Commentary on Proposition X.96

Back to Proposition X.96.

[Folio 0215 | p. 211]

We have to find and classify

$$\sqrt{\rho(\sqrt{k}.\rho - \sqrt{\lambda}.\rho)}.$$

Put, as usual,

$$\left. \begin{aligned} u + v &= \sqrt{k}.\rho \\ uv &= \frac{1}{4}\lambda\rho^2 \end{aligned} \right\} \dots\dots\dots\dots\dots\dots\dots\dots(1),$$

and, $u$, $v$ being thus found, let

$$\left. \begin{aligned} x^2 &= \rho u \\ y^2 &= \rho v \end{aligned} \right\} \dots\dots\dots\dots\dots\dots\dots\dots(2).$$

Then, as before, $(x - y)$ is the square root required. For, from (1), by x. 18,

$$u \mathbin{\smile} v,$$

whence

$$\rho u \mathbin{\smile} \rho v,$$

or

$$x^2 \mathbin{\smile} y^2,$$

and $x, y$ are incommensurable in square.

Next, $x^2 + y^2 = \rho(u + v) = \sqrt{k}.\rho^2$, which is a medial area. Also $2xy = \sqrt{\lambda}.\rho^2$, which is again a medial area. Lastly, $\sqrt{k}.\rho$, $\sqrt{\lambda}.\rho$ are by hypothesis $\sim$, so that

$$\sqrt{k}.\rho \mathbin{\smile} \sqrt{\lambda}.\rho,$$

whence

$$\sqrt{k}.\rho^2 \mathbin{\smile} \sqrt{\lambda}.\rho^2,$$

or

$$(x^2 + y^2) \mathbin{\smile} 2xy.$$

Thus $(x - y)$ is the “side” of a medial, minus a medial, area [x. 78].

Algebraical solution gives

$$u = \frac{\rho}{2}(\sqrt{k} + \sqrt{k - \lambda}),$$

$$v = \frac{\rho}{2}(\sqrt{k} - \sqrt{k - \lambda}),$$

whence

$$x - y = \rho\sqrt{\frac{1}{2}(\sqrt{k} + \sqrt{k - \lambda})} - \rho\sqrt{\frac{1}{2}(\sqrt{k} - \sqrt{k - \lambda})}.$$

This, as explained in the note on x. 59, is the lesser positive root of the equation

$$x^4 - 2\sqrt{k}.\rho^2 x^2 + (k - \lambda)\rho^4 = 0.$$