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next polygon of double the number of sides. According to what has been demonstrated, b' is a mean proportional between a and b, and d' is a mean proportional between a and "+6

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; so that we have

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hence the radii a and b of one polygon being known, the radii a', b', of the following polygon are easily deduced; and we may proceed in this manner till the difference between the two radii shall become insensible; then either of these radii may be taken for the radius of a circle equivalent to the proposed square or polygon.

This method may be readily applied to lines, since it consists in finding successive mean proportionals between known lines; but it succeeds still better by means of numbers, and it is one of the most convenient, that elementary geometry can furnish, for finding expeditiously the approximate ratio of the circumference of a circle to its diameter. Let the side of the square be equal to 2, the first inscribed radius CA will be 1, and the first circumscribed radius CB will be/2 or 1,4142136. Putting then a = = 1, and b = 1,4142136, we shall have

b = √ax b=1X 1,4142136 1,1892071;

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These numbers may be used in calculating the succeeding ones according to the law of continuation.

See the result of this calculation extended to seven or eight figures by means of a table of common logarithms.

Radii of the circumscribed circles. Radii of the inscribed circles.

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The first half of the figures being now the same in both, we can take the arithmetical instead of the geometrical means, since they do not differ from each other except in the remoter decimals (Alg. 102). The operation is thus greatly abridged, and the results are,

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Hence 1,1283792 is very nearly the

radius of a circle equal From this it is easy to

in surface to a square whose side is 2.
find the ratio of the circumference of a circle to its diameter; for
it has been demonstrated that the surface of a circle is equal to
the square of the radius multiplied by the number; therefore
if we divide the surface 4 by the square of 1,1283792, we shall
have the value of л equal to 3,1415926 &c., as determined by
the other method.

Fig. 172.

Appendix to the Fourth Section.


298. AMONG quantities of the same kind that which is greatest is called a maximum; and that which is smallest a minimum.

Thus the diameter of a circle is a maximum among all the straight lines drawn from one point of the circumference to another, and a perpendicular is a minimum among all the straight lines drawn from a given point to a given straight line.

299. Those figures which have equal perimeters are called isoperimetrical figures.


300. Among triangles of the same base and the same perimeter that is a maximum in which the two undetermined sides are equal. Demonstration. Let AC CB (fig. 172), and


the isosceles triangle ACB will be greater than the triangle AMB of the same base and the same perimeter.


From the point C, as a centre, and with the radius CA CB, describe a circle meeting CA produced in D; join DB; and the angle DBA, inscribed in a semicircle is a right angle (128). Produce the perpendicular DB towards N, and make MN = MB, and join AN. From the points M and C let fall upon DN the


perpendiculars MP and CG. Since CBCD, and MN MB, AC+CBAD, and AM+MBAM + MN. But


consequently AD=AM+MN; therefore AD> AN.

Now, if the oblique line AD is greater than the oblique line AN, it must be at a greater distance from the perpendicular AB (52); hence DB > BN, and BG the half DB is greater than BP the half BN. But the triangles ABC, ABM, which have the same base AB, are to each other as their altitudes BG, BP; therefore, since BG > BP, the isosceles triangle ABC is greater than the triangle ABM of the same base and same perimeter which is not isosceles.


301. Among polygons of the same perimeter and of the same number of sides, that is a maximum which has its sides equal.

Demonstration. Let ABCDEF (fig. 173) be the maximum Fig. 173. polygon; if the side BC is not equal to CD, make, upon the base BD, an isosceles triangle BOD, having the same perimeter as BCD, the triangle BOD will be greater than BCD (300), and consequently the polygon ABODEF will be greater than ABCDEF; this last then will not be a maximum among all those of the same perimeter and the same number of sides, which is contrary to the supposition. Hence BC must be equal to CD; and, for the same reason, we shall have CD=DE, DE=EF, &c.; therefore all the sides of the maximum polygon are equal to each other.


302. Of all triangles formed with two given sides making any angle at pleasure with each other, the maximum is that in which the two given sides make a right angle.


