4

The golden ratio $\frac{1+\sqrt 5}2$ is found in the regular pentagon with side length 1 as the length of the 1st diagonal.

The silver ratio $\frac{2+\sqrt 8}2$ is found in the regular octagon with side length 1 as the length of the 2nd diagonal.

The bronze ratio $\frac{3+\sqrt{13}}2 = 2 \cos \frac{\pi}{13}\left( \sin \frac{2\pi}{13} \csc \frac{3\pi}{13} + 1 \right) \approx 3.3027$ is not found in the regular tridecagon with side length 1 as the length $L$ of the 3rd diagonal. (Editor's note: Actually, length $L =\cos{\frac{\color{red}5\pi}{26}} \csc{\frac \pi{13}} \approx 3.43891,$ fixing a small typo by the OP.)

Is there a more complex geometric expression of the bronze ratio that can be found in a regular tridecagon?

Tito Piezas III
  • 60,745
Alexey
  • 199

2 Answers2

3

I found an answer in Polygons, Diagonals, and the Bronze Mean by Antonia Redondo Buitrago:

Consequently, we can conclude: There exists no regular polygon in which the ratio of the length of the diagonal to the side of the polygon is equal to the Bronze Mean.

and in https://tellerm.com/math/metal_ngons/, which identified 1708 various combinations of diagonal segments in a regular 13-gon that are equal to the bronze mean.

Alexey
  • 199
2

The OP asks for "...a more complex geometric expression" and accepts in the comments that it may "...possibly something other than a diagonal".

With these more relaxed conditions, by using more than one diagonal, then the bronze ratio can certainly can be found in the regular $13$-gon. Assume one with unit side length,

Polygon with 13 sides

Given a common vertex $v_0$ which, going clockwise, connects to vertices $v_1, v_3, v_4, v_5$. Then the sum of the three blue lines minus the green line is the bronze ratio,

$$D(1)+D(3)+D(4)-D(5) = \frac{3+\sqrt{13}}2$$

or using their exact trigonometric values,

$$\left(\frac{\sin\frac{1\pi}{13}}{\sin\frac{\pi}{13}}\right)+\left(\frac{\sin\frac{3\pi}{13}}{\sin\frac{\pi}{13}}\right)+\left(\frac{\sin\frac{4\pi}{13}}{\sin\frac{\pi}{13}}\right)-\left(\frac{\sin\frac{5\pi}{13}}{\sin\frac{\pi}{13}}\right)= \frac{3+\sqrt{13}}2$$

In fact, more metallic ratios can be found in $n$-gons in a similar manner as discussed in this post about the $17$-gon and others.

P.S. The OP mentions in his answer there are 1708 various combinations of diagonal segments in a $13$-gon that are equal to the bronze ratio. I am not aware of these plethora of solutions, but the simple one given here seems enough.

Tito Piezas III
  • 60,745