Commentary on Proposition X.98

Back to Proposition X.98.

[Folio 0221 | p. 217]

In this case we have to find and classify

$$\frac{(k^{\frac{1}{4}}\rho \sim k^{\frac{3}{4}}\rho)^2}{\sigma}.$$

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

$$\left.\begin{aligned} \sigma x &= k^{\frac{1}{2}}\rho^2 \\ \sigma y &= k^{\frac{3}{2}}\rho^2 \\ \sigma \cdot 2z &= 2k\rho^2 \end{aligned}\right\} \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$

$(\alpha)$ Now $k^{\frac{1}{2}}\rho^2$, $k^{\frac{3}{2}}\rho^2$ are medial areas; therefore $\sigma (x + y)$ is medial, whence $(x + y)$ is rational and $\smile \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2).$

But $2k\rho^2$, and therefore $\sigma \cdot 2z$, is rational, whence $2z$ is rational and $\sim \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3).$

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

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

or

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

Hence $(x + y)$, $2z$ are rational straight lines commensurable in square only, and therefore $(x + y) - 2z$ is an apotome.

$(\beta)$ We prove, as before, that

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

Also $k^{\frac{1}{2}}\rho^2 \sim k^{\frac{3}{2}}\rho^2$, or $\sigma x \sim \sigma y$, so that

$$x \sim y\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(5).$$

[This step is omitted in P, and Heiberg accordingly brackets it. The result is, however, assumed.]

Therefore [(4), (5)], by x. 17,

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

And $2z \sim \sigma$. Therefore $(x + y) - 2z$ is a second apotome.

Obviously

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