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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[51.] PROBLEM X.
[52.] PROBLEM XI.
[53.] PROBLEM XII.
[54.] PROBLEM XIII.
[55.] PROBLEM XIV.
[56.] PROBLEM XV.
[57.] Synopſis of the PROBLEMS.
[58.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCC LXXII.
[59.] ADVERTISEMENT.
[60.] EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II.
[61.] THE PREFACE.
[62.] PROBLEMS CONCERNING DETERMINATE SECTION. PROBLEM I.
[63.] LEMMA I.
[64.] LEMMA II.
[65.] LEMMA III.
[66.] PROBLEM II.
[67.] LEMMA IV.
[68.] LEMMA V.
[69.] PROBLEM III.
[70.] PROBLEM IV.
[71.] DETERMINATE SECTION. BOOK I. PROBLEM I. (Fig. 1.)
[72.] PROBLEM II. (Fig. 2 and 3.)
[73.] PROBLEM III. (Fig. 4. and 5.)
[74.] PROBLEM IV. (Fig. 6. 7. and 8.)
[75.] PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
[76.] PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
[77.] THE END OF BOOK I.
[78.] DETERMINATE SECTION. BOOK II. LEMMA I.
[79.] LEMMA II.
[80.] LEMMA III.
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          <p>
            <s xml:id="echoid-s1052" xml:space="preserve">
              <pb o="[2]" file="0072" n="79"/>
            but if it only touches, then only of one; </s>
            <s xml:id="echoid-s1053" xml:space="preserve">if it neither touches nor cuts, it is
              <lb/>
            then impoſſible.</s>
            <s xml:id="echoid-s1054" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1055" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s1056" xml:space="preserve">Let a point of interſection then be O, and join O Y
              <lb/>
            and OR. </s>
            <s xml:id="echoid-s1057" xml:space="preserve">The angles AYO and IOR are equal, the angle AOY being the
              <lb/>
            complement of each of them to a right one, and hence the triangles AOY and
              <lb/>
            IOR are ſimilar.</s>
            <s xml:id="echoid-s1058" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1059" xml:space="preserve">Hence AY = AE: </s>
            <s xml:id="echoid-s1060" xml:space="preserve">AO:</s>
            <s xml:id="echoid-s1061" xml:space="preserve">: OI: </s>
            <s xml:id="echoid-s1062" xml:space="preserve">IR = AI
              <lb/>
            And by div . </s>
            <s xml:id="echoid-s1063" xml:space="preserve">or comp . </s>
            <s xml:id="echoid-s1064" xml:space="preserve">EO: </s>
            <s xml:id="echoid-s1065" xml:space="preserve">AO:</s>
            <s xml:id="echoid-s1066" xml:space="preserve">: AO: </s>
            <s xml:id="echoid-s1067" xml:space="preserve">AI
              <lb/>
            And
              <emph style="ol">AO</emph>
              <emph style="sub">2</emph>
            = EO x AI
              <lb/>
            Therefore
              <emph style="ol">AO</emph>
              <emph style="sub">2</emph>
            (= EO x AI): </s>
            <s xml:id="echoid-s1068" xml:space="preserve">EO x AU:</s>
            <s xml:id="echoid-s1069" xml:space="preserve">: AI: </s>
            <s xml:id="echoid-s1070" xml:space="preserve">AU:</s>
            <s xml:id="echoid-s1071" xml:space="preserve">: R: </s>
            <s xml:id="echoid-s1072" xml:space="preserve">S
              <lb/>
            Q. </s>
            <s xml:id="echoid-s1073" xml:space="preserve">E. </s>
            <s xml:id="echoid-s1074" xml:space="preserve">D.</s>
            <s xml:id="echoid-s1075" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1076" xml:space="preserve">This Problem admits of two Caſes. </s>
            <s xml:id="echoid-s1077" xml:space="preserve">The 1ſt determinate or limited, the 2d
              <lb/>
            unlimited.</s>
            <s xml:id="echoid-s1078" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1079" xml:space="preserve">
              <emph style="sc">Case</emph>
            I. </s>
            <s xml:id="echoid-s1080" xml:space="preserve">Is when OE the co-efficient of the given external line AU is part of
              <lb/>
            AO the ſide of the required ſquare [fig. </s>
            <s xml:id="echoid-s1081" xml:space="preserve">1. </s>
            <s xml:id="echoid-s1082" xml:space="preserve">2.</s>
            <s xml:id="echoid-s1083" xml:space="preserve">] and here the
              <emph style="sc">Llmitation</emph>
            is,
              <lb/>
            that AI muſt not be given leſs than four times AE, as appears from fig. </s>
            <s xml:id="echoid-s1084" xml:space="preserve">2.
