Commentary on Proposition X.34

Back to Proposition X.34.

In this case we take [x. 31, 2nd part] the medial straight lines

$$\frac{\rho}{(1 + k^2)^{\frac{1}{4}}}, \quad \frac{\rho}{(1 + k^2)^{\frac{3}{4}}}.$$

Solve the equations

$$\left. \begin{aligned} x + y &= \frac{\rho}{(1 + k^2)^{\frac{1}{4}}} \\ xy &= \frac{\rho^2}{4(1 + k^2)^{\frac{3}{2}}} \end{aligned} \right\} \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$

Take $u$, $v$ such that, if $x$, $y$ be the result of the solution,

$$\left. \begin{aligned} u^2 &= \frac{\rho}{(1 + k^2)^{\frac{1}{4}}} \cdot x \\ v^2 &= \frac{\rho}{(1 + k^2)^{\frac{1}{4}}} \cdot y \end{aligned} \right\} \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2),$$

and $u$, $v$ are straight lines satisfying the given conditions.

Euclid’s proof is similar to the preceding.

($\alpha$) From (1) it follows [x. 18] that

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

whence

$$u^2 \mathbin{\smile} v^2,$$

and $u$, $v$ are thus incommensurable in square.

($\beta$) $u^2 + v^2 = \frac{\rho^2}{\sqrt{1 + k^2}}$, which is a medial area.

($\gamma$)

$$\begin{aligned} uv &= \frac{\rho}{(1 + k^2)^{\frac{1}{4}}} \cdot \sqrt{xy} \\ &= \frac{1}{2} \cdot \frac{\rho^2}{1 + k^2}, \text{ which is a rational area.} \end{aligned}$$

Therefore $uv$ is rational.

To find the actual form of $u$, $v$, we have, by solving the equations (1) (if $x > y$),

$$x = \frac{\rho}{2(1 + k^2)^{\frac{3}{4}}} (\sqrt{1 + k^2} + k),$$

$$y = \frac{\rho}{2(1 + k^2)^{\frac{3}{4}}} (\sqrt{1 + k^2} - k);$$

and hence

$$u = \frac{\rho}{\sqrt{2}(1 + k^2)^{\frac{1}{2}}} \sqrt{\sqrt{1 + k^2} + k},$$

$$v = \frac{\rho}{\sqrt{2}(1 + k^2)^{\frac{1}{2}}} \sqrt{\sqrt{1 + k^2} - k}.$$

Bearing in mind the forms which $\frac{\rho}{(1 + k^2)^{\frac{1}{4}}}$, $\frac{\rho}{(1 + k^2)^{\frac{3}{4}}}$ may take (see note on x. 31), we shall find that $u$, $v$ may have any of the forms

$$(1) \quad \sqrt{\frac{(a + \sqrt{B})\sqrt{a^2 - B}}{2}}, \quad \sqrt{\frac{(a - \sqrt{B})\sqrt{a^2 - B}}{2}};$$

$$(2) \quad \sqrt{\frac{(\sqrt{A} + \sqrt{B})\sqrt{A - B}}{2}}, \quad \sqrt{\frac{(\sqrt{A} - \sqrt{B})\sqrt{A - B}}{2}};$$

$$(3) \quad \sqrt{\frac{(\sqrt{A} + b)\sqrt{A - b^2}}{2}}, \quad \sqrt{\frac{(\sqrt{A} - b)\sqrt{A - b^2}}{2}}.$$