PROPOSITION IL THEOREM.

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

F

B

For, let ABCDEF be the maximum polygon. If the side BC is not equal to CD, construct upon the base BD an isosceles triangle BOD, which shall be isoperimetrical with BCD; this triangle BOD will (Prop. I. Ap.) be greater than BCD, and consequently the polygon ABODEF will be greater than ABCDEF; E 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.

PROPOSITION III. THEOREM.

D

Of all the triangles, having two sides given in length and containing an angle which is not given, the maximum is that triangle in which the two given sides contain a right angle.

Let BAC, BAD be two triangles, in

E

which the side AB is common, and the ong
side AC=AD; if the angle BAC is
right, the triangle BAC will be greater
than the triangle BAD, of which the
angle A is acute or obtuse.

A

For, the base AB being the same, the two triangles BAC, BAD are to each other as their altitudes AC, DE; E but the perpendicular DE is shorter than the oblique line AD or its equal o AC; hence the triangle BAD is less than the triangle BAC.

B

PROPOSITION IV. THEOREM.

Among polygons formed of sides which are all given but one, the maximum is such that all its angles can be inscribed in a semicircle, of which the unknown side is the diameter.

triangle ADF (Prop. 3. Ap.); and consequently augment the whole polygon, because the parts ABCD, DEF would continue exactly as they are. But this polygon, being already a maximum, cannot be augmented; hence the angle ADF is no other than a right angle. The same are ABF, ACF, AEF; hence all the angles A, B, C, D, E, F of the maximum polygon are in a semicircumference, of which the indeterminate side AF is the diameter.

B

Scholium. This proposition gives rise to a question: Whether there be more ways than one of forming a polygon with sides which are all given, except the last side which is unknown and is to form the diameter of the semicircle wherein all the others are inscribed? Before deciding this question, it will be necessary to observe, that if the same chord AB subtend two arcs described with different radii AC, AD, the central angle standing upon this chord, will be smaller in the circle whose radius is greater; thus ACB ADB. For (27. I.), the angle ADO ACD+CAD; hence ACDADO, and doubling both, ACB ADB.

PROPOSITION V. THEOREM.

There is but one way of forming a polygon ABCDEF with sides which are all given, except the last side, which is unknown, and is to form the diameter of the semicircle wherein all the others are inscribed.

B

D

For, suppose we have found one circle which satisfies the conditions of the problem: if we take a greater circle, the chords AB, BC, CD, &c. will lie opposite to angles at the centre, which are smaller. Hence the sum of these central angles will be less than two right angles; hence the extremities of the given sides will not fall at the extremities of a diameter. The contrary error will arise, if we assume a smaller circle: hence the polygon in question can only be inscribed in one circle.

A

Scholium. The order of the sides AB, BC, CD may be altered at will, the diameter of the circumscribed circle, as well as the area of the polygon, always continuing the same; for whatever be the order of the arcs AB, BC, &c., it is enough if their sum be a semicircumference, and the polygon will always have the same area, being always equal to the semicircle minus the segments AB, BC &c. the sum of which is the same in any order.

PROPOSITION VI. THEOREM.

Of all the polygons formed with given sides, the maximum is the one which can be inscribed in a circle.

Let ABCDEFG be the inscribed polygon, and abcdefg the polygon which cannot be inscribed, both having equal sides, AB=ab, BC=bc, &c.; the inscribed polygon will be greater than the other.

Ďraw the diameter EM; join AM, MB; upon ab AB, construct the triangle abm equal to ABM, and join em.

By Prop. 4. of this Appendix, the polygon EFGAM, is greater than efgam, unless this efgam can be inscribed in a semicircle, of which the side em is the diameter, in which case the two polygons would be equal, by the last Proposition. For the same reason, the polygon EDCBM is greater than edcbm, saving a similar exception, by means of which they would be equal. Hence the whole polygon EFGAMBCDE is greater than efgambcde, unless they are equal in all respects: but they are not equal in all respects, because the one is inscribed

in a circle, and the other cannot be inscribed; hence the inscribed polygon is greater. Take away from both respectively the equal triangles ABM, abm; there will remain the inscribed polygon ABCDEFG greater than abcdefg, which cannot be inscribed.

