In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular
be the given circle; thus it is required to inscribe in the circle ABCD
a fifteenangled figure which shall be both equilateral and equiangular.
In the circle ABCD
let there be inscribed a side AC
of the equilateral triangle inscribed in it, and a side AB
of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD
, there will be five in the circumference ABC
which is one-third of the circle, and there will be three in the circumference AB
which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments.
be bisected at E
; [III. 30
] therefore each of the circumferences BE
is a fifteenth of the circle ABCD
If therefore we join BE
and fit into the circle ABCD
straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it. Q. E. F.
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle a fifteen-angled figure which is equilateral and equiangular.
And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteenangled figure and circumscribe one about it. Q. E. F.