math and physics play

Inscribed Triangle in circle problem

December 25, 2022 math and physics play , , , , , ,

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In the LinkedIn Pre-University Geometric Algebra group, James presents a problem from the MindYourDecisions youtube channel Impossible Viral Problem, as a candidate for solution using geometric algebra.

I tried this out and found a couple ways to solve it. One of those I’ll detail here. I have to admit that part of the reason that I wanted to solve this is that the figure in the beginning of the video really bugged me. The triangle that was inscribed in the circle didn’t have any of the length properties from the problem. I could do much better with a sloppy freehand sketch, but to do a good figure, you have to actually solve for the vertexes of the triangle (once you do that, the area is easy to figure out.)

Formulating the problem.

Having solved the problem, the geometry of the problem is illustrated in fig. 1.

fig. 1. Inscribed triangle in circle.

fig. 1. Inscribed triangle in circle.

I set up the problem so that the \( A,C \) triangle vertices were symmetric with respect to the x-axis, and the \(B \) vertex located elsewhere. I can describe those algebraically as
\begin{equation}\label{eqn:inscribedTriangleProblem:20}
\begin{aligned}
\BA &= r \Be_1 e^{i\theta} \\
\BC &= r \Be_1 e^{-i\theta} \\
\BB &= r \Be_1 e^{i\phi},
\end{aligned}
\end{equation}
where the radius \( r \) and two angles \( \theta, \phi \) are to be determined, and \( i = \Be_1 \Be_1\) the pseudoscalar for the \(x-y\) plane.
The vector pointing to the midpoint of the upper triangular face is given by the average of the \( \BA, \BB \) vectors, which can be seen from
\begin{equation}\label{eqn:inscribedTriangleProblem:40}
\BA + \frac{\BB – \BA}{2} = \frac{\BA + \BB}{2},
\end{equation}
and similarly, the midpoint of the lower face is found at
\begin{equation}\label{eqn:inscribedTriangleProblem:60}
\BC + \frac{\BB – \BC}{2} = \frac{\BB + \BC}{2},
\end{equation}
The problem tells us that the respective lengths of those vectors from the origin are \( r-2, r – 3\) respectively, so
\begin{equation}\label{eqn:inscribedTriangleProblem:80}
\begin{aligned}
r – 2 &= \inv{2} \Abs{ \BA + \BB } \\
r – 3 &= \inv{2} \Abs{ \BB + \BC },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:100}
\begin{aligned}
(r – 2)^2 &= \frac{r^2}{4} \lr{ \Be_1 e^{i\theta} + \Be_1 e^{i\phi} }^2 \\
(r – 3)^2 &= \frac{r^2}{4} \lr{ \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} }^2 \\
\end{aligned}
\end{equation}
Finally, since the midpoint of the right edge is found at \( (r-1)\Be_1 \), it is clear that
\begin{equation}\label{eqn:inscribedTriangleProblem:120}
\frac{r-1}{r} = \cos\theta,
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:140}
r = \inv{1 – \cos\theta}.
\end{equation}
This leaves us with three equations and three unknowns. Unfortunately, these are rather non-linear equations. In the video, a direct method of solving equivalent equations was demonstrated, but I picked the lazy route, and used Mathematica’s NSolve routine, solving for \( r,\theta, \phi\) numerically. Since NSolve has intrinsic complex number support, I made the following substitutions:
\begin{equation}\label{eqn:inscribedTriangleProblem:160}
\begin{aligned}
z &= e^{i\theta} \\
w &= e^{i\phi},
\end{aligned}
\end{equation}
and then plugged those into our relations above, after expanding the squares, to find
\begin{equation}\label{eqn:inscribedTriangleProblem:180}
\begin{aligned}
\lr{ \Be_1 e^{i\theta} + \Be_1 e^{i\phi} }^2
&=
2 + \Be_1 e^{i\theta} \Be_1 e^{i\phi} + \Be_1 e^{i\phi} \Be_1 e^{i\theta} \\
&=
2 + e^{-i\theta} \Be_1^2 e^{i\phi} + e^{-i\phi} \Be_1^2 e^{i\theta} \\
&=
2 + e^{-i\theta} \Be_1^2 e^{i\phi} + e^{-i\phi} \Be_1^2 e^{i\theta} \\
&=
2 + \frac{w}{z} + \frac{z}{w},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:inscribedTriangleProblem:200}
\begin{aligned}
\lr{ \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} }^2
&=
2 + \Be_1 e^{i\phi} \Be_1 e^{-i\theta} + \Be_1 e^{-i\theta} \Be_1 e^{i\phi} \\
&=
2 + e^{-i\phi} e^{-i\theta} + e^{ i\theta} e^{i\phi} \\
&=
2 + w z + \inv{w z}.
\end{aligned}
\end{equation}
This gives us
\begin{equation}\label{eqn:inscribedTriangleProblem:220}
\begin{aligned}
4 \lr{ \frac{r – 2 }{r} }^2 &= 2 + \frac{w}{z} + \frac{z}{w} \\
4 \lr{ \frac{r – 3 }{r} }^2 &= 2 + w z + \inv{w z},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eqn:inscribedTriangleProblem:240}
r = \inv{1 – \inv{2}\lr{ z + \inv{z}}}.
\end{equation}

