Commentary on Proposition X.56

Back to Proposition X.56.

[Folio 0129 | p. 125]

This proposition in like manner is the equivalent of finding the square root of the product of $\rho$ and the third binomial [X. 50], i.e. of the expression

$$\rho(\sqrt{k}\cdot\rho + \sqrt{k}\cdot\rho\sqrt{1-\lambda^2}).$$

As before, put

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

Next, $u$, $v$ being found, let

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

then $(x+y)$ is the square root required and is a second bimedial straight line. [X. 38]

For, as in the last proposition, it is proved that $(x+y)$ is the square root, and $x$, $y$ are medial and $\sim$.

Again, $xy = \frac{1}{2}\sqrt{k}\cdot\rho^2\sqrt{1-\lambda^2}$, which is medial.

Hence $(x+y)$ is a second bimedial straight line.

By solving equations (1), we find

$$\begin{aligned} u &= \frac{1}{2}(\sqrt{k}\cdot\rho + \lambda\sqrt{k}\cdot\rho), \\ v &= \frac{1}{2}(\sqrt{k}\cdot\rho - \lambda\sqrt{k}\cdot\rho), \end{aligned}$$

and

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

The corresponding second apotome of a medial found in X. 93 is the same thing with a minus sign between the terms, and the two are the roots (cf. note on X. 50) of the biquadratic equation

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