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Commentary on Proposition X.63¶

We have to prove that [cf. x. 39]

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

is a fourth binomial straight line. For brevity we must call this expression

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

Take $x$, $y$, $z$ such that

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

wherein it has to be remembered [x. 39] that $u$, $v$ are incommensurable in square, $(u^2 + v^2)$ is rational, and $uv$ is medial. ($\alpha$) $(u^2 + v^2)$, and therefore $\sigma(x + y)$, is rational ; therefore $(x + y)$ is rational and $\sim \sigma$ …………………………………(1). $2uv$, and therefore $\sigma \cdot 2z$, is medial ; therefore $2z$ is rational and $\smile \sigma$ ……………………………………(2). Thus $(x + y)$, $2z$ are rational and $\sim$ ……………………………..(3), so that $(x + y) + 2z$ is a binomial straight line. ($\beta$) As before, $x + y > 2z$, and $xy = z^2$. Now, since $u^2 \smile v^2$, $\sigma x \smile \sigma y$, or $x \smile y$. Hence [x. 18] $\sqrt{(x + y)^2 - (2z)^2} \smile (x + y)$ ……………………(4). And $(x + y) \sim \sigma$, by (1). Therefore [(3), (4)] $(x + y) + 2z$ is a fourth binomial straight line. It is of course

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