fell above MN, the sum A'B + A'C would be still greater. There fore the altitude of the triangle ABC is greater than that of A'BC, and hence also its area is the greater.

54. Corollary. Of all isoperimetric triangles, that which is equilat eral is the maximum. For, the maximum triangle having a given perimeter must be isosceles whichever side is taken as the base.

PROPOSITION XX.-THEOREM.

55. Of all triangles formed with the same two given sides, that in which these sides are perpendicular to each other is the maximum. Let ABC, A'BC, be two triangles having the

sides AB, BC, respectively equal to A'B, BC; tnen, if the angle ABC is a right angle, the area of the triangle ABC is greater than that of the triangle A'BC.

Α

B

D

For, taking BC as the common base, the altitude AB of the triangle ABC is evidently greater than the altitude A'D of the triangle A'BC.

PROPOSITION XXI.-THEOREM.

56. Of all isoperimetric plane figures, the circle is the maximum. 1st. With a given perimeter, there may be constructed an infinite number of figures of different forms and various areas. The area may be made as small as we please (IV. 35), but obviously cannot be increased indefinitely. Therefore, among all the figures of the same perimeter there must be one maximum figure, or several maximum figures of different forms and equal areas.

2d. Every closed figure of given perimeter containing a maximum area must necessarily be convex, that is, such that any straight lire joining two points of the perimeter lies wholly within the figure.

Let ACBNA be a non-convex figure, the

straight line AB, joining two of the points in its perimeter, lying without the figure; then, if the re-entrant portion ACB be revolved about the line AB into the position AC'B, the figure AC'BNA has the same perimeter as the first

N

A

figure, but a greater area. Therefore, the non-convex figure cannot be a maximum among figures of equal perimeters.

3d. Now let ACBFA be a maximum figure formed with a given

FI

E'

B

E

perimeter; then we say that, taking any point A in its perimeter and drawing AB so as to divide the perimeter into two equal parts, this line also divides the area of the figure into two equal parts. For, if the area of one of the parts, as AFB, were greater than that of the other part, ACB, then, if the part AFB were revolved upon the line AB into the position AF'B, the area of the figure AF'BFA would be greater than that of the figure ACBFA, and yet would have the same perimeter; thus the figure ACBFA would not be a maximum.

F

G

Hence also it appears that, ACBFA being a maximum figure, AF'BFA is also one of the maximum figures, for it has the same perimeter and area as the former figure. This latter figure is symmetrical with respect to the line AB, since by the nature of the revolution about AB, every line FF' perpendicular to AB, and terminated by the perimeter, is bisected by AB (I. 140). Hence Fand F ́ being any two symmetrical points in the perimeter of this figure, the triangles AFB and AF'B are equal.

Now the angles AFB and AF'B must be right angles; for if they were not right angles the areas of the triangles AFB and AF'B could be increased without varying the lengths of the chords AF, FB, AF', F'B (55), and then (the segments AGF, FEB, AG'F', F'E'B, still standing on these chords), the whole figure would have its area increased without changing the length of its perimeter; consequently the figure AF'BFA would not be a maximum. Therefore, the angles Fand F' are right angles. But Fis any point in the curve AFB; therefore, this curve is a semi-circumference (II. 59, 97).

Hence, if a figure ACBFA of a given perimeter is a maximum, its half AFB, formed by drawing AB from any arbitrarily chosen point A in the perimeter, is a semicircle. Therefore the whole figure is a circle.*

*This demonstration is due to STEINER, Crelle's Journal für die reine und angewandle Mathematik, vol. 24. (Berlin, 1842.)

mar 28

PROPOSITION XXII.-THEOREM.

57. Of all plane figures containing the same area, the circle has the minimum perimeter.

Let C be a circle, and A any other figure having the same area as C; then, the perimeter of C is less than that of A.

For, let B be a circle having the same perimeter as the figure A; then, by the preceding theorems AB, or CB. Now, of two circles, that which has the less

area has the less perimeter; there-

B

A

fore, the perimeter of C is less than that of B, or less than that of A.

PROPOSITION XXIII.-THEOREM.

58. Of all the polygons constructed with the same given sides, that is the maximum which can be inscribed in a circle.

S

Let P be a polygon constructed with the sides a, b, c, d, e, and inscribed in a circle S, and let P' be any other polygon constructed with the same sides and not inscriptible in a circle; then, P> P'.

e

a

S'

a

P!

P

d

d

For, upon the sides a, b, c, etc., of the polygon P' construct circular segments equal to those standing on the corresponding sides of P. formed has the same perimeter as the circle S; therefore, area of S> area of S' (56); subtracting the circular segments from both, we have P> P'.

The whole figure S' thus

PROPOSITION XXIV.-PROBLEM.

59. Of all isoperimetric polygons having the same number of sides, the regular polygon is the maximum.

B

B'

C

P

1st. The maximum polygon P, of all the isope-' rimetric polygons of the same number of sides must have its sides equal; for if two of its sides, as AB', B'C, were unequal, we could, by (53), substitute for the triangle AB'C the isosceles triangle ABC having the same perimeter as AB'C' and a greater area, and thus the area of the whole polygon could be increased with. out changing the length of its perimeter or the number of its sides.

2d. The maximum polygon constructed with the same number of equal sides must, by (58), be inscriptible in a circle; therefore it must be a regular polygon.

PROPOSITION XXV.-THEOREM.

60. Of all polygons having the same number of sides and the same area, the regular polygon has the minimum perimeter.

Let P be a regular polygon, and M any irregular polygon having the same number of sides and the same area as P; then, the perimeter of Pis less than that of M.

For, let N be a regular polygon having the same perimeter and the same number of sides as M; then, by (59), MN, or PN. But of two regular polygons having the same number of sides, that which has the

M

P

N

less area has the less perimeter; therefore the perimeter of P is less than that of N, or less than that of M.

15

PROPOSITION XXVI.-THEOREM.

61. If a regular polygon be constructed with a given perimeter, its area will be the greater, the greater the number of its sides.

Let P be the regular polygon of three sides, and the regular polygon of four sides, constructed with the same given perimeter. In any side AB of P take any arbitrary point D; the polygon P may be regarded as an irregular poly

A

gon of four sides, in which the sides AD, DB, make an angle with each other equal to two right angles (I. 16); then, the irregular polygon P of four sides is less than the regular isoperimetric polygon Qof four sides (59). In the same manner it follows that is less than the regular isoperimetric polygon of five sides, and so on.

PROPOSITION XXVII.-THEOREM.

62. If a regular polygon be constructed with a given area, its perimeter will be the less, the greater the number of its sides.

Let P and Q be regular polygons having the same area, and let Q have the greater number of sides; then, the perimeter of P will be greater than that of Q.

For, let R be a regular polygon having the same perimeter as Q and the same number of sides as P; then, by (61),

P

R

> R, or P > R; therefore the perimeter of P is greater than

that of R, or greater than that of Q.