Commentary on Proposition X.95

Back to Proposition X.95.

[Folio 0212 | p. 208]

Here the problem is to find and classify

$$\sqrt{\rho(k\rho\sqrt{1+\lambda} - k\rho)}.$$

As usual, we put

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

and, $u$, $v$ being found, we take

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

Then $(x - y)$ so found is our required square root. This fact is proved as before, and, as before, we see that

$$xy = \frac{1}{2} k\rho^2.$$

Now 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 $(x^2 + y^2) = \rho(u + v) = k\rho^2 \sqrt{1+\lambda}$, which is a medial area. And $2xy = k\rho^2$, which is a rational area. Hence $(x - y)$ is the “side” of a medial, minus a rational, area. [X. 77]

Algebraical solution gives

$$u = \frac{k\rho}{2}(\sqrt{1+\lambda} + \sqrt{\lambda}),$$

$$v = \frac{k\rho}{2}(\sqrt{1+\lambda} - \sqrt{\lambda}),$$

and therefore

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

which is, as explained in the note to x. 58, the lesser positive root of the equation

$$x^4 - 2k\rho^2\sqrt{1+\lambda} \cdot x^2 + \lambda k^2\rho^4 = 0.$$