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THE side of an Icosaëdrum(α) 1.1 is equal in power to the side of a Pentagon in a circle containing only five sides of the Icosaëdrum; and the semi diameter of this circle is in power sub∣quintuple of the Diam. of the sphere of the circumscribed Icosa∣ëdrum.
Both these are evident from the genesis of the Icosaëdrum in Def. 22. The first immediately hence, because all the other
sides of the triangles (Fig. 99.) Aa, Ba, &c. are made equal to the side of the Pentagon AB by Consect. 4. Prop. 51. The latter from this inference; if for OA the radius of the circle you put a (since the side of the Pentagon, which here is also the side of the Icosaëdrum, it will be equal in power to the radius and side of the Decagon taken together by the aforesaid Prop.) the altitude OG will be the side of the Decagon = 〈 math 〉〈 math 〉 by Consect. 1. Prop. 54. to which the equal in∣ferior part oH being added, and the intermediate altitude Oo =a, you'l have the whole diameter of the circumscribed sphere GH= 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 i e. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 and so the square of the diameter of the sphere will be 5aa: Therefore the square of the diameter of the sphere is to the square of the semi-diam. of the circle containing the five sides of the Icosaëdrum as 5 to 1. Q. E. D.
IT is also evident that a sphere described on the diameter GH will pass thro' the other angles of this Icosaëdrum; for assuming the center between O and o the radius FG will be = 〈 math 〉〈 math 〉. But FA is also = 〈 math 〉〈 math 〉; for the □ of FO is =¼aa, and the □ AO=aa: Therefore the sum is =〈 math 〉〈 math 〉aa = □ FA. Q. E. D.
THerefore, if the radius of the circle ABCDE remain a, you'l have the altitude OG 〈 math 〉〈 math 〉, and the side of the Icosaëdrum 〈 math 〉〈 math 〉, by Cons. 1. and 2. Prop. 54. and the diam of the circumscribed Sphere 2 〈 math 〉〈 math 〉, as is evident from the Demonstration.
IF AB (Fig. 100) be the diameter of a sphere(α) 1.2 divided in D so that AD shall be ⅓ AB, then (having erected the perpendicular DF) BF will be the side of the Tetraëdrum by Prop. 56. and AF the side of the Hexaëdrum by Prop. 58. Cons. 2. and BE or AE (erecting from the center the perpendicular CE) will be the side of the Octaëdrum by Prop. 57. Now if AF be cut in mean and extreme reason in O, you'l have AO the side of the Dodecaëdrum by Prop. 59. Lastly, if you erect BG double of CB, HI will be double of CI, and the □ of HI=4 □ of CI; consequently the □ CH or CB=5 □ CI. Therefore the □ of AB (double of CH) is also=to 5 □ of HI (which is double of CI) therefore HI is the radius of the circle circumscribing the Pentagon of the Icosaëdrum, and IB the side of the Decagon inscribed in the same circle, and HB the side of the Pentagon, and also the side of the Ico∣saëdrum, by Prop. 60.
Eucl. Prop. 16. Co∣roll. lib. 13.
Eucl. Prop. 18. and los••, lib. 13.