# Finding angles of non-overlapping adjacent triangles whose vertices lie on the edges of an annulus

## The problem

I have two concentric circles with radii $$r$$ and $$R$$, where $$R \geq r$$. These circles forms an annulus. On the outer circle $$n$$ equidistant points are placed, denote them by $$(p_0 = p_n, p_1, p_2, \dots, p_n)$$. I want to find $$n$$ equidistant points $$\{q_1, \dots, q_n\}$$ on the inner circle such that the triangles $$\{\Delta(p_i, p_{i+1}, q_i) \mid 1 \leq i \leq n \}$$ do not overlap and are adjacent to each other.

See the following picture for an illustration where $$n = 5$$ and $$R = 2*r$$: , and a larger example: The problem comes down to essentially finding the angle $$\alpha$$ indicated in the first picture.

## What I've tried

Using the fact that the angle of one corner of a regular $$n$$-gon is $$\frac{(n-2)*\pi}{n}$$ I could derive one angle of the coloured triangles. Similarly, the sides of the outer and inner polygon are found using $$2R\sin{\frac{\pi}{n}}$$ and $$2r\sin{\frac{\pi}{n}}$$. Also the area of one colored triangle can be calculated by $$\frac{1}{2}\sin{(\frac{2\pi}{n})}(R^2 - r^2)$$.

The visuals linked previously were made by manually finding the correct angle, for $$n=5$$ and $$R=2*r$$ the angle $$\alpha \approx 0.527$$ radians or around $$30$$ degrees.

Could someone give a hint on how to continue? This is not a homework question or anything; I want to make an animation and for that I need the angle $$\alpha$$.

For outer radius $$r$$, inner radius $$s$$, and $$\theta := \pi/n$$ (ie, half the central angle of the $$n$$-gon), we have ... $$s\cos\theta = r\cos(\theta+\alpha) \quad\to\quad \alpha = \arccos\left(\frac{s}{r}\,\cos\theta\right)-\theta$$
For $$n=5$$ and $$r=2s$$, this yields $$\alpha = 0.52603\ldots$$ (aka, $$30.1396\ldots^\circ$$).