4
# construct the Gram matrix
G = array( [
[dot(u, u), dot(u, v), dot(u, w)],
[dot(v, u), dot(v, v), dot(v, w)],
[dot(w, u), dot(w, v), dot(w, w)] ])
return sqrt(linalg.det(G))/6
p0 = array([ 0, 0, 0, 0])
p1 = array([11, 0, 0, 0])
p2 = array([ 0, 3, 0, 0])
p3 = array([ 0, 0, 25, 0])
p4 = array([ 0, 0, 0, 7])
V0 = volume(p1, p2, p3, p4)
V1 = volume(p0, p2, p3, p4)
V2 = volume(p0, p1, p3, p4)
V3 = volume(p0, p1, p2, p4)
V4 = volume(p0, p1, p2, p3)
print(V0**2 - V1**2 - V2**2 - V3**2 - V4**2)
This prints -9.458744898438454e-11. The result is 0, modulo the limitations of floating point arithmetic.
Floating point arithmetic is generally accurate to 15 decimal places. Why is the numerical error above relatively large? The loss of precision is the usual suspect: subtracting nearly equal numbers. We have
V0 = 130978.7777777778
and
V1**2 + V2**2 + V3**2 + V4**2 = 130978.7777777779
Both results are correct to 16 decimal places, but when we subtract them we lose all precision.
The post Tetrahedral analog of the Pythagorean theorem first appeared on John D. Cook.2025-11-03 06:12:53
As I’ve written about several times lately, the Smith chart is the image of a rectangular grid in the right half-plane under the function
f(z) = (z − 1)/(z + 1).
What would the image of a grid in the left half-plane look like?
For starters, since f maps the right half-plane to the interior of the unit circle, it must map the left-half plane to the exterior of the unit circle.
As we said before, the function f is a Möbius transformation, and so it takes generalized circles, i.e. either a circle or a line, to generalized circles. So the grid lines in the left half-plane are either mapped to lines or circles. Which is it?
The function f has a singularity at −1 and so the image of any line (or circle) through z = −1 is unbounded, i.e. a line, not a circle. Any line not passing through −1 has a bounded image, which must be a circle.
The line Re(z) = −1 in the z plane is mapped to the line Re(w) = 1 in the w plane. Otherwise a vertical line crossing the real axis at x is mapped to a circle passing through w = (x − 1)/(x + 1). The circle also passes through w = 1 because f(∞) = 1. The circle is symmetric about the real axis, and so this is enough information to uniquely determine the circle.
Note that (x − 1)/(x + 1) > 1 when x < −1 and so vertical lines with real part less than −1 are mapped to circles to the right of w = 1. When −1 < x < 0, vertical lines are mapped to circles to the left of w = 1.
The images of horizontal lines we’ve looked at before. These are all circles passing through w = 1 and tangent to the circles that are images of vertical lines. But this time instead of taking the portion of the circles inside the unit circle, we take the portion outside the unit circle.
And without further ado, we present the anti-Smith chart, the image of a grid in the left half plane.
The post The anti-Smith chart first appeared on John D. Cook.
2025-11-02 21:10:50
A few days ago I wrote two posts about how to create a Smith chart, a graphical device used for impedance calculations. Then someone emailed me to point out the connection between the Smith chart and triangular numbers.
The Smith chart is the image of a rectangular grid in the right half-plane under the function
f(z) = (z − 1)/(z + 1).
If you subtract the values of f at consecutive integers, you get the reciprocal of a triangular number.
f(n) − f(n − 1) = 2/(n(n + 1)) = 1 / Tn
Or to put it another way,
f(n) − f(n − 1) = 1 / (1 + 2 + 3 + … + n).
In the first post on the Smith chart we showed that the function f maps vertical lines
in the z plane to circles in the w plane all touching at w = 1.
The circles are symmetric about the real axis and the diameter runs from f(n) to 1. The separation between the circles on the left side is thus
f(n) − f(n − 1) = 1 / Tn.
