Commentary on Proposition XIII.11

Back to Proposition XIII.11.

Here we require certain definitions and propositions of Book X. First we require the definition of an apotome [see X. 73], which is a straight line of the form $(\rho \sim \sqrt{k} \cdot \rho)$, where $\rho$ is a “rational” straight line and $k$ is any integer or numerical fraction, the square root of which is not integral or expressible in integers. The lesser of the straight lines $\rho$, $\sqrt{k} \cdot \rho$ is the annex.

Next we require the definition of the fourth apotome [X. Deff. III. (after X. 84)], which is a straight line of the form $(x - y)$, where $x$, $y$ are (being both rational and commensurable in square only) are also such that $\sqrt{x^2 - y^2}$ is incommensurable with $x$, while $x$ is commensurable with a given rational straight line $\rho$. As shown on X. 88 (note), the fourth apotome is of the form

$$(k\rho - \frac{k\rho}{\sqrt{1 + \lambda}}).$$

Lastly the minor (straight line) is the irrational straight line defined in

    1. It is of the form $(x - y)$, where $x$, $y$ are incommensurable in square, and $(x^2 + y^2)$ is ‘rational,’ while $xy$ is ‘medial.’ As shown in the note on

  1. 76, the minor irrational straight line is of the form

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

The proposition may be put as follows. ABCDE being a regular pentagon inscribed in a circle, AG, BH the diameters through A, B meeting CD in L and AC in M respectively, FK is made equal to $\frac{1}{4}$ AF. Now, the radius AF ($r$) being rational, so are FK, BK. The arcs CG, GD are equal; hence $\angle\text{s}$ at L are right, and CD $= 2$ CL. Similarly $\angle\text{s}$ at M are right, and AC $= 2$ CM.

We have to prove

  1. that BM is an apotome,

  2. that BM is a fourth apotome,

  3. that BA is a minor irrational straight line.

Remembering that, if CA is divided in extreme and mean ratio, the greater segment is equal to the side of the pentagon [XIII. 8], and that accordingly [XIII. 1] $($ CD $+ \frac{1}{2}$ CA $)^2 = 5(\frac{1}{2}$ CA $)^2$, we work towards a proportion containing the ratio $($ CD $+ $ CM $)^2 : $ CM $^2$, thus.

The $\Delta\text{s}$ ACL, AFM are equiangular and therefore similar. Therefore LC $:$ CA $= $ MF $:$ FA, and accordingly $2$ LC $:$ CA $= $ MF $: \frac{1}{2}$ FA; thus $2$ LC $: \frac{1}{2}$ CA $= $ MF $: \frac{1}{4}$ FA, or DC $:$ CM $= $ MF $:$ FK; whence, componendo, and squaring, $($ DC $+ $ CM $)^2 : $ CM $^2 = $ MK $^2 : $ KF $^2$. But $($ DC $+ $ CM $)^2 = 5$ CM $^2$; therefore MK $^2 = 5$ KF $^2$.

[This means that MK $^2 = \frac{5}{16}r^2$, or MK $= \frac{\sqrt{5}}{4}r$.] It follows that, KF being rational, MK $^2$, and therefore MK, is rational.

  1. To prove that BM is an apotome and MK its annex. We have BF $= 4$ FK; therefore BK $= 5$ FK, BK $^2 = 25$ FK $^2$ $= 5$ MK $^2$, from above; therefore BK $^2$ has not to MK $^2$ the ratio of a square number to a square number; therefore BK, MK are incommensurable in length. They are therefore rational and commensurable in square only; accordingly BM is an apotome. [BK $^2 = 5$ MK $^2 = \frac{25}{16}r^2$, and BK $= \frac{5}{4}r$. Consequently BK $- $ MK $= (\frac{5}{4}r - \frac{\sqrt{5}}{4}r)$.]

  2. To prove that BM is a fourth apotome. First, since KF, FB are commensurable, BK, BF are commensurable, i.e. BK is commensurable with BH, a given rational straight line. Secondly, if N $^2 = $ BK $^2 - $ KM $^2$, since BK $^2 : $ KM $^2 = 5 : 1$, it follows that BK $^2 : $ N $^2 = 5 : 4$, whence BK, N are incommensurable. Therefore BM is a fourth apotome.

  3. To prove that BA is a minor irrational straight line. If a fourth apotome form a rectangle with a rational straight line, the side of the square equivalent to the rectangle is minor [X. 94]. Now BA $^2 = $ HB $\cdot$ BM, HB is rational, and BM is a fourth apotome; therefore BA is a minor irrational straight line. [BA $= r\sqrt{2} \cdot \sqrt{\frac{5}{4} - \frac{\sqrt{5}}{4}} = \frac{r}{2}\sqrt{10 - 2\sqrt{5}}$. If this is separated into the difference between two straight lines, we have BA $= \frac{r}{2}\sqrt{5 + 2\sqrt{5}} - \frac{r}{2}\sqrt{5 - 2\sqrt{5}}$.]