The NSolve gave me some garbage solutions (like \(\theta = 0 \)) that must have been valid numerically, but did not encode the geometry of the problem, so I added a few additional constraints to the problem, namely
\begin{equation}\label{eqn:inscribedTriangleProblem:260}
\begin{aligned}
z \bar{z} &= 1 \\
w \bar{w} &= 1 \\
\inv{2} \lr{ z + \inv{z} } &\ne 1 \\
1/(1 – (1/2) \textrm{Re}(z + 1/z)) &> 3.
\end{aligned}
\end{equation}
This provided exactly two solutions, but when plotted, they turn out to just be mirror images of each other. After back substitution, the solution illustrated above was given by
\begin{equation}\label{eqn:inscribedTriangleProblem:280}
\begin{aligned}
r &= 3.87939 \\
\theta &= 42.078 \\
\phi &= 164.125,
\end{aligned}
\end{equation}
where these angles are in degrees, not radians.

The triangular area.

There are probably lots of formulas for the area of a triangle (that I have forgotten), but we can compute it easily by doubling the triangle, forming a parallelogram, to find
\begin{equation}\label{eqn:inscribedTriangleProblem:300}
\textrm{Area} = \inv{2} \Abs{ \lr{ \BA – \BC } \wedge {\BC – \BB } },
\end{equation}
or
\begin{equation}\label{eqn:inscribedTriangleProblem:320}
\begin{aligned}
\textrm{Area}^2
&= \frac{-1}{4} \lr{ \lr{ \BA – \BC } \wedge \lr{\BC – \BB } }^2 \\
&= \frac{-1}{4} \lr{ \BA \wedge \BC – \BA \wedge \BB + \BC \wedge \BB }^2 \\
&= \frac{-r^4}{4} \lr{\gpgradetwo{ \Be_1 e^{i\theta} \Be_1 e^{-i\theta} – \Be_1 e^{i\theta} \Be_1 e^{i\phi} + \Be_1 e^{-i\theta} \Be_1 e^{i\phi} }}^2 \\
&= \frac{-r^4}{4} \lr{\gpgradetwo{ e^{-2 i \theta} – e^{i \phi -i\theta} + e^{i\theta + i \phi} }}^2,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:inscribedTriangleProblem:340}
\textrm{Area} = \frac{r^2}{2} \Abs{ -\sin( 2 \theta ) – \sin(\phi- \theta) + \sin(\theta + \phi)}.
\end{equation}
Plugging in \( r, \theta, \phi \), we find
\begin{equation}\label{eqn:inscribedTriangleProblem:360}
\textrm{Area} = 17.1866.
\end{equation}
After computing this value, I then finally watched the original video to compare my answer, and was initially disturbed to find that this wasn’t even one of the possible values. However, that was because the problem itself, as originally stated, didn’t include the correct answer, and my worry that I’d made a mistake was unfounded, as the value I computed matched what was computed in the video (it also looks “about right” visually.)

