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arrived at this point, we shall infer that the last result expresses the surface of the circle, which, since it must always lie between the inscribed and the circumscribed polygon, and since those polygons agree as far as a certain place of decimals, must also agree with both as far as the same place.
We have subjoined the computation of those polygons, carried on till they agree as far as the seventh place of decimals.
The surface of the circle, we infer therefore, is equal to 3.1415926. Some doubt may exist perhaps about the last decimal figure, owing to errors proceeding from the parts omitted; but the calculation has been carried on with an additional figure, that the final result here given might be absolutely correct even to the last decimal place.
Since the surface of the circle is equal to half the circumference multiplied by the radius, the half circumference must be 3.1415926, when the radius is 1; or the whole circumferrence must be 3.1415926, when the diameter is 1: hence the ratio of the circumference to the diameter, formerly expressed by, is equal to 3.1415926.
PROPOSITION XV. LEMMA.
The triangle CAB is equivalent to the isosceles triangle DCE, which has the same angle C, and one of its equal sides CE or CD a mean proportional between CA and CB. And if the angle CAB is right, the perpendicular CF, drawn to the base of the isosceles triangle, will be a mean proportional between the side CA and half the sum of the sides CA, CB.
First. Because of the common angle C, the triangle ABC is to the isosceles triangle DCE as ACX CB is to DCXCE or DC2 (24. III.): hence those triangles will be equivalent, if DC2 ACX CB, or if DC is a mean proportional between AC and CB.
Secondly. Because the perpendicular CGF bisects the angle ACB, we shall have AG: GB:: AC: CB (17. III.); and therefore, componendo, AG: AG+GB or AB:: AC: AC+CB; but AG is to AB as the triangle ACG is to the triangle ACB, or 2 CDF;
besides if the angle A is right, the right-angled triangles ACG, CDF must be similar, and give ACG: ČDF:: AC2: CF2; hence,
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); 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.
PROPOSITION XVI. PROBLEM.
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 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 compress those limits.
Take CA and CE, each equal 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 CF, CA+CB a mean proportional between CA and > 2 ed 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.
will be inscrib
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.
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 between a and
a+b ; so that b'a.b, and a'=√a. : hence a and b the 2 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 become 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 proportionals 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 ciphers 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.
I. 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 given straight line. II. Isoperimetrical figures are such as have equal perimeters.
PROPOSITION I. THEOREM.
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 (15. II.). Pro- A duce 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 ADAN: and since
the oblique line AD is greater than the oblique line AN, it must be farther from the perpendicular AB; therefore DBBN; therefore BG, which (12. I.) 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 BGBP, the isosceles triangle ABC is greater than ABM, which is not isosceles, and has the same base and the same perimeter.