# What is the minimum size of a circle that circumscribes N vertices of equilateral triangles?

This question was inspired by a meme I saw floating around:

I got kicked out of a flat Earth group for asking if this 6 ft social distance rule pushed anyone over the edge yet

Chuckling to myself at the joke, I attempted to solve the problem by modeling it assuming that the "flat Earth" is a uniformly flat, featureless disk, and all people are restricted to distancing themselves in the 2-d plane of the disk. I conjectured that the supposedly flat Earth forces the social-distancers into tessellating the disk as the vertices of equilateral triangles. Given that, how large would the disk need to be for all humans to fit on it with appropriate social distance? I quickly found myself stumped as I couldn't figure out how to determine how many triangles are produced by $$N$$ vertices, given a circular packing.

My general question is then:

Given a set of $$N$$ vertices of unit equilateral triangles, such that no two vertices of different triangles are less than 1 unit distance from each other, what is the radius of the smallest possible circle that circumscribes all of the vertices?

Trivial cases for $$N = 3$$ and $$N = 7$$ (a circumcircle of a single equilateral triangle and a hexagon, respectively) are $$\frac{1}{\sqrt 3}$$ and $$1$$, respectively, but I'm finding it difficult to generalize to large $$N$$.

(Editors feel free to reformulate the question if there is a better way to state it; my math lingo is rusty.)

• Does this answer your question? max points in circle given radius and min spacing between points Commented Aug 3, 2020 at 19:15
• Such packing problems have the unpleasant habit of creating irregular solutions and resisting to analyis.
– user65203
Commented Aug 3, 2020 at 19:19
• @DonThousand no, that's related, but working from the other direction. It gave me some ideas for avenues to pursue here though. Commented Aug 3, 2020 at 20:52

This is just the problem of circle packing in a circle (Wikipedia). If the radius of the smallest possible circle enclosing $$n$$ circles, each of radius $$1$$, is $$k$$, then for just the vertices to be enclosed, the radius can just be reduced by $$\frac{1}{2}$$. Wikipedia uses the diameter $$d$$, so the solution to your problem is $$\frac{d}{2} - \frac{1}{2} = \frac{1}{2}(d-1)$$.