Apollonius <Pergaeus>; Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

Table of contents

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[11.] PROBLEM IV.
Page: 15 ((3))
[12.] PROBLEM V.
Page: 16 ((4))
[13.] The general Solution.
Page: 16 ((4))
[14.] PROBLEM VI.
Page: 17 ((5))
[15.] The general Solution.
Page: 17 ((5))
[16.] PROBLEM VII.
Page: 18 ((6))
[17.] LEMMA I.
Page: 19 ((7))
[18.] PROBLEM VIII.
Page: 19 ((7))
[19.] Mr. Simpſon conſtructs the Problem thus.
Page: 19 ((7))
[20.] PROBLEM IX.
Page: 19 ((7))
[21.] LEMMA II.
Page: 20 ((8))
[22.] LEMMA III.
Page: 20 ((8))
[23.] PROBLEM X.
Page: 20 ((8))
[24.] PROBLEM XI.
Page: 21 ((9))
[25.] PROBLEM XII .
Page: 22 ((10))
[26.] LEMMA IV.
Page: 23 ((11))
[27.] LEMMA V.
Page: 23 ((11))
[28.] PROBLEM XIII.
Page: 23 ((11))
[29.] PROBLEM XIV.
Page: 24 ((12))
[30.] SUPPLEMENT. PROBLEM I.
Page: 26 ((14))
[31.] PROBLEM II.
Page: 26 ((14))
[32.] PROBLEM III.
Page: 26 ((14))
[33.] PROBLEM IV.
Page: 28 ((16))
[34.] PROBLEM V.
Page: 28 ((16))
[35.] PROBLEM VI.
Page: 29 ((17))
[36.] General Solution.
Page: 29 ((17))
[37.] A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I.
Page: 31 ((19))
[38.] PROBLEM II.
Page: 32 ((20))
[39.] PROBLEM III.
Page: 33 ((21))
[40.] PROBLEM IV.
Page: 33 ((21))
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page |< < ([32]) of 161 > >|
109[32] CU as IO is to BO; and ſo, by compound ratio, the rectangle AO, UO
is to the rectangle BO, CU as the rectangle EO, IO is to the rectangle
BO, CE; by permutation, the rectangle AO, UO is to the rectangle EO,
IO, as the rectangle BO, CU is to the rectangle BO, CE; or (Eu. V. 15.)
as CU is to CE; that is, by conſtruction as R to S.
Q. E. D.
Scholium. In enumerating the ſeveral Caſes of this Problem I ſhall en-
deavour to follow the method which I conceive Apollonius did: and there-
fore, notwithſtanding the preceding Analyſis and Conſtruction are general
for the whole, divide it into three Problems, each Problem into three Epi-
tagmas, or general Caſes, and theſe again into their ſeveral particular ones.
PROBLEM I. (Fig. 32 to 45.)
Here O is ſought between the two mean points of the four given ones:
and the three Epitagmas are, firſt, when A and U, the points which bound
the ſegments containing the antecedent rectangle, are one an extreme, and
the other an alternate mean; ſecondly, when thoſe points are one an ex-
treme and the other an adjacent mean; thirdly, when they are both means,
or both extremes.
Epitagma I, Conſiſts of eight Caſes, viz. when the order of the given
points is A, I, U, E; U, E, A, I; A, E, U, I; or U, I, A, E, and the
given ratio of a leſs to a greater, and four others wherein the order of the
points is the ſame as in thoſe, but the ratio of R to S, the ratio of a greater
to a leſs.
Case I. Let the order of the given points be A, I, U, E, and the given
ratio of a leſs to a greater; and the Conſtruction will be as in Fig. 32, where
B is made to fall beyond A, with reſpect to I, and C beyond U with re-
ſpect to E, and DH is drawn through F, the center of the circle on BC.
That O, when this conſtruction is uſed, will fall between I and U is
plain, becauſe CO is to CU as IB is to BO; and therefore if CU be
greater than CO, BO will be greater than IB, and if leſs, leſs; but this,
it is manifeſt, cannot be the Caſe if O falls either beyond I or U, and
therefore it falls between them.

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