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A history of Greek mathematics - Wilbourhall.org

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THE COLLECTION. BOOK III 365<br />

Pappus does not seem to have seen this, for he observes<br />

that the geometer in question, though saying that DF is<br />

a harmonic mean, does not say how it is a harmonic mean<br />

or between what straight lines.<br />

In the next chapters (pp. 84-104) Pappus, following Nicomachus<br />

and others, defines seven more means, three <strong>of</strong> which<br />

were ancient and the last four more modern, and shows how<br />

we can form all ten means as linear functions <strong>of</strong> oc, ft, y, where<br />

a, ft, y are in geometrical progression. The expositiop has<br />

already been described (vol. i, pp. 86-9).<br />

Section (3).<br />

The 'Paradoxes' <strong>of</strong> Erycinus.<br />

The third section <strong>of</strong> Book III (pp. 104-30) contains a series<br />

<strong>of</strong> the same sort, which are curious rather<br />

<strong>of</strong> propositions, all<br />

than geometrically important. They appear to have been<br />

taken direct from a collection <strong>of</strong> Paradoxes by one Erycinus. 1<br />

The first set <strong>of</strong> these propositions (Props. 28-34) are connected<br />

with Eucl. I. 21, which says that, if from the extremities<br />

<strong>of</strong> the base <strong>of</strong> any triangle two straight lines be drawn meeting<br />

at any point within the triangle, the straight lines are together<br />

less than the two sides <strong>of</strong> the triangle other than the base,<br />

but contain a greater angle. It is pointed out that, if the<br />

straight lines are allowed to be drawn from points in the base<br />

other than the extremities, their sum may be greater than the<br />

other two sides <strong>of</strong> the triangle.<br />

The first case taken is that <strong>of</strong> a right-angled triangle ABC<br />

right-angled at B. Draw AD to any point D on BC. Measure<br />

on it BE equal to AB, bisect AE<br />

in F, and join FC. Then shall A<br />

DF+FC be > BA + AC.<br />

For EF+FC=AF + FC> AC.<br />

Add BE and AB respectively,<br />

and we have<br />

BF+FC> BA + AC.<br />

More elaborate propositions are next proved, such as the<br />

following.<br />

1 . In<br />

any triangle, except an equilateral triangle or an isosceles<br />

1<br />

Pappus, iii, p. 106. 5-9.

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