Prove the smallest circle containing an equilateral triangle has radius d/sqrt(3)

Question

Consider an equilateral triangle in $A \subset \mathbb{R}^2$ with vertices $x_1$, $x_2$, and $x_3$. I would like to show that the smallest circle which contains $A$ has radius $\frac{d}{\sqrt{3}}$, where $d$ is the length of each side.

I've been able to show that a (closed) ball centered at $(x_1 + x_2 + x_3)/3$ with radius $\frac{d}{\sqrt{3}}$ contains A. To complete the proof, I need to show that no smaller ball can contain $A$. My strategy so far has been as follows. Pick a point $x \in \mathbb{R}^2$. If $d(x_1,x) \leq \frac{d}{\sqrt{3}}$ and $d(x_2,x) \leq \frac{d}{\sqrt{3}}$ with at least one equality strict, then $d(x_3,x) > \frac{d}{\sqrt{3}}$ (where $d(\cdot, \cdot)$ is the Euclidean distance function). I have not been able to find a simple way to verify this statement (either directly or by contradiction). Is is possible to show this? Or is another strategy better?

Answer 1

Choose coordinate system so that the centroid of the equilateral triangle of side $d$ is the origin. i.e $$x_1 + x_2 + x_3 = 0$$ We will have $|x_1| = |x_2| = |x_3| = \frac{d}{\sqrt{3}}$.

Given any circle of radius $r$ centered at $x$. If it contains the three point $x_1, x_2, x_3$, it will satisfies: $$3r^2 \ge |x_1 - x|^2 + |x_2 - x|^2 + |x_3 - x|^2 = |x_1|^2 + |x_2|^2 + |x_3|^2 + 3|x|^2 \ge 3\left(\frac{d}{\sqrt{3}}\right)^2$$ This leads to $r \ge \frac{d}{\sqrt{3}}$.

Answer 2

The circumcenter of a triangle is the intersection of the perpendicular bisector of each edge. A right triangle B can be formed between the bisector of an edge, a vertice of the same edge and the circumcenter. Other right triangles can be formed at each edge and they will be congruent because triangle A is equilateral. The hypothenuse of each right triangle will therefore be of the same length. A circle can be formed at the circumcenter with a radius the length of the hypothenuse. The circle will touch each edge and will therefore be the smallest circle.

To calculate the hypothenuse $h$ of triangle B given any side $x$, consider that the perpendiculars intersect the center of each edge in triangle A by definition and that the angles of the equilateral triangle are halved because triangle A is equilateral:

$$ cos(\pi/6) = \frac{x/2}{h}\\ h = \frac{x/2}{cos(\pi/6)}\\ h = \frac{x/2}{\sqrt{3}/2}\\ h = \frac{x}{\sqrt{3}} $$

This hopefully proves what you already know.