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A history of Greek mathematics - Wilbourhall.org

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48 ARCHIMEDES<br />

Draw QMN through M parallel<br />

meet KG produced in F.<br />

to AK. Produce K'M to<br />

By similar triangles,<br />

FA:AM = K'A':A'M, or FA:h = a:h' y<br />

whence FA = AH (k, suppose).<br />

Similarly # A'E — A'H' (Jc, suppose).<br />

Again, by^similar triangles,<br />

(FA + AM) :<br />

(A'K' + A'M) = AM: A'M<br />

= (AK + AM):(EA' + A'M),<br />

or (k + h):(a + h') = (a + h):(¥ + h') }<br />

i. e. (k + h) (k' + h') = (a + h)(a + h f ). (1)<br />

Now, by hypothesis,<br />

m:n = (k + h):(k' + hf)<br />

= (k + h)(k' + h'):(k' + h') 2<br />

= (a + h) (a + hf) :<br />

{¥ + h!f [by (I)]. (2)<br />

Measure AR, A f R! on AA f produced both ways equal to a.<br />

Draw RP, R'P' at right angles to RR /<br />

Measure along MJS T the length MV equal to<br />

draw PP' through V, A' to meet RP, R'P'.<br />

Then QV=k' + h', P'V= V2 (a + h'),<br />

PV= V2(a + h),<br />

whence PV.P'V = 2 (a + h) {a + h!) ;<br />

and, from (2) above,<br />

as shown in the figure.<br />

MA' or h\ and<br />

2m:n=2(a + h) (a + h') :<br />

(¥ + h'f<br />

= PV.P'V: QV 2 .<br />

(3)<br />

Therefore Q is on an ellipse in which PP' is a diameter, and<br />

Q V is an ordinate to it.<br />

Again, GQNK is equal to AA'K'K, whence<br />

GQ.QN= AA'. A'K' =(h + h!) a = 2ra, (4)<br />

and therefore Q is on the rectangular hyperbola with KF,<br />

KK' as asymptotes and passing through A'.

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