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Commentary on Proposition X.93¶

Here we are to find and classify the irrational straight line

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

Following the same method, we put

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

Next, $u, v$ being found, let

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

then $(x - y)$ is the square root required and is a second apotome of a medial straight line.

That $(x - y)$ is the square root required and that $x^2, y^2$ are medial areas, so that $x, y$ are medial straight lines, is proved exactly as in the last proposition.

The rectangle $xy$, being equal to $\frac{1}{2}\sqrt{k}.\rho^2\sqrt{1-\lambda^2}$, is also medial.

$$\begin{aligned} &\text{Now, from (1), by x. 17,} & u &\mathbin{\frown} v, \\ &\text{whence} & u + v &\mathbin{\frown} u. \\ &\text{But} & (u + v)\text{, or }\sqrt{k}.\rho &\mathbin{\smile} \frac{1}{2}\sqrt{k}.\rho\sqrt{1-\lambda^2}; \\ &\text{therefore} & u &\mathbin{\smile} \frac{1}{2}\sqrt{k}.\rho\sqrt{1-\lambda^2}, \\ &\text{and consequently} & \rho u &\mathbin{\smile} \frac{1}{2}\sqrt{k}.\rho^2\sqrt{1-\lambda^2}, \\ &\text{or} & x^2 &\mathbin{\smile} xy, \\ &\text{whence} & x &\mathbin{\smile} y. \\ &\text{And, since } u \mathbin{\frown} v\text{,} & \rho u &\mathbin{\frown} \rho v, \\ &\text{or} & x^2 &\mathbin{\frown} y^2. \end{aligned}$$

Thus $x, y$ are medial straight lines commensurable in square only. And $xy$ is a medial area. Therefore $(x - y)$ is a second apotome of a medial straight line.

Its actual form is found by solving equations (1), (2); thus

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

and

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

As explained in the note on x. 56, this is the lesser positive root of the equation

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