Commentary on Proposition X.92

Back to Proposition X.92.

There is an evident flaw in the text in the place (Heiberg, p. 282, ll. 17—20: translation p. 196 above) where it is said that “since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore :g:`LP <0199.g1:/ L P />`, :g:`PN <0199.g1:/ P N />` are also medial straight lines commensurable in square only.” It is not till the last lines of the proposition (Heiberg, p. 284, ll. 17, 18) that it is proved that LP, PN are incommensurable in length. What should have been proved in the former passage is that the squares on LP, PN are commensurable, so that LP, PN are commensurable in square (not commensurable in square only). I have supplied the step in the note below: “Also $x^2 \mathbin{\frown} y^2$, since $u \mathbin{\frown} v$.” Theon seems to have observed the omission and to have put “and commensurable with one another” after “medial” in the passage quoted, though even this does not show why the squares on LP, PN are commensurable. One MS. (V) also has “only” ($\mu\acute{o}\nu\omicron\nu$) erased after “commensurable in square.”

This proposition amounts to finding and classifying

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

The method is that of the last proposition. Euclid solves, first, the equations

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

Then, using the values of $u, v$ so found, he puts

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

and $(x - y)$ is the square root required.

That

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

is proved in the same way as is the corresponding fact in x. 91.

From (1)

$$u : \frac{1}{2}k\rho = \frac{1}{2}k\rho : v,$$

so that

$$\rho u : \frac{1}{2}k\rho^2 = \frac{1}{2}k\rho^2 : \rho v.$$

But

$$x^2 : xy = xy : y^2,$$

whence, by (2),

$$xy = \frac{1}{2}k\rho^2 \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3).$$

Therefore

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

Next, we have to prove that $(x - y)$ is a first apotome of a medial straight line.

From (1) it follows, by x. 17, that

$$u \mathbin{\frown} v, \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(4),$$

therefore $u, v$ are both $\mathbin{\frown} (u + v)$.

But [(1)] $(u + v)$ is rational and $\mathbin{\smile} \rho$; therefore $u, v$ are both rational and $\mathbin{\smile} \rho \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(5)$.

Therefore $\rho u, \rho v$, or $x^2, y^2$, are both medial areas, and $x, y$ are medial straight lines $\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(6)$.

Also $x^2 \mathbin{\frown} y^2$, since $u \mathbin{\frown} v$ [(4)] $\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(7)$.

Now $xy$, or $\frac{1}{2}k\rho^2$, is a rational area; therefore

$$xy \mathbin{\smile} y^2,$$

and

$$x \mathbin{\smile} y.$$

Hence [(6), (7), (3)] $x, y$ are medial straight lines commensurable in square only and containing a rational rectangle; therefore $(x - y)$ is a first apotome of a medial straight line.

Algebraical solution of the equations gives

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

and

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

As explained in the note on x. 55, this is the lesser positive root of the equation

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