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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            <s xml:id="echoid-s1453" xml:space="preserve">
              <emph style="sc">Case</emph>
            VI. </s>
            <s xml:id="echoid-s1454" xml:space="preserve">Let the given ratio of EL to LI be inæqualitatis majoris, and
              <lb/>
            let the point ſought be required to lie beyond either extreme. </s>
            <s xml:id="echoid-s1455" xml:space="preserve">Here we muſt
              <lb/>
            uſe the IIId
              <emph style="sc">Case</emph>
            of the IId
              <emph style="sc">Problem</emph>
            ; </s>
            <s xml:id="echoid-s1456" xml:space="preserve">and the
              <emph style="sc">Determination</emph>
            is that
              <lb/>
            UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
              <lb/>
            IE + EL + √4 IEL*.</s>
            <s xml:id="echoid-s1457" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1458" xml:space="preserve">
              <emph style="sc">Case</emph>
            VII. </s>
            <s xml:id="echoid-s1459" xml:space="preserve">Let the ſituation of O be required the ſame as in the two laſt
              <lb/>
            Caſes, but let the given ratio be that of equality, which was there ſuppoſed
              <lb/>
            of inequality. </s>
            <s xml:id="echoid-s1460" xml:space="preserve">Here the IId
              <emph style="sc">Problem</emph>
            will be of no uſe, and this Caſe
              <lb/>
            requires a particular conſtruction.</s>
            <s xml:id="echoid-s1461" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1462" xml:space="preserve">Let then the three Points be A, I, E, and I the middle one; </s>
            <s xml:id="echoid-s1463" xml:space="preserve">and let it
              <lb/>
            be required to find a fourth O beyond E, ſuch that AO x OE may equal
              <lb/>
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1464" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1465" xml:space="preserve">
              <emph style="sc">Construction</emph>
            . </s>
            <s xml:id="echoid-s1466" xml:space="preserve">Upon AE diameter deſcribe a circle,, and let another YS
              <lb/>
            cut the former at right angles. </s>
            <s xml:id="echoid-s1467" xml:space="preserve">Join SI, and continue it to meet the circum-
              <lb/>
            ference in R. </s>
            <s xml:id="echoid-s1468" xml:space="preserve">From R draw a tangent to meet the given line in O, and I
              <lb/>
            ſay O is the point required.</s>
            <s xml:id="echoid-s1469" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1470" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s1471" xml:space="preserve">Joining YR, the triangles SUI and SYR will be ſimi-
              <lb/>
            lar, and the angle UIS or RIO = SYR. </s>
            <s xml:id="echoid-s1472" xml:space="preserve">But the angle IRO made by the
              <lb/>
            tangent and ſecant = SYR in the alternate ſegment. </s>
            <s xml:id="echoid-s1473" xml:space="preserve">Therefore RIO = IRO,
              <lb/>
            and OR = OI. </s>
            <s xml:id="echoid-s1474" xml:space="preserve">But by the property of the circle AOE =
              <emph style="ol">OR</emph>
              <emph style="sub">2</emph>
            . </s>
            <s xml:id="echoid-s1475" xml:space="preserve">And
              <lb/>
            therefore AOE =
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1476" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1477" xml:space="preserve">Q. </s>
            <s xml:id="echoid-s1478" xml:space="preserve">E. </s>
            <s xml:id="echoid-s1479" xml:space="preserve">D.</s>
            <s xml:id="echoid-s1480" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1481" xml:space="preserve">The
              <emph style="sc">Determination</emph>
            is that AI muſt be greater than IE.</s>
            <s xml:id="echoid-s1482" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1483" xml:space="preserve">
              <emph style="sc">Case</emph>
            VIII. </s>
            <s xml:id="echoid-s1484" xml:space="preserve">Whereas in the Iſt and IId
              <emph style="sc">Cases</emph>
            the given ratio was that of
              <lb/>
            inequality, let us now ſuppoſe it that of equality; </s>
            <s xml:id="echoid-s1485" xml:space="preserve">and let the three points be
              <lb/>
            A, E, I, and E the middle one; </s>
            <s xml:id="echoid-s1486" xml:space="preserve">and let a fourth O be ſought between E and
              <lb/>
            I, ſuch that AOE may equal
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1487" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1488" xml:space="preserve">The
              <emph style="sc">Construction</emph>
            and
              <emph style="sc">Demonstration</emph>
            of this Caſe is in every reſpect
              <lb/>
            the ſame as that of the preceeding, as will appear by comparing the figures.</s>
            <s xml:id="echoid-s1489" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div74" type="section" level="1" n="70">
          <head xml:id="echoid-head83" xml:space="preserve">PROBLEM IV.</head>
          <p>
            <s xml:id="echoid-s1490" xml:space="preserve">To cut a given indefinite right line ſo in one point that, of the four ſeg-
              <lb/>
            ments intercepted between the ſame and four points given in the indefinite
              <lb/>
            line, the rectangle under any two aſſigned ones may be to the rectangle under
              <lb/>
            the two remaining ones in a given ratio.</s>
            <s xml:id="echoid-s1491" xml:space="preserve"/>
          </p>
        </div>
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