Irish Mathematical Olympiad

Let the lines $DA$ and $CB$ intersect at $E$. We have to distinguish two possibilities: either $|CD| < |AB|$ or $|CD| > |AB|$, see diagram below.

Since $AD = BC$, by symmetry we get $DE = CE$ and $AB \parallel DC$. Since $\angle AEB = 60^\circ$ this implies that $\triangle ABE$ and $\triangle DCE$ are equilateral. Thus $\angle ABC = 60^\circ$ if $|CD| < |AB|$ and $\angle ABC = 120^\circ$ if $|CD| > |AB|$. The Cosine Rule for the triangle $ABC$ now gives

$$|CA|^2 = |AB|^2 + |BC|^2 - 2 \cdot |AB| \cdot |BC| \cdot \cos \angle ABC.$$

Using $\cos 60^\circ = -\cos 120^\circ = 1/2$, we get $|CA|^2 = 16 + 4 \pm 8$, hence $|CA| = 2\sqrt{3}$ if $|CD| < |AB|$ and $|CA| = 2\sqrt{7}$ if $|CD| > |AB|$.

The radius $R$ of the circle is the circumradius of $\triangle ABC$ which is given by the Sine Rule for this triangle

$$R = \frac{|CA|}{2 \sin \angle ABC} = \frac{|CA|}{\sqrt{3}}.$$

Therefore, $R = 2$ if $|CD| < |AB|$ and $R = 2\sqrt{\frac{7}{3}}$ if $|CD| > |AB|$.