Commentary on Proposition X.99

Back to Proposition X.99.

[Folio 0224 | p. 220]

We have to find and classify

$$\frac{1}{\sigma}\left(k^{\frac{1}{4}}\rho \sim \frac{\sqrt{\lambda}\cdot\rho}{k^{\frac{1}{4}}}\right)^2.$$

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

$$\left.\begin{aligned} \sigma x &= \sqrt{k}\cdot\rho^2 \\ \sigma y &= \frac{\lambda}{\sqrt{k}}\cdot\rho^2 \\ \sigma \cdot 2z &= 2\sqrt{\lambda}\cdot\rho^2 \end{aligned}\right\}.$$

$(\alpha)$ Then $\sigma(x + y)$ is a medial area, whence $(x + y)$ is rational and $\smile \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$ Also $\sigma \cdot 2z$ is medial, whence $2z$ is rational and $\smile \sigma\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2).$

Again

$$k^{\frac{1}{4}}\rho \smile \frac{\sqrt{\lambda}\cdot\rho}{k^{\frac{1}{4}}},$$

whence

$$\sqrt{k}\cdot\rho^2 \smile \sqrt{\lambda}\cdot\rho^2.$$

And

$$\sqrt{k}\cdot\rho^2 \sim \left(\sqrt{k}\cdot\rho^2 + \frac{\lambda}{\sqrt{k}}\rho^2\right),$$

while

$$\sqrt{\lambda}\cdot\rho^2 \sim 2\sqrt{\lambda}\cdot\rho^2;$$

therefore

$$\left(\sqrt{k}\cdot\rho^2 + \frac{\lambda}{\sqrt{k}}\rho^2\right) \smile 2\sqrt{\lambda}\cdot\rho^2,$$

or

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

and

$$(x + y) \smile 2z\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3).$$

[Folio 0225 | p. 221]

Thus [(1), (2), (3)] $(x + y)$, $2z$ are rational and $\sim$, so that $(x + y) - 2z$ is an apotome.

$(\beta)$ $\sigma x \sim \sigma y$, so that $x \sim y$.

And, as before,

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

Therefore [x. 17]

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

And neither $(x + y)$ nor $2z$ is $\sim \sigma$.

Therefore $(x + y) - 2z$ is a third apotome.

It is of course equal to

$$\frac{\rho^2}{\sigma}\left\{\frac{k + \lambda}{\sqrt{k}} - 2\sqrt{\lambda}\right\}.$$