We already know that Pappus of Alexandria (around the year 320) was interested in the shape of honeycombs. He thought that the hexagonal prism was an optimal
solution for the bees. It is to say that this hexagonal structure minimized the amount of wax that the bees needed to build the cells. These cells contain more
honey in relation with the surface area.

Humankind has always been fascinated by how bees build their honeycombs. Kepler related honeycombs with a polyhedron called Rhombic Dodecahedron.

Pappus wrote:

'Bees, then, know just this fact which is of service to themselves, that the hexagon is greater than the square and the triangle and will hold more honey for
the same expenditure of material used in constructing the different figures'
(You can read more about Pappus in note 16 of the book 'The Six Cornered Snowflake' or in
Sir Thomas Heath's book 'A History of Greek Mathematics' [Dover, 1981]).

Pappus considered only the hexagonal prism but not the rhombic bottom.

The problem that a hexagonal prism is optimal or isoperimetric was known as the Honeycomb Conjeture. This conjeture remainded unproven until the middle of the
twentieth century (Laszlo Fejes Toth, Thomas Hale). And Toth proved that the bases of the cells are not optimal shapes.

In this page we are going to study a much more simple optimization problem: We want to close a hexagonal prism as bees do, using three equal rhombi.

You can play with the applet to see different ways to do that:

If the shape changes the surface area is different.

We want to minimize this surface area.

We can notice that the volume of these different cells is always the same.

You can play with the applet, opening and closing the keel, to see the reason:

To close the cell we cut three pyramids from the hexagonal prism and then we put these three pyramids together to form the keel, then the volume is always the same.

The surface area needed to close the prism changes when we change the shape.

We can calculate the lateral hexagonal prims surface area but we will not need this calculation. Our problem does not depend on the height of the prism and we can suppose
that it is large enough. Our problem is related with the three rhombi that close the prism, and we want to optimize this surface area.

*R* is the radius of the circumcircle of the hexagon and *h* is the height to the rhombus vertex from one vertex of the hexagon. *h* is our independen variable.
Changing *h* we change the shape and the surface area needed to close the prism (but no the volume of the body).

We can make our calculations considering only one rhombus because the three of them are equal.

If we start with the hexagonal prism lateral surface area the first thing that we can do is to cut two right triangles. These triangles *t* have base *R* and height *h*.

Then the area of one of these triangles *t* is:

To close the prism we add one rhombus made of two triangles *r*. They have base *d* and height *a*.

The diagonal *d* is easy to calculate, because it is one diagonal of the hexagon:

We need *s* that is related with *R*, the radius of the circumcircle:

Now we can apply the Phytagorean Theorem to calculate *a*:

Then the surface area of the triangle *r* is:

We can simplify a little:

The surface area of the rhombus that we add is twice the area of this triangle.

Now we can calculate the surface area of our body. Taking into consideration only one rhombus, we cut two triangles *t* and we add two triangles *r*.

The function that we want to optimize is:

To obtain the minimun of this function we can use the derivative:

Then the minimum is:

We can check the relation between the two diagonals of this rhombus:

This rhombus is a very special one. With twelve of these rhombi we can build a polyhedra called rhombic dodecahedron.

Johannes Kepler was the first mathematician who wrote about the rhombic dodecahedron and thought that bees close their cells with rhombi that are rhombi of a rhombic dodecahedron.

"Intrigued by these rhombi [the three equal rhombi that form the keel of a cell in a beehive], I began looking into geometry to see whether a body resembling
the five regular solids and the fourteen Archimedean solids could be constructed exclusively from rhombi. I discovered two, one related to the cube
and the octahedron, the other to the dodecahedron and the icosahedron. (The cube itself can be considered a third, since it is like two tetrahedra
joined together)." (Johannes Kepler, 'De Nive Sexangula'). [Note: There are only thirteen Archimedean solids, and not fourteen as Kepler said]

You can play with the transparency using the next applet:

REFERENCES

Johannes Kepler - The Six Cornered Snowflake: a New Year's gif - Paul Dry Books, Philadelphia, Pennsylvania, 2010. English translation of Kepler's book 'De Nive Sexangula'.
With notes by Owen Gingerich and Guillermo Bleichmar and illustrations by the spanish mathematician Capi Corrales Rodrigáñez.

D'Arcy Thompson - On Growth And Form - Cambridge University Press, 1942.

Hugo Steinhaus - Mathematical Snapshots - Oxford University Press - Third Edition.

Magnus Wenninger - 'Polyhedron Models', Cambridge University Press.

Peter R. Cromwell - 'Polyhedra', Cambridge University Press, 1999.

H.Martin Cundy and A.P. Rollet, 'Mathematical Models', Oxford University Press, Second Edition, 1961.

W.W. Rouse Ball and H.S.M. Coxeter - 'Matematical Recreations & Essays', The MacMillan Company, 1947.

MORE LINKS

Material for a sesion about polyhedra (Zaragoza, 9th May 2014). Simple techniques to build polyhedra like the tetrahedron, octahedron, the cuboctahedron and the rhombic dodecahedron. We can build a box that is a rhombic dodecahedron.

There is a standarization of the size of the paper that is called DIN A. Successive paper sizes in the series A1, A2, A3, A4, and so forth, are defined by halving the preceding paper size along the larger dimension.

A cuboctahedron is an Archimedean solid. It can be seen as made by cutting off the corners of a cube.

A cuboctahedron is an Archimedean solid. It can be seen as made by cutting off the corners of an octahedron.

The volume of an octahedron is four times the volume of a tetrahedron. It is easy to calculate and then we can get the volume of a tetrahedron.

The truncated octahedron is an Archimedean solid. It has 8 regular hexagonal faces and 6 square faces. Its volume can be calculated knowing the volume of an octahedron.

These polyhedra pack together to fill space, forming a 3 dimensional space tessellation or tilling.

Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the truncated octahedron.

Leonardo da Vinci made several drawings of polyhedra for Luca Pacioli's book 'De divina proportione'. Here we can see an adaptation of the cuboctahedron.

You can chamfer a cube and then you get a polyhedron similar (but not equal) to a truncated octahedron. You can get also a rhombic dodecahedron.

A Cube can be inscribed in a Dodecahedron. A Dodecahedron can be seen as a cube with six 'roofs'. You can fold a dodecahedron into a cube.

If you fold the six roofs of a regular dodecahedron into a cube there is an empty space. This space can be filled with an irregular dodecahedron composed of identical irregular pentagons (a kind of pyritohedron).