Commentary on Proposition X.33

Back to Proposition X.33.

Euclid takes the straight lines found in x. 30, viz. $\rho$, $\dfrac{\rho}{\sqrt{1+k^2}}$.

He then solves geometrically the equations

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

If $x, y$ are the values found, he takes $u, v$ such that

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

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

Solving algebraically, we get (if $x > y$)

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

whence

$$\left.\begin{aligned} u &= \frac{\rho}{\sqrt{2}}\sqrt{1 + \frac{k}{\sqrt{1+k^2}}} \\ v &= \frac{\rho}{\sqrt{2}}\sqrt{1 - \frac{k}{\sqrt{1+k^2}}} \end{aligned}\right\} \text{ .........................................(3).}$$

Euclid’s proof that these straight lines fulfil the requirements is as follows.

($\alpha$) The constants in the equations (1) satisfy the conditions of x. 18 ; therefore

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

But $x : y = u^2 : v^2$. Therefore

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

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

($\beta$) $u^2 + v^2 = \rho^2$, which is rational.

($\gamma$) By (1),

$$\sqrt{xy} = \frac{\rho}{2\sqrt{1+k^2}}.$$

By (2),

$$\begin{aligned} uv &= \rho\ .\ \sqrt{xy} \\ &= \frac{\rho^2}{2\sqrt{1+k^2}}. \end{aligned}$$

But $\dfrac{\rho^2}{\sqrt{1+k^2}}$ is a medial area, therefore $uv$ is medial.

Since $\rho$, $\dfrac{\rho}{\sqrt{1+k^2}}$ may have any of the three forms (1) $a$, $\sqrt{a^2-B}$, (2) $\sqrt{A}$, $\sqrt{A-B}$, (3) $\sqrt{A}$, $\sqrt{A-b^2}$, $u, v$ may have any of the forms

$$\begin{array}{lcc} (1) & \sqrt{\dfrac{a^2 + a\sqrt{B}}{2}}, & \sqrt{\dfrac{a^2 - a\sqrt{B}}{2}} ; \\ (2) & \sqrt{\dfrac{A + \sqrt{AB}}{2}}, & \sqrt{\dfrac{A - \sqrt{AB}}{2}} ; \\ (3) & \sqrt{\dfrac{A + b\sqrt{A}}{2}}, & \sqrt{\dfrac{A - b\sqrt{A}}{2}}. \end{array}$$