# The Archimedean Solids

by Johannes Kepler## Proposition 28

There are 13 solid congruences which are perfect in an inferior degree.

From these 13 we obtain the Archimedean solids.

For congruences in this degree figures of different kinds are combined, thus by proposition XXI either two or three kinds of figure will be involved.

Cases involving 2 kinds either will or will not include trigons.

Thus, trigons and tetragons will make three solids which .satisfy definition

IX. This definition rules out three ways of constructing a solid angle, namely by using one tetragon angle and either one or three trigon angles or by using two tetragon angles and one trigon angle. For in the first case the congruence includes only one tetragon and we obtain half an octahedron, a figure whose solid angles are not all alike, while in the second case we have only two tetra gons^^ and in the third only two trigons.^^ So all these congruences are imper fect, by definition X. There remain the following methods of constructing a solid angle from plane ones. First, by using four trigon angles and one tetragon angle. For they add up to less than four right angles. Thus six tetragons and thirty-two (that is twenty and twelve) trigons fit together to make a triacontaoctahedral figure which I call a snub cube.^’^ It is shown in the diagram below, numbered Oo. I Snub cube. 12 .

Five trigon angles and one tetragon angle are more than four right angles, whereas to form a solid angle they would need to be less than four right angles, by Part 16.

The same is true of four trigon angles and two tetragon angles. In fact, three trigon angles and two tetragon angles make four right angles.

Second, two trigon angles and two tetragon angles are less than four right angles. Thus eight trigons and six tetragons fit together to form a tessareshae- decahedron, which I call a cuboctahedron.^^ It is shown here with the number eight. Two trigon angles and three tetragon angles are more than four right angles. Third, one trigon angle and three tetragon angles come to less than four right angles. Therefore eight triangles and eighteen (that is, twelve and six) squares join up to make an icosihexahedron, which I call a truncated cuboctahedral rhombus or a rhombicuboctahedron.^^ It is shown on this page, numbered 10.

In these three figures we have tetragons combined with trigons. In what follows we shall combine each of them separately with pentagons.

Five trigon angles will not combine with a pentagon angle, since they will not even combine with a tetragon angle, which is smaller.

Four trigon angles and one pentagon angle make less than four right angles, and eighty (that is 20 and 60) trigons will fit together with 12 pentagons to make an enenecontahaedyhedron, which I call a snub do decahedron.

It is shown here numbered 13. In this series of snubfigures the icosahedron could make a third, since it is like a snub tetrahedron.

If you combine 3 trigon angles with one pentagon angle the result is as described above, namely that the solid formed includes only 2 pentagons.

If you combine 2 trigon angles with one pentagon angle the solid includes only 1 pentagon.

The former case gives a zone or central column and the latter gives a pyramid, both parts of an icosahedron.

The solid angles of the second body are not all the same, since one is surrounded by 5 trigon angles, as in an icosahedron. We have now dealt with all the cases involving only one pentagon angle.

Three trigon angles with two pentagon angles make more than four right angles. So we have dealt with all the cases in which three trigon angles are combined with pentagon angles.

Two trigon angles with two pentagon angles make less than four right angles.

Thus 20 trigons and twelve pentagons fit together to make a triacontakaedyhedron, which I call an icosidodecahedron.’’^ It is shown here numbered 9.

Since we have already rejected the case in which two trigon angles are joined up with one pentagon angle we have now dealt with all the cases involving two trigons.

One trigon angle added to three pentagon angles makes more than four right angles, and if it is joined up with two pentagon angles it cannot make a regular solid, by XXIII, since the pentagon has an odd number of sides.

So we have now dealt with all the cases involving pentagons combined with trigons.

Four trigon angles with one hexagon angle, and two trigon angles with two hexagon angles fill the plane around a point; three trigon angles with two hexagon angles are greater than four right angles, and with only one hexagon angle they give a figure which contains only two hexagons.

So we must reject cases involving 3 trigon angles. Two trigon angles are equal to one hexagon angle, so this case is also rejected, by XXII.

It remains to unite one trigon angle with two hexagon angles. Thus four trigons and four hexagons fit together to make an octahedron, which I call a truncated tetrahedron.

It is shown as number 2 vi Truncated on the next page. tetrahedron.

Four trigon angles with one heptagon angle, or a larger angle, come to more than four right angles, so we need not discuss cases involving four trigon angles, nor cases involving three, for reasons already given.

In fact, two trigon angles with 2 plane angles larger than those of the hexagon come to more than 4 right angles, so we do not need to discuss cases involving two trigon angles joined with two plane angles of a figure larger than the hexagon, rwr cases involving two trigon angles joined with one plane angle of a larger figure, because such an angle is larger than two trigon angles, so the case is rejected by axiom 22.

It remains for us to examine the case in which one trigon angle is united with two plane angles of a figure larger than a hexagon. The case in which these are two heptagon angles is rejected by XXIII, as are all the cases involving two angles of a figure with an odd number of sides. With two octagon angles we obtain a solid in which eight trigons join up with six octagons to make a tessarakaedecahedron, which I call a truncated cube.

