Maximum of Isoperimetrical Triangles of the saine Base.

or, taking away AB,

AC+BC

AD+BD,

and suppose ACB isosceles, or AC = CB.

We are to prove that

the triangle ACB > the triangle ADB.

But, since these triangles have the same base AB, they are to each other as their altitudes CE and DF; so that we need only prove

CE > DF.

=

Produce AC to H, making CH=CB=AC. Join BH; and if a semicircle is described upon AH as a diameter with the radius AC CH, it will pass through the point B; and ABH, being inscribed in it, must be a right angle.

Produce BH towards L, and take DL-DB. Join AL, and we have

AD+DL=AD+DB=AC+CB-AC+CH—AH.

Now, letting fall the perpendiculars CI and DM upon

BH and BL, we have

whence

304. Theorem.

BH-BICE,

BL-BM=DF ;

CE DF.

The maximum of isoperimetrical

polygons of the same number of sides is equilateral.

Proof. Let ABCD &c. (fig. 142) be the maximum of isoperimetrical polygons of any given number of sides.

Maximum of Polygous formed of sides all given but one.

Join AC. The triangle ABC must be the maximum of all the triangles which are formed upon AC, and with a perimeter equal to that of ABC. Otherwise a greater triangle AFC could be substituted for ABC, without changing the perimeter of the polygon, which would be inconsistent with the hypothesis that ABCD &c. is the maximum polygon.

Therefore, by the preceding article,

305. Theorem. 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.

Proof. Let ABC, ADC (fig. 143) be triangles, formed with the side AC common and the side AB- AD, and suppose BAC to be a right angle.

As these triangles have the same base AC, they are to each other as their altitudes AB and DE.

But

and

the triangle ABC> the triangle ADC.

306. Theorem. The maximum of polygons formed of sides, all given but one, can be inscribed in a semicircle having the undetermined side for its diameter.

Proof. Let ABCD &c. (fig. 144) be the maximum polygon formed of the given sides AB, BC, CD &c.

Draw from either vertex, as D, to the extremities A

Maximum of Polygons formed of given Sides.

and S of the side not given, the lines DA, DS. The triangle ADS must be the maximum of all triangles formed with the sides A and S; otherwise, either by increasing or else by diminishing the angle ADS, the triangle ADS would be enlarged, while the rest of the polygon ABCD, DEF &c. would be unchanged; so that the polygon would be enlarged, which is inconsistent with the hypothesis that it is the maximum polygon. The angle ADS is, therefore, a right angle by the preceding article, and is inscribed in the semicircle which has AS for its diameter.

307. Theorem. The maximum of all polygons formed of given sides can be inscribed in a circle.

Proof. Let ABCD &c. (fig. 145) be a polygon which can be inscribed in a circle, and A'B'C'D' &c. one which cannot be inscribed in a circle, but equilateral with respect to ABCD &c.

Draw the diameter AM. Join EM, MF. Upon E'F', equal to EF, construct the triangle E'M'F', equal to EMF, and join A'M'.

The polygon ABCDEM, which is inscribed in the semicircle having AM for its diameter is, by the preceding article, greater than A'B'C' DE'M' formed of the same sides but one, and which cannot be so inscribed. In the same way

the polygon AMFG &c. > A'M'F'G' &c.

Hence, the entire polygon ABCDEMF &c. > A'B'C'D' EMF &c., and, subtracting the triangle EMF EMF the polygon ABCD &c. > A'B'C'D' &c.

308. Theorem. The maximum of isoperimetrical polygons of the same number of sides is regular.

Greatest of Isoperimetrical Regular Polygons.

Proof. For, by § 304, it is equilateral; and, by the preceding article, it can be inscribed in a circle; so that, by 202, it is regular.

309. Theorem. Of isoperimetrical regular polygons that is the greatest which has the greatest number of sides.

Proof. Let ABCD &c., A'B'C'D' &c. (fig. 146) be two isoperimetrical regular polygons, of which ABCD &c. has the greater number of sides.

Denote the area of ABCD &c. by S, and the radius OH of its inscribed circle by R; and denote the area of A'B'C'D' &c. by S, and the radius O'H' of its inscribed circle by R'; also the common perimeter of the two polygons by P.

Then we have, by § 277,

S: SPR:1PX R',

or, striking out the common factor † P,

Upon A'B', as a side, describe a polygon A'B'C"D" &c. similar to ABCD &c. ; denote its perimeter by P", and the radius O'M' of its inscribed circle by R".

Join A'O' and 'O"; describe the arc MN with the radius R', and the arc M'N' with the radius R".

The half side A'M' is, evidently, the same part of the perimeter P, which the arc M'N is of its circumference, which circumference is, by § 237, equal to 2 л × R'; that is,

AM: PMN: 2π x R',

Greatest of Isoperimetrical Regular Polygons.

and in the same way,

P: AM2 л × R" : MN,

π

the product of these two proportions is, by striking out the factors common to the terms of each ratio,

P": P=R" × MN: R' × M'N'.

But, by 233,

P": PR": R,

and, on account of the common ratio P": P,
R": R=R" × MN: R x MN,
which, multiplied by the identical proportion

gives, by striking out the common factors, R': RMN: MN,

so that we need only prove

in order to prove

MN MN,

RR'.

Now, the angle A'O'M' is obtained by dividing 360° by twice the number of sides of the polygon A'B'C'D', &c., and the angle A'O'M' is obtained by dividing 360° by twice the number of sides of the polygon A'B'C"D" &c., but the second number of sides was supposed to be greater than the first, and, therefore,

the angle A' O'M' < the angle A'O'M' ;

and, therefore, as the angle O"A'M' is, by § 69, the remainder after subtracting the angle A'O"M' from 90°, it is greater than the angle O'A'M' which remains after subtracting A'O'M' from 90°; and O"AM' includes O'A'M' ; so that the radius MO" is greater than M'O', and the circle described with M'O" as a radius includes the circle described with M'O' as a radius.