Commentary on Proposition XIII.18¶
Back to Proposition XIII.18.
for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [XI. 21]
For the same reason, neither can a solid angle be constructed by more than six plane angles.
By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.
By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.
Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.
LEMMA.¶
Let ABCDE be an equilateral and equiangular pentagon, let the circle ABCDE be circumscribed about it, let its centre F be taken, and let FA, FB, FC, FD, FE be joined.
We have seen in the preceding notes that, if $r$ be the radius of the sphere circumscribing the five solid figures,
$$\begin{aligned} (\text{edge of tetrahedron}) &= \tfrac{2}{3} \sqrt{6} \cdot r, \\ (\text{edge of octahedron}) &= \sqrt{2} \cdot r, \\ (\text{edge of cube}) &= \tfrac{2}{3} \sqrt{3} \cdot r, \\ (\text{edge of icosahedron}) &= \tfrac{r}{5} \sqrt{10(5-\sqrt{5})}, \\ (\text{edge of dodecahedron}) &= \tfrac{r}{3} (\sqrt{15}-\sqrt{3}). \end{aligned}$$
Euclid here exhibits the edges of all the five regular solids in one figure.
Make AD equal to $2$ DB. Thus BA $= \tfrac{3}{2}$ AD, and BA $:$ AD $=$ BA $^2 :$ AF $^2$; therefore BA $^2 = \tfrac{3}{2}$ AF $^2$. Thus AF $= \sqrt{\tfrac{2}{3}} \cdot 2r = \tfrac{2}{3} \sqrt{6} \cdot r = (\text{edge of tetrahedron})$.
AB $^2 :$ BF $^2 =$ AB $:$ BD $= 3 : 1$. Therefore BF $^2 = \tfrac{1}{3}$ AB $^2$, or BF $= \frac{2}{\sqrt{3}} \cdot r = \tfrac{2}{3} \sqrt{3} \cdot r = (\text{edge of cube})$.
AB $^2 = 2$ BE $^2$. Therefore BE $= \sqrt{2} \cdot r = (\text{edge of octahedron})$.
Draw AG perpendicular and equal to AB. Join GC, meeting the semicircle in H, and draw HK perpendicular to AB. Then GA $= 2$ AC; therefore, by similar triangles, HK $= 2$ KC. Hence HK $^2 = 4$ KC $^2$, and therefore $5$ KC $^2 =$ HK $^2 +$ KC $^2$ $= $ HC $^2$ $= $ CB $^2$.
Again, since AB $= 2$ CB, and AD $= 2$ DB, by subtraction, BD $= 2$ DC, or BC $= 3$ DC.
Therefore $9$ DC $^2 =$ BC $^2$ $= 5$ KC $^2$.
Hence KC $>$ CD.
Make CL equal to KC, draw LM at right angles to AB, and join AM, MB.
Since CB $^2 = 5$ KC $^2$, AB $^2 = 5$ KL $^2$.
It follows that KL ($= \sqrt{\tfrac{4}{5}} \cdot r$) is the radius of, or the side of the regular hexagon in, the circle containing the pentagonal sections of the icosahedron. [XIII. 16]
And, since $2r = (\text{side of hexagon}) + 2 (\text{side of decagon in same circle})$ [XIII. 16, Por.] AK $=$ LB $= (\text{side of decagon in the said circle})$.
But LM $=$ HK $=$ KL $= (\text{side of hexagon in circle})$.
Therefore LM $^2 +$ LB $^2 (=$ BM $^2) = (\text{side of pentagon in circle})^2$ [XIII. 10] $= (\text{edge of icosahedron})^2$, and BM $= (\text{edge of icosahedron})$.
[More shortly, HK $= 2$ KC, whence HK $^2 = 4$ KC $^2$, and $5$ KC $^2 =$ HC $^2 = r^2$.
Also AK $= r -$ CK $= r \left( 1 - \frac{1}{\sqrt{5}} \right)$.
Thus BM $^2 =$ HK $^2 +$ AK $^2$ $= \tfrac{4}{5} r^2 + r^2 \left( 1 - \frac{1}{\sqrt{5}} \right)^2$ $= r^2 \left( \frac{10}{5} - \frac{2}{\sqrt{5}} \right)$ $= \frac{r^2}{5} (10 - 2\sqrt{5})$, and BM $= \frac{r}{5} \sqrt{10(5-\sqrt{5})} = (\text{edge of icosahedron})$.]
Cut BF (the edge of the cube) in extreme and mean ratio at N. Then, if BN be the greater segment, BN $= (\text{edge of dodecahedron})$. [XIII. 17]
[Solving, we obtain BN $= \frac{\sqrt{5}-1}{2} \cdot$ BF $= \frac{\sqrt{5}-1}{2} \cdot \frac{2}{\sqrt{3}} \cdot r$ $= \frac{r}{3} (\sqrt{15}-\sqrt{3})$ $= (\text{edge of dodecahedron})$.]
If $t, o, c$ are the edges of the tetrahedron, octahedron and cube respectively,
$$4r^2 = \tfrac{3}{2} t^2 = 2o^2 = 3c^2.$$
If each of these equals is put equal to $X$,
$$\begin{aligned} 4r^2 &= X, \\ t^2 &= \tfrac{2}{3} \cdot X, \\ o^2 &= \tfrac{1}{2} \cdot X, \\ c^2 &= \tfrac{1}{3} \cdot X, \end{aligned}$$
whence
$$4r^2 : t^2 : o^2 : c^2 = 6 : 4 : 3 : 2,$$
and the ratios between $2r, t, o, c$ are all rational (in Euclid’s sense). The ratios between these and the edges of the icosahedron and the dodecahedron are irrational.