              <lb/>
            </s>
            <s xml:id="echoid-s1085" xml:space="preserve">for AE: </s>
            <s xml:id="echoid-s1086" xml:space="preserve">AO:</s>
            <s xml:id="echoid-s1087" xml:space="preserve">: OI: </s>
            <s xml:id="echoid-s1088" xml:space="preserve">AI; </s>
            <s xml:id="echoid-s1089" xml:space="preserve">and here OI being the half of AI, AE will be
              <lb/>
            the half of AO, or the fourth part of AI. </s>
            <s xml:id="echoid-s1090" xml:space="preserve">In this Caſe the Homotactical Con-
              <lb/>
            ſtruction is uſed.</s>
            <s xml:id="echoid-s1091" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1092" xml:space="preserve">
              <emph style="sc">Case</emph>
            II. </s>
            <s xml:id="echoid-s1093" xml:space="preserve">Is when AO the ſide of the required ſquare is part of OE the co-
              <lb/>
            efficient of the given external line AU, [fig. </s>
            <s xml:id="echoid-s1094" xml:space="preserve">3.</s>
            <s xml:id="echoid-s1095" xml:space="preserve">] and this is unlimited, for here
              <lb/>
            the Anlitactical Conſtruction is uſed. </s>
            <s xml:id="echoid-s1096" xml:space="preserve">Or if O be required between A and E,
              <lb/>
            this is effected by the ſame Conſtruction.</s>
            <s xml:id="echoid-s1097" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div67" type="section" level="1" n="63">
          <head xml:id="echoid-head76" xml:space="preserve">LEMMA I.</head>
          <p>
            <s xml:id="echoid-s1098" xml:space="preserve">If from the extremes of any diameter perpendiculars be let fall upon any
              <lb/>
            Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs
              <lb/>
            are equal, and moreover alſo the ſegments of the Chords themſelves.</s>
            <s xml:id="echoid-s1099" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1100" xml:space="preserve">That YO is equal to IU may be thus ſhewn. </s>
            <s xml:id="echoid-s1101" xml:space="preserve">Having joined YI, the
              <lb/>
            angle IYE is a right one, being in a ſemicircle, and the angle at O is right by
              <lb/>
            hypotbeſis; </s>
            <s xml:id="echoid-s1102" xml:space="preserve">hence YI is parallel to the Chord, and YOUI is a parallelogram,
              <lb/>
            and the oppoſite ſides YO and IU will be equal. </s>
            <s xml:id="echoid-s1103" xml:space="preserve">In the ſame manner OE is
              <lb/>
            proved equal to UL. </s>
            <s xml:id="echoid-s1104" xml:space="preserve">And as to the ſegments of the Chord, it is thus ſhewn.
              <lb/>
            </s>
            <s xml:id="echoid-s1105" xml:space="preserve">By Euc. </s>
            <s xml:id="echoid-s1106" xml:space="preserve">III. </s>
            <s xml:id="echoid-s1107" xml:space="preserve">35. </s>
            <s xml:id="echoid-s1108" xml:space="preserve">and 36, the rect. </s>
            <s xml:id="echoid-s1109" xml:space="preserve">EOY = rect. </s>
            <s xml:id="echoid-s1110" xml:space="preserve">SOR, and rect. </s>
            <s xml:id="echoid-s1111" xml:space="preserve">LUI = rect. </s>
            <s xml:id="echoid-s1112" xml:space="preserve">
              <lb/>
            SUR. </s>
            <s xml:id="echoid-s1113" xml:space="preserve">But, by what has been juſt proved, rect. </s>
            <s xml:id="echoid-s1114" xml:space="preserve">EOY = rect. </s>
            <s xml:id="echoid-s1115" xml:space="preserve">LUI; </s>
            <s xml:id="echoid-s1116" xml:space="preserve">hence
              <lb/>
            rect. </s>
            <s xml:id="echoid-s1117" xml:space="preserve">SOR = rect. </s>
            <s xml:id="echoid-s1118" xml:space="preserve">SUR, and the ſegments SO and OR are reſpectively equal
              <lb/>
            to the ſegments UR and SU.</s>
            <s xml:id="echoid-s1119" xml:space="preserve"/>
          </p>
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