Hart’s theorem

Posted on 25 June 2026 by John

Hart’s theorem says

If a triangle be formed by the arcs of three circles, the inscribed and the three escribed circles are all tangent to a new circle or line.

Here “triangle” means a three-sided figure whose sides are portions of a circle. The inscribed circle is the largest circle that can fit inside the three-sided figure.

The “escribed” circles are analogous to the excircles in the previous post: you extend two sides and find a circle that is tangent to the triangle side and the extended side. The difference here being that the side extensions are now circles.

Incircles and Excircles of Pythagorean triangles

Posted on 25 June 2026 by John

This post will reveal the connection between my two previous posts: one on the Star Trek lemma and one on Pythagorean triples.

In the process of writing the latter, I looked at the Wikipedia article on Pythagorean triples and noticed this curious paragraph.

In every Pythagorean triangle, the radius of the incircle and the radii of the three excircles are positive integers. Specifically, for a primitive triple the radius of the incircle is r = n(m − n), and the radii of the excircles opposite the sides m² − n², 2mn, and the hypotenuse m² + n² are respectively m(m − n), n(m + n), and m(m + n).

The citation for the paragraph above was the book by my former officemate, which led to the post on the Star Trek lemma. The passage in Arthur Baragar’s book that Wikipedia cites is Exercise 15.3.

Let ΔABC be a right angle triangle with sides of integer length. Prove that the inradius r and the exradii r_a, r_b, and r_c are all integers.

I don’t know whether Arthur discovered this theorem, but I’ll call it Baragar’s theorem for this post.

Incircles and excircles

To unpack Baragar’s theorem, let’s start by saying what incircles and excircles are. Incircles are more familiar. The incircle of a triangle is the largest circle that can be inscribed inside the triangle, and the radius of this circle is the inradius.

Since an incircle is an inscribed circle, you might expect an excircle to be a circumscribed circle, but that’s not it. There are three excircles, one for each side. To find the excircle for a side, extend the other two sides and find the circle tangent to the side and the two extensions. The radius of an excircle is its exradius.

Proof

Baragar’s theorem follows directly from Euclid’s formula for Pythagorean triples mentioned in the previous post

$\begin{align*} a &= m^2 - n^2 \\ b &= 2mn \\ c &= m^2 + n^2 \end{align*}$

and formulas for the inradius r and the exradii r_a, r_b, and r_c.

$\begin{align*} r &= \frac{K}{s} \\ r_a &= \frac{K}{s - a} \\ r_b &= \frac{K}{s - b} \\ r_c &= \frac{K}{s - c} \\ \end{align*}$

Here K is the area of the triangle, which in our case is ab/2, and s is the semiperimeter, half the perimeter.

Expressing the radii in terms of m and n gives the values cited by Wikipedia above.

Illustrating the theorem

I’d like to write a Python script to illustrate the theorem, and knowing the radii of the circles help, but we also need to know the centers of the circles.

The center of the incircle is the weighted average of the vertices, with weights given by the lengths of the opposite sides. That is, if the vertices are A, B, and C, and the sides opposite these vertices are a, b, and c, the the incenter is

$I = \frac{aA + bB + cC}{a + b + c}$

The centers for the excircles have remarkably similar expressions. For the incenter of the circle opposite a vertex, flip the sign of the corresponding side.

$\begin{align*} I_a = \frac{-aA + bB + cC}{-a + b + c} \\ I_a = \frac{aA - bB + cC}{a - b + c} \\ I_a = \frac{aA + bB - cC}{a + b - c} \\ \end{align*}$

Python code

Putting it all together, here’s an illustrate the theorem.

And here’s the code that produced it. Note that everything in this section works for right triangles in general, not just Pythagorean triangles.

import numpy as np
import matplotlib.pyplot as plt

def connect(A, B):
    plt.plot([A[0], B[0]], [A[1], B[1]], "C0")

def draw_circle(c, r, color):
    t = np.linspace(0, 2*np.pi)
    plt.plot(r*np.cos(t) + c[0], r*np.sin(t) + c[1], color=color)

a, b, c, = 3, 4, 5

A = np.array([0, b])
B = np.array([-a, 0])
C = np.array([0, 0])

s = (a + b + c)/2
K = a*b/2
r = K/s
ra = K/(s - a)
rb = K/(s - b)
rc = K/(s - c)
I = (a*A + b*B + c*C)/(a + b + c)
Ia = (-a*A + b*B + c*C)/(-a + b + c)
Ib = (a*A - b*B + c*C)/(a - b + c)
Ic = (a*A + b*B - c*C)/(a + b - c)

draw_circle(I, r, "C1")
draw_circle(Ia, ra, "C2")
draw_circle(Ib, rb, "C3")
draw_circle(Ic, rc, "C4")

plt.plot([-2*rc, 2*rb], [0, 0], "C0")
plt.plot([0, 0], [-2*ra, 2*rc], "C0")
plt.plot([(-2*ra - b)*a/b, 2*rb], [-2*ra, 2*rb*b/a + b], "C0")

