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-s1606" xml:space="preserve">
              <pb o="[13]" file="0083" n="90"/>
            ſuppoſe it that of equality, and they may all be ſolved by one conſtruction, viz.
              <lb/>
            </s>
            <s xml:id="echoid-s1607" xml:space="preserve">the rectangle AOU made equal to the rectangle EOI.</s>
            <s xml:id="echoid-s1608" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1609" xml:space="preserve">Let it be made as UI: </s>
            <s xml:id="echoid-s1610" xml:space="preserve">AE:</s>
            <s xml:id="echoid-s1611" xml:space="preserve">: UO: </s>
            <s xml:id="echoid-s1612" xml:space="preserve">EO</s>
          </p>
          <p>
            <s xml:id="echoid-s1613" xml:space="preserve">Then by permutation UO: </s>
            <s xml:id="echoid-s1614" xml:space="preserve">UI:</s>
            <s xml:id="echoid-s1615" xml:space="preserve">: EO: </s>
            <s xml:id="echoid-s1616" xml:space="preserve">AE</s>
          </p>
          <p>
            <s xml:id="echoid-s1617" xml:space="preserve">And by comp. </s>
            <s xml:id="echoid-s1618" xml:space="preserve">or diviſ. </s>
            <s xml:id="echoid-s1619" xml:space="preserve">UO: </s>
            <s xml:id="echoid-s1620" xml:space="preserve">OI:</s>
            <s xml:id="echoid-s1621" xml:space="preserve">: EO: </s>
            <s xml:id="echoid-s1622" xml:space="preserve">AO</s>
          </p>
          <p>
            <s xml:id="echoid-s1623" xml:space="preserve">Hence AOU = EOI.</s>
            <s xml:id="echoid-s1624" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1625" xml:space="preserve">
              <emph style="sc">Lemma</emph>
            VI. </s>
            <s xml:id="echoid-s1626" xml:space="preserve">Let there be two ſimilar triangles IAE, UAO, having their
              <lb/>
            baſes IE and UO parallel; </s>
            <s xml:id="echoid-s1627" xml:space="preserve">I ſay Iſt when they are right-angled, that the ex-
              <lb/>
            ceſs of the rectangle EAO, under the greater ſides of each, above the rect-
              <lb/>
            angle IAU, under the leſſer ſides of each, will be equal to the rectangle
              <lb/>
            IE x OU, under their baſes. </s>
            <s xml:id="echoid-s1628" xml:space="preserve">IIdly, When they are obtuſe-angled, that the
              <lb/>
            ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum
              <lb/>
            of the diſtances of the angles at the baſe of the other from the perpendicular,
              <lb/>
            viz. </s>
            <s xml:id="echoid-s1629" xml:space="preserve">EI x
              <emph style="ol">OS + US</emph>
            . </s>
            <s xml:id="echoid-s1630" xml:space="preserve">IIIdly, When they are acute-angled, that then the ſaid
              <lb/>
            exceſs will be equal to the rectangle under the baſe of one and the difference
              <lb/>
            of the ſegments of the baſe of the other made by the perpendicular, viz.
              <lb/>
            </s>
            <s xml:id="echoid-s1631" xml:space="preserve">OU x EL.</s>
            <s xml:id="echoid-s1632" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1633" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s1634" xml:space="preserve">Since EA: </s>
            <s xml:id="echoid-s1635" xml:space="preserve">AO:</s>
            <s xml:id="echoid-s1636" xml:space="preserve">: IA: </s>
            <s xml:id="echoid-s1637" xml:space="preserve">AU:</s>
            <s xml:id="echoid-s1638" xml:space="preserve">: EI: </s>
            <s xml:id="echoid-s1639" xml:space="preserve">OU, the rect-
              <lb/>
            angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles
              <lb/>
            are right-angled EAO = IAU + EI x OU by Euc. </s>
            <s xml:id="echoid-s1640" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s1641" xml:space="preserve">19. </s>
            <s xml:id="echoid-s1642" xml:space="preserve">and I. </s>
            <s xml:id="echoid-s1643" xml:space="preserve">47. </s>
            <s xml:id="echoid-s1644" xml:space="preserve">But if
              <lb/>
            they be oblique-angled, draw the perpendicular YAS. </s>
            <s xml:id="echoid-s1645" xml:space="preserve">Then IIdly, in caſe
              <lb/>
            they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt; </s>
            <s xml:id="echoid-s1646" xml:space="preserve">and IAU =
              <lb/>
            YAS + IY x US by the ſame. </s>
            <s xml:id="echoid-s1647" xml:space="preserve">And therefore EAO - IAU = EY x OS -
              <lb/>
            IY x US =
              <emph style="ol">EY - IY</emph>
            or EI x
              <emph style="ol">OS + US</emph>
            . </s>
            <s xml:id="echoid-s1648" xml:space="preserve">But if IIIdly they be acute-angled,
              <lb/>
            and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR
              <lb/>
            which will be equal and ſimilarly divided to IAU. </s>
            <s xml:id="echoid-s1649" xml:space="preserve">Then by part IId EAO
              <lb/>
            - LAR, i. </s>
            <s xml:id="echoid-s1650" xml:space="preserve">e. </s>
            <s xml:id="echoid-s1651" xml:space="preserve">EAO - IAU = EL x
              <emph style="ol">OS + RS</emph>
            = EL x OU.</s>
            <s xml:id="echoid-s1652" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1653" xml:space="preserve">Q. </s>
            <s xml:id="echoid-s1654" xml:space="preserve">E. </s>
            <s xml:id="echoid-s1655" xml:space="preserve">D.</s>
            <s xml:id="echoid-s1656" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1657" xml:space="preserve">
              <emph style="sc">Lemma</emph>
            VII. </s>
            <s xml:id="echoid-s1658" xml:space="preserve">If a right line VY, joining the tops of two perpendiculars
              <lb/>
            drawn from two points of the diameter of a circle E and I to the circum-
              <lb/>
            ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and
              <lb/>
            A and U be the extremes of the ſaid diameter, I ſay that the ratio of the
              <lb/>
            rectangle AOU to the rectangle EOI is a Minimum.</s>
            <s xml:id="echoid-s1659" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1660" xml:space="preserve">But if VY joins the tops of two perpendiculars from E and I drawn on
              <lb/>
            the ſame ſide of the diameter, and conſequently meets the diameter produced
              <lb/>
            in O, that then the ratio of AOU to EOI is a Maximum.</s>
            <s xml:id="echoid-s1661" xml:space="preserve"/>
          </p>
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