Canonical bivectors in spacetime algebra.

December 5, 2022 math and physics play , , , , ,

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I’ve been enjoying XylyXylyX’s QED Prerequisites Geometric Algebra: Spacetime YouTube series, which is doing a thorough walk through of [1], filling in missing details. The last episode QED Prerequisites Geometric Algebra 15: Complex Structure, left things with a bit of a cliff hanger, mentioning a “canonical” form for STA bivectors that was intriguing.

The idea is that STA bivectors, like spacetime vectors can be spacelike, timelike, or lightlike (i.e.: positive, negative, or zero square), but can also have a complex signature (squaring to a 0,4-multivector.)

The only context that I knew of that one wanted to square an STA bivector is for the electrodynamic field Lagrangian, which has an \( F^2 \) term. In no other context, was the signature of \( F \), the electrodynamic field, of interest that I knew of, so I’d never considered this “Canonical form” representation.

Here are some examples:
\begin{equation}\label{eqn:canonicalbivectors:20}
\begin{aligned}
F &= \gamma_{10}, \quad F^2 = 1 \\
F &= \gamma_{23}, \quad F^2 = -1 \\
F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\
F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\
F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = -15 \\
F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 I \\
F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = -3 + 4 I.
\end{aligned}
\end{equation}
You can see in this table that all the \( F \)’s that are purely electric, have a positive signature, and all the purely magnetic fields have a negative signature, but when there is a mix, anything goes. The idea behind the canonical representation in the paper is to write
\begin{equation}\label{eqn:canonicalbivectors:40}
F = f e^{I \phi},
\end{equation}
where \( f^2 \) is real and positive, assuming that \( F \) is not lightlike.

The paper gives a formula for computing \( f \) and \( \phi\), but let’s do this by example, putting all the \( F^2 \)’s above into their complex polar form representation, like so
\begin{equation}\label{eqn:canonicalbivectors:60}
\begin{aligned}
F &= \gamma_{10}, \quad F^2 = 1 \\
F &= \gamma_{23}, \quad F^2 = 1 e^{\pi I} \\
F &= 4 \gamma_{10} + \gamma_{13}, \quad F^2 = 15 \\
F &= \gamma_{10} + \gamma_{13}, \quad F^2 = 0 \\
F &= \gamma_{10} + 4 \gamma_{13}, \quad F^2 = 15 e^{\pi I} \\
F &= \gamma_{10} + \gamma_{23}, \quad F^2 = 2 e^{(\pi/2) I} \\
F &= \gamma_{10} – 2 \gamma_{23}, \quad F^2 = 5 e^{ (\pi – \arctan(4/3)) I}
\end{aligned}
\end{equation}

Since we can put \( F^2 \) in polar form, we can factor out half of that phase angle, so that we are left with a bivector that has a positive square. If we write
\begin{equation}\label{eqn:canonicalbivectors:80}
F^2 = \Abs{F^2} e^{2 \phi I},
\end{equation}
we can then form
\begin{equation}\label{eqn:canonicalbivectors:100}
f = F e^{-\phi I}.
\end{equation}

If we want an equation for \( \phi \), we can just write
\begin{equation}\label{eqn:canonicalbivectors:120}
2 \phi = \mathrm{Arg}( F^2 ).
\end{equation}
This is a bit better (I think) than the form given in the paper, since it will uniformly rotate \( F^2 \) toward the positive region of the real axis, whereas the paper’s formula sometimes rotates towards the negative reals, which is a strange seeming polar form to use.

