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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[41.] PROBLEM V.
[42.] PROBLEM VI.
[43.] PROBLEM VII.
[44.] LEMMA I.
[45.] LEMMA II.
[46.] LEMMA III.
[47.] LEMMA IV.
[48.] LEMMA V.
[49.] PROBLEM VIII.
[50.] PROBLEM IX.
[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.
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          <p>
            <s xml:id="echoid-s1150" xml:space="preserve">
              <pb o="[4]" file="0074" n="81"/>
            two ſolutions; </s>
            <s xml:id="echoid-s1151" xml:space="preserve">if it only touches, then but of one; </s>
            <s xml:id="echoid-s1152" xml:space="preserve">if it neither cuts nor
              <lb/>
            touches, it is then impoſſible.</s>
            <s xml:id="echoid-s1153" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1154" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s1155" xml:space="preserve">Let the point of interſection then be O or o. </s>
            <s xml:id="echoid-s1156" xml:space="preserve">We ſhall
              <lb/>
            have, by Lemma I. </s>
            <s xml:id="echoid-s1157" xml:space="preserve">AO x ON = MN x NK = MN x AY = AI x AE, by
              <lb/>
            conſtruction. </s>
            <s xml:id="echoid-s1158" xml:space="preserve">Let now from N be ſet off NL = AI in the ſame direction as A
              <lb/>
            is from I; </s>
            <s xml:id="echoid-s1159" xml:space="preserve">then by what has been demonſtrated NL: </s>
            <s xml:id="echoid-s1160" xml:space="preserve">ON:</s>
            <s xml:id="echoid-s1161" xml:space="preserve">: AO: </s>
            <s xml:id="echoid-s1162" xml:space="preserve">AE.</s>
            <s xml:id="echoid-s1163" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1164" xml:space="preserve">And by Diviſion or Compoſition OL: </s>
            <s xml:id="echoid-s1165" xml:space="preserve">ON:</s>
            <s xml:id="echoid-s1166" xml:space="preserve">: OE: </s>
            <s xml:id="echoid-s1167" xml:space="preserve">AE
              <lb/>
            And by Permutation OL: </s>
            <s xml:id="echoid-s1168" xml:space="preserve">OE:</s>
            <s xml:id="echoid-s1169" xml:space="preserve">: ON: </s>
            <s xml:id="echoid-s1170" xml:space="preserve">AE
              <lb/>
            But by what has been proved ON: </s>
            <s xml:id="echoid-s1171" xml:space="preserve">AE:</s>
            <s xml:id="echoid-s1172" xml:space="preserve">: AI: </s>
            <s xml:id="echoid-s1173" xml:space="preserve">AO
              <lb/>
            Therefore by Equality OL: </s>
            <s xml:id="echoid-s1174" xml:space="preserve">OE:</s>
            <s xml:id="echoid-s1175" xml:space="preserve">: AI: </s>
            <s xml:id="echoid-s1176" xml:space="preserve">AO
              <lb/>
            And by Diviſion or Compoſition LE: </s>
            <s xml:id="echoid-s1177" xml:space="preserve">OE:</s>
            <s xml:id="echoid-s1178" xml:space="preserve">: OI: </s>
            <s xml:id="echoid-s1179" xml:space="preserve">AO
              <lb/>
            And LE x AO = OE x OI
              <lb/>
            But LE = NU. </s>
            <s xml:id="echoid-s1180" xml:space="preserve">for NL was put = AI, and IU = AE
              <lb/>
            Hence NU x AO = S x AO = OE x OI
              <lb/>
            And R: </s>
            <s xml:id="echoid-s1181" xml:space="preserve">S:</s>
            <s xml:id="echoid-s1182" xml:space="preserve">: R x AO: </s>
            <s xml:id="echoid-s1183" xml:space="preserve">S x AO or OE x OI
              <lb/>
            Q. </s>
            <s xml:id="echoid-s1184" xml:space="preserve">E. </s>
            <s xml:id="echoid-s1185" xml:space="preserve">D.</s>
            <s xml:id="echoid-s1186" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1187" xml:space="preserve">This Problem may be conſidered as having two
              <emph style="sc">Epitacmas</emph>
            , the firſt,
              <lb/>
