Measurementsby Johannes Kepler
The surfaces of the Decagon and the Pentagon belong to more remote Degrees of knowledge, as does the side of the Icosigon and the remaining [sides] of figures of this class.
For the side of the Pentagon FH, multiplied by AN makes twice [the area of triangle] FAH, a fifth part of the Pentagonal Area. Now, FH is an Elasson, and AN is such that in square it is equal to the [square of the] Expressible line AF, less the square of the Elasson FN.
Now, if the square of an Elasson is subtracted from the square of an Expressible line, the result is a new kind of line which, in square, is equal to this remainder.
The rectangle contained between a line of this new type and an Elasson will be of a still more remote kind; in fact, the area of the Pentagon will be commensurable with it, namely being in the ratio of five to two, so that it [the area of the pentagon] too will thus be of a more remote kind. Thus the side of the Decagon FG, multiplied by its perpendicular distance from the center, makes twice [the area of triangle] FAG, 1/10 of the Decagon Surface, that is, [the product is] onefifth.
FG is a fourth Apotome; and the perpendicular to it from the center, squared, is equal to a quarter of its [i.e. FG’s] square less than the square of the semidiameter.
But if the square of an Apotome is subtracted from the square of an Expressible line, the line which, in square, is equal to the remainder, is of a new kind beyond those listed so far; and if such a line were to make a rectangle with an Apotome, it [i.e. the area of the rectangle] would be of a still more remote kind, and with it also 5 times it, that is the Area of the Decagon.
Finally, since half the side of the Decagon is a fourth Apotome and the square of the Apotome, stretched out to the length of the Diameter (which is Expressible in length), gives a width which is a first Apotome, that is to say the sagitta of the tenth part of the Circle^’^’^: indeed, the side of the Icosigon, squared, is equal to the sum of the squares of half the side of the Decagon, a fourth Apotome, and of this sagitta, a first Apotome.
The surface composed from [i.e. with sides equal to] Apotomes of different kinds, and thus incommensurable with one another, will not be equal to the square of any line like those listed already; but [will be equal to the square of] some line of a completely new kind: and thus also of lower degree (ignobilior).
How much more will this apply to the forty-sided polygon (Tessaracontagon) and the others of this class'?
The sides of the Pentekaedecagon and its stars, namely the chords subtending 2, 4, or 7 15ths [of the circle], do have a Geometrical description, but not apart from the circle.
In the circle, also, not through their angles, thus [the geometrical description] is not intrinsic (impropriam) and the knowledge is of a different kind, of a more remote degree than that of all the preceding sides.
The triacontagon and the remaining shapes of this class are of even more remote degree.
For it is described from figures before it, the relevant ones having a number of sides that is not obtainable by doubling, because 15 is an odd number, half of it not being a [whole] numbeF that is from the Triangle BCD and the Pentagon BIFHK, each star ting from the same point B. For if you subtract 1/3 of a circle, the arc BC, from 2/5, L [the arc]
BIF, that is 5 fifteenths from 6 fifteenths, A / the remainder is CF, 1 fifteenth.
So joining the angles
F gives the line
CF as the side of the pentekaedecagon.
Here neither the size of the angle J m nor the number of Angles in thefigure is concerned in the process of description; nor do I construct any triangle in accordance with this number, as was done for the previous figures.
But neither can it [this figure] be described in any other way.
Thus knowledge of it is also of a remote and low kind. For since FH, a side of the Pentagon, is parallel to CD, a side of the Trigon, because each figure has an odd number of sides and begins at the same point B: therefore let there be drawn from F a perpendicular to [CD, meeting it in the point[ L, andfrom B a diameter through the center A, cutting the lines [CD, FH and the circumference] in the points
Therefore the side CF, squared, will be equal to the sum of the squares of
FL-y but CL is the magnitude by which CE, Expressible in Square, exceeds FN, that is LE, an Elasson: so CL is [a line] of a completely new kind.
On the other hand, AN is a line which, squared, is equal to the remainder when
a surface whose side is an Elasson is subtracted from an Expressible one: so
it is of a new kind.
But EN is what remains of this new [kind of line] after subtracting AE, Expressible in length. So EN is two steps more remote.
Finally CF, the side of the Pentekaedecagon, squared, is equal to the sum of the squares of CL and EN, both of new kinds; so in the former case it [CF] is twice, and in the latter case three times [more remote], and is thus [in all]five times more remote.
Furthermore, the properties of different classes, those of the Trigon and of the Pentagon, are combined, so that knowledge [of the polygon] is of a different kind. What should the decision now be about the sides of the Triacontagon?
Since the degree of remoteness always increases with the doubling of the number of sides of an earlier figure.
But the chord subtending seven fifteenths, that is 14 Thirtieths, uses the side of the Triacontagon, and is posterior to iO
The chord subtending 7 Thirtieths is obtained from it by bisection [of the arc]: the same procedure generates the chord subtending 8 Thirtieths, that is 4 fifteenths, from which the chord subtending 2 fifteenths can also be obtained by bisection [of the arc].
However, this last also has another origin; for example, the chord subtended by the arc MF, squared, is equal to the sum of the square of CF, the side of the Pentehaedecagon, and the Rectangle contained by the same CF and FI, the side of the PentagonJ^^^ In either way it [i.e. the chord subtending two fifteenths] is posterior to [i.e. of lower degree than] the previous figures.