Final Round of National Olympiad

(i) Let the hexagon be $ABCDEF$. It suffices to show that $|AB| + |BC| = |DE| + |EF|$.
Let $K$ be the intersection point of rays $FA$ and $CB$ and let $L$ be the intersection point of rays $FE$ and $CD$ (Fig. 8). The size of every internal angle of the hexagon is $120^\circ$, whence triangles $KAB$ and $LDE$ are equilateral. The quadrilateral $FKCL$ is a parallelogram since its opposite sides are parallel. This implies $|KC| = |LF|$ or $|KB| + |BC| = |LE| + |EF|$, which together with $|KB| = |AB|$ and $|LE| = |DE|$ implies the desired equality $|AB| + |BC| = |DE| + |EF|$.

(ii) Take a parallelogram with side lengths 7 and 5 and internal angles $60^\circ$ and $120^\circ$, and cut off equilateral triangles with side lengths 1 and 2 at its acute angles. This gives rise to a hexagon with all internal angles having size $120^\circ$ and side lengths 1, 4, 5, 2, 3, 6 (Fig. 9).

Figure 8

Figure 9

(i) Note that the external angles of the hexagon have size $60^\circ$. Any two opposite sides of the hexagon are parallel, as they are separated by exactly three external angles.
Consider a line $s$ perpendicular to opposite sides $CD$ and $FA$ of the hexagon $ABCDEF$ (Fig. 10). As all other sides form the same angle $30^\circ$ with line $s$, the lengths of these sides are proportional to the lengths of the projections of the sides to line $s$. The sum of the lengths of the projections of sides $AB$ and $BC$ is equal to the distance between the parallel lines $CD$ and $FA$ and the same holds also for the opposite sides $DE$ and $EF$. Therefore the sum of the lengths of the projections of sides $AB$ and $BC$ is equal to that of sides $DE$ and $EF$, whence the sums of the lengths of the sides are equal as well.

Figure 10

(ii) Figure 11 shows a hexagon in a triangular grid with distance between neighbouring nodes being 1. The side lengths of the hexagon are 1, 4, 5, 2, 3 and 6.

Figure 11

Consider two types of transformations on hexagons that maintain the property that all internal angles are of the same size.
(1) Prolonging two opposite sides by the same quantity $x$ (Fig. 12); this causes the side lengths to change according to the template
$$(a, b, c, d, e, f) \longleftrightarrow (a+x, b, c, d+x, e, f).$$
(2) Prolonging the two neighbouring sides of one particular side by the same quantity $x$ (Fig. 13); this causes the side lengths to change according to the template
$$(a, b, c, d, e, f) \longleftrightarrow (a+x, b-x, c+x, d, e, f)$$
A straightforward check shows that both transformations maintain the desired property, no matter of in which direction the transformations are applied.

(i) As the internal angles of all regular hexagons are equal, it suffices to show that an arbitrary hexagon with all internal angles equal can be turned into a regular hexagon by a finite sequence of the transformations above.
Indeed, let the side lengths of a given hexagon with all internal angles equal be $(a, b, c, d, e, f)$. Assume w.l.o.g. that $d \ge a$ and $f \ge c$. Choose a quantity $s$ such that $s \ge \max(a, b, c)$; by applying the transformation (1) thrice, we can obtain a hexagon with three consecutive sides having the same length:
$$(a, b, c, d, e, f) \xrightarrow{(1)} (s, b, c, d', e, f) \xrightarrow{(1)} (s, s, c, d', e', f) \xrightarrow{(1)} (s, s, s, d', e', f').$$
By the assumption made above we have $d' \ge s$ and $f' \ge s$. W.l.o.g., assume also $d' \le f'$. The transformation
$$(s, s, s, d', e', f') \xrightarrow{(2)} (s, s, s, s, e'', f'')$$
leads to a hexagon with four consecutive sides of equal length. But this must be regular since all of its internal angles are equal (Fig. 14).

(ii) Such a hexagon can be obtained from a regular hexagon with side length 1 by the following transformations:
$$(1, 1, 1, 1, 1, 1) \xrightarrow{(1)} (1, 4, 1, 1, 4, 1) \xrightarrow{(1)} (1, 4, 5, 1, 4, 5) \xrightarrow{(2)} (1, 4, 5, 2, 3, 6).$$

Figure 12

Figure 13

Figure 14