# From Arbitrary Pentagon to Regular One

### What Might This Be About?

### Problem

Let $A_0A_1A_2A_3A_4$ be an arbitrary pentagon. For $i=0,\ldots,4,$ form linear combinations (where indices are taken modulo $5)$:

$\displaystyle B_{i}=\frac{A_{i}+\phi A_{i+1}+A_{i+2}}{2+\phi}.$

Then the centers of regular polygons erected either internally or externally on the sides of $B_{0}B_{1}B_{2}B_{3}B_{4}$ form a regular pentagon.

### Solution

We shall use complex numbers and identify them with the points in the plane. The proof is based on the assertion that all four pentagons involved - $A_0A_1A_2A_3A_4,$ $B_{0}B_{1}B_{2}B_{3}B_{4},$ and the two allegedly regular ones - have the same barycenter.

This is easily seen for the first two:

$\begin{align}\displaystyle \frac{1}{5}\sum_{k=0}^{4}B_{k} &= \frac{1}{5(2+\phi )}\sum_{k=0}^{4}(A_{k}+\phi A_{k+1}+A_{k+2})\\ &= \frac{1}{5(2+\phi )}\bigg(\sum_{k=0}^{4}A_{k}+\phi\sum_{k=0}^{4}A_{k+1}+\sum_{k=0}^{4}A_{k+2}\bigg)\\ &= \frac{1}{5(2+\phi )}\bigg(\sum_{k=0}^{4}A_{k}+\phi\sum_{k=0}^{4}A_{k}+\sum_{k=0}^{4}A_{k}\bigg)\\ &= \frac{1}{5(2+\phi )}(2+\phi )\sum_{k=0}^{4}A_{k}\\ &= \frac{1}{5}\sum_{k=0}^{4}A_{k}. \end{align}$

The same derivation works for any polygon obtained from $A_0A_1A_2A_3A_4$ by a linear polygonal transformation, of which $B_{0}B_{1}B_{2}B_{3}B_{4}$ is one example. It will be seen that this is how the two allegedly regular polygons are obtained from $B_{0}B_{1}B_{2}B_{3}B_{4}.$

It is convenient to assume that $A_0A_1A_2A_3A_4$ and $B_{0}B_{1}B_{2}B_{3}B_{4}$ have the origin as the barycenter by this convention, $\displaystyle\sum_{k=0}^{4}A_{k}=0.$

Given two points $X$ and $Y,$ there are two regular pentagons that have $XY$ as one of the sides. The centers of these pentagons can be found as

$Q=(1-\tau )X+\tau Y,$

where $\displaystyle \tau=\frac{1}{2}(1\pm i\cdot \tan 54^{\circ}).$ In what follows I'll use the sign $+$ since the other case may be also obtained by simply exchanging $X$ and $Y.$

Define now $\displaystyle\omega=e^{i\frac{2\pi}{5}},$ multiplication by which causes a rotation by $72^{\circ}$ around the origin.

**Statement**

Assume $\displaystyle\sum_{k=0}^{4}A_{k}=0.$ Define $Q_{i}=(1-\tau )B_{i}+\tau B_{i+1},$ $i \mod 5.$ Then

$Q_{i+1}=wQ_{i},$ $i \mod 5.$

Observe that this exactly says that $Q_0Q_1Q_2Q_3Q_4$ is regular.

**Proof**

Let's express $\{Q_{i}\}$ explicitly in terms of $\{A_{i}\}.$

$\begin{align}\displaystyle Q_{i}&=(1-\tau )B_{i}+\tau B_{i+1}\\ &=(1-\tau)A_{i}+[(1-\tau )\phi + \tau]A_{i+1}+[(1-\tau )+\tau\phi ]A_{i+2}+\tau A_{i+3}. \end{align}$

It is simpler to collect the coefficients by $A_0,A_1,A_2,A_3,A_4$ (sic!), into a vector such that we may rewrite the above as

