# Pentagons with Rational Trigonometry

The other day I wanted to work out the properties of a pentagon, for reasons which I might go into in a future post. In particular I wanted to know the horizontal and vertical dimensions of a pentagon with unit length “spokes”.

A started scratching out diagrams and end up with a mess of lines and stupid values like $sin\left(\frac{\pi}{5}\right)$. It all quickly because a mess and gave me no precise numbers. Everything was in terms of trig functions and pi. Don’t get me wrong, pi’s a great number to work with if you’ve got circles, but I’m just dealing with straight lines here, I shouldn’t need transcendental numbers!

So, I reverted back to good old rational trigonometry. Rational trigonometry uses the concept of spread to measure the angle between lines, and quadrance to measure the extent between points. If the angle between two lines is $\theta$ then the spread, $s$, is equal to $\sin^2\theta$. The quadrance between two points is just the distance squared. It turns out that using these measures makes calculations much simpler than in traditional trigonometry.

The diagram below shows a regular pentagon with unit length (and thus unit quadrance) spokes. We shall use rational trigonometry to compute exact values for all the labels shown.

The first (and most complex) step is to compute $s_1$ and $s_2$. We note that $s_1$ is one fifth of a full revolution and $s_2$ is one fifth of a half revolution. Both a half and a full revolution represent a spread of zero. Therefore $s_1$ and $s_2$ are both solutions of the 5th order spread polynomial,

$S_5(s) = s(16s^2 - 20s + 5)^2 = 0$.

We use the quadratic equation to find the two non-zero solutions to this equation and find

$s_1 = \frac{5 + \sqrt{5}}{8}\\ s_2 = \frac{5 - \sqrt{5}}{8}$.

A corollary of the triple spread law when applied to right angled triangles gives

$s_3 = 1 - s_2 = \frac{3 + \sqrt{5}}{8}$.

By the spread law we can now directly find

$Q_2 = s_2 = \frac{5 - \sqrt{5}}{8}\\ Q_3 = s_3 = \frac{3 + \sqrt{5}}{8}\\ Q_4 = s_1 = \frac{5 + \sqrt{5}}{8}\\ Q_5 = 1 - s_1 = \frac{3 - \sqrt{5}}{8}$.

Applying the triple quad formula to a bisected line we get

$Q_1 = 4Q_2 = \frac{5 - \sqrt{5}}{2}$.

Pythagoras’ theorem now gives us

$Q_6 = Q_1 - Q_4 = \frac{15 - 5\sqrt{5}}{8} = 5Q_5$.

Once again applying the spread law we get

$Q_7 = \frac{s_3}{s_1} = \frac{5 + \sqrt{5}}{10} = \frac{4}{5}Q_4\\ Q_8 = \frac{1 - s_1}{s_1} = \frac{5 - 2\sqrt{5}}{5}$.

So, there we have it, all the interesting dimensions of a pentagon without once having to resort to transcendental numbers or functions.

If you want to know more about rational geometry I strongly suggest watching this series of short lectures by Norman Wildberger.

## 2 thoughts on “Pentagons with Rational Trigonometry”

1. Hey that’s cool! I’m starting to watch the Norman Wildberger videos now.

Incidentally, when I first saw the picture, the first thing that went through my head was “Calculate the current flowing throw Q7″ …