Commentary on Proposition X.94

Back to Proposition X.94.

[Folio 0209 | p. 205]

We have here to find and classify the straight line

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

As usual, we find $u$, $v$ from the equations

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

and then, giving $u$, $v$ their values, we put

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

Then $(x - y)$ is the required square root. This is proved in the same way as before, and, as before, it is proved that

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

Now, from (1), by x. 18,

$$u \mathbin{\sim\kern-0.6em\raisebox{0.1ex}{/}} v;$$

therefore

$$\rho u \mathbin{\sim\kern-0.6em\raisebox{0.1ex}{/}} \rho v,$$

or

$$x^2 \mathbin{\sim\kern-0.6em\raisebox{0.1ex}{/}} y^2,$$

so that $x$, $y$ are incommensurable in square. And $x^2 + y^2$, or $\rho(u + v)$, is a rational area ($k\rho^2$). But $2xy = \frac{k\rho^2}{\sqrt{1+\lambda}}$, which is a medial area. Hence [x. 76] $(x - y)$ is the irrational straight line called minor.

[Folio 0210 | p. 206]

Algebraical solution gives

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

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

whence

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

As explained in the note on x. 57, this is the lesser positive root of the equation

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