Let’s compute \( f \) for \( F = \gamma_{10} – 2 \gamma_{23} \), using
\begin{equation}\label{eqn:canonicalbivectors:140}
2 \phi = \pi – \arctan(4/3).
\end{equation}
The exponential expands to
\begin{equation}\label{eqn:canonicalbivectors:160}
e^{-\phi I} = \inv{\sqrt{5}} \lr{ 1 – 2 I }.
\end{equation}

Multiplying each of the bivector components by \(1 – 2 I\), we find
\begin{equation}\label{eqn:canonicalbivectors:180}
\begin{aligned}
\gamma_{10} \lr{ 1 – 2 I}
&=
\gamma_{10} – 2 \gamma_{100123} \\
&=
\gamma_{10} – 2 \gamma_{1123} \\
&=
\gamma_{10} + 2 \gamma_{23},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:canonicalbivectors:200}
\begin{aligned}
– 2 \gamma_{23} \lr{ 1 – 2 I}
&=
– 2 \gamma_{23}
+ 4 \gamma_{230123} \\
&=
– 2 \gamma_{23}
+ 4 \gamma_{23}^2 \gamma_{01} \\
&=
– 2 \gamma_{23}
+ 4 \gamma_{10},
\end{aligned}
\end{equation}
leaving
\begin{equation}\label{eqn:canonicalbivectors:220}
f = \sqrt{5} \gamma_{10},
\end{equation}
so the canonical form is
\begin{equation}\label{eqn:canonicalbivectors:240}
F = \gamma_{10} – 2 \gamma_{23} = \sqrt{5} \gamma_{10} \frac{1 + 2 I}{\sqrt{5}}.
\end{equation}

It’s interesting here that \( f \), in this case, is a spatial bivector (i.e.: pure electric field), but that clearly isn’t always going to be the case, since we can have a case like,
\begin{equation}\label{eqn:canonicalbivectors:260}
F = 4 \gamma_{10} + \gamma_{13} = 4 \gamma_{10} + \gamma_{20} I,
\end{equation}
from the table above, that has both electric and magnetic field components, yet is already in the canonical form, with \( F^2 = 15 \). The canonical \( f \), despite having a positive square, is not necessarily a spatial bivector (as it may have both grades 1,2 in the spatial representation, not just the electric field, spatial grade-1 component.)

References

[1] Justin Dressel, Konstantin Y Bliokh, and Franco Nori. Spacetime algebra as a powerful tool for electromagnetism. Physics Reports, 589:1–71, 2015.

Verifying dimensions of Planck length

October 31, 2022 math and physics play

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I’m reading [1], which has problems, despite being a sort of pop-sci book. The first such problem is showing that the particular constant
\begin{equation*}
\sqrt{ \frac{h G}{c^3} }
\end{equation*}
has dimensions of length.

My first thought for this was that we have lots of ways of expressing energy in ways that bring in some, but not all of those constants. Examples are
\begin{equation*}
m c^2
,\quad
h \nu
,\quad
i \,\hbar \PD{t}{}
,\quad
– \frac{\hbar^2}{2m} \PDSq{x}{}
,\quad
– \frac{G m M}{r^2}.
\end{equation*}

Some of these are identical with respect to dimensions, for example:
\begin{equation*}
[h\nu] = [i \,\hbar \PD{t}{}] = [h]/T.
\end{equation*}
Let’s use the fact that the dimensions of a particle’s rest energy match that of the photon energy, to find a way to eliminate mass from the dimensions of the gravitation potential energy, that is
\begin{equation*}
[ m c^2 ] = [m] \frac{L^2}{T^2} = [h]/T,
\end{equation*}
or
\begin{equation*}
M L^2/T^2 = [h]/T,
\end{equation*}
so
\begin{equation*}
M
= [h] \frac{T}{L^2}
= [h/c] \inv{L}.
\end{equation*}

Now we can relate the photon energy dimensions with the dimensions of gravitational potential energy, to find

\begin{equation*}
\begin{aligned}
\frac{[h]}{T}
&=
\frac{[G] M^2}{L} \\
&=
\frac{[G]}{L}
[h^2/c^2] \inv{L^2},
\end{aligned}
\end{equation*}
or
\begin{equation*}
[h G/c^3] = L^2.
\end{equation*}
so, we see that the root of this odd combination of units, does, as claimed, have dimensions of length.

References

[1] Carlo Rovelli. General Relativity: The Essentials. Cambridge University Press, 2021.

Angular momentum bivector in cylindrical and spherical bases.

September 15, 2022 math and physics play , , , , , , ,

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Motivation

In a discord thread on the bivector group (a geometric algebra group chat), MoneyKills posts about trouble he has calculating the correct expression for the angular momentum bivector or it’s dual.