$Q_{i}=\bigg((1-\tau),\,(1-\tau )\phi + \tau,\,(1-\tau )+\tau\phi,\,\tau ,\,0\bigg).$

Next

$\omega Q_{i}=\bigg(\omega (1-\tau),\,\omega (1-\tau )\phi + \omega \tau,\,\omega (1-\tau )+\omega \tau\phi,\,\omega \tau ,\,0\bigg).$

Finally, replace $A_{0}$ with $\displaystyle A_{0}=-\sum_{k=1}^{4}A_{k}:$

$\begin{align}\omega Q_{i}=\bigg(0,&\,-\omega (1-\tau)+\omega (1-\tau )\phi + \omega \tau,\\ &-\omega (1-\tau)+\omega (1-\tau )+\omega \tau\phi,\,-\omega (1-\tau)+\omega \tau ,\,-\omega (1-\tau)\bigg). \end{align}$

We may simplify that a little:

$\omega Q_{i}=\bigg(0,\,\omega (2\tau -1)+\omega (1-\tau )\phi,\omega\tau\phi,\,\omega (2\tau -1),\,\omega (\tau -1)\bigg).$

To remind

$Q_{i+1}=\bigg(0,\,(1-\tau),\,(1-\tau )\phi + \tau,\,(1-\tau )+\tau\phi,\,\tau\bigg).$

In order to prove that $\omega Q_{i}=Q_{i+1}$ we need to establish four identities:

$\begin{cases} I: & 1-\tau = \omega (2\tau -1)+\omega (1-\tau )\phi,\\ II: & (1-\tau )\phi + \tau = \omega\tau\phi,\\ III: & (1-\tau )+\tau\phi = \omega (2\tau -1),\\ IV: & \tau = \omega (\tau -1). \end{cases}$

The four identities are not independent. E.g., substituting $\omega (2\tau -1)=(1-\tau )+\tau\phi$ from (III) into (I) we obtain (IV) times $\phi.$ Also, if we multiply (III) by $\phi,$ use $\phi^{2}=1+\phi$ and substitute into that $(1-\tau )\phi=\omega\tau\phi-\tau$ from (II) we again obtain (IV) times $\phi.$ So, at best, only two of these equations are independent - say, (II) and (IV).

Now $\omega (II)+\phi (IV)$ leads to $\omega-\phi=\omega^{2}\phi.$

Several trigonometric identities will serve well to help finish the proof:

$ \displaystyle\omega = \frac{\sqrt{5}-1}{4}+i\sqrt{\frac{5+\sqrt{5}}{8}},\\ \displaystyle\omega^{2}= -\frac{\sqrt{5}+1}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}},\\ \displaystyle\sin 54^{\circ}=\frac{\sqrt{5}+1}{4},\\ \displaystyle\cos 54^{\circ}=\sqrt{\frac{5-\sqrt{5}}{8}}. $

All that needs to be verified now is:

$\omega-\phi=\omega^{2}\phi$ and

$\tau = \omega (\tau -1),$

where $\displaystyle\phi=\frac{\sqrt{5}+1}{2}$ and $\displaystyle \tau=\frac{1}{2}(1+ i\cdot \tan 54^{\circ}).$ Let's check the second one:

$\displaystyle\begin{align} \frac{\tau}{\tau -1} &= \frac{\cos 54^{\circ} +i\sin 54^{\circ}}{-\cos 54^{\circ}+i\sin 54^{\circ}}\\ &=\frac{(\cos 54^{\circ} +i\sin 54^{\circ})^{2}}{1}\\ &=(\sin^{2} 54^{\circ}-\cos^{2}54^{\circ})+2i\cos 54^{\circ}\sin 54^{\circ}\\ &=-\cos 108^{\circ}+i\sin 108^{\circ}\\ &=\cos 72^{\circ}+i\sin 72^{\circ}\\ &=\omega. \end{align}$

Verifying the other identity is left as an exercise.

### Acknowledgment

The above construction of the regular pentagon is by Dao Thanh Oai (Vietnam) who posted it at the CutTheKnotMath facebook page.

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