Commentary on Proposition X.62

Back to Proposition X.62.

[Folio 0143 | p. 139]

We have to prove that [cf. x. 38]

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

is a third binomial straight line. Take $x, y, z$ such that

$$\begin{aligned} \sigma x &= k^{\frac{1}{2}}\rho^2,\\ \sigma y &= \frac{\lambda\rho^2}{k^{\frac{1}{2}}},\\ \sigma \cdot 2z &= 2\sqrt{\lambda}\cdot \rho^2. \end{aligned}$$

($\alpha$) Now $k^{\frac{1}{4}}\rho, \frac{\lambda^{\frac{1}{2}}\rho}{k^{\frac{1}{4}}}$ are medial straight lines commensurable in square only and containing a medial rectangle. [x. 38] The sum of the squares on them, or $\sigma(x + y)$, is medial; therefore

$$(x + y)\text{ is rational and } \smile \sigma \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$

And $\sigma \cdot 2z$ being medial also,

$$2z\text{ is rational and } \smile \sigma \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2).$$

Now

$$k^{\frac{1}{4}}\rho : \frac{\lambda^{\frac{1}{2}}\rho}{k^{\frac{1}{4}}} = (k^{\frac{1}{4}}\rho)^2 : k^{\frac{1}{4}}\rho \cdot \frac{\lambda^{\frac{1}{2}}\rho}{k^{\frac{1}{4}}}$$

$$= \sigma x : \sigma z,$$

whence $\sigma x \smile \sigma z.$ But $(k^{\frac{1}{4}}\rho)^2 \sim \left\{(k^{\frac{1}{4}}\rho)^2 + \left(\frac{\lambda^{\frac{1}{2}}\rho}{k^{\frac{1}{4}}}\right)^2\right\},$ or $\sigma x \sim \sigma(x + y),$ and $\sigma z \sim \sigma \cdot 2z;$ therefore

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

or

$$(x + y) \smile 2z \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3).$$

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

($\beta$) As before,

$$(x + y) > 2z,$$

and

$$x \sim y.$$

Also

$$xy = z^2.$$

Therefore [x. 17]

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

And [(1), (2)] neither $(x + y)$ nor $2z$ is $\sim \sigma.$ Therefore $(x + y) + 2z$ is a third binomial straight line. Obviously

$$(x + y) + 2z = \frac{\rho^2}{\sigma}\left\{\frac{k + \lambda}{\sqrt{k}} + 2\sqrt{\lambda}\right\}.$$