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therefore, if the angle A is a right angle, the perpendicular CF is a mean proportional between the side AC and half the sum of the sides AC, CB.

PROBLEM.

296. To find a circle which shall differ as little as we please from a given regular polygon.

Solution. Let there be given, for example, the square BMNP Fig. 171. (fig. 171); let fall from the centre C the perpendicular CA upon the side MB, and join CB.

The circle described upon the radius CA is inscribed in the square, and the circle described upon the radius CB is circumscribed about this square; the first will be less than the square, and the second will be greater; it is proposed to reduce these limits.

Take CD and CE, each equal to a mean proportional between CA and CB, and join ED; the isosceles triangle CDE will be equivalent to the triangle CAB (295); let the same be done with respect to each of the eight triangles which compose the square, and there will be formed a regular octagon equivalent to the square BMNP. The circle described upon CF, a mean proportional between CA and CA CB, will be inscribed in the octa

+

2

gon; and the circle described upon CD, as a radius, will be circumscribed about it. Thus the first will be less, and the second greater, than the given square.

If we change, in the same manner, the right-angled triangle CDF into an equivalent isosceles triangle, we shall form in this way a regular polygon of sixteen sides equivalent to the proposed square. The circle inscribed in this polygon will be less than the square, and the circle circumscribed about it will be greater.

We can proceed in this manner till the ratio between the radius of the inscribed circle and that of the circumscribed circle shall differ as little as we please from equality. Then either of these circles may be regarded as equivalent to the proposed

square.

297. Scholium. To exhibit the result of this investigation of the successive radii, let a be the radius of the circle inscribed in one of the polygons, and b the radius of the circle circumscribed about the same polygon; and let a', b', be similar radii to the

next polygon of double the number of sides. According to what has been demonstrated, b' is a mean proportional between a and b, and a' is a mean proportional between a and +; so that we have

b' = √a xb, and a' =

a+b

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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 = √1 x 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.

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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 deci mals (Alg. 102). The operation is thus greatly abridged, and the results are,

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2.

radius of a circle equal From this it is easy to

in surface to a square whose side is
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; there-
fore, if we divide the surface 4 by the square of 1,1283792, we
shall have the value of equal to 3,1415926, &c., as determined
by the other method.

Fig. 172.

Appendix to the Fourth Section.

DEFINITIONS.

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.

THEOREM.

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

AM+MBAC + CB;

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 CB = CD, and MN = MB, AC+ CB AD, and AM+MB= AM + MV. But

=

AC+ CB AM + MB;

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.

THEOREM.

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.

PROBLEM.

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 Fig. 174. AC=AD; 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

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

Fig. 175.

THEOREM.

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 differFig. 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 ACBADB. For ADO=ACD+ CAD (63); therefore ACDADO, and, each being doubled, we have ACB <ADB.

THEOREM.

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

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