Commentary on Proposition X.55

Back to Proposition X.55.

A second binomial straight line being [x. 49] of the form

$$\frac{k\rho}{\sqrt{1 - \lambda^2}} + k\rho,$$

the present proposition is equivalent to finding the square root of the expression

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

As in the last proposition, Euclid finds $u$, $v$ from the equations

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

then finds $x$, $y$ from the equations

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

and then proves ($\alpha$) that

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

and ($\beta$) that $(x + y)$ is a first bimedial straight line [x. 37].

The steps in the proof are as follows. For ($\alpha$) reference to the corresponding part of the previous proposition suffices.

($\beta$) By (1) and x. 17,

$$u \sim v;$$

therefore $u$, $v$ are both rational and $\cap (u + v)$, and therefore $\cup \rho$ [by (1)] $\ldots(3).$ Hence $\rho u$, $\rho v$, or $x^2$, $y^2$, are medial areas, so that $x$, $y$ are also medial $\ldots\ldots\ldots\ldots\ldots\ldots(4).$ But, since $u \sim v,$

$$x^2 \sim y^2 \ldots\ldots\ldots\ldots\ldots\ldots(5).$$

Again $(u + v)$, or $\frac{k\rho}{\sqrt{1 - \lambda^2}}$, $\cup\ k\rho,$ so that

$$\begin{aligned} u &\cup \frac{1}{2}k\rho, \\ \text{whence} \quad \rho u &\cup \frac{1}{2}k\rho^2, \\ \text{or} \quad x^2 &\cup xy, \\ \text{and} \quad x &\cup y \ldots\ldots\ldots\ldots\ldots\ldots(6). \end{aligned}$$

Thus [(4), (5), (6)] $x$, $y$ are medial and $\sim$. Lastly, $xy = \frac{1}{2}k\rho^2$, which is rational. Therefore $(x + y)$ is a first bimedial straight line. The actual straight lines obtained from (1) are

$$\left. \begin{aligned} u &= \frac{1}{2} \cdot \frac{1 + \lambda}{\sqrt{1 - \lambda^2}} k\rho \\ v &= \frac{1}{2} \cdot \frac{1 - \lambda}{\sqrt{1 - \lambda^2}} k\rho \end{aligned} \right\},$$

so that

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

The corresponding first apotome of a medial straight line found in x. 92 being the same thing with a minus sign between the terms, the two expressions are the roots of the biquadratic

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

being the equation in $x^2$ corresponding to that in $x$ in x. 49.