Hex
1 out of 3
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
Problem
Explain how the thirteen pieces making up the regular hexagon shown in the diagram can be re-assembled to form three smaller regular hexagons congruent to each other.
Click here for a poster of this problem.
Getting Started
If the original hexagon has sides of length 3 units what is the length of the sides of the smaller hexagons?
Student Solutions
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: $$\cos30 = {1.5\over H}$$ $$H = {1.5\over \cos 30}$$ $$H = 1.73$$
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: $$l^2 = (a + a)^2 - a^2$$ $$l^2 = 4a^2 - a^2$$ $$l^2 = 3a^2$$ $$l = a \sqrt{3}$$
If the length of the side of the great hexagon is 3 units long $l = 3$
Therefore $a = \sqrt{3}$ 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:
| $6{l^2 \sqrt3\over 4} = {l^2 3\sqrt3\over2}$ | $ | (1)$ |
The 3 smaller hexagons of side "a" have a total area of:
| $3{a^2 3\sqrt3\over 2} = {a^2 9\sqrt3\over2}$ | $(2)$ |
Since, $(1)$ and $(2)$ represent the same area, $3a^2 = l$
which is the same as we found above.