Malcolm Findlay from Madras College in St Andrews, Scotland has solved the first part of this problem:

Izumi Tomioka, Carol Chow and Priscilla Luk from The Mount School in York solved the second part of the problem:

The original hexagon has sides of length 3 units and we need to work out x, the lengths of the sides of the smaller hexagons.

The hexagon below has been made from 6 of the shaded triangles above.
We need to find the length of one of the sides of this hexagon.

Splitting the shaded triangle into half gives us a right angled triangle such that:

Andrei Lazanu (aged 12) from School 205 in Bucharest, Romania, solved both parts of the problem.

This is how he tackled the second part:

To calculate the lengths of the sides of the smaller hexagons I used the following notations:

l for the length of side of the great hexagon
a for the length of side of the small hexagon

I used the following notation:

Triangle ACE is equilateral, because its sides are congruent.
So, angle EAC is 60°.

Angle FAB is 120°, since each angle of a regular hexagon is 120°.

Triangle AEF is congruent with triangle ACB, having all sides congruent. They are also isosceles triangles.
This means that each of the angles FAE and CAB is 30°.

Therefore angle EAB is 90°.

Triangle AMN is also equilateral, because it has a 60° angle (MAN) and AM = AN.
Triangle ANB is isosceles, so AN and NB are congruent.

Therefore, in the right angled triangle MAB,
MA = MN = NB = a

AB has length l

Applying the Pythagorean Theorem:

l ² = (a + a) ² - a ²
l ² = 4a ² - a ²
l ² = 3a ²
l = a √3

If the length of the side of the great hexagon is 3 units long
l = 3

Therefore
a =√3 units

An alternative way of calculating "a" takes into account the first part of the problem:
the area of the great hexagon is three times the area of the small one.

For a hexagon of side l, the area is calculated as 6 times the area of an equilateral triangle of side l,
this means that the area of the great hexagon is:

The 3 smaller hexagons of side "a" have a total area of:

Since, (1) and (2) represent the same area,
3a ² = l ²

which is the same as we found above.