Commentary on Proposition X.100

Back to Proposition X.100.

[Folio 0227 | p. 223]

We have to find and classify

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

We will call this, for brevity,

$$\frac{1}{\sigma}(u-v)^2.$$

Take $x, y, z$ such that

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

where it has to be remembered that $u^2, v^2$ are incommensurable, $(u^2+v^2)$ is rational, and $2uv$ medial.

[Folio 0228 | p. 224]

It follows that $\sigma(x+y)$ is rational and $\sigma \cdot 2z$ medial, so that $(x+y)$ is rational and $\sim \sigma$ ………………………………(1), while $2z$ is rational and $\smile \sigma$ ………………………………….(2), and $\sigma(x+y) \smile \sigma \cdot 2z$, so that $(x+y) \smile 2z$ …………………………………………(3).

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

Next, since

$$u^2 \smile v^2,$$

$$\sigma x \smile \sigma y,$$

or

$$x \smile y.$$

And it is proved, as usual, that

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

Therefore [X. 18] $\sqrt{(x+y)^2 - (2z)^2} \smile (x+y)$.

But $(x+y) \sim \sigma$, therefore $x+y-2z$ is a fourth apotome.

Its value is of course $\frac{\rho^2}{\sigma}\left(1 - \frac{1}{\sqrt{1+k^2}}\right)$.