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A history of Greek mathematics - Wilbourhall.org

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ON THE CUTTING-OFF OF A RATIO 179<br />

Further, if we put for A the ratio between the lengths <strong>of</strong> the<br />

two fixed tangents, then if h, k be those lengths,<br />

k<br />

h<br />

y<br />

h + x-2\/hx<br />

which can easily be reduced to<br />

©*+©-'<br />

the equation <strong>of</strong> the parabola referred to the two fixed tangents<br />

as axes.<br />

•(/?) On the cutting-<strong>of</strong>f <strong>of</strong> an area (\coptov ccttoto/jltj),<br />

two Books.<br />

This work, also in two Books, dealt with a similar problem,<br />

with the difference that the intercepts on the given straight<br />

lines measured from the given points are required, not to<br />

have a given ratio, but to contain a given rectangle. Halley<br />

included an attempted restoration <strong>of</strong><br />

this work in his edition<br />

<strong>of</strong> the De sectione rationis.<br />

The general case can here again be reduced to the more<br />

special one in which one <strong>of</strong> the fixed points is at the intersection<br />

<strong>of</strong> the two given straight lines. Using the same<br />

figure as before, but with D taking the position shown by (D)<br />

in the figure, we take that point such that<br />

or<br />

OC .<br />

AD — the given rectangle.<br />

We have then to draw ON'M through<br />

B'N' .AM=OC.AD,<br />

B'N':OC=AD:AM.<br />

But, by parallels, B''N' : OC = B'M: CM;<br />

therefore<br />

AM :CM=AD: B'M<br />

= MD:B'C,<br />

such that<br />

so that<br />

B'M .MD = AD. B'C.<br />

Hence, as before, the problem is reduced to an application<br />

<strong>of</strong> a rectangle in the well-known manner. The complete<br />

n 2

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