The Construction Of Regular Figures

The 16-sided figure (hekkaedecagon) has a Geometrical description from its angles, but knowledge of the side takes us far afield into degrees lower than all the preceding ones: and even more so for the sides of its stars, whether they subtend three, five, or 7/16 [of the circle].

Because 2 8s are 16, this 16-sided polygon can be described by working via the side of the Octagon and thus using the same principles that were employed before in deriving the side of the Octagon via the side of the Tetragon.

Let QO be the side not, as before, of the Tetragon, but of the Octagon. QT, TO now are the sides of the 16-angle.

Let QP be the side of the star Octagon: before, the first [i.e. QO] was a Mizon: therefore LM, which is half of it, was a Mizon.

So the rectangle contained by ST, which is Expressible, and LM, which is Mizon, is of a completely new kind, not mentioned among the degrees discussed above, which were of higher kinds.

A new area of this kind, subtracted from that contained by the lines LT, TS, which are Expressible in length, again leaves something of a more distant kind, namely the rectangle contained by MT, TS which is equal to the square of TQ the side of the 16-sided polygon (Hekkaedecagon).

This holds even more for the figures of this Class with more angles, such as 32, 64, 128, and so on.

Since this holds for one side, the chord subtended by 1/16 [of the circle], its square, subtracted from that of the diameter of the circle, leaves [the square of] the chord subtended by 7/16 of the circle, so the latter is of more distant degree.

The chord subtended by three sixteenths [of the circle] is de­ rived from that subtended by three eighths by bisection so the former is of a more distant degree [i.e. lower degree] than the latter. And the square of the chord subtended by three sixteenths, subtracted from the square of the diameter, leaves the square of the chord subtended by five sixteenths. So this again is of a more distant [i.e. lower] degree.

Proposition 38

The sides of the Trigon and the Hexagon have a geometrical description, from the angles of the figures; and when they are described in a circle, they are knowable, the former in the third, the latter in the second degree; the surfaces or areas of the figures are Medials, and are in the ratio T 2 .

The construction of a Trigon independent of the circle is very easy, by Euclid 127 Yhe most expeditious way of inscribing it in a circle (and I pass over other methods in silence) is to use the side of the Hexagon. Because half of six is three. And the description and inscription of the Hexagon are given in Euclid IV.13.

But it remains to show how the magnitude of the side follows, from the properties (rationibus) of the angles. Let the Hexagon be BHCGDE So since there are 6 angles, the surface of the Hexagon will be divided into six triangles, with their vertices meeting at the center A: one of which [triangles] is CAG.

Since there are four Right angles surrounding the center A, their sum, divided among the six vertices, gives for the single vertical angle CAG [the value] four sixths or two thirds of a Right angle. But all three angles of triangle CAG add up to two right angles, or six thirds of a Right angle; so subtracting the angle at A, 2 thirds, from the sum of 6 thirds, there remain, for the two angles at C and G, the sum of 4 thirds: now, all the angles are equal; so for each of the angles at C and at G there remain 2 thirds of a Right angle, no less than for the vertical angle at A. But if the three angles are equal, the 3 sides of the Triangle must be equal also. So CG, which is at the same time the side of the Hexagon and of the triangle that IS one sixth of it, is equal to the semidiameter of the circle, CA or AG.

Thus the side of the Hexagon is expressible in length, namely half the diameter of the circle.

So the former belongs to Degree II by XIII of this book.

Let us consider a triangle [inscribed in the circle] such as triangle BCD.

Its side BC links two sides of the hexagon CH, HB, which meet at H. Thus since BHC is two thirds of the semicircle and CG one third, therefore the arc BCG is a semicircle, and BG a diameter, passing through A. Therefore the angle BCG, the angle in it [sc. the semicircle[ is a Right angle, by Euclid III.31.

So [the sum of] the squares of BC and CG is equal to the square of BG, by Euclid 1.47.

But CG is a semidiameter, and its square is a quarter of that of the diameter; so subtracting a quarter from the square of the diameter BG gives as remainder the square of the side of the triangle BC. So this square is Expressible: but because its ratio to the square of BG is not equal to that of a square number to a square number, but is as 3 to 4, therefore BC is expressible only in square.

It therefore belongs to the Third degree, by section XIV above.

Because BC, BD are equal, and the angles BCD, BDC are equal; therefore BE, the perpendicular dropped [from B] to CD, will cut it in E into equal lines CE, ED. The complete line CD was Expressible only in square; so the same is true of half of it, CE.

Therefore the rectangle contained by CE, AG, lines commensurable only in square, the latter being Expressible in length, is a Medial.

But this Rectangle is equal to the [sum of the] areas of two triangles, each equal to the triangle CGA (one of the six triangles that make up the Hexagon), and is thus one third of the Area of the Hexagon.

So the area of the Hexagon is a Medial surface. And because the triangles BCA and BCH have sides BA and BH, CA and CH equal, and one common side BC: they therefore have equal areas.

But BCH, BDF, CDG are parts of the Hexagonal area, the parts by which it exceeds the Triangular area BCD, which is equal to the sum of the triangles BAC, CAD, DAB.

Therefore the Hexagonal area is twice the Triangular area. Thus the Triangular area is also Medial, because it is commensurable with, that is twice, the Hexagonal [area], which was Medial.

Proposition 39

The sides of the Dodecagon and of the star of the same name, namely the chords subtended by five twelfths of the circle, can be described Geometrically, and when they are inscribed in the same circle they [sc. the sides] are individually knowable in the eighth degree of nobility of knowledge (nobilioris cognitionis), taken together [they are knowable] in the fifth degree; in fact the surface of the Dodecagon is Expressible.*

Let the dodecagon be BMHLCKGQDPFN and the star Dodecagon BKFLDMGNCPHQB.

So, because twice six are twelve, these figures can be described by working via the side of the Hexagon, using the same principles as were employed before in deriving the side of the Octagon via the side of the Tetragon,^^’^ [namely by] drawing from A, the center of the circle, a line perpendicular to HC, the side of the hexagon, to cut the side in O and the circle in L and P, and joining the points L, Hfor the side of the Dodecagon, and the points H, Pfor the side of the star.

So since HC, the side of the hexagon (sexanguli) is Expressible in length; so too will be its half HO, but AC, equul to HC itself, is, when squared, equal to the square of its half OC plus the square of AO,^^^ therefore the ratio of the suare of AO to the square of AC or AP is that of 3 to 4, not the ratio of a square number to a square numberA^^

Therefore PA, AO are commensurable with one another only in square, as are LA and AO.

CA, that is PA or AL, the greater, which is Expressible is, in square, greater than the square of the smaller, AO, by [the square of] a magnitude which is commensurable with CO itself Therefore, by the definitions before Euclid the compound line PO is a Binomial, and by the definition before [Euclid X.]83, the remainder OL is an Apotome, each designated First

The Terms are AP, Expressible in length, and AO, Expressible only in square.

But, by Euclid X.54,^"^^ LLP, in square equal to the rectangle contained by OP, a first Binomial, and PL, which is Expressible, is a Binomial, and by 91 of the same, the side of the Dodecagon, in square equal to the rectangle contained by OL, a first Apotome and LP, which is expressible, is an Apotome.

Thus the sides belong in the 8th degree of nobility of knowledge.

The Terms of this compound line PH, and of the diminished line HL, are PS and And since HB is the side of the Hexagon (sexangulij, KP of the Triangle, [and] BP of the tetragon fquadranguli), the square of the first is equal to twice the square of the smaller Term, that is it is equal to the square of HS plus [that of] SB, the square of the second is equal to twice the square of the greater [Term], that is it is equal to the square of KS plus [that of] SP;

while the square of the last is equal to the sum of the square of both [Terms], each taken once, that is [the squares] of BS and SP.^^"^ PH is also a binomial composed of PR, the side of the square, and RH, the side of the Dodecagon; but it is not called a Binomial on account of this composition; because, by Euclid there is only one point, here the point S, which can divide it into its Terms.

Since HO, LP are Expressible in length, the rectangle they contain, that is the rectangle contained by LH, HP will be Expressible,and the sum of the squares of LH and HP is similarly Expressible, being indeed equal to the square of LP itself.

Therefore on this basis LH and HP taken together are in the fifth degree of knowledge. Neither do they produce anything new when they are combined, nor do they produce a Binomial or an Apotome again; for adding LH and HP gives a line Expressible only in square, that is the line whose square is three halves of the square of LP: while subtracting LH, or HR, from HP again produces a line Expressible in square, [namely] PR, the side of a square, whose square is half the square of LP.'

And since the Area of the Dodecagon is made up of 12 Triangles, one of which is LAC, four of these [the triangles] might be contained in the Expressible rectangle LHPD, that is [it has] a Third of the total Area, therefore the total Area is also Expressible, namely of quantity equal to three times the product of HO and LP; so the Area is Three Quarters of the square of the diameter, or the Arithmetic Mean between the Tetragon circumscribing the circle and the Tetragon inscribed in the circle; just as the Area of the Octagon (Octanguli) is the Geometric Mean between them.