Geometry - Equilateral triangle covered with five circles

Question

I have to cover an equilateral triangle (whose sides are 1m long) with 5 identical circles: what's the minimum radius of the circles?

Answer 1

As a lower bound, consider circles that perfectly cover the triangle (which is impossible) and thus have the same combined area as the triangle. The area of the triangle is $\frac {\sqrt{3}}{4}$, so each circle would need to have an area of $\frac {\sqrt{3}}{20}$, or a radius of $\sqrt \frac {\sqrt{3}}{20 \pi} \approx 0.166$.

As an upper bound, consider three copies of the inscribed circle, each moved halfway to each corner of the triangle (the placement of the fourth and fifth circles is unimportant in this solution). The radius of those circles is $\frac {\sqrt{3}}{6}\approx 0.289$.

Thus the minimum radius to solve your problem is $0.166 \lesssim r \lesssim 0.289$.

Answer 2

Here is a solution where $r \approx 0.251$. This is not optimal, but is very close. Finding a precise, or non-numerical, solution of this sort is an exercise in trigonometry and algebra.

Answer 3

The minimum radius $r$ of the circle is $1/4$.

Hans Melissen showed that $r\le 1/4$.

Now let $A$ be a subset of the triangle, consisting of its vertices and the middles of its edges. Then the distance between each pair of different points of $A$ is at least $1/2$. Since in each cover of the triangle by five circles there is a circle covering two different points of $A$, its radius is at least $1/4$.