# A pentagon, hexagon, and decagon walk into a bar …

The new book A Panoply of Polygons cites a theorem Euclid (Proposition XIII.10) saying

If a regular pentagon, a regular hexagon, and a regular decagon are inscribed in congruent circles, then their side lengths form a right triangle.

This isn’t exactly what Euclid said, but it’s an easy deduction from what he did say.

Here’s an illustration. Update: Ralph Corderoy suggested in the comments that it would be good to add the congruent circles through the polygon vertices. Here is a graphic with the circles added. ## Connection with golden ratio

I learned about this theorem from the book cited above, which arrived in the mail yesterday. The figure looked familiar because Cliff Pickover had posted essentially the same illustration on Twitter recently, saying that not only is there a right triangle in the middle of the diagram, this triangle is half of a golden rectangle.

The length of the side of a regular polygon with n sides is 2R sin(π/n) where R is the radius of the circle through the vertices. So to prove that the ratio between the side of the hexagon and the side of the decagon is the golden ratio φ, it suffices to prove

sin(π/6) / sin(π/10) = φ

You could verify this in Mathematica by seeing that the following returns `True`.

`    Sin[Pi/6]/ Sin[Pi/10] == GoldenRatio`

## Python code

Here’s the code I used to create the image above.

```    from numpy import exp, pi, arange
import matplotlib.pyplot as plt

# Vertices of a hexagon
hexa = exp(2j*pi*arange(6)/6)

# Vertices of a decagon, shifted and rotated
# to have a side perpendicular to the hexagon
deca = exp(2j*pi*(0.5+arange(10))/10)
deca += hexa - deca

# Shift and rotate a pentagon to share one vertex
# with the hexagon and one vertex with the decagon
omega = exp(2j*pi/5)
c = (hexa*omega - deca)/(omega - 1)
penta = c + (hexa - c)*exp(2j*pi*arange(5)/5)

def connect(p, q, c):
plt.plot([p.real, q.real], [p.imag, q.imag], '-', color=c)

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

for n in range(-1, 5):
connect(hexa[n], hexa[n+1], 'blue')
for n in range(-1, 9):
connect(deca[n], deca[n+1], 'green')
for n in range(-1, 4):
connect(penta[n], penta[n+1], 'red')

plt.show()
```

The ability to use -1 to as the index to the last element of an array simplifies the code that connects the vertices. In a language like C you’d have to write an extra statement after your loop to join the last vertex to the first.

When I first saw the illustration in Panoply of Pentagons I though that the top of the pentagon was parallel to the top of the hexagon. When I wrote the Python code first assuming this orientation of the pentagon, things didn’t line up. Writing the code made me pay closer attention, as is often the case.

## 6 thoughts on “A pentagon, hexagon, and decagon walk into a bar …”

1. Ralph Corderoy

It would be nice to see the ‘congruent circles’ in the diagram.

2. Thanks for the suggestion. I updated the post accordingly.

3. And the punch line is?

4. Arthur

I think in language without index wraparound you can do something like

connect(hexa[n], hexa[(n+1)%6], ‘blue’)

Looping from 0 to 5 inclusive.

5. An exercise left to the reader. :)

6. Victor

It is possible to draw two congruent exterior angles at each vertex of a polygon, but only one is considered when talking of the exterior angle at a specific vertex.