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

The Star Trek lemma

Posted on 24 June 2026 by John

I was reading an article this evening and saw a footnote to a book by Arthur Baragar [1]. This caught my eye because he was my officemate at UT for a year.

I found his book on Archive.org and was surprised to see “The Star Trek Lemma” in the table of contents. What could this be?

It’s a theorem that goes back to Euclid that applies to an angle formed by connecting a point to two other points on a circle. The theorem says “The measure of an inscribed angle is half the measure of the arc it subtends.” But why call it the Star Trek lemma? Quoting Arthur:

In the spirit of Euclid, we will refer to this theorem as the Star Trek lemma because of the figure associated with the statement of the theorem. … Before Star Trek, as far as I know, this theorem had no name, though some might call it Euclid III.20, which is its proposition number in Euclid’s Elements (Book III, Proposition 20).

Here is my reconstruction of the figure given in the book.

Baragar's illustration of the Star Trek lemma

The lemma says that ∠BAC is half of ∠BOC.

[1] Baragar, Arthur (2001), A Survey of Classical and Modern Geometries: With Computer Activities, Prentice Hall

Circular arc approximation

Posted on 28 April 2026 by John

Suppose you have an arc a, a portion of a circle of radius r, and you know two things: the length c of the chord of the arc, and the length b of the chord of half the arc, illustrated below.

Here θ is the central angle of the arc. Then the length of the arc, rθ, is approximately

a = rθ ≈ 12 b²/(c + 4b).

If the arc is moderately small, the approximation is very accurate.

This approximation is simple, accurate, and not obvious, much like the one in this post

Derivation

Let φ = θ/4. Then the angle between the chords b and c is φ. This follows from the inscribed angle theorem, illustrated below.

There are two right triangles in the diagram above that have an angle φ: a smaller triangle with hypotenuse b and a larger triangle with hypotenuse 2r. From the smaller triangle we learn

cos(φ) = c / 2b

and from the larger triangle we learn

sin(φ) = b / 2r.

Now expand in power series.

c / 2b = cos(φ) = 1 − φ²/2! + φ⁴/4! − …
2b / a = sin(φ) / φ = 1 − φ²/3! + φ⁴/5! − …

If we multiply 2b / a by 3 and subtract c / 2b then the φ² terms cancel out and we get

6b / a − c / 2b = 2 − φ⁴/60 + …

and so

6b / a − c / 2b ≈ 2

to a very high degree of accuracy when φ is small. The approximation follows by solving for a.

Example

Let θ = π/3 and so φ = 0.26…, not a particularly small value of φ, but small enough for the approximation to work well.

Set r = 1 so a = θ. Then

b = 2 sin(π/12) = 0.51764

and

c = 2b cos(π/12) = 1.

Now in application, we know b and c, not θ, and so pretend we measured b = 0.51764 and c = 1. Then we would approximate a by

12b²/(c + 4b) = 1.04718

while the exact value is 1.04720. Unless you can measure lengths to more than four significant figures, the approximation may has well be exact because approximation error would be less than measurement error.

[1] J. M. Bruce. Approximation to a Circular Arc. The American Mathematical Monthly. Vol. 49, No. 3 (March 1942), pp. 184–185

Intersecting spheres and GPS

Posted on 14 April 2026 by John

If you know the distance d to a satellite, you can compute a circle of points that passes through your location. That’s because you’re at the intersection of two spheres—the earth’s surface and a sphere of radius d centered on the satellite—and the intersection of two spheres is a circle. Said another way, one observation of a satellite determines a circle of possible locations.

If you know the distance to a second satellite as well, then you can find two circles that contain your location. The two circles intersect at two points, and you know that you’re at one of two possible positions. If you know your approximate position, you may be able to rule out one of the intersection points.

If you know the distance to three different satellites, now you know three circles that you’re standing on, and the third circle will only pass through one of the two points determined by the first two satellites. Now you know exactly where you are.

Knowing the distance to more satellites is even better. In theory additional observations are redundant but harmless. In practice, they let you partially cancel out inevitable measurement errors.