There is a diagram of vii Truncated it numbered 1 on the following page. With two decagon angles we obtain a solid in which twenty trigons join up with twelve decagons to make a triacontakaedy- hedron which I call a truncated dodecahedron.

This is shown below as number viii Truncated 3. With two dodecagon angles the plane is filled, so no solid angle can be made dodecahedron. with these or any still larger angles. We have thus entirely finished with cases involving trigons together with any other single kind of figure.

Since the 2 kinds of plane figure will no longer include trigons the smallest figure involved will now be the tetragon. Three tetragon angles with one larger angle come to more than 4 right angles, and by definition IX we know we cannot combine two tetragon angles with one larger angle, since only 2 of the larger figures will occur in the resultant solid.

The case of one tetragon angle combined with two pentagon angles is rejected, by XXIII, but one tetragon angle will go with two hexagon angles, and six tetragons and eight hexagons will fit together to make a tessarakaedecahedron which I call a truncated ix Truncated octahedron.

t is shown as number 5 in the diagram below. The case of one tetragon angle combined with two heptagon angles is rejected because the heptagon has an odd number of sides, that is by XXIII. With two octagon angles the plane is filled. With larger angles the sum exceeds four right angles and no solid angle can be formed.

So we have dealt with all the cases involving the tetragon, since there must be only 2 kinds of plane figure.

Two pentagon angles combined with one hexagon angle or the angle of any other figure will not form a congruence, by XXIII, so these cases must be rejected, just as we earlier rejected the cases involving a trigon angle or a tetragon angle combined with two pentagon angles.

Moreover, two pentagon angles combined with one decagon angle cover the plane, so neither with this angle nor with a larger one will they form a solid angle.

Now one pentagon angle with two hexagon angles comes to less than 4 right angles, and twelve pentagons and twenty hexagons willfit together to make a triacontakaedyhedron, which I call a truncated icosahedron.

It is shown numbered 4. We cannot expect any more congruences from the pentagon. For one pentagon angle combined with two heptagon angles is already larger than 4 right angles.

One hexagon angle with 2 others fills the plane, and with two larger angles the sum exceeds four right angles.

So we have now dealt with all the cases in which two kinds of figure are combined.

Turning to cases in which 3 kinds of face may fit together to make a solid angle, we first note that two plane angles, one from a tetragon and one from a pentagon, add up to more than 2 right angles, and larger angles add up to even more, so since 3 trigon angles come to two right angles, it is clear that the two plane angles we have mentioned will not fit together with three trigon angles, for the total sum would be more than 4 right angles. The cases in which two trigon angles are combined with one tetragon angle and one pentagon angle, or, instead of the pentagon angle, a hexagon angle or some larger one, all these cases are to be rejected, by proposition XXIII, because they would require the trigon, which has an odd number of sides,

to be surrounded by tetragons and either pentagons or, instead of the pentagons, hexagons, etc.

One trigon angle, two tetragon angles and one pentagon angle add up to xi Rhombico- less than four right angles, and twenty trigons, thirty tetragons and twelve penta- s> 4 «decahedron. gons will fit together to make a hexacontadyhedron which I call a rhombicosido- decahedron or a truncated icosidodecahedral rhombus:’^ It is shown as number 11 on the previous page.

One trigon angle and two tetragon angles combined with one hexagon angle come to 4 right angles; with a larger angle they come to more; so they do not form a solid angle.

Let us therefore dismiss cases involving two tetragon angles.

One trigon angle, one tetragon angle and two pentagon angles add up to more than four right angles, and they add up to even more with two larger angles.

So we have finished with cases in which four plane angles are put together to form a solid angle, and also with cases in which the trigon angle is one of the 3 kinds of angle involved.

For the case of one trigon angle and one tetragon angle and either one pentagon angle or any other angle is to be rejected, by XXIV, since the trigon has an odd number of sides.

In fact, since we are now combining only three plane angles, no figure with an odd number of sides can be admitted, by XXIV again.

A tetragon angle with a hexagon angle and an octagon angle, the smallest xii Truncated admissible angles, comes to less than four right angles; and twelve tetragons, cuboctahedron. eight hexagons and six octagons will fit together to make an icosihexahedron which I call a truncated cuboctahedron: not because it can be formed by trun cation but because it is like a cuboctahedron that has been truncated.’^^ It is shown numbered 6.

A tetragon angle with a hexagon angle and a decagon angle comes to less xiii Truncated than four right angles; and thirty tetragons, twenty hexagons and twelve deca- »cosidodeca- hedron.

gons fit together to make a hexacontadyhedron which I call a truncated icosido- decahedron, for reasons similar to those given in the previous case.^^ It is shown numbered 7.

If we replace the decagon angle by a dodecagon angle the sum is 4 right angles and we cannot make a solid angle; also, if we replace the hexagon angle by an octagon angle and take as our third angle any angle larger than an octagon angle we have more than four right angles; nor is the sum less if we set aside the tetragon angle and instead join up three angles from larger figures with an even number of sides.

Therefore the whole family of Archimedean solids numbers 13, as was to be shown.