Nov 19 2024
Reuleaux Triangles
r r
Problem
Given two reuleaux triangles with distance from the center to a vertex \(r\) and \(R\), such that \( R \geq r \): find the angles \( \theta \) between them such that the corners of the smaller triangle intersect the larger one as shown in the diagram.
Solution
The key insight was to notice that a circle inscribed around the verticies of the inner reuleaux triangle would always intersect with the arc of each side, at the point where the rotated vertex should lie.
you can then solve for the intersection of the inscribed circle, and the circle that one side's arc belongs to. (shown beside)
The solutions I found are: $$\theta= \pm \left(\arctan{\left( {\sqrt{p^2 - a^2} \over a} \right)} - {\pi \over 3} \right) $$
and $$ a = { {p^2-2} \over 2} $$
Where:
- \(r\) is the distance from the center of the inner shape to a vertex
- \(R\) is the same distance for the outer triangle
- \(p\) is the ratio \(r \over R \)