295.

LEMMA.

A triangle is equivalent to an ososceles triangle, when one of its angles is equal to the verticle angle of the ososceles triangle, and the product of the sides containing this angle equal to the square of one of the equal sides of the isosceles triangle. And if the third side of the first triangle, is perpendicular to either of the other sides, then the perpendicular let fall from the vertex on the base of the isosceles triangle, will be a mean proportional between the less of these other sides, and half the ir sum.

First. Because of the common angle C, the triangle ABC is to the isosceles triangle DCE,as ACX CB is to DCX CE or DC2 (216.): hence those triangles will be equivalent, if DC2=ACXCB, or if DC is a mean proportional between AC and CB.

D

Secondly. Because the perpendicular CGF bisects the angle ACB, we shall have AG: GB:: AC: CB (201.); and therefore, by composition, AG: AG+GB or AB :: AC : AC+CB; but AG is to AB as the triangle ACG is to the triangle ACB, or 2CDF; besides if the angle A is right, the right-angled triangles ACG, CDF must be similar, and give ACG: CDF:: AC: CF2; or ACG: 2CDF:: AC2:2CF2; therefore, AC2: 2CF2 :: AC: AC+CB.

Multiply the second pair by AC; the antecedents will be equal, and consequently we shall have 2CF2=AC. (AC+CB), or CF2=AC.

AC+CB
2

+CB)

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

PROBLEM.

296. To find a circle differing as little as we please from a given regular polygon.

Let the square BMNP be the proposed polygon. From the M centre C, draw CA perpendicular to MB, and join CB.

The circle described with the radius CA is inscribed in the square, and the circle described with the radius CB circumscribes this same square; the first will in consequence be less than it, the second greater: it is now required to reduce those limits.

Take CD and CE, each equal N

D

F

B

A

E

P

to the mean proportional between CA and CB, and join ED; the isosceles triangle CDE will, by the last proposition, be equivalent to the triangle CAB. Perform the same operation on each of the eight triangles which compose the square you will thus form a regular octagon equivalent to the square BMNP. The circle described with the radius CA+CB CF, a mean proportional between CA and

2

will be

inscribed in this octagon, and the circle whose radius is CD will circumscribe it. The first of them will therefore be less than the given square, the second greater.

If the right-angled triangle CDF be, in like manner, changed into an equivalent isosceles triangle, we shall by this means form a regular polygon of 16 sides, equivalent to the proposed square. The circle inscribed in this polygon will be less than the square; the circumscribed circle will be greater.

The same process may be continued, till the ratio between the radius of the inscribed and that of the circumscribed circle, approach as near to equality as we please. In that case, both circles may be regarded as equivalent to the square.

297. Scholium. The investigation of the successive radii is reduced to this. Let a be the radius of the circle inscribed

in one of the polygons, b the radius of the circle circumscribing the same polygon; let a and b' be the corresponding radii for the next polygon, which is to have twice the number of sides. From what has been demonstrated, b' is a mean proportional between a and b, and a is a mean proportional a+b

a+b
2

between a and ; so that b√a.b, and a'=' a.

2

hence a and b the radii of one polygon being known, we may easily discover the radii a' and b' of the next polygon; and the process may be continued till the difference between the two radii becomes insensible; then either of those radii will be the radius of the circle equivalent to the proposed square or polygon.

This method is easily practised with regard to lines; for it implies nothing but the finding of successive mean propor tionals between lines which are given it is still more easily practised with regard to numbers, and forms one of the most commodious plans which elementary geometry can furnish, for discovering speedily the approximate ratio of the circumference to the diameter. Let the side of the square be 2; the first inscribed radius CA will be 1, and the first circumscribed radius CB will be 2 or 1.4142136. Hence, putting a=1, b=1.4142136, we shall find b=1.1892071, and a'=1.0986841. These numbers will serve for computing the rest, the law of their combination being known.

Since the first half of these figures is now become the same on both sides, it will occasion little error to assume the arithmetical means instead of the mean proportionals or geometrical means, which differ from the former only in their last figures. By this method, the operation is greatly abridged; the results are:

Thus 1.1283792 is very nearly the radius of a circle equal in surface to the square whose side is 2. From this, it is easy to find the ratio of the circumference to the diameter: for it has already been shewn that the surface of the circle is equal to the square of its radius multiplied by the number ; hence if the surface 4 be divided by the square of 1.1283792 the radius, we shall get the value of , which by this computation is found to be 3.1415926, &c., as was formerly determined by another method.

APPENDIX TO BOOK IV.

Definitions.

298. A maximum is the greatest among all the quantities of the same species; a minimum is the least.

Thus the diameter of a circle is a maximum among all the lines joining two points in the circumference; the perpendicular is a minimum among all the lines drawn from a given point to a straight line.

299. Isoperimetrical figures are such as have equal peri

meters.

THEOREM.

300. Of all the triangles having the same base and the same perimeter, the maximum is that triangle of which the two undetermined sides are equal.

Suppose AC=CB, and AM+MB =AC+CB; then is the isosceles triangle ACB greater than the triangle AMB, which has the same base and the same perimeter.

From the centre C, with a radius CA=CB, describe a circle meeting CA produced in D; join DB; the angle DBA, inscribed in a semicircle, will be right (128.). Produce the perpendicular DB towards N, make MN=MB, and join AN. Lastly, from the points M and C, draw MP and CG perpendicular to DN.

Since CB-CD, and MN=MB, we have AC+CB=AD, and AM+MB=AM+MN. But AC+CB=AM+MB ; therefore AD=AM+MN; therefore AD7AN: and since the oblique line AD is greater than the oblique line AN, it must be further from the perpendicular AB (52.); therefore DB7 BN; therefore BG, which is half of BD, will be greater than BP which is half of BN. But the triangles ABC, ABM, having the same base AB, are to each other as their altitudes BG, BP; therefore, since BG7BP, the isosceles triangle ABC is greater than ABM, which is not isosceles, and has the same base and the same perimeter.

THEOREM.

301. Of all the isoperimetrical polygons having a given number of sides, the maximum is the one which has its sides equal.

B

For, let ABCDEF be the maximum polygon. If the side BC is not equal to CD, construct upon the base BD an F isosceles triangle BOD, which shall be isoperimetrical with BCD; this triangle BOD will (300.) be greater E than BCD, and consequently the polygon ABODEF will be greater than ABCDEF; hence the latter is not the maximum of all the polygons having the same perimeter and the same number of sides, which contradicts the hypothesis. BC must therefore be equal to CD: for the same reasons must CD=DE, DE= EF, &c.; hence all the sides of the maximum polygon are equal.

D