Commentary on Proposition X.71

Back to Proposition X.71.

[Folio 0158 | p. 154]

A rational area being of the form $k\rho^2$, and a medial area of the form $\sqrt\lambda . \rho^2$, the problem is to classify

$$\sqrt{k\rho^2 + \sqrt\lambda . \rho^2}$$

according to the different possible relations between $k, \lambda$.

Put

$$\begin{aligned} \sigma u &= k\rho^2, \\ \sigma v &= \sqrt\lambda . \rho^2. \end{aligned}$$

Then, since the former rectangle is rational, the latter medial, $u$ is rational and $\frown \sigma$, $v$ is rational and $\smile \sigma$.

Also the rectangles are incommensurable; so that

$$u \smile v.$$

Hence $u, v$ are rational and $\sim$; whence $(u + v)$ is a binomial straight line.

[Folio 0159 | p. 155]

The possibilities now are as follows :

1. $u > v$. Then either

1. $\sqrt{u^2 - v^2} \frown u$, or (2) $\sqrt{u^2 - v^2} \smile u$, while in both cases $u \frown \sigma$.

In case (1) $(u + v)$ is a first binomial straight line, and in case (2) $(u + v)$ is a fourth binomial straight line.

Thus $\sqrt{\sigma (u + v)}$ is either (1) a binomial straight line [X. 54] or (2) a major irrational straight line [X. 57].

2. $v > u$. Then either

1. $\sqrt{v^2 - u^2} \frown v$, or (2) $\sqrt{v^2 - u^2} \smile v$, while in both cases $v \smile \sigma$, but $u \frown \sigma$.

Hence, in case (1), $(v + u)$ is a second binomial straight line, and, in case (2), $(v + u)$ is a fifth binomial straight line.

Thus $\sqrt{\sigma (v + u)}$ is either (1) a first bimedial straight line [X. 55], or (2) a side of a rational plus a medial area [X. 58].