PROBLEM II.

To describe about a given circle a regular polygon having a given number of sides.

E

T

Q B

A

This problem can be solved if the Oce of a can be divided into the given number of equal arcs.

Let the Oce of the given be divided into a number of equal arcs PQ, QR, RS &c.;

and let tangents be drawn at P, Q, R, S &c. intersecting in A, B, C &c.

Then shall ABCD be a regular polygon circumscribing the given .

For if the figure be turned about its centre 0,

so that P may coincide with the former position of Q: then will Q, R, S &c. coincide with the former positions of R, S, T &c.;

and the tangents at P, Q, R, S &c. with those of the tangents at Q, R, S, T &c.

Hence the polygon will coincide with its former position; .. AB=BC= CD= DE &c. ;

.. the polygon is equilateral and also equiangular, i.e. it is a regular polygon and it is circumscribed about the given circle.

COROLLARY. Hence we may describe about a circle regular

figures of 3, 4, 5, 6, 8 &c. sides.

(III. G. H. K.)

PROBLEM 12.

To describe a circle about a given regular figure.

Let ABCDE be the given regular figure.

It is required to describe a

about it.

Bisects EAB, ABC by AO, BO intersecting in O,

and with centre O at distance OA describe a O.
This shall be the required.

Join OC, OD, &c.

Then.. EA, AO and ▲ EAO are respectively

L

= BA, AO and BAO;

.. LOEA = L OBA (1. 1) = half of one of the 4s of the regular polygon;

.. LAED is bisected.

Similarly it may be proved that all the 4s of the polygon are bisected.

Then . OAB= LOBA, .. OA=OB: (1.4)

... the circle passes through B, C, D, &c.,

and is.. described about the given polygon.

PROBLEM 13.

To inscribe a circle in a given regular figure.

E

B

Let ABCD &c. be the given regular polygon.
It is required to inscribe a circle within it.

Bisect the s ABC, BCD by BO, CO intersecting in O.

Then the

From O let fall OP1 on AB.

described with centre O at the distance OP

shall be the one required.

Join OD, OE, &c., and draw OQ, OR, &c. to BC, CD, &c.

Then it may be shewn as in the previous problem that thes of the polygon are bisected by OD, OE, &c. Also ... LS. OPB, OBP and OB are respectively =LS OQB, OBQ and OB ;

and it will touch AB, BC, CD, &c., . the 4 s at P, Q, R, &c.

are rights;

.. it is inscribed in the given polygon.

(III. 6)

CIRCUMSCRIBED POLYGONS.

PROBLEM 14.

To find the difference between the areas of two regular polygons having the same number of sides, one inscribed in, and the other circumscribed about a given circle.

H

Let ABC... be a regular polygon inscribed in a circle and HIJ... one of the same number of sides circumscribed about it,

formed by drawing tangents to the circle at A, B, C....... Join OA, OB, OC, &c. OH, OI, OJ, &c.

From OH cut off OS = AN, and draw XSY || to AB.

Then ▲ OSX is = ▲ ANH in all respects,

so AOSY=ABNH;

.. ΔΟΧΥ=ΔΑΗΒ;

(1. 25)

Then as OXY, O YZ, &c. are all equal to one another,

as are also AHB, BIC, &c.

Hence the difference between the polygons ABC..., and HIJ...is equal to polygon XYZ....

COR. I. If the polygons are formed by the continual bisections of the circumference, then the polygon XYZ is less than a square whose side is ST (which is = AB).

Hence the difference between the polygons ABC, HIJ, and.. the difference between either polygon and the circle ABC, is less than the square on a side of the inscribed polygon.

COR. 2. By continual bisections, the circumference of a circle can be divided into arcs each less than any assignable magnitude;

and .. the chords will each be less than any assignable magnitude;

.. the ce may be subdivided till the squares on the subtended chords shall each be less than any assignable

area.

.. a regular polygon can be inscribed in a circle whose area shall differ from that of the circle by less than any assignable

area:

so also can one be circumscribed about it, &c.