Containing All

Note that the equilateral triangle of length $1$ has diameter $1$ and the smallest circle containing it is its circumcircle which has radius $\frac 1{\sqrt3}$ . This shows $r_{\min}\ge \frac 1{\sqrt 3}$ . On the other hand, the following theorem shows $r_{\min}\le \frac 1{\sqrt 3}$ .

Theorem. If a plane figure $F$ has diameter $1$ , then it can be contained in a circle of radius $\frac 1{\sqrt3}$ .

Lemma. Suppose that $\mathcal G$ is a family of at least three convex subsets of $\mathbb R^2$ and that $C$ is a convex subset of $\mathbb R^2$ . Suppose also that either $\mathcal G$ has only finitely many members or that $C$ and all of members of $\mathcal G$ are compact. If for each three members of $\mathcal G$ , there is a translate of $C$ that contains the three members, then there is a translate of $C$ that contains all members of $\mathcal G$ .

Proof. For each $A\in\mathcal G$ , let $A'=\{x\in\mathbb R^2\mid A\subseteq (x+C)\}$ . It's easy to prove that each $A'$ is convex by the convexity of $C$ . It follows from the hypotheses of the lemma that every three of the sets $A'$ have a point in common. Therefore, Helly's theorem implies that there is a point $x_0$ common to all of the sets $A'$ , and this means that $A\subseteq x_0+C$ for each $A\in\mathcal G$ . $_\square$

Proof of the Theorem. If every three points of $F$ is contained in a circle of radius $r$ , then by the lemma, the entire figure is contained in a circle of radius $r$ . As a consequence, we need only to prove that every three points that form a set of diameter $1$ can be contained in a circle of radius $\frac 1{\sqrt3}$ . In other words, we have to prove that a triangle whose longest side is of length $1$ has a circumradius no greater than $\frac 1{\sqrt3}$ .

A glance at part (1) of the following figure makes this obvious. if the side $[a,b]$ is of length $1$ , then the third vertex of the triangle must be within the shade area, and as shown in part (2) of the following figure, the shaded area is contained in a circle of radius $\frac 1{\sqrt3}$ . $_\square$