Squaring lunes; and a maxim on natural wit

Two concentric circles, a sector and portions, with:

d e c – b a. These are double circles, and the irrational portion a b is one half of the irrational portion c d. Therefore half of the larger portion, which is c e, will be equal to the whole portion a b — which was our intent. But to return to the experiment above, with the help of this we shall find how to remove a portion, or half a portion, of c equal to the portion made from the arc of a above; and then we shall subtract the portion made from the arc d from the remainder of the arc c with its portion, if it is capable; and if the arc of c were not capable, we shall subtract the arc of c from the arc of d.

Left column, lower part.

To give an irrational portion of a half-sized circle equal to a half irrational portion of a circle double the first. And I shall do thus:

Left column.

Here it is proved that c is equal to d, and consequently a remains equal to b. And we shall say thus: having set the lune c b equal to the sector a d, and likewise the half-sector c a equal to the whole sector a d, I shall remove the part a from the half-sector c a, and also remove the same a from the portion a d; there will remain c equal to d, as I had promised to prove to you. Therefore the remainder of the one and the other surface remain of equal value between them, that is a b, which was our proposition. In the second demonstration three surfaces are given equal in magnitude or value, that is the portion a b, the lune b c d, and the half-portion b c e. I remove the part b from each of them; there remains a equal to c d, and there remains a equal to the rectilinear c e. I shall put the rectilinear c e back into the base of the lune a and remove the triangle c from the one and the other surface, and there will remain a o equal to e; and I shall lower the portion o e and join it with a, and I shall have made a surface of three sides, of which two are convex of equal curvature and the third side is concave, of a curvature of a circle double that of the two convex curvatures, which concave curvature is worth the two convex ones.

Upper left corner.

Always the little lune of two double portions is in itself squarable by exchanging its parts; and squarable is the remainder of the lune.

Left half. Centre of the upper margin:

He who is foolish by nature and wise by accident, when he speaks or acts naturally, always seems foolish, and seems wise in what is accidental.

At the bottom, a circumscribed octagon, with:

b a e – c. Square the triangle a b c and double the squaring, and you will have the triangle e a b c — or you may call it a quadrangle — which contains the eighth part of the circle and has in itself the angle of the octagon, which is given similar to the square d; and the remainder is a pyramid, or rectilinear triangle, equal to the sector that contains the eighth part of a circle. And what you do with the eighths of circles you can do with the sixths of those circles; because if the beginning of this operation is made with the quarters of circles double one to the other, you will do this second operation with thirds of circles double one to the other; and if that of the quarters of divided circles gives me eighths, this second will give me sixths.

Little column at the centre.

To every lune squarable in itself, a square will be given equal to that lune, outside of it. It is proved: in the lune a the arc will be divided into two equal parts, and to these the chords will be drawn, and two portions will be made equal to the portion b of double arc; and likewise in the lune c d the lower arc will be divided into two equal parts, and two portions will be made equal to the portion of the upper arc, which is of an arc double the lower arc; and so will all of them do. And by this a squaring is given outside the squaring that is generated in them in themselves. No square will be given equal to any surface of a single curved side, whether it be convex or concave.

Two figures, with:

o – f c – m n – r d; S – c o d. Since above c d is equal, I shall lend f to c, opening it into c o, and likewise I shall lend as much to d, and they will remain equal. Then I shall place one over the other, as you see here, and the two portions n m will be removed from the one surface and the other, and the remainder of d will remain a square equal to the remainder of c, with four sides, of which three are curved — one convex which is worth the two curved ones.

Three similar figures, with:

o – e c a; o a; e. Here one may lend to the triangle and to the lune a rectilinear part.

Beneath the larger figure and along the central fold:

Here the arc of c is worth the arc of a d, and if I subtract the arc of d from that of c, the remainder of c will be equal to the arc of a. But because the arc of d includes a portion larger than is fitting — perhaps as much larger as a takes less than is fitting to make a portion of its whole arc (which I do not believe, but experience will show us not to believe it) — and this one below {there follows a figure of two triangles in a semicircle, with n – d – c a}, since the portions a b and b c are equal and similar, I then make two others in a d and d c, which are neither similar nor equal; and if you take the value of the two portions a b and b c, and take the value of the portions a d and d c, you will find the second portions to be worth much more than the first. Therefore my doubt is resolved, and for this I shall return to the first proposition.

On the right, triangles within a portion:

To give a portion of a smaller circle equal to the portion of a larger circle.

Lune and overlaid portions, with:

a – d b – c – e.

Two lunes, a triangle, with:

a; d c; c e.

