Filling Up the Planeby Johannes Kepler
Proposition 20: There are 4 ways in which the plane can be filled by the congruence of plane angles of 3 kinds.
Here, we cannot use 3 or more trigon angles at each meeting-point This is because 3 trigon angles make 2 right angles. This leaves a gap which is less than the sum of the angles of the next 2 polygons, the tetragon and the pentagon.
This is why we cannot employ 2 trigon angles with 2 tetragon angles, or with larger ones, since they do not leave enough space for the angle of the third kind of figure.
- So if we have two trigon angles and one tetragon angle, the dodecagon angle will join up with them. However, this pattern cannot be continued. See letters Cc, Dd, and Ee, which show three forms, all belonging to the first case of the proposition.
Here, as above, we must consider the star dodecagon. For four trigon angles, one tetragon angle and one point of the star will fill the plane. See forms F f Gg, and Hh.
If a pentagon angle is joined up with two trigon angles the remainder will be incongruent, being twenty-two twenty-fifths of a right angle, for there is no angle of a regular figure which is eleven twenty-fifths. If one hexagon angle is added to two trigon angles the remainder will also be a hexagon angle, and the form will be one of those described above. So there are no more plane con gruences involving two trigons.
So let there be one trigon angle. Three tetragon angles cannot be added to it, because then the sum is too large and there will not be enough space left for the angle of the third kind.
- To one trigon angle let us add two tetragon angles. When we subtract the sum from four right angles the remainder is the angle of a hexagon.
This pattern takes two forms: the one shown at li can he continued, the one shown at
Kk cannot be continued without additional figures. This is the second case of the proposition.
A single trigon angle cannot join up with two pentagon angles, because the gap left is fourteen fifteenths of a right angle, an angle not found in regular figures.
Nor can a single trigon angle join up with one pentagon angle, because the gap left would be thirty-two fifteenths of a right angle and no regular figure has an angle of 16 fifteenths.
Nor can a trigon angle join up with one hexagon angle, because they come to two right angles, and no single angle is that size; and half of the gap is the size of the tetragon angle, which we have already dealt with.
Nor can a trigon angle join up with one angle of a heptagon or of an octagon or of an enneagon, for the gap left for the third kind of angle would be 40 21s of a right angle or 11 16ths or 16 9ths, none of which is the angle of a regular figure.
A trigon angle combined with a decagon angle leaves a gap of 26 15ths of a right angle, which is the angle of the pentehaedecagon.
These figures do form a congruence, but a limited one. For the pentehaedecagon has an odd number of sides, so by XVII the meeting-points at the angles of the figure will not all be the same.
Since the decagon has an even number of sides it can be surrounded by alternate trigons and pentekaedecagons, but two of the pentekaedecagons immediately run up against one another and prevent the pattern from being continued.
Further, a trigon cannot be joined with a hendecagon, for this would leave fifty-six thirty-thirds of a right angle, which is not the angle of any regular figure.
Next, a trigon angle taken together with a dodecagon angle leaves a gap the size of a decagon angle, a form we have already discussed.
If a trigon angle is subtracted from four right angles, the space left is not great enough to fit in angles of two other kinds, which together would come to more than two right angles. Ll. Mm. Nn.
- One tetragon angle joined up with one pentagon angle leaves a gap the sizeof an icosigon angle.
An icosigon can therefore fit together with two such angles at every one of its own angles, forming a true congruence, but this pattern cannot be continued outwards. It is therefore an imperfect congruence. See figure Ll. This is the third case of the proposition.
- A tetragon angle joined up with a hexagon angle leaves a gap the size of a dodecagon angle. See figure Mm. This is the fourth and last case of the proposition.
Here we must consider the star decagon which can be filled out with 12 trigons.
Four angles thus fit together to fill the space: two trigon angles, one tetragon angle, one hexagon angle, and a point of the star. See figure Nn.
If a tetragon angle is added to a heptagon angle it leaves a gap of 11 sevenths of a right angle, an angle not found in any regular figure.
Added to an octagon angle, a tetragon angle leaves a gap the size of an octagon angle.
We have discussed this form above. We have therefore dealt with all the cases involving a tetragon angle.
A pentagon angle together with a hexagon angle leaves a gap of 22 15ths of a right angle. Taken together with a heptagon angle it leaves forty-eight thirty-fifths; with an octa^n angle thirteen tenths, while no such angle is to be found in any regular figure.
Taking a pentagon angle with angles of figures having more sides than an octagon] the gap left starts to be less than an octagon angle, which is fifteen tenths of a right angle. But we have already dealt with congruences involving smaller angles. We have therefore dealt with all the cases involving a pentagon angle.
3 hexagon angles fill the plane, so a hexagon angle cannot be combined with two angles larger than itself.
We have therefore dealt with all cases involving figures of 3 different kinds.