TRACT 15 . GEOMETRICAL DIVISION OP THE CIRCLE. 259

Solution .—Put the radius ad or db = r, de = ar, the arcbg = g, gh = y, and df = i>. Now, if z to v be a constant

But, by similar triangles, eh = ar + */(i' 7 '—y' 1 ) : gh = y : :

ed = ar : dp =--—-- = v ; the fluxion of which is

ar + ^{r'—y*)

i* ..K strray

ratio. Therefore there is no such fixed point e, as that men-tioned in the problem.

Corollary 1.—Hence then it appears, that the commonmethod of finding the side of a polygon inscribed in a circle,by drawing a line from a certain fixed point e, through f andg, making af to ac as 2 is to the number of sides of the poly-gon, is not generally true.

Corol. 2.—But such a point e maybe found, as shall renderthat construction at least nearly true, in the following man-ner. Suppose the line efg to revolve about e, from B to a:at b, the arc bg and the line df arise in the ratio of be toDE; and at a they are in the ratio of ba to ad or db ; there-fore make these two ratios equal to each other, and it willdetermine the point e, so as that the ratios in all the inter-mediate points, or situations, will be nearly equal: thus then,BE : DE : : BA : ad : : p : 2, making p = 3T416 ; or ed : DE:: p — 2:2; hence de = x BD = T752 bd = t bdvery nearly. If, therefore, de be taken to da as 7 to 4; thenany line drawn from e, to cut the diameter ac, and the semi-circumference abc, it will very nearly cut them proportion**ally. Therefore, if a polygon is to be inscribed, or if thewhole circumference is to be divided into any number ofequal parts j first divide the diameter into the same number

£ 2

ratio, then k to v will also be constant; and the contrary.

v ; putting — y‘) = oh ; aisu

10 (ar + 10)

Hence then k : v

V(r*-S-)

x = y xr'-yarw

w(ar + w)*

1 : ar x

x - g; which is evidently a variable

(ar 4 - w)