MAGIC POLYHEDRONS

Close your eyes and imagine that you are connecting the midpoint of a cube with its vertices by line segments, creating in this way six congruent square pyramids, which will completely fill this cube.

Now duplicate each of these pyramids by reflecting each of them on the plane given by its base. You get now 6 square piramids positioned onto the faces of the cube outside. The cube together with these six pyramids perform a new polyhedron.

Draw this polyhedron in that way you can imagine it. Then answer the following questions:

How many vertices, faces, edges does have this new polyhedron?
Which kind of polygonal shapes are its faces?
Are its faces congruent?
Is this polyhedron a regular one?
What’s its volume? (Compare the volume of this polyhedron with the volume of the cube in regard with the method you did create it.)

The new solid is a polyhedron with 14 vertices. 8 of them belong to the cube, and the remaining 6 you did create additionally. The polyhedron has 24 edges, none of them, as you can see, belongs to the base cube.

The edges of the base cube are now integrated into the faces of the new polyhedron and they are identical with the shorter diagonals of the rhombuses.
Of course, the polyhedron is not regular although the faces are congruent, but it does have incongruent corners: either three of four rhombuses built a corner.
From numbers and shape of its faces the name of the polyhedron is derived (fig.1) rhombus dodecahedron.


fig 1

After creation of the rhombus dodecahedron in our imagination we use now materialized models for fitting the 6 suitable square pyramides to the sides of a cube using ashesive tape.

Then we cut them partially off along the edges with a knife or a razon blade and we get a model enabling us to visualize the fact that the volume of the rhombus dodecahedron is twice the volume of the cube from which it was generated.

Below, we illustrated the phases of transforming the rhombus dodecahedron into two cubes with a common side. First, we wind out two pyramids from the opposite sides of the cube which is located inside the dodecahedron.....


fig 2a


fig 2b

...Then we detach succesively the remainibng pyramids and put them one by one in the line....

fig 3a

fig. 3b

....Finally we fit them together to the second cube..

fig  4a/ 4b and 5a/ 5b

It is easy to recognize that the edge of the rhombus dodecahedron is one half of a spatial diagonal line of the cube.

This means that for the edge „a” of the cube, the edge of the rhombus dodecahedron is b = .
This gives to us an easy derivation of the formula for the volume of the rhombus dodecahedron with the edge b:

The reader is able to calculate in the mind the surface of thw rhombus dodecahedron with side “b”.

The fact that one rhombus dodecahedron with the edge „b” may be transforrned into two cubes of the edge „a” which allows us to make further deducations.

From eight cubes, each of the edge a we may create another cube with an edge of double length (because 23 = 8). Based on the preceeding reasoning, the eight cubes may be transformed into four rhombus dodecahedron. This means that into the cube of the edge “2a” we may pack four rhombus dodecahedrons of the edge b =

But how to do that? The easiest way would be to make reverse transformation in to the cubes and square pyramides, however it is not as convenient as it might be supposed.

Let’s try something different. First put one such dodecahedron into the cube of the double length edge of the cube, from which it was created. Recognize, that the distance between the opposite situated vertices of the rhombus dodecahedron is the height of the cube, into which we are going to fill in. This suggest how it should be done - fig. 6.

fig. 6

 

We are aware that the dodecahedron dissected into four congruent heptahedrons could be positioned onto the bottom of the cube. One of its edges equals the cube’s edge with the length „2a”.


fig. 7

Now is clear, that to each of the heptahedrons belong: four faces which are isocsceles triangles with the base of length of , side of length of /2 and the hight of a/2, the rhombus with the length diagonal and a ,

two rectangular isosceles triangles with the hypotenuse 2a and the leg of right angled of length .

It’s interesting, that four such heptahedrons create one rhombus dodecahedron, the same one, that has been put into the cube.

fig. 8

Do you know how many of these heptahedrons you can place in the cube with the edge „2a”? Try to figure it out: the volume of this cube is 8a3, so you can place inside 8 cubes with the side „a”. Each couple of them does have the same volume of the one adequate dodecahedron. It means, that inside the cube which you are just begining to fill you can place four rhombus dodecahedrons that are exactly sixteen heptahedrons.

fig. 9

 Try to perform it and assembly this cube.


1 step


2 step


3 step


4 step

fig. 10

At last we show the net the “magic” heptahedron including its maeasurement.

fig. 11