Commentary on Proposition X.64

Back to Proposition X.64.

[Folio 0147 | p. 143]

To prove that [cf. x. 40]

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

is a fifth binomial straight line. For brevity denote it by $\frac{1}{\sigma} (u + v)^2$, and put

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

Remembering that [x. 40] $u^2 \smile v^2$, $(u^2 + v^2)$ is medial, and $2uv$ is rational, we proceed thus. ($\alpha$) $\sigma(x + y)$ is medial ; therefore $(x + y)$ is rational and $\smile \sigma$ ……………………………(1). Next, $\sigma \cdot 2z$ is rational ; therefore $2z$ is rational and $\sim \sigma$ …………………………………(2). Thus $(x + y)$, $2z$ are rational and $\sim$ …………………………..(3), so that $(x + y) + 2z$ is a binomial straight line. ($\beta$) As before,

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

and

$$ x \smile y. $$

Therefore [x. 18] $\sqrt{(x + y)^2 - (2z)^2} \smile (x + y)$ …………………..(4). Hence [(2), (3), (4)] $(x + y) + 2z$ is a fifth binomial straight line. It is of course

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