            when the ſegment aſſigned for the coefficient of the given external line R is
              <lb/>
            terminated by an extreme point of the three given ones and the point ſought;
              <lb/>
            </s>
            <s xml:id="echoid-s1188" xml:space="preserve">and this again admits of three Caſes. </s>
            <s xml:id="echoid-s1189" xml:space="preserve">The other is when the aforeſaid ſeg-
              <lb/>
            ment is terminated by the middle point of the three given ones and the
              <lb/>
            point ſought.</s>
            <s xml:id="echoid-s1190" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1191" xml:space="preserve">
              <emph style="sc">Epitagma</emph>
            I. </s>
            <s xml:id="echoid-s1192" xml:space="preserve">
              <emph style="sc">Case</emph>
            I. </s>
            <s xml:id="echoid-s1193" xml:space="preserve">Let the aſſigned points be A, E, I. </s>
            <s xml:id="echoid-s1194" xml:space="preserve">A an extreme
              <lb/>
            and E the middle one. </s>
            <s xml:id="echoid-s1195" xml:space="preserve">And let the point O ſought (ſuch that AO x R: </s>
            <s xml:id="echoid-s1196" xml:space="preserve">OE
              <lb/>
            x OI:</s>
            <s xml:id="echoid-s1197" xml:space="preserve">: R: </s>
            <s xml:id="echoid-s1198" xml:space="preserve">S) be required to lie between A and E, or elſe beyond I, which
              <lb/>
            will ariſe from the ſame conſtruction.</s>
            <s xml:id="echoid-s1199" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1200" xml:space="preserve">Here the Homotactical conſtruction is uſed, and IU as likewiſe UN is ſet off
              <lb/>
            in the ſame direction as AI. </s>
            <s xml:id="echoid-s1201" xml:space="preserve">And ſince AO: </s>
            <s xml:id="echoid-s1202" xml:space="preserve">AE:</s>
            <s xml:id="echoid-s1203" xml:space="preserve">: AI: </s>
            <s xml:id="echoid-s1204" xml:space="preserve">ON, and AO + ON
              <lb/>
            is greater than AE + AI or AU, by
              <emph style="sc">Lemma</emph>
            II. </s>
            <s xml:id="echoid-s1205" xml:space="preserve">AO and ON will be the
              <lb/>
            leaſt and greateſt of all; </s>
            <s xml:id="echoid-s1206" xml:space="preserve">and AO will therefore be leſs than AE, as likewiſe
              <lb/>
            Ao (being equal ON by
              <emph style="sc">Lemma</emph>
            I.) </s>
            <s xml:id="echoid-s1207" xml:space="preserve">greater than AI. </s>
            <s xml:id="echoid-s1208" xml:space="preserve">This Caſe is
              <lb/>
            unlimited.</s>
            <s xml:id="echoid-s1209" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1210" xml:space="preserve">
              <emph style="sc">Case</emph>
            II. </s>
            <s xml:id="echoid-s1211" xml:space="preserve">Let the aſſigned points be in the ſame poſition as before, and let
              <lb/>
            the point O ſought be required between E and I.</s>
            <s xml:id="echoid-s1212" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1213" xml:space="preserve">Here the conſtruction is Homotactical, and UN is ſet off the contraty way, viz.
              <lb/>
            </s>
            <s xml:id="echoid-s1214" xml:space="preserve">in the direction IA. </s>
            <s xml:id="echoid-s1215" xml:space="preserve">And ſince AO: </s>
            <s xml:id="echoid-s1216" xml:space="preserve">AE:</s>
            <s xml:id="echoid-s1217" xml:space="preserve">: AI: </s>
            <s xml:id="echoid-s1218" xml:space="preserve">ON, and AO + ON is leſs
              <lb/>
            than AE + AI or AU, by
              <emph style="sc">Lemma</emph>
            II. </s>
            <s xml:id="echoid-s1219" xml:space="preserve">AE and AI will be the leaſt </s>
          </p>
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