Demonstration. Let there be the two triangles BAC, BAD (fig. 174), which have the side AB common, and the side ACAD; if the angle BAC is a right angle, the triangle BAC will be greater than the triangle BAD, in which the angle A is acute or obtuse.

For, the base AB being the same, the two triangles BAC, BAD, are as their altitudes AC, DE. But the perpendicular

Fig. 174.

Fig. 175.

DE is less than the oblique line AD or its equal AC; therefore the triangle BAD is less than BAC.


303. Of all polygons formed of given sides and one side to be taken of any magnitude at pleasure, the maximum must be such that all the angles may be inscribed in a semicircle of which the unknown side shall be the diameter.

Demonstration. Let ABCDEF (fig. 175), be the greatest of the polygons formed of the given sides AB, BC, CD, DE, EF, and the side AF taken at pleasure; draw the diagonals AD, DF. If the angle ADF is not a right angle, we can, by preserving the parts ABCD, DEF, as they are, augment the triangle ADF, and consequently the entire polygon by making the angle ADF a right angle, according to the preceding proposition; but this polygon can no longer be augmented, since it is supposed to have attained its maximum; therefore the angle ADF is already a right angle. The same may be said of the angles ABF, ACF, AEF; hence all the angles A, B, C, D, E, F, of the maximum polygon are inscribed in a semicircle of which the undetermined side AF is the diameter.

304. Scholium. This proposition gives rise to a question, namely, whether there are several ways of forming a polygon with given sides and one unknown side, the unknown side being the diameter of the semicircle in which the other sides are inscribed. Before deciding this question it is proper to observe that, if the same chord AB subtends arcs described upon differ Fig. 176. ent radii AC, AD (fig. 176), the angle at the centre subtended by this chord will be least in the circle of the greatest radius ; thus ACB <ADB. For ADO= ACD + CAD (78); therefore ACD < ADO, and, each being doubled, we have ACB<ADB.


305. There is but one way of forming a polygon ABCDEF, Fig. 175. (fig. 175) with given sides and one side unknown, the unknown side being the diameter of the semicircle in which the others are inscribed.

Demonstration. Let us suppose that we have found a circle which satisfies the question; if we take a greater circle, the chords AB, BC, CD, &c., answer to angles at the centre that are

smaller. The sum of the angles at the centre will accordingly be less than two right angles; thus the extremities of the given sides will not terminate in the extremities of a diameter. The contrary will occur if we take a smaller circle; therefore the polygon under consideration can be inscribed in only one circle.

306. Scholium. We can change at pleasure the order of the sides AB, BC, CD, &c., and the diameter of the circumscribed circle will always be the same as well as the surface of the polygon; for, whatever be the order of the arcs AB, BC, CD, &c., it is sufficient that their sum makes a semicircumference, and the polygon will always have the same surface, since it will be equal to the semicircle minus the segments AB, BC, CD, &c., the sum of which is always the same.


307. Of all polygons formed of given sides the maximum is that which can be inscribed in a circle.

Demonstration. Let ABCDEFG (fig. 177) be an inscribed Fig. 177. polygon, and abcdefg one that does not admit of being inscribed, having its corresponding sides equal to those of the former, namely, a b = AB, bc= BC, cd = CD, &c.; the inscribed polygon will be greater than the other.

Draw the diameter EM, and join AM, MB; upon ab= AB construct the triangle a b m equal to ABM, and join e m.

According to what has just been demonstrated (303), the polygon EFGAM is greater than efg am, unless this last can also be inscribed in a semicircle having em for its diameter, in which case the two polygons would be equal (305). For the same reason the polygon EDCBM is greater than edc bm, with the exception of the case in which they are equal. Hence the entire polygon EFGAMBCDE is greater than efg ambcde, unless they should be in all respects equal; but they are not so (161), since one is inscribed in a circle, and the other does not admit of being inscribed; therefore the inscribed polygon is greater than the other. Taking from them respectively the equal triangles ABM, abm, we have the inscribed polygon ABCDEFG greater than the polygon not inscribed abcdefg.

308. Scholium. It may be shown, as in art. 305, that there is only one circle and consequently only one maximum polygon

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