Can the bronze ratio be found in a regular tridecagon?

Question

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

The silver ratio $\frac{2+\sqrt 8}2$ is found in the regular octagon with side length 1 as the length of the second 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 of the third diagonal, which is $\cos{\frac{3\pi}{26}} \csc{\frac \pi{13}} \approx 3.43891$

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

Answer 1

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.