Geometry

Random hexagon fractal

Posted on 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.

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Hyperbolic version of Napier’s mnemonic

Posted on 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 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 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.

A triangle whose interior angles sum to zero

Posted on by John

Spherical geometry

In spherical geometry, the interior angles of a triangle add up to more than π. And in fact you can determine the area of a spherical triangle by how much the angle sum exceeds π. On a sphere of radius 1, the area equals the triangle excess

Area = E = interior angle sum − π.

Small triangles have interior angle sum near π. But you could, for example, have a triangle with three right angles: put a vertex on the north pole and two vertices on the equator 90° longitude apart.

Hyperbolic geometry

In hyperbolic geometry, the sum of the interior angles of a triangle is always less than π. In a space with curvature −1, the area equals the triangle defect, the difference between π and the angle sum.

Area = D = π − interior angle sum.

Again small triangles have an interior angle sum near π. Both spherical and hyperbolic geometry are locally Euclidean.

The interior angle sum can be any value less than π, and so as the angle sum goes to 0, the triangle defect, and hence the area, goes to π. Since the minimum angle sum is 0, the maximum area of a triangle is π.

The figure below has interior angle sum 0 and area π in hyperbolic geometry.

Strictly speaking this is an improper triangle because the three hyperbolic lines (i.e. half circles) don’t intersect within the hyperbolic plane per se but at ideal points on the real axis. But you could come as close to this triangle as you like, staying within the hyperbolic plane.

Note that the radii of the (Euclidean) half circles doesn’t change the area. Any three semicircles that intersect on the real line as above make a triangle with the same area. Note also that the triangle has infinite perimeter but finite area.

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A circle in the hyperbolic plane

Posted on by John

Let ℍ be the upper half plane, the set of complex real numbers with positive imaginary part. When we measure distances the way we’ve discussed in the last couple posts, the geometry of ℍ is hyperbolic.

What is a circle of radius r in ℍ? The same as a circle in any geometry: it’s the set of points a fixed distance r from a center. But when you draw a circle using one metric, it may look very different when viewed from the perspective of another metric.

Suppose we put on glasses that gave us a hyperbolic perspective on ℍ, draw a circle of radius r centered at i, then take off the hyperbolic glasses and put on Euclidean glasses. What would our drawing look like?

In the previous post we gave several equivalent expressions for the hyperbolic metric. We’ll use the first one here.

d(z_1, z_2) = 2\, \text{arcsinh}\left( \frac{|z_1 - z_2|}{2\, \sqrt{\Im z_1\, \Im z_2}} \right)

Here the Fraktur letter ℑ stands for imaginary part. So the set of points in a circle of radius r centered at i is

\{ x + iy \mid d(x + iy, i) = r \}

Divide the expression for d(x + iy, i) by 2, apply sinh, and square. This gives us

\sinh^2\left(\frac{r}{2}\right) = \frac{x^2 + (y-1)^2}{4y}

which is an equation for a Euclidean circle. If we multiply both sides by 4y and complete the square, we find that the center of the circle is (0, cosh(r)) and the radius is sinh(r).

Summary so far

So to recap, if we put on our hyperbolic glasses and draw a circle, then switch out these glasses for Euclidean glasses, the figure we drew again looks like a circle.

To put it another way, a hyperbolic viewer and a Euclidean viewer would agree that a circle has been draw. However, the two viewers would disagree where the center of the circle is located, and they would disagree on the radius.

Both would agree that the center is on the imaginary axis, but the hyperbolic viewer would say the imaginary part of the center is 1 and the Euclidean viewer would say it’s cosh(r). The hyperbolic observer would say the circle has radius r, but the Euclidean observer would say it has radius sinh(r).

Small circles

For small r, the hyperbolic and Euclidean viewpoints nearly agree because

cosh(r) = 1 + O(r²)

and

sinh(r) = r + O(r³)

Big circles

Note that if you asked a Euclidean observer to draw a circle of radius 100, centered at (0, 1), he would say that the circle will extend outside of the half plane. A hyperbolic observer would disagree. From his perspective, the real axis is infinitely far away and so he can draw a circle of any radius centered at any point and stay within the half plane.

Moving circles

Now what if we looked at circles centered somewhere else? The hyperbolic metric is invariant under Möbius transformations, and so in particular it is invariant under

z ↦ x0 + y0 z.

This takes a circle with hyperbolic center i to a circle centered at x0 + i y0 without changing the hyperbolic radius. The Euclidean center moves from cosh(r) to y0 cosh(r) and the radius changes from sinh(r) to y0 sinh(r).

Hyperbolic metric

Posted on by John

One common model of the hyperbolic plane is the Poincaré upper half plane ℍ. This is the set of points in the complex plane with positive imaginary part. Straight lines are either vertical, a set of points with constant imaginary part, or arcs of circles centered on the real axis. The real axis is not part of ℍ. From the perspective of hyperbolic geometry these are ideal parts, infinitely far away, and not part of the plane itself.

We can define a metric on ℍ as follows. To find the distance between two points u and v, draw a line between the two points, and let a and b be the ideal points at the end of the line. By a line we mean a line as defined in the geometry of ℍ, what we would see from our Euclidean perspective as a half circle or a vertical line. Then the distance between u and v is defined as the absolute value of the log of the cross ratio (u, v; a, b).

d(u, v) = |\log (u, v; a, b) | = \left| \log \frac{|a - u|\,|b - v|}{|a - v|\,|b - u|} \right|
Cross ratios are unchanged by Möbius transformations, and so Möbius transformations are isometries.

Another common model of hyperbolic geometry is the Poincaré disk. We can use the same metric on the Poincaré disk because the Möbius transformation

z \mapsto \frac{z - i}{z + i}

maps the upper half plane to the unit disk. This is very similar to how the Smith chart is created by mapping a grid in the right half plane to the unit disk.

Update: See the next post for four analytic expressions for the metric, direct formulas involving u and v but not the ideal points a and b.

Four generalizations of the Pythagorean theorem

Posted on by John

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.

a^2 + b^2 = 2(m^2 + h^2)

2. Edsgar Dijkstra’s extension of the Pythagorean theorem for general triangles.

\text{sgn}(\alpha + \beta - \gamma) = \text{sgn}(a^2 + b^2 - c^2)

3. A generalization of the Pythagorean theorem to tetrahedra.

V_0^2 = \sum_{i=1}^n V_i^2

4. A unified Pythagorean theorem that covers spherical, plane, and hyperbolic geometry.

A(c) = A(a) + A(b) - \kappa \frac{A(a) \, A(b)}{2\pi}

Analog of Heron’s formula on a sphere

Posted on by John

The area of a triangle can be computed directly from the lengths of its sides via Heron’s formula.

A = \sqrt{s(s-a)(s-b)(s-c)}

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).

\tan^2 \frac{S}{4} = \tan \frac{s}{2} \tan \frac{s-a}{2} \tan \frac{s-b}{2} \tan \frac{s-c}{2}

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.

Japanese polygon theorem

Posted on by John

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.

Glossary

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.

Equations for inradius and incenter

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

\frac{aA + bB + cC}{a + b + c}

where A is the vertex opposite side a, B is the vertex opposite side b, and C is the vertex opposite side c.

Python code

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.

 

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