vraag met Aziatische origine

Call the pentagon $OABCD$, where $\angle O= 60^{\circ}$ and the other angles equal $120^{\circ}.$

Let $P= BC \cap OA$ and $Q= BC \cap OD.$

Note that $ABP, CDQ$ and $OPQ$ are equilateral triangles.

Hence $|OA|+|AB| = |AB|+|BC|+|CD| = |CD|+|DO|.$

Set the lengths of five segment of this pentagon as $x, x+1, x+2, x+3$ and $x+4.$

Note that $|AB|+|BC|+|CD| \ge 3x+3$ and $|OA|+|AB| = |CD|+|DO| =\frac{5x+10-|BC|}2\le \frac{4x+10}{2}=2x+5$.
Hence $2x+5 \ge 3x+3,$ i.e. $x \le 2$, so $x \in \{1,2\}$ and one can also note that $|BC|=x$ or $|BC|=x+2$ (but then $x=1$).

Now there are only a few cases where we have to check the possible values:

When $x=1$ and $|BC|=x+2$, one gets $|BC|=3, |CD|=2,|OD|=4, |OA|=5$ and $|AB|=1$ or the symmetric version.

When $|BC|=x=2$, one gets $ |OA|=6, |AB|=3, |BC|=2, |CD|=4$ and $|DO|=5$ and the symmetric variant.
When $|BC|=x=1$, we get $ |OA|=5, |AB|=2, |BC|=1, |CD|=4$ and $|DO|=3$ and the symmetric variant.

One simply checks that these values indeed result in pentagons which satisfy the conditions.

So there are $3$ non-congruent pentagons satisfying the condition.