At the bottom, a circumscribed octagon, with:

b a e – c. Square the triangle a b c and double the squaring, and you will have the triangle e a b c — or you may call it a quadrangle — which contains the eighth part of the circle and has in itself the angle of the octagon, which is given similar to the square d; and the remainder is a pyramid, or rectilinear triangle, equal to the sector that contains the eighth part of a circle. And what you do with the eighths of circles you can do with the sixths of those circles; because if the beginning of this operation is made with the quarters of circles double one to the other, you will do this second operation with thirds of circles double one to the other; and if that of the quarters of divided circles gives me eighths, this second will give me sixths.

Double figure with: d n

c o. Since I cannot subtract the arc of c from the chord of d — the chord of d being shorter than that same arc of c — I shall lend to d the triangle n, and lend as much to c with the triangle o, and these surfaces likewise remain equal, and the chord of n d is capable of receiving within itself the arc of o c.

(d K n); (c h o); (o b); (L); (o); (2); (4); m d; L o; m; d; L; o. Once such surfaces are made equal, I place them one upon the other and remove the superimposed part, which is the portion n, which is double because the parts of these two surfaces are superimposed; and by this equal diminution the remainders are equal, and there remains the lune d equal to the c below. This done, I wish to square the lune in itself by taking the upper arc and rendering the lower value in n. But because the upper arc of the lune is smaller than the lower — which is 1/8 of the arc of a circle double the outer circle — the inner arc would need 1/4 of the circle half the outer circle, and thus the values would be equal; and for this I shall lend to the outer arc as much arc as will suffice to bring it to the sum of the quarter of a circle, and I shall render as much value of arc to the surface c, and the two surfaces will be equal — that is, in the lune d there will be the addition m, and in the other surface the addition o, equal to m; and this equality will be made by the fifth of the first, set opposite in the opposite face, in the two circles that are one within the other. Besides this, I shall square the lune m d, equal to c o, and I shall square it in one of the two ways figured below, by filling the concave arc and casting away as much value of the convex arc; and so I shall have made the value of the lune m d, which I shall reduce in part. There remains to us, then, the surface c o, equal to the squared d n. I shall give an angle of the octagonal surface to the arc o c and place it below in o L, and I shall take the squarable part of that L o, which will be L, and take it from the square m d placed beside it, which is the squaring of the lune equal to L o. Behold, then, L o separated into two parts, which I place below, o separated from L; which L, being squarable, I subtract from the lune reduced into the square d m. Therefore I shall divide that square into two parts, of which one is m and the other d; and because m is equal to the squarable L, d remains equal to the unsquarable o.

On the right margin, with reference to the figures distinguished by:

x y. This equality gives a fallacy, because it is difficult to give it, unless x y were already to the purpose, drawing it away then at the end. And by x y is understood each to be composed of a triangle with one convex side and a portion, that is {there follow four little figures of triangles and portions}.

Upper right corner.

Double figure with:

d n; c o. Since I cannot subtract the arc of c from the chord of d — the chord of d being shorter than that same arc of c — I shall lend to d the triangle n, and lend as much to c with the triangle o, and these surfaces likewise remain equal, and the chord of n d is capable of receiving within itself the arc of o c.

The same figures overlaid, with:

d – n – c.

Decomposition and separation of the preceding, with:

d m; x; c – o; y.

Large double portion with:

b c – d a.

Due cerchi concentrici, settore e porzioni, con:

Colonna sinistra, parte inferiore.

Colonna sinistra.

Angolo superiore sinistro.

Metà sinistra. Centro del margine superiore:

In basso, ottagono circoscritto, con:

Colonnina al centro.

Due figure, con:

Tre simili figure, con:

Sotto la maggior figura e lungo la piega mediana:

A destra, triangoli entro porzione:

Falcata e porzioni sovrapposte, con:

Due falcate, un triangolo, con:

In basso, ottagono circoscritto, con:

Doppia figura con: d n

x y Questa equalità dà fallacia, perché difficile a darla, se già x y non fussi al proposito, traendola poi via del fine. E esso x y s'intende ciascuna essere composta d'un triangolo d'un lato convesso ed una porzione, cioè (seguono quattro figurine di triangoli e porzioni).

Doppia figura con:

Le stesse figure sovrapposte, con:

Scomposizione e separazione della precedente, con:

Grande doppia porzione con:

Two concentric circles, a sector and portions, with:

Left column, lower part.

Left column.

Upper left corner.

Left half. Centre of the upper margin:

At the bottom, a circumscribed octagon, with:

Little column at the centre.

Two figures, with:

Three similar figures, with:

Beneath the larger figure and along the central fold:

On the right, triangles within a portion:

Lune and overlaid portions, with:

Two lunes, a triangle, with:

At the bottom, a circumscribed octagon, with:

Double figure with: d n

Double figure with:

The same figures overlaid, with:

Decomposition and separation of the preceding, with:

Large double portion with:

On this page

A square outside every self-squarable lune

Squaring a sector through a circumscribed octagon

Irrational portions of double and half circles

Maxim: foolish by nature, wise by accident