Commentary on Proposition X.97

Back to Proposition X.97.

Here begins the hexad of propositions solving the problems which are the converse of those in the hexad just concluded. Props. 97 to 102 correspond of course to Props. 60 to 65 relating to the binomials etc.

We have in X. 97 to prove that, $(\rho - \sqrt{k} \cdot \rho)$ being an apotome,

$$\frac{(\rho - \sqrt{k} \cdot \rho)^2}{\sigma}$$

is a first apotome, and we have to find it geometrically.

Euclid’s procedure may be represented thus.

Take $x$, $y$, $z$ such that

$$\left.\begin{aligned} \sigma x &= \rho^2 \\ \sigma y &= k\rho^2 \\ \sigma \cdot 2z &= 2\sqrt{k} \cdot \rho^2 \end{aligned}\right\} \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$

Thus

$$(x + y) - 2z = \frac{(\rho - \sqrt{k} \cdot \rho)^2}{\sigma},$$

and we have to prove that $(x + y) - 2z$ is a first apotome.

$(\alpha)$ Now $\rho^2 + k\rho^2$, or $\sigma (x + y)$, is rational; therefore $(x + y)$ is rational and $\sim \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2).$

And $2\sqrt{k} \cdot \rho^2$, or $\sigma \cdot 2z$, is medial: therefore $2z$ is rational and $\smile \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3).$

But, $\sigma (x + y)$ being rational, and $\sigma \cdot 2z$ medial,

$$\sigma (x + y) \smile \sigma \cdot 2z,$$

whence

$$(x + y) \smile 2z.$$

Therefore, since $(x + y)$, $2z$ are both rational [(2), (3)], $(x + y)$, $2z$ are rational and $\sim\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(4).$

Hence $(x + y) - 2z$ is an apotome.

$(\beta)$ Since $\sqrt{k} \cdot \rho^2$ is a mean proportional between $\rho^2$, $k\rho^2$, $\sigma z$ is a mean proportional between $\sigma x$, $\sigma y$ [by (1)].

That is,

$$\sigma x : \sigma z = \sigma z : \sigma y,$$

or

$$x : z = z : y,$$

and

$$xy = z^2\text{, or }\frac{1}{4}(2z)^2\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(5).$$

And, since $\rho^2 \sim k\rho^2$,

$$\begin{aligned} \sigma x &\sim \sigma y, \\ \text{or } \quad x &\sim y \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(6). \end{aligned}$$

Hence [(5), (6)], by x. 17,

$$\sqrt{(x + y)^2 - (2z)^2} \sim (x + y).$$

And [(4)] $(x + y)$, $2z$ are rational and $\sim$, while [(2)] $(x + y) \sim \sigma$; therefore $(x + y) - 2z$ is a first apotome.

The actual value of $(x + y) - 2z$ is of course

$$\frac{\rho^2}{\sigma}\{(1 + k) - 2\sqrt{k}\}.$$