If you’re not on the earth’s surface, you’re still at the intersection of n spheres if you know the distance to n satellites. If you’re in an airplane, or on route to the moon, the same principles apply.

Errors and corrections

How do you know the distance to a satellite? The satellite can announce what time it is by its clock, then when you receive the announcement you compare it to the time by your clock. The difference between the two times tells you how long the radio signal traveled. Multiply by the speed of light and you have the distance.

However, your clock will probably not be exactly synchronized with the satellite clock. Observing a fourth satellite can fix the problem of your clock not being synchronized with the satellite clocks. But it doesn’t fix the more subtle problems of special relativity and general relativity. See this post by Shri Khalpada for an accessible discussion of the physics.

Numerical computation

Each distance measurement gives you an equation:

|| x – s_i || = d_i

where s_i is the location of the ith satellite and d_i is your distance to that satellite. If you square both sides of the equation, you have a quadratic equation. You have to solve a system of nonlinear equations, and yet there is a way to transform the problem into solving linear equations, i.e. using linear algebra. See this article for details.

Finding a parabola through two points with given slopes

Posted on 14 April 2026 by John

The Wikipedia article on modern triangle geometry has an image labeled “Artzt parabolas” with no explanation.

A quick search didn’t turn up anything about Artzt parabolas [1], but apparently the parabolas go through pairs of vertices with tangents parallel to the sides.

The general form of a conic section is

ax² + bxy + cy² + dx + ey + f = 0

and the constraint b² = 4ac means the conic will be a parabola.

We have 6 parameters, each determined only up to a scaling factor; you can multiply both sides by any non-zero constant and still have the same conic. So a general conic has 5 degrees of freedom, and the parabola condition b² = 4ac takes us down to 4. Specifying two points that the parabola passes through takes up 2 more degrees of freedom, and specifying the slopes takes up the last two. So it’s plausible that there is a unique solution to the problem.

There is indeed a solution, unique up to scaling the parameters. The following code finds parameters of a parabola that passes through (x_i, y_i) with slope m_i for i = 1, 2.

def solve(x1, y1, m1, x2, y2, m2):
    
    Δx = x2 - x1
    Δy = y2 - y1
    λ = 4*(Δx*m1 - Δy)*(Δx*m2 - Δy)/(m1 - m2)**2
    k = x2*y1 - x1*y2

    a = Δy**2 + λ*m1*m2
    b = -2*Δx*Δy - λ*(m1 + m2)
    c = Δx**2 + λ
    d =  2*k*Δy + λ*(m1*y2 + m2*y1 - m1*m2*(x1 + x2))
    e = -2*k*Δx + λ*(m1*x1 + m2*x2 - y1 - y2)
    f = k**2 + λ*(m1*x1 - y1)*(m2*x2 - y2)

    return (a, b, c, d, e, f)

[1] The page said “Artz” when I first looked at it, but it has since been corrected to “Artzt”. Maybe I didn’t find anything because I was looking for the wrong spelling.

Random hexagon fractal

Posted on 9 April 2026 by John

I recently ran across a post on X describing a process for creating a random fractal. First, pick a random point c inside a hexagon.

Then at each subsequent step, pick a random side of the hexagon and create the triangle formed by that side and c. Update c to be the center of the new triangle and plot c.

Note that you only choose a random point inside the hexagon once. After that you randomly choose sides.

Now there are many ways to define the center of a triangle. I assumed the original meant barycenter (centroid) when it said “center”, and apparently that was correct. I was able to create a similar figure.

But if you define center differently, you get a different image. For example, here’s what you get when you use the incenter, the center of the largest circle inside the triangle.

Hyperbolic version of Napier’s mnemonic

Posted on 2 April 2026 by John

I was looking through an old geometry book [1] and saw a hyperbolic analog of Napier’s mnemonic for spherical trigonometry. In hindsight of course there’s a hyperbolic analog: there’s a hyperbolic analog of everything. But I was surprised because I’d never thought of this before. I suppose the spherical version is famous because of its practical use in navigational calculations, while the hyperbolic analog is of more theoretical interest.

Napier’s mnemonic is a clever way to remember 10 equations in spherical trig. See the linked post for the meanings of the variables.

sin a = sin A sin c = tan b cot B
sin b = sin B sin c = tan a cot A
cos A = cos a sin B = tan b cot c
cos B = cos b sin A = tan a cot c
cos c = cot A cot B = cos a cos b

The hyperbolic analog replaces every circular function of a, b, or c with its hyperbolic counterpart.

sinh a = sin A sinh c = tanh b cot B
sinh b = sin B sinh c = tanh a cot A
cos A = cosh a sin B = tanh b coth c
cos B = cosh b sin A = tanh a coth c
cosh c = cot A cot B = cosh a cosh b

[1] D. M. Y. Sommerville. The Elements of Non-Euclidean Geometry. 1919.

