Commentary on Proposition X.102

Back to Proposition X.102.

[Folio 0233 | p. 229]

We have to find and classify

$$\frac{1}{\sigma}\left\{\frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1+\frac{k}{\sqrt{1+k^2}}} - \frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1-\frac{k}{\sqrt{1+k^2}}}\right\}^2.$$

Call this $\frac{1}{\sigma}(u-v)^2$, and put

$$\left.\begin{aligned} \sigma x &= u^2, \\ \sigma y &= v^2, \\ \sigma \cdot 2z &= 2uv. \end{aligned}\right.$$

Here $u^2, v^2$ are incommensurable, $(u^2+v^2), 2uv$ are both medial areas, and

$$(u^2+v^2) \smile 2uv.$$

Since $\sigma(x+y), \sigma \cdot 2z$ are medial and incommensurable, $(x+y)$ is rational and $\smile \sigma$, $2z$ is rational and $\smile \sigma$, and

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

Hence $(x+y), 2z$ are rational and $\sim$, so that $(x+y) - 2z$ is an apotome.

Again, since $u^2, v^2$, or $\sigma x, \sigma y$, are incommensurable,

$$x \smile y.$$

And, as before,

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

Therefore [X. 18]

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

And neither $(x+y)$ nor $2z$ is $\frown \sigma$; therefore $(x+y) - 2z$ is a sixth apotome.

It is of course

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