Scholium. It might be shewn, as in the foregoing Proposition, that there can be only one circle, and therefore only one maximum polygon that will satisfy the conditions of the problem; and this polygon will always have the same area, in whatever order we arrange its sides.

PROPOSITION VII. THEOREM.

The regular polygon is the greatest of all the polygons which have the same perimeter and the same number of sides.

For, by the second of these Theorems, the maximum polygon has all its sides equal; and by the last Theorem, it can be inscribed in a circle : hence it is a regular polygon.

PROPOSITION VIII. LEMMA.

Two angles at the centre, measured in two different circles, are to each other as their included arcs divided by their radii.

AB

have C O::AB: FG; hence C: 0 :: :

G

E

AC FO

of the similar arcs FG, DE, (11. IV.) we have FG: DE::FO: DO;

therefore the quotient is equal to the quotient

FG
FO

DE

and con

DO'

PROPOSITION IX. THEOREM.

Of two isoperimetrical regular polygons, the one having the greater number of sides is the greater.

Let DE be a half-side of one of those polygons, O the centre, OE the apothem let AB be a half-side of the other polygon, C the centre, CB the apothem. Suppose the centres O and C to be situated at any distance OC, and the apothems OE, CB, in the direction OC: thus

DOE and ACB will be half-angles at the centres of the polygons; and because these angles are not equal, the lines CA, OD, if produced, will meet in some point F; from this point let fall the perpendicular OC

on FG; from the points O and C as centres, describe the arcs GI, GH, terminated by the sides OF, CF.

GI GH

Now, by the preceding lemma, we have O: C:: OG: CG; but DE is to the perimeter of the first polygon as the angle O is to four right angles; and AB is to the perimeter of the second polygon as C is to four right angles; therefore, since the perimeters of the polygons are GI GH

equal, DE: AB:: O: C, or DE:AB::

OG CG

Multiply the antecedents by OG, and the consequents by CG; we shall have DE.OG: AB.CG:: GI: GH. But the similar triangles ODE, OFG give OE:OG::DE: FG, whence DE.OG=OE.FG; in like manner, we should find AB.CG=CB.FG; therefore OE.FG: CB.FG:: GI: GH, or OE: CB:: GI: GH. Hence, if we can shew that the arc GI is greater than the arc GH, it will follow that the apothem OE is greater than CB.

On the other side of CF, construct a figure CKx entirely equal to the figure CGx, so that CK=CG, the angle HCK=HCG, and the arc KaxG; the curve KG will envelope the arc KHG, and (9. IV.) be greater than it. Therefore Ga, half of the curve, is greater than GH half of the arc; therefore still more is GI greater than GH.

It follows, therefore, that the apothem OE is greater than CB; but (7. IV.) the two polygons, having the same perimeter, are to each other as their apothems; hence the polygon which has DE for a half-side is greater than the polygon which has AB for its half-side: the first has more sides, because its central angle is smaller; hence of two isoperimetrical regular polygons, the one having more sides is greater.

PROPOSITION X. THEOREM.

The circle is greater than any polygon of the same perimeter.

We have already shewn, that of all the isoperimetrical polygons having the same number of sides, the regular polygon is the greatest; therefore we need only to compare the circle with some regular polygon of the same perimeter. Let AI be the half-side of this polygon; C its centre. In the isoperimetrical circle,

?

I

D

the

let the angle DOE be equal to ACI, and onsequently, the arc DE be equal to the half-side AI. The polygon P is to the circle C as triangle ACI is to the sector ODE; hence P:C::AI.CI : {DE.OE :! CI: OE. From the point Edraw a tangent EG meeting OD produced in G: the similar triangles ACI, GOE will give the proportion CI: OE: AL or DE: GE; hence P: C:: DE GE, or as DEOE,