Superphysics
Part 45b

# Algebra

by Kepler
##### 8 minutes  • 1647 words

So no Regular Heptagon (Septangulum) has ever been constructed by anyone knowingly and deliberately, and working as proposed.

Nor can it be constructed as proposed; but it can well be constructed fortuitously; yet it is, all the same, [logically] necessary that it cannot be known whether the figure has been constructed or no.

Here it might be suggested that I should use the Analytic art called Algebra after the Arab Geber, its Italian name being Cossa: for in this art the sides of all kinds of Polygon seem to be determinable.

For example, for the Heptagon the following procedure is adopted byjost Biirgi, Instrument maker (Mechanicus) to the Emperor and to the Land grave of Hesse who is noted for his very ingenious and surprising achievements in this matter.

First he assigns the value 2 to BP, the diameter of the circle, so that AB shall be a complete unit, which will be divided into parts by an indefinite [procedure of] subdivision, and these parts will give a numerical value for the length of the side BC. Then he assumes that the ratio of AB to BC is known, though this ratio is in fact what we are required to find.

He sets up the series of ratios so that the ratio of AB [taken as] 1 to BC [taken as] I!^, is equal to the ratio of iij to U, and U to and ic^ to m , and nh to IJ and so on for ever, which we shall express in a more convenient notation using Roman numerals, thus: 1, Ij, lij, liij, liiij, Iv, Ivj, Ivij, and so on having made these assumptions we first consider the quadrilateral BEDC.

So, since it has been proved by Ptolemy, Copernicus, Regiomontanus, Pitiscus, and others who have written on the theory of sines; that in any cyclic quadrilateral the single rectangle contained by the Diagonals, CE, DB, is equal to the sum of the two rectangles contained by the [pairs of] opposite sides, namely, that of DC and ED, and that of CB and DE’^^ f- And again since it is known from Geometry that the sum of the squares of CO, half the chord CH, and OB, the sagitta, is equal to the square of the side CB.

Therefore let BP be equal to 2 [units] and CB be equal to Ij, so that its square is lij, divide this by BP, it gives BOf^^^ namely lij divided by 2 [units], squared [this is] liiij divided by 4, subtract this from the square of CB, lij, the remainder is 4ij - liiij [all] divided by 4, [which is] the square of CO. Now since CH is twice the line CO, the square of the line CH is 16ij - 4iiij [all] divided by 4, that is 4ij - liiij.

Therefore, since we wish to have the square of CH or CB, that is the rectangle contained by BC and CEN’^ multiply CB into DE, so that the rectangle contained by these lines is lij, subtract this from the rectangle contained by BD, CE, which is 4ij - liiij, there remains the rectangle contained by CD, BE, which is 3ij - liiij, divide this by Ij, that is by CD, the result will be BE, 3j - liij.

Further, we turn to the Quadrilateral DBHE. And because BE is 3j - liij, the rectangle contained by BE, DE, that is the square of the line BE, will be 9ij - 6iiij + Ivj: subtract the rectangle contained by BH, DE, [which is] lij, there will remain the rectangle contained between BD, EH, which is 8ij - 6iiij + Ivj, divide this by EH, which is 3j - liij, the result will be BD, 8ij - 6iiij + Ivj [all] divided by 3j - liij: its square [Le. BD^] will be 64iiij - 96vj + 32viij - 12x + Ixij [all] divided by 9ij - 6iiij + Ivj, which was [earlier found to be] 4ij - liiij: multiply this [value] by the denominator [of the previous expression] and we have 36iiij - 33vj + lOviij - lx equals 64iiij - 96vj + 52iij - 12x + Ixij therefore also 63vj + llx equals 28iiij + 42viij + lxij3^^ therefore also 63ij + llvj equals 28 + 42iiij + lviij3^^ This equation gives the quantity of the side of the Heptagon. Or we turn, further, to BD, EG. Now the square DC, gij _

• Ivj. But the square DB, EG^’^^ is 4ij - liiij, subtract this latter from the former, the rectangle contained by DE, BG will be 5ij - 5iiij + Ivj, divide this by DE, that is Ij, BG will be 5j - 5iij + Iv, whose square is 25ij - 50iiij + 35vj - lOviij + lx, which earlier was [found to be] 4ij - liiij. So 49iiij + lOviij equals 2 lij + 33vj + lx

Therefore also 49ij + lOvj equals 21 + 33iiij + Iviij.

