THE CONGRUENCE OF REGULAR FIGURES
by Johannes KeplerProposition 19: There are 6 ways in which the plane can be filled around a point by 2 kinds of shapes:
 2 ways use 5 angles
 1 way uses 4 angles
 3 ways using 3 angles
6 plane figures cannot fit together since the angle of one of them must be larger than the angle of the trigon.
The angle of the trigon, the first of the polygons, is 2/3 of a right angle, so taking it 6 times gives 12 thirds, or 4 right angles.
So if one of the 6 angles were larger, that is if it were the angle of a higher polygon, the sum would be more than four right angles, so the plane is not covered, by XVI of this book.

5 shapes fit together if 4 trigon angles are combined with another angle equal to 2 trigon angles, that is a hexagon angle. The form is shown in L.

5 figures fit if 3 trigon angles are combined with 2 tetragon angles, because the last two add up to three more trigon angles.
The form is as shown in M
, or N
, both of which forms can be extended uniformly, or it is as shown in O
, a form which cannot be extended uniformly.
But if you take two trigon angles with three tetragon angles they will come to more than four right angles. The sum is even larger if you add two^^ larger angles to two trigon angles.
 Four figures of two kinds fit together if two trigon angles are combined with 2 hexagon angles.
The form is either as shown in P or as shown in R.
Whatever other four angles you fit together you always get more or less than 4 right angles so you do not fill the plane. If we join up three angles, taking care not to use more than two kinds of angle, we may begin by ruling out cases which use two trigon angles or 2 tetragon angles, for these do not come to more than two right angles and the gap they leave for the third angle is therefore too large for any one angle to fill on its own.
 Now if we assume one of the three angles to be a trigon angle, we obtain a congruence with two dodecagon angles.
This pattern can be continued without involving any different kind of meetingpoint. The result in the plane is as shown in S.
Here we must consider the star dodecagon, because its reentrant angles are equal to a trigon angle, so that a dodecagon can be divided up into a star and twelve trigons. Therefore, five trigon angles and two points of two stars will fit together. The form, which can be continued, is as seen marked with the letter T.
If a tetragon angle is taken as one of the three, there is a congruence with two octagon angles, and this form too may be continued.
It is seen marked with the letter V.
Here we must consider the star octagon, because its reentrant angles are equal to a tetragon angle, so that an octagon can be divided up into a star and eight rightangled triangles, two of which make a tetragon.
Thus 3 tetragon angles and 2 points of 2 stars fill the plane.
The mixed form is as shown marked with X
, or otherwise, again mixed, marked Y
.
 Having dealt with sets of three angles which include trigon angles and tetragon angles, if we now come to the pentagon angle we may take two of them, because together they come to more than two right angles; and a decagon angle fits into the space they leave. The decagon is encircled by ten pentagons, but this pattern z. cannot be continued in its pure form. See the inner part of diagram Z. Here we must consider the star pentagon, since we can fit together three pentagon angles and one point of a star, because the reentrant angle of the star takes one pentagon angle while, no less, the gap left by fitting together three pentagon angles takes the point of the star. See the outer part of the same z. diagram Z.
However, this pattern cannot be continued indefinitely, for the domain it builds up is unsociable^^ and when it has added to its size a little it builds fortifications. You may see a different arrangement of these two forms marked Aa.
with the letters Aa.
If you really wish to continue the pattern, certain irregularities must be admitted, two decagons must be combined, two sides being removed from each of them.
As the pattern is continued outwardsfivecorneredforms appear repeatedly: in the first and smallest of the fivecornered ranks there are five decagons with no intermediate irregularity, in the second and wider rank lines of single decagons lie between decagons joined in pairs, in the third rank the corners are taken by pairs of joined decagons and between two such pairs there lies a simple decagon, in the fourth rank we again have simple decagons in the corners and on the side between them there are two more decagons, spaced at equal intervals, in the fifth rank the corners are marked by the tips of the outermost points of stars and the sides each contain two simple decagons between which there are two pairs of combined decagons. So as it progresses this fivecornered pattern con tinually introduces something new. The structure is very elaborate and intricate. Aa.
See the diagram marked Aa.
Here we must also consider the star decagon, whose reentrant angle fits around the angle of a pentagon.
In this way two of the points, each one 3 tenths of a right angle, join up with two pentagon angles to fill the plane around a point. This pattern takes in pentagons of a different size. It can be continued.
ut the continuation includes incomplete open decagons. The pattern is shown marked with the letters Bb. Bb.
We cannot take a single pentagon as one of the three plane figures which are to form a congruence, for its angle is six fifths of a right angle, by Book 1 Part 33, so the remaining 2 angles would be left with 14 fifths of a right angle, that is each would be 7 fifths, which is not the angle of any regular figure.
Nor can we take 2 hexagons, for the remainder is also the angle of a hexagon and we shall obtain theform of congruence already described above, and we are now looking for structures involving two kinds offigure, not only one kind.
As for the higher polygons, whose angles are greater than that of the hexagon, when two such angles are subtracted from four right angles the remainder is less than the angle of a hexagon; taking one angle from four right angles, what is left for the remaining two angles of the proposed congruence is less than two hexagon angles.
We have already dealt with those figures whose angles are smaller than that of the hexagon, so we have dealt with all the ways of covering the plane with three figures at each meetingpoint.