Number the circles starting from the outermost as 0, 1, 2, …. Then the maximum distance between circle n and circle n − 1 is 1 / Tn. You can see in the graph above that the distance between circle 0 and circle 1 is 1. It’s a little harder to see that the distance between circle 1 and circle 2 is 1/3. It looks like the distance between circles 2 and 3 is about half of that between circles 1 and 2, so it would be 1/6.
2025-11-01 22:28:24
The cross ratio of four points A, B, C, D is defined by
where XY denotes the length of the line segment from X to Y.
The idea of a cross ratio goes back at least as far as Pappus of Alexandria (c. 290 – c. 350 AD). Numerous theorems from geometry are stated in terms of the cross ratio. For example, the cross ratio of four points is unchanged under a projective transformation.
The cross ratio of four (extended [1]) complex numbers is defined by
The absolute value of the complex cross ratio is the cross ratio of the four numbers as points in a plane.
The cross ratio is invariant under Möbius transformations, i.e. if T is any Möbius transformation, then
This is connected to the invariance of the cross ratio in geometry: Möbius transformations are projective transformations on a complex projective line. (More on that here.)
If we fix the first three arguments but leave the last argument variable, then
is the unique Möbius transformation mapping z1, z2, and z3 to ∞, 0, and 1 respectively.
Suppose (a, b; c, d) = λ ≠ 1. Then there are 4! = 24 permutations of the arguments and 6 corresponding cross ratios:
Viewed as functions of λ, these six functions form a group, generated by
This group is called the anharmonic group. Four numbers are said to be in harmonic relation if their cross ratio is 1, so the requirement that λ ≠ 1 says that the four numbers are anharmonic.
The six elements of the group can be written as
When I was looking at the six possible cross ratios for permutations of the arguments, I thought about where I’d seen them before: the linear transformation formulas for hypergeometric functions. These are, for example, equations 15.3.3 through 15.3.9 in A&S. They relate the hypergeometric function F(a, b; c; z) to similar functions where the argument z is replaced with one of the elements of the anharmonic group.
I’ve written about these transformations before here. For example,
There are deep relationships between hypergeometric functions and projective geometry, so I assume there’s an elegant explanation for the similarity between the transformation formulas and the anharmonic group, though I can’t say right now what it is.
[1] For completeness we need to include a point at infinity. If one of the z equals ∞ then the terms involving ∞ are dropped from the definition of the cross ratio.
The post Cross ratio first appeared on John D. Cook.2025-10-31 23:44:10
Let’s make a QR code out of a sentence two ways: mixed case and upper case. We’ll use Python with the qrcode library.
>>> import qrcode
>>> s = "The quick brown fox jumps over the lazy dog."
>>> qrcode.make(s).save("mixed.png")
>>> qrcode.make(s.upper()).save("upper.png")
Here are the mixed case and upper case QR codes.
The QR code creation algorithm interprets the mixed case sentence as binary data but it interprets the upper case sentence as alphanumeric data.
Alphanumeric data, in the context of QR codes, comes from the following alphabet of 44 characters:
0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ $%*+-.:
Since 44² = 1936 < 2048 = 211 two alphanumeric characters can be encoded in 11 bits. If text contains a single character outside this alphabet, such as a lower case letter, then the text is encoded as ISO/IEC 8859-1 using 8 bits per character.
Switching from mixed-case text to upper case text reduces the bits per character from 8 to 5.5, and so we should expect the resulting QR code to require about 30% fewer pixels. In the example above we go from a 33 × 33 grid down to a 29 × 29 grid, from 1089 pixels to 841.
Bech32 encoding uses an alphabet of 32 characters while Base58 encoding uses an alphabet of 58 characters, and so the former needs about 17% more characters to represent the same data. But Bech32 uses a monocase alphabet, and base 58 does not, and so Bech32 encoding requires fewer QR code pixels to represent the same data as Base58 encoding.
(Bech32 encoding uses a lower case alphabet, but the letters are converted to upper case before creating QR codes.)