This equation too gives the quantity of the Heptagon side: but Biirgi turns his attention away from the complete circle and considers it only as an arc that is to be divided into 7 [equal] parts.

So since the chord subtending 2 parts can be found by this algebraic procedure {cossice), he seeks the chord subtending 4 parts, and finds it (by the same method as above) to be the Root of 16ij - 20iiij + 8 vj - Iviij. He now makes use of the Diagonal in a new quadrilateral, [two of] whose sides are chords subtending three sevenths, so that the Rec­ tangle they contain is 9iJ - 6 iiij + Ivj, which, subtracted from the Rectangle 16ij - 20iiij + 8 vj - Iviij, leaves, as the rectangle of the remaining [two] sides, 7ij - 14iiij -I- 7vj - Iviij. He makes use of this chord, comparing it either with the number that expresses the chord subtended by the arc that is to be divided into seven parts, or with the figure zero, if the whole circle is to be divided into seven, as here: and then either that number or the figure zero is equal to the quantities 7j - 14iij + 7v - Ivij or 7 - 14ij + 7iiij - lvj .^^2 Then he deduces from the equation, which he solves mechani- cally,223 not one value for the root, but two for the Pentagon, three for the Heptagon, four for the Nonagon, and so on: for one value is BC, the second BD and the third B E .224

This type of investigation of the sides of the figure has absolutely nothing in common with the Definitions we gave in Parts 1-3.

What does the algebraic chord of Biirgi’s signify?

It signifies that if 7 lines are constructed in continuous proportion, the proportion being that between the side of the heptagon and the semidiameter of the circle, and the first proportional is made equal to the side of the heptagon: then seven lines equal to the first proportional plus seven equal to the fifth will add up to the same as 14 lines equal to the third proportional plus one line equal to the 7th.

This statement is Geometrical and can be demonstrated like when we showed that the surface of the Octagon was Medial, or the side of the Dodecagon was an Apotome of some line.

For there, something was being stated about the surface or line, here something is stated about the proportion between lines.

But just as it is not enough for me, for knowing and measuring a surface, to know that it is a Medial, and not enough for measuring a line to know that it is an Apotome of some line:

Since there are many quantities of such a type, and there is no construction [to be deduced] from this general remark, and no precise and certain quantity for the plane or line may be elicited from it, but these properties only follow from quantities previously constructed and described.

So here also, it is not enough for me to know what would happen once the 7 lines in continuous proportion, according to the proportion that I require, have been set up: but since I do not yet have that proportion described by geometrical means: therefore I waited for someone to explain to me how to set up that proportion first.

For thus for all previous figures the procedure was [in the order]: description, inscription in a circle, determination of a precise quantity, and a precise Geometrical means by which this determination might be carried out; finally there followed the knowledge of the properties which permitted comparison of figures one with another.

To make the distinction in this matter clearer, let us look at the side of the Pentagon, whose mode of description, described above, was that, having combined two squares, one [whose side was] the semidiameter, the other [with side equal to] half of it, to make a square shape, we subtracted from the side of this square half of the semidiameter; the square of the line that Remains was combined again with the square of the semidiameter, and [the result] made into a square shape, and the side of this square would be the side of the Pentagon.

All this is possible and easier to do than to explain in words, as anyone knows who is used to handling compasses. For what is easier than to construct a right angle GAM, and to take on the lines enclosing it any length AM and double that length AG, and having placed one point of the compasses in M, and opened the compasses so that the other reaches to G, draw the circle GP, extend the line MA to P, then take the length GP with the compasses and transfer it into another circle^‘^^ whose diameter is GA?