Commentary on Proposition X.101¶

We have to find and classify

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

Call this $\frac{1}{\sigma}(u-v)^2$, and take $x, y, z$ such that

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

In this case $u^2, v^2$ are incommensurable, $(u^2+v^2)$ is a medial area and $2uv$ a rational area.

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

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

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

Again, as before,

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

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

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

$$x \smile y.$$

Hence [X. 18]

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

And $2z \frown \sigma$.

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

It is of course equal to

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