This blog post is a more long winded answer than my bivector response and includes this calculation using both cylindrical and spherical coordinates.

Cylindrical coordinates.

The position vector for any point on a plane can be expressed as
\begin{equation}\label{eqn:amomentum:20}
\Br = r \rcap,
\end{equation}
where \( \rcap = \rcap(\phi) \) encodes all the angular dependence of the position vector, and \( r \) is the length along that direction to our point, as illustrated in fig. 1.

fig. 1. Cylindrical coordinates position vector.

The radial unit vector has a compact GA representation
\begin{equation}\label{eqn:amomentum:40}
\rcap = \Be_1 e^{i\phi},
\end{equation}
where \( i = \Be_1 \Be_2 \).

The velocity (or momentum) will have both \( \rcap \) and \( \phicap \) dependence. By chain rule, that velocity is
\begin{equation}\label{eqn:amomentum:60}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
where
\begin{equation}\label{eqn:amomentum:80}
\begin{aligned}
\dot{\rcap}
&= \Be_1 i e^{i\phi} \dot{\phi} \\
&= \Be_2 e^{i\phi} \dot{\phi} \\
&= \phicap \dot{\phi}.
\end{aligned}
\end{equation}
It is left to the reader to show that the vector designated \( \phicap \), is a unit vector and perpendicular to \( \rcap \) (Hint: compute the grade-0 selection of the product of the two to show that they are perpendicular.)

We can now compute the momentum, which is
\begin{equation}\label{eqn:amomentum:100}
\Bp = m \Bv = m \lr{ \dot{r} \rcap + r \dot{\phi} \phicap },
\end{equation}
and the angular momentum bivector
\begin{equation}\label{eqn:amomentum:120}
\begin{aligned}
L
&= \Br \wedge \Bp \\
&= m \lr{ r \rcap } \wedge \lr{ \dot{r} \rcap + r \dot{\phi} \phicap } \\
&= m r^2 \dot{\phi} \rcap \phicap.
\end{aligned}
\end{equation}

This has the \( m r^2 \dot{\phi} \) magnitude that the OP was seeking.

Spherical coordinates.

In spherical coordinates, our position vector is
\begin{equation}\label{eqn:amomentum:140}
\Br = r \lr{ \Be_1 \sin\theta \cos\phi + \Be_2 \sin\theta \sin\phi + \Be_3 \cos\theta },
\end{equation}
as sketched in fig. 2.

fig. 2. Spherical coordinates.

We can factor this into a more compact representation
\begin{equation}\label{eqn:amomentum:160}
\begin{aligned}
\Br
&= r \lr{ \sin\theta \Be_1 (\cos\phi + \Be_{12} \sin\phi ) + \Be_3 \cos\theta } \\
&= r \lr{ \sin\theta \Be_1 e^{\Be_{12} \phi } + \Be_3 \cos\theta } \\
&= r \Be_3 \lr{ \cos\theta + \sin\theta \Be_3 \Be_1 e^{\Be_{12} \phi } }.
\end{aligned}
\end{equation}

It is useful to name two of the bivector terms above, first, we write \( i \) for the azimuthal plane bivector sketched in fig. 3.

Spherical coordinates, azimuthal plane.

\begin{equation}\label{eqn:amomentum:180}
i = \Be_{12},
\end{equation}
and introduce a bivector \( j \) that encodes the \( \Be_3, \rcap \) plane as sketched in fig. 4.

Spherical coordinates, “j-plane”.

\begin{equation}\label{eqn:amomentum:200}
j = \Be_{31} e^{i \phi}.
\end{equation}

Having done so, we now have a compact representation for our position vector
\begin{equation}\label{eqn:amomentum:220}
\begin{aligned}
\Br
&= r \Be_3 \lr{ \cos\theta + j \sin\theta } \\
&= r \Be_3 e^{j \theta}.
\end{aligned}
\end{equation}

This provides us with a nice compact representation of the radial unit vector
\begin{equation}\label{eqn:amomentum:240}
\rcap = \Be_3 e^{j \theta}.
\end{equation}