Visualizing orbital velocity

Posted on 16 February 2026 by John

The shape of a planet’s orbit around a star is an ellipse. To put it another way, a plot of the position of a planet’s orbit over time forms an ellipse. What about the velocity? Is its plot also an ellipse? Surprisingly, a plot of the velocity forms a circle even if a plot of the position is an ellipse.

If an object is in a circular orbit, it’s velocity vector traces out a circle too, with the same center. If the object is in an elliptical orbit, it’s velocity vector still traces out a circle, but one with a different center. When the orbit is eccentric, the hodograph, the figure traced out by the velocity vector, is also eccentric, though the two uses of the word “eccentric” are slightly different.

The eccentricity e of an ellipse is the ratio c/a where c is the distance between the two foci and a is the semi-major axis. For a circle, c = 0 and so e = 0. The more elongated an ellipse is, the further apart the foci are relative to the axes and so the greater the eccentricity.

The plot of the orbit is eccentric in the sense that the two foci are distinct because the shape is an ellipse. The hodograph is eccentric in the sense that the circle is not centered at the origin.

The two kinds of eccentricity are related: the displacement of the center of the hodograph from the origin is proportional to the eccentricity of the ellipse.

Imagine the the orbit of a planet with its major axis along the x-axis and the coordinate of its star positive. The hodograph is a circle shifted up by an amount proportional to the eccentricity of the orbit e. The top of the circle corresponds to perihelion, the point closest to the star, and the bottom corresponds to aphelion, the point furthest from the star. For more details, see the post Max and min orbital speed.

Solving spherical triangles

Posted on 4 December 2025 by John

This post is a side quest in the series on navigating by the stars. It expands on a footnote in the previous post.

There are six pieces of information associated with a spherical triangle: three sides and three angles. I said in the previous post that given three out of these six quantities you could solve for the other three. Then I dropped a footnote saying sometimes the missing quantities are uniquely determined but sometimes there are two solutions and you need more data to uniquely determine a solution.

Todhunter’s textbook on spherical trig gives a thorough account of how to solve spherical triangles under all possible cases. The first edition of the book came out in 1859. A group of volunteers typeset the book in TeX. Project Gutenberg hosts the PDF version of the book and the TeX source.

I don’t want to duplicate Todhunter’s work here. Instead, I want to summarize when solutions are or are not unique, and make comparisons with plane triangles along the way.

SSS and AAA

The easiest cases to describe are all sides or all angles. Given three sides of a spherical triangle (SSS), you can solve for the angles, as with a plane triangle. Also, given three angles (AAA) you can solve for the remaining sides of a spherical triangle, unlike a plane triangle.

SAS and SSA

When you’re given two sides and an angle, there is a unique solution if the angle is between the two sides (SAS), but there may be two solutions if the angle is opposite one of the sides (SSA). This is the same for spherical and plane triangles.

There could be even more than two solutions in the spherical case. Consider a triangle with one vertex at the North Pole and two vertices on the equator. Two sides are specified, running from the pole to the equator, and the angles at the equator are specified—both are right angles—but the side of the triangle on the equator could be any length.

ASA and AAS

When you’re given two angles and a side, there is a unique solution if the side is common to the two angles (ASA).

If the side is opposite one of the angles (AAS), there may be two solutions to a spherical triangle, but only one solution to a plane triangle. This is because two angles uniquely determine the third angle in a plane triangle, but not in a spherical triangle.

The example above of a triangle with one vertex at the pole and two on the equator also shows that an AAS problem could have a continuum of solutions.

Summary

$\begin{tabular}{|l|c|c|} \hline \textbf{Case} & \textbf{Plane} & \textbf{Spherical} \\ \hline SSS & 1 & 1 \\ SAS & 1 & 1 \\ SSA & 1 or 2 & 1 or 2 \\ AAS & 1 & 1 or 2 \\ ASA & 1 & 1 \\ AAA & $\infty$ & 1 \\ \hline \end{tabular}$

Note that spherical triangles have a symmetry that plane triangles don’t: the spherical column above remains unchanged if you swap S’s and A’s. This is an example of duality in spherical geometry.

Geometry