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PROBLEM IJ. To describe about a given circle a regular polygon having a given number of sides.
This problem can be solved if the o ce of a O can be divided into the given number of equal arcs.
Let the Oce of the given o 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 o.
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.;
also LABC= BCD= . CDE, &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.
(111. G. H. K.)
To describe a circle about a given regzular figure.
Let ABCDE be the given regular figure.
It is required to describe a o about it. BisectZs EAB, ABC by AO, BO intersecting in 0, and with centre o at distance OA describe a O.
This shall be the required.
Join OC, OD, &c.
Then :: EA, AO and 2 EAO are respectively
= BA, AO and _ BAO; .. LOEA = 1 OBA (r. 1) = half of one of the 2s of the regular polygon;
.. LAED is bisected. Similarly it may be proved that all the as of the polygon are bisected. Then ... _OAB = OBA, .. OA = OB: (1. 4)
similarly OB = OC, &c.;
To inscribe a circle in a given regular figure.
Let ABCD &c. be the given regular polygon.
It is required to inscribe a circle within it.
Bisect the 2s ABC, BCD by BO, CO intersecting in 0.
From O let fall OP 1 on AB. Then the o described with centre ( at the distance OP shall be the one required.
Join OD, OE, &c., and draw OQ, OR, &c. I to BC, CD, &c.
Then it may be shewn as in the previous problem that the zs of the polygon are bisected by OD, OE, &c.
Also :: LS. OPB, OBP and OB are respectively = LS OQB, OBQ and OB; .. OP=OQ.
(1. 25) Similarly OQ = OR = OS &c.
.. the O will pass through Q, R, S &c.; and it will touch AB, BC, CD, &c.,' the 2 sat P, Q, R, &c. are right 2S;
(III. 6) .:. it is inscribed in the given polygon.
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.
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, O1, OJ, &c. From OH cut off OS = AN, and draw XSY || to AB. Then s OSX is = A ANH in all respects, (1. 25)
SO A OSY = A BNH; .:. Δ ΟΧΥ= Δ ΑΗΒ;
cut off OZ, &c. = OX or O Y. 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....
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 oce may be subdivided till the squares on the subtended chords shall each be less than any assignable
.. a regular polygon can be inscribed in a circle whose area shall differ from that of the circle by less than any assignable
so also can one be circumscribed about it, &c.