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

< >
[81.] LEMMA IV.
Page: 105 ([28])
[82.] LEMMA V.
Page: 106 ([29])
[83.] PROBLEM VII. (Fig. 32, 33, 34, &c.)
Page: 108 ([31])
[84.] PROBLEM I. (Fig. 32 to 45.)
Page: 109 ([32])
[85.] PROBLEM II. (Fig. 46 to 57.)
Page: 113 ([36])
[86.] PROBLEM III.
Page: 116 ([39])
[87.] THE END.
Page: 117 ([40])
[88.] A SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS Have hitherto been in Print.
Page: 133
[89.] By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT. ROCHESTER:
Page: 133
[90.] MDCCLXXIII. [Price One Shilling.]
Page: 133
[91.] ADVERTISEMENT.
Page: 135
[92.] AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS.
Page: 137
[93.] INDEX OF THE Authors refered to in the SYNOPSIS.
Page: 139 ([i])
[94.] Lately was publiſhed by the ſame Author; [Price Six Shillings in Boards.]
Page: 140 ([ii])
[95.] SYNOPSIS.
Page: 141 ([1])
[96.] Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length.
Page: 149 ([9])
[97.] SYNOPSIS
Page: 152 ([12])
[98.] FINIS.
Page: 156 ([16])
< >
page |< < ([4]) of 161 > >|
81[4] two ſolutions; if it only touches, then but of one; if it neither cuts nor
touches, it is then impoſſible.
Demonstration. Let the point of interſection then be O or o. We ſhall
have, by Lemma I. AO x ON = MN x NK = MN x AY = AI x AE, by
conſtruction. Let now from N be ſet off NL = AI in the ſame direction as A
is from I; then by what has been demonſtrated NL: ON: : AO: AE.
And by Diviſion or Compoſition OL: ON: : OE: AE
And by Permutation OL: OE: : ON: AE
But by what has been proved ON: AE: : AI: AO
Therefore by Equality OL: OE: : AI: AO
And by Diviſion or Compoſition LE: OE: : OI: AO
And LE x AO = OE x OI
But LE = NU. for NL was put = AI, and IU = AE
Hence NU x AO = S x AO = OE x OI
And R: S: : R x AO: S x AO or OE x OI
Q. E. D.
This Problem may be conſidered as having two Epitacmas, the firſt,
when the ſegment aſſigned for the coefficient of the given external line R is
terminated by an extreme point of the three given ones and the point ſought;
and this again admits of three Caſes. The other is when the aforeſaid ſeg-
ment is terminated by the middle point of the three given ones and the
point ſought.
Epitagma I. Case I. Let the aſſigned points be A, E, I. A an extreme
and E the middle one. And let the point O ſought (ſuch that AO x R: OE
x OI: : R: S) be required to lie between A and E, or elſe beyond I, which
will ariſe from the ſame conſtruction.
Here the Homotactical conſtruction is uſed, and IU as likewiſe UN is ſet off
in the ſame direction as AI. And ſince AO: AE: : AI: ON, and AO + ON
is greater than AE + AI or AU, by Lemma II. AO and ON will be the
leaſt and greateſt of all; and AO will therefore be leſs than AE, as likewiſe
Ao (being equal ON by Lemma I.) greater than AI. This Caſe is
unlimited.
Case II. Let the aſſigned points be in the ſame poſition as before, and let
the point O ſought be required between E and I.
Here the conſtruction is Homotactical, and UN is ſet off the contraty way, viz.
in the direction IA. And ſince AO: AE: : AI: ON, and AO + ON is leſs
than AE + AI or AU, by Lemma II. AE and AI will be the leaſt

Text layer

  • Dictionary

Text normalization

  • Original

Search


  • Exact
  • All forms
  • Fulltext index
  • Morphological index
Impressum