Commentary on Proposition X.65

Back to Proposition X.65.

[Folio 0149 | p. 145]

To prove that [cf. X. 41]

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

is a sixth binomial straight line.

Denote it by $\frac{1}{\sigma} (u + v)^2$, and put

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

Now, by X. 41, $u^2 \smile v^2$, $(u^2 + v^2)$ is medial, $2uv$ is medial, and $(u^2 + v^2) \smile 2uv$.

($\alpha$) In this case $\sigma(x + y)$ is medial ; therefore $(x + y)$ is rational and $\smile \sigma$ ……………………………(1).

In like manner, $2z$ is rational and $\smile \sigma$ ……………………….(2).

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

$$ (x + y) \smile 2z \text{ .........................................(3).} $$

Therefore $(x + y) + 2z$ is a binomial straight line.

($\beta$) As before,

$$ \begin{aligned} x + y &> 2z, \\ xy &= z^2, \\ x &\smile y ; \end{aligned} $$

therefore [X. 18] $\sqrt{(x + y)^2 - (2z)^2} \smile (x + y)$ …………………..(4).

Hence [(1), (2), (3), (4)] $(x + y) + 2z$ is a sixth binomial straight line.

It is obviously

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