Just as was the case in cylindrical coordinates, our azimuthal plane unit vector is
\begin{equation}\label{eqn:amomentum:280}
\phicap = \Be_2 e^{i\phi}.
\end{equation}

Now we want to compute the velocity vector. As was the case in cylindrical coordinates, we have
\begin{equation}\label{eqn:amomentum:300}
\Bv = \dot{r} \rcap + r \dot{\rcap},
\end{equation}
but now we need the spherical representation for the \( \rcap \) derivative, which is
\begin{equation}\label{eqn:amomentum:320}
\begin{aligned}
\dot{\rcap}
&=
\PD{\theta}{\rcap} \dot{\theta} + \PD{\phi}{\rcap} \dot{\phi} \\
&=
\Be_3 e^{j\theta} j \dot{\theta} + \Be_3 \sin\theta \PD{\phi}{j} \dot{\phi} \\
&=
\rcap j \dot{\theta} + \Be_3 \sin\theta j i \dot{\phi}.
\end{aligned}
\end{equation}
We can reduce the second multivector term without too much work
\begin{equation}\label{eqn:amomentum:340}
\begin{aligned}
\Be_3 j i
&=
\Be_3 \Be_{31} e^{i\phi} i \\
&=
\Be_3 \Be_{31} i e^{i\phi} \\
&=
\Be_{33112} e^{i\phi} \\
&=
\Be_{2} e^{i\phi} \\
&= \phicap,
\end{aligned}
\end{equation}
so we have
\begin{equation}\label{eqn:amomentum:360}
\dot{\rcap}
=
\rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi}.
\end{equation}

The velocity is
\begin{equation}\label{eqn:amomentum:380}
\Bv = \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} }.
\end{equation}

Now we can finally compute the angular momentum bivector, which is
\begin{equation}\label{eqn:amomentum:400}
\begin{aligned}
L &=
\Br \wedge \Bp \\
&=
m r \rcap \wedge \lr{ \dot{r} \rcap + r \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } } \\
&=
m r^2 \rcap \wedge \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } \\
&=
m r^2 \gpgradetwo{ \rcap \lr{ \rcap j \dot{\theta} + \sin\theta \phicap \dot{\phi} } },
\end{aligned}
\end{equation}
which is just
\begin{equation}\label{eqn:amomentum:420}
L =
m r^2 \lr{ j \dot{\theta} + \sin\theta \rcap \phicap \dot{\phi} }.
\end{equation}

I was slightly surprised by this result, as I naively expected the cylindrical coordinate result. We have a \( m r^2 \rcap \phicap \dot{\phi} \) term, as was the case in cylindrical coordinates, but scaled down with a \( \sin\theta \) factor. However, this result does make sense. Consider for example, some fixed circular motion with \( \theta = \mathrm{constant} \), as sketched in fig. 5.

fig. 5. Circular motion for constant theta

The radius of this circle is actually \( r \sin\theta \), so the total angular momentum for that motion is scaled down to \( m r^2 \sin\theta \dot{\phi} \), smaller than the maximum circular angular momentum of \( m r^2 \dot{\phi} \) which occurs in the \( \theta = \pi/2 \) azimuthal plane. Similarly, if we have circular motion in the “j-plane”, sketched in fig. 6.

fig. 6. Circular motion for constant phi.

where \( \phi = \mathrm{constant} \), then our angular momentum is \( L = m r^2 j \dot{\theta} \).

Interior angles of a regular n-sided polygon: a strange way to find them.

August 8, 2022 math and physics play , , , , , ,

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Fig 1. Regular polygon, interior angles.

For reasons that I can’t explain, I woke up this morning dreaming about the interior angles of regular polygons. i.e. the angles \( \pi – \theta \), as illustrated in fig. 1.

The logical way to calculate that angle would be to slice the polygon up into triangles from the center, since each slice would have an interior angle would be \( 2 \pi / N \), and then the problem is just trigonometric. However, in my dream, I was going around the outside, each time rotating by a constant angle, until reaching the original starting point. This was a vector algebra problem, instead of a trigonometric problem, as illustrated in
fig. 2.

fig. 2. Polygon vertex iteration.

