Here are four theorems that generalize the Pythagorean theorem. Follow the links for more details regarding each equation.
1. Theorem by Apollonius for general triangles.
2. Edsgar Dijkstra’s extension of the Pythagorean theorem for general triangles.
3. A generalization of the Pythagorean theorem to tetrahedra.
4. A unified Pythagorean theorem that covers spherical, plane, and hyperbolic geometry.
The area of a triangle can be computed directly from the lengths of its sides via Heron’s formula.
Here s is the semiperimeter, s = (a + b + c)/2.
Is there an analogous formula for spherical triangles? It’s not obvious there should be, but there is a formula by Simon Antoine Jean L’Huilier (1750–1840).
Here we denote area by S for surface area, rather than A because in the context of spherical trigonometry A usually denotes the angle opposite side a. The same convention applies in plane trigonometry, but the potential for confusion is greater in L’Huilier’s formula since the area appears inside a tangent function.
Now tan θ ≈ θ for small θ, and so L’Huilier’s formula reduces to Heron’s formula for small triangles.
Imagine the Earth as a sphere of radius 1 and take a spherical triangle with one vertex at the north pole and two vertices on the equator 90° longitude apart. Then a = b = c = π/2 and s = 3π/4. Such a triangle takes of 1/8 of the Earth’s surface area of 4π, so the area S is π/2. You can verify that L’Huilier’s formula gives the correct area.
It’s not a proof, but it’s a good sanity check that L’Huilier’s formula is correct for small triangles and for at least one big triangle.
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(r*cos(t) + center[0], r*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.
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 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.
Apollonius of Perga (c. 262 BC – c. 190 BC) discovered a theorem that generalizes the Pythagorean theorem but isn’t nearly as well known.
Let ABC be a general triangle, and let D be the midpoint of the segment AB. Let a be the length of the side opposite A and b the length of the side opposite B. Let m be the length of AD and h the length of the mediant, the line CD.
Apollonius’s theorem says
a² + b² = 2(m² + h²).
To see that this is a generalization of the Pythagorean theorem, apply Apollonius’ theorem to an isosceles triangle. Now a = b and ACD is a right triangle.
Apollonius’ theorem says
2b² = 2m² + 2h²
which is the Pythagorean theorem applied to ACD with each term doubled.
Exactly one month ago I wrote about sampling points from a triangle. This post will look at the three dimensional analog, sampling from a tetrahedron. The generalization to higher dimensions works as well.
In the triangle post, I showed that a naive approach doesn’t work but a variation does. If the vertices of the triangle are A, B, and C, then the naive approach is to generate three uniform random numbers, normalize them, and use them to form a linear combination of the vertices. That is, generate
r1, r2, r3
uniformly from [0, 1], then define
r1′ = r1/(r1 + r2 + r3)
r2′ = r2/(r1 + r2 + r3)
r3′ = r3/(r1 + r2 + r3)
and return as a random sample the point
r1′ A + r2′ B + r3′ C.
This oversamples points near the middle of the triangle and undersamples points near the vertices. But everything above works if you replace uniform samples with exponential samples.
The analogous method works for a tetrahedron with vertices A, B, C, and D. Generate exponential random variables ri for i = 1, 2, 3, 4. Then normalize, defining each ri′ to be ri divided by the sum of all the ri s. Then the random sample is
r1′ A + r2′ B + r3′ C + r4′ D.
In the case of the triangle, it was easy to visualize that uniformly generated rs did not lead to uniform samples of the triangle, but that exponentially generated rs did. This is harder to do in three dimensions. Even if the points were uniformly sampled from a tetrahedron, they might not look uniformly distributed from a given perspective.
So how might we demonstrate that our method works? One approach would be to take a cube inside a tetrahedron and show that the proportion of samples that land inside the cube is what we’d expect, namely the ratio of the volume of the cube to the volume of the tetrahedron.
This brings up a couple interesting subproblems.
To find the volume of a tetrahedron, form a matrix M whose columns are the vectors from three of the vertices to the remaining vertex. So we could take A − D, B − D, and C − D. The volume of the tetrahedron is the determinant of M divided by 6.
A tetrahedron is convex, and so the line segment joining any two points inside the tetrahedron lies entirely inside the tetrahedron. It follows that if all the vertices of a cube are inside the a tetrahedron, the entire cube is inside.
So this brings us to the problem of testing whether a point is inside a tetrahedron. We can do this by converting the coordinates of the point to barycentric coordinates, then testing whether all the coordinates are non-negative and add to 1.
