Irish Mathematical Olympiad

Let $EF$ be the crease such that $E$ is on $AB$ and $F$ on $BC$. Join $H$ to $C$, $F$ and $E$.

The crease $EF$ lies along the perpendicular bisector of $HB$ and the triangles $\triangle EHF$ and $\triangle EBF$ are congruent. If we let $x = |AE|$ and $y = |CF|$, then $|EH| = |EB| = 4-x$ and $|HF| = |BF| = 4-y$.

In triangle $EHA$ we have $|AH| = 2$ and $\angle HAE = 120^\circ$, thus the cosine rule reads
$$(4 - x)^2 = 2^2 + x^2 - 4x \cos(120^\circ) = x^2 + 2x + 4$$
and so we have $10x = 12$, i.e. $x = 6/5$, thus $|EB| = 4 - x = 14/5$.

In triangle $HDC$ we have $|HD| = 2$, $|DC| = 4$ and $\angle HDC = 60^\circ$, and the cosine rule gives
$$|HC|^2 = 4 + 16 - 16 \cos(60^\circ) = 12.$$
In particular, $|HC|^2 + |HD|^2 = |CD|^2$ and so $\angle FCH = \angle CHD = 90^\circ$. Therefore we have $|HF|^2 = |HC|^2 + |CF|^2$, i.e. $(4 - y)^2 = 12 + y^2$ and so $4 = 8y$, that is $y = 1/2$. This implies $|BF| = 4 - y = 7/2$.

The cosine rule for triangle $BEF$ now gives
$$\begin{aligned} |EF|^2 &= |BF|^2 + |EB|^2 - 2|BF| \cdot |EB| \cos(60^\circ) \\ &= \left(\frac{7}{2}\right)^2 + \left(\frac{14}{5}\right)^2 - 2 \cdot \frac{7}{2} \cdot \frac{14}{5} \cdot \frac{1}{2} = \frac{21 \cdot 49}{100}, \quad \text{and so} \\ |EF| &= \frac{7}{10}\sqrt{21} \end{aligned}$$