I didn’t have the computational power in my dream to solve the problem, and had to write it down when I woke up, to do so. The problem has the structure of a recurrence relation:
\begin{equation}\label{eqn:regularNgon:20}
\Bp_k = \Bp_{k-1} + a \Be_1 \lr{ e^{i\theta} }^{k-1},
\end{equation}
where
\begin{equation}\label{eqn:regularNgon:40}
\Bp_N = \Bp_0.
\end{equation}

We can write these out explicitly for the first few \( k \) to see the pattern
\begin{equation}\label{eqn:regularNgon:60}
\begin{aligned}
\Bp_2
&= \Bp_{1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ e^{i\theta} }^{1-1} + a \Be_1 \lr{ e^{i\theta} }^{2-1} \\
&= \Bp_{0} + a \Be_1 \lr{ 1 + \lr{ e^{i\theta} }^{2-1} },
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:80}
\Bp_k = \Bp_{0} + a \Be_1 \lr{ 1 + e^{i\theta} + \lr{ e^{i\theta} }^{2-1} + \cdots + \lr{ e^{i\theta} }^{k-1} },
\end{equation}
so the equation to solve (for \(\theta\)) is
\begin{equation}\label{eqn:regularNgon:100}
\Bp_N = \Bp_0 + a \Be_1 \lr{ 1 + \cdots + \lr{ e^{i\theta} }^{N-1} } = \Bp_0,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:120}
1 + \cdots + \lr{ e^{i\theta} }^{N-1} = 0.
\end{equation}
The LHS is a geometric series of the form
\begin{equation}\label{eqn:regularNgon:140}
S_N = 1 + \alpha + \cdots \alpha^{N-1}.
\end{equation}
Recall that the trick to solve this is noting that
\begin{equation}\label{eqn:regularNgon:160}
\alpha S_N = \alpha + \cdots \alpha^{N-1} + \alpha^N,
\end{equation}
so
\begin{equation}\label{eqn:regularNgon:180}
\alpha S_N – S_N = \alpha^N – 1,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:200}
S_N = \frac{\alpha^N – 1}{\alpha – 1}.
\end{equation}
For our polygon, we seek a zero numerator, that is
\begin{equation}\label{eqn:regularNgon:220}
e^{N i \theta} = 1,
\end{equation}
and the smallest \( \theta \) solution to this equation is
\begin{equation}\label{eqn:regularNgon:240}
N \theta = 2 \pi,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:260}
\theta = \frac{2 \pi}{N}.
\end{equation}
The interior angle is the complement of this, since we are going around the outside edge. That is
\begin{equation}\label{eqn:regularNgon:280}
\begin{aligned}
\pi – \theta &= \pi – \frac{2 \pi}{N} \\ &= \frac{ N – 2 }{N} \pi,
\end{aligned}
\end{equation}
and the sum of all the interior angles is
\begin{equation}\label{eqn:regularNgon:300}
N \lr{ \pi – \theta } = \lr{N – 2 } \pi.
\end{equation}

Plugging in some specific values, for \( N = 3, 4, 5, 6 \), we find that the interior angles are \( \pi/3, \pi/2, 3 \pi/5, 4 \pi/6 \), and the respective sums of these interior angles for the entire polygons are \( \pi, 2 \pi, 3 \pi, 4 \pi \).

Like, I said, this isn’t the simplest way to solve this problem. Instead, we could solve for \( 2 \mu \) with respect to interior triangle illustrated in
fig. 3, where

fig. 3. Polygon interior geometry.

\begin{equation}\label{eqn:regularNgon:320}
2 \mu + \frac{ 2\pi}{N} = \pi,
\end{equation}
or
\begin{equation}\label{eqn:regularNgon:340}
2 \mu = \frac{N – 2}{N} \pi,
\end{equation}
as found the hard way. The hard way was kind of fun though.

The toughest problem to solve would be “why on earth was my brain pondering this in the early morning?” I didn’t even go to bed thinking about anything math or geometry related (we finished the night with the brain-dead activity of watching an episode of “Stranger things”.)