I made up four points as vertices of a tetrahedron.
import numpy as np A = [0, 0, 0] B = [5, 0, 0] C = [0, 7, 0] D = [0, 1, 6] A, B, C, D = map(np.array, [A, B, C, D])
For my test cube I chose the cube whose vertex coordinates are all combinations of 1 and 2. So the vertex nearest the origin is (1, 1, 1) and the vertex furthest from the origin is (2, 2, 2). You can verify that all the vertices are inside the tetrahedron, and so the cube is inside the tetrahedron.
The tetrahedron has volume 35, and the cube has volume 1, so we expect 1/35th of the random points to land inside the cube. Here’s the code to sample the tetrahedron and calculate the proportion of points that land in the cube.
from scipy.stats import expon
def incube(v):
return 1 <= v[0] <= 2 and 1 <= v[1] <= 2 and 1 <= v[2] <= 2
def good_sample(A, B, C, D):
r = expon.rvs(size=4)
r /= sum(r)
return r[0]*A + r[1]*B + r[2]*C + r[3]*D
def bad_sample(A, B, C, D):
r = np.random.random(size=4)
r /= sum(r)
return r[0]*A + r[1]*B + r[2]*C + r[3]*D
N = 1_000_000
badcount = 0
goodcount = 0
for _ in range(N):
v = bad_sample(A, B, C, D)
if incube(v):
badcount += 1
v = good_sample(A, B, C, D)
if incube(v):
goodcount += 1
print(badcount/N, goodcount/N, 1/35)
This produced
0.094388 0.028372 0.028571…
The good (exponential) method match the expected proportion to three decimal places but the bad (uniform) method put about 3.3 times as many points in the cube as expected.
Note that three decimal agreement is about what we’d expect via the central limit theorem since 1/√N = 0.001.
If you’re porting the sampler above to an environment where you don’t a function for generating exponential random samples, you can roll your own by returning −log(u) where u is a uniform sample from [0. 1].
After writing the post about sampling from triangles, I wrote a followup post about sampling from polygons. In a nutshell, divide your polygon into triangles, then randomly select a triangle in proportion to its area, then sample with that triangle. You can do the analogous process in three dimensions by dividing a general volume into tetrahedra.
John Conway discovered a right triangle that can be partitioned into five similar triangles. The sides are in proportion 1 : 2 : √5.
You can make a larger similar triangle by making the entire triangle the central (green) triangle of a new triangle.
Here’s the same image with the small triangles filled in as in the original.
Repeating this process creates an aperiodic tiling of the plane.
The tiling was discovered by Conway, but Charles Radin was the first to describe it in a publication [1]. Radin attributes the tiling to Conway.
It would be easiest to illustrate the tiling if we were standing together in a room and placing new triangles on the floor, watching the tiling expand. Given the limitations of a screen, it may be easier to visualize subdividing the triangle rather than tiling the plane.
Imagine the smallest triangles are a constant size and at each step we’re viewing the process from further away. We see a constant size outer triangle at each step, but the triangle is growing and covering the plane.
Here’s an animated GIF of the process.
[1] Charles Radin. “The Pinwheel Tilings of the Plane.” Annals of Mathematics, vol. 139, no. 3, 1994, pp. 661–702.
Yesterday I wrote about a triangle inequality discovered by Paul Erdős.
Let P be a point inside a triangle ABC. Let x, y, z be the distances from P to the vertices and let p, q, r, be the distances to the sides. Then Erdős’ inequality says
x + y + z ≥ 2(p + q + r).
Using the same notation, here are four more triangle inequalities discovered by Oppenheim [1].
[1] A. Oppenheim. The Erdös Inequality and Other Inequalities for a Triangle. The American Mathematical Monthly. Vol. 68, No. 3 (Mar., 1961), pp. 226–230
Ted Dunning left a comment on my post on random sampling from a triangle saying you could extend this to sampling from a polygon by dividing the polygon into triangles, and selecting a triangle each time with probability proportional to the triangle’s area.
To illustrate this, let’s start with a irregular pentagon.
To pick a point inside, I used the centroid, the average of the vertices. Connecting the centroid to each of the vertices splits the pentagon into triangles. (Here I implicitly used the fact that this pentagon is convex. The centroid of a non-convex polygon could be outside the polygon.)
We can find the area of the triangles using Heron’s rule.
Here’s what we get for random samples.
