Commentary on Proposition X.86

Back to Proposition X.86.

[Folio 0185 | p. 181]

Take, as before, $k\rho$ commensurable in length with $\rho$. Let $m^2, n^2$ be again square numbers, but $(m^2 - n^2)$ not square. Take $x$ such that

$$(m^2 - n^2) : m^2 = k^2\rho^2 : x^2 \tag{1},$$

whence

$$x = k\rho \frac{m}{\sqrt{m^2 - n^2}}$$

$$= \frac{k\rho}{\sqrt{1 - \lambda^2}}, \text{ say.}$$

Thus $x$ is greater than $k\rho$. Then $x - k\rho$, or $\frac{k\rho}{\sqrt{1 - \lambda^2}} - k\rho$, is a second apotome.

For ($\alpha$), as before, $x$ is rational and $\sim k\rho$. ($\beta$) If $x^2 - k^2\rho^2 = y^2$, we have, from (1),

$$m^2 : n^2 = x^2 : y^2.$$

Thus $y$, or $\sqrt{x^2 - k^2\rho^2}$, is commensurable in length with $x$. And $k\rho \frown \rho$. Therefore $x - k\rho$ is a second apotome.

As explained in the note on x. 49, the second apotome

$$\frac{k\rho}{\sqrt{1 - \lambda^2}} - k\rho$$

is the lesser root of the equation

$$x^2 - \frac{2k\rho}{\sqrt{1 - \lambda^2}} \cdot x + \frac{\lambda^2}{1 - \lambda^2} k^2\rho^2 = 0.$$