Part 33-36


by Johannes Kepler Icon

Proposition 33

If from twice the number of angles of the figure you subtract 4, you will obtain the Numerator of the parts of a right angle which give you the angle of the figure

The Denominator of the parts is the number of angles itself.

So, for the Triangle, twice 3 is 6, subtract 4, which leaves 2.

Thus the angle of the Triangle is equal to two thirds of a Right angle.

Similarly, for the Icosigon, twice 20 is 40, subtract 4. Thus the angle of the Icosigon is 36 twentieths or 9 fifths of one Right angle. For the angles of each figure are distributed among a number of triangles which is equal to the number of sides of the figure, less two.

But the angles of any triangle add up to two Right angles: therefore the angles of any figure add up to twice as many Right angles as the figure has sides, less four. This number of Right angles is to be divided by the number of angles of the figure, therefore the former number is the de­ nominator and the latter the numerator of the parts of one Right angle [in each angle of the figure].

Proposition 34

A circle is cut by Geometrical description into 2 equal parts.

The line bisecting it is known by first degree knowledge: for it is itself the Diameter.

For the basis^^ for inscribing figures in a circle is to draw a straight line through a specified point, producing it as far as necessary.

A straight line bisecting the circle is a diameter, that is, drawn through the center, because the greatest of the equal parts into which a circle may be divided is a semicircle, so the line cutting it into two semicircles is the longest, by Euclid III.14, and so it is the diameter, by 15, and by definition.

Now the diameter is itself the Expressible line proposed as the Measure of others; it is equal to itself, and its own perfect Measure, the basis^’^ of Geometrical knowledge.

Proposition 35

The side of the Tetragon derives its Geometrical description [method of drawing] from its angles, if it is drawn independently of a circle {extra circulum)

If it is inscribed in a circle this description is of the third degree of kno wledge,the description of its square is of the second degree, and so is that of the area of the figure.

Let the Tetragon be OQPR, its angle, by XXXIII of this book, is a right angle, so by Euclid 1.46^^^ it is easy, given the side, to describe (draw) the Tetragon.

Since it has 4 angles it has the same number of sides; thus two sides that meef^^ cut off two quarters of the P Circle, that is half the circle. So by XXXIV of this book the end-points of contiguous sides lie on a diameter of the B 1 Xt circle.

As QO, QP, which form the right angle OQP in the semicircle OQP, have their end-points 0,P lying on the di­ O * ameter OLP of the circle. So by Eiwlid 1.47 the [sum of the] squares of the two sides OQ QP is equal to the square of the diameter.

If [an area equal to] a half of the square of the diameter is redrawn in the shape of a square, by Euclid 11.14 the side of this square will be the side of the Tetragon.

So the square of the side is Expressible. And because the ratio of the square OP to the square OQ is 2 to 1, not the ratio of a square number to a square Number; [and] OP is in fact Expressible in length: therefore the side OQ is Expressible only in square, by Euclid X.9.

The area of the tetragon is the same in this figure as the Square of the side, so the area of the Figure is also Expressible.

Proposition 36

The side of the Octagon has a Geometrical description from its Angles, as equally does the side of the star Octagon, being the chord subtending three eighths of a circle

These sides are of the 8th degree of knowledge, individually. The former is an Elasson, the latter a Mizon.

When combined, they [the sides] are of the sixth degree [of knowledge], and bear a particular proportion one to another.

In short, the area is inexpressible, in fact a Medial.

Let the octagon be UQTX)XRSP, and the star UOSQXPTRU^^^: therefore when a pair of lines, say QT, TO, containing the ti octagon angle QTO, have a line drawn through their other ends Q, O, the connecting line is the side of a Tetragon, because half of eight is four.

Therefore, after a Tetragon has been constructed in the circle (to leave out other ways of constructing the Octagon), let there he drawn from the center L a perpendicular to its side 0(^ to cut the side in M and the arc in T, by using [the construction given in] Euclid 7 . 7 2 , Then, by Euclid 111.30, the 2 parts of the quarter circle OQ namely the arcs OT, TQ, will be equal. Joining the points O and T will give the line OT as the side of the Octagon, and joining O, S will give OS as the side of the star [octagon].

Joining the center L with Qy because QML is a right angle, therefore QL [which is] Expressible in length, is, [when] squared, equal to the sum of the squares of (JM and ML.

Moreover, the semidiameter QL, squared, is equal to twice the square of QM, half the side of the square. Therefore QM and ML are equal, and each is Expressible only in square, by section XXXV above.^^^

Thus the square of LQ exceeds that of LM by the square of the line MQ which in length is incommensurable with the line LQ But LQ and LS, and LT are equal.

Therefore the compound line SM will be a fourth Binomial, whose Terms are SL and LM, by the definitions before Euclid X.48.’^^ The remainder M T will be a fourth Apotome, whose Terms are TL and LM, by the definition before Euclid X. 5 5 . 1'2

And because MS is a fourth binomial, and ST is expressible, therefore, by Euclid the line QS, which when squared is equal to the Rectangle contained by them,^^^ is a Mizon: thus because TM is a fourth Apotome, and TS is expressible: therefore TQ the side of the Octangle, which when squared is equal to the rectangle contained by MT, TS, is an Elasson, by Euclid X.94.

The elements of these lines are shown in the diagram as PA, the greater one, and AT, the smaller one. For adding AT to PA gives PT, the side of the star: and on the other hand taking TA away from either PA or YT leaves AY, that is QU the side of the Octagon. That is to say, the Elasson TQ squared, is twice the square of the Prosharmozusa TA; and the side of the Tetragon QP, squared, is equal to the sum of the squares of the elements PA and AQ that is AT.

The ratio of PX, a Mizon, to the greater of the Elements PA, is the same as the ratio of TQ an Elasson, to the smaller of the Elements TA, and in turn the ratio of the greater of the elements PA to the smaller AT is equal to the ratio of the Mizon PX to the Elasson TQ As the greater is to the smaller; so the sum is to the difference.

These sides SQ QT are not only Mizon and Elasson. They are also lines such that other such lines can be made from them by addition or subtraction.

For, first, they are incommensurable with one another, second, the sum of the squares of TQ QS is equal to the square of the expressible line TS.

Third, the Rectangle contained by TQ QS is a Medial for it is equal to the rectangle contained by QM, half the side of the Tetragon, expressible only in square, and by TS, expressible in length: from which [it follows] that they [the sides of the 2 figures are also of the sixth degree of knowledge when compounded.

Therefore by [Euclid] X.39 when they are compounded into one line TQS they make a Mizon,^^^ and by X.76 when TQ that is QZ, is subtracted from QS, the remainder, ZS, is an Elasson.^^^ Thus it can come about that an Elasson and a Mizon, belonging to one pair, become the Elements of another pair, and the Elasson, subtracted from its Mizon, leaves the Elasson of the other [pair].

As for the area of the Octagon, the figure is made up of eight triangles like LQT. But the rectangle QTRS is made up of four such [triangles]; therefore it has half the area: and it is a Medial, as just proved; therefore twice this area, that is the area of the Octagon, will also be a Medial, by a porism to Euclid

Hence CLAVIUS proves in his Geometria Practica, Book VIII, Proposition 31, that its area is a mean proportional between the area of the inscribed Tetragon and the area of the circumscribed Tetragon, which are in the ratio 1:2, and the [method of] determination of this definite quantity [i.e. that of the area of the octagon] implies that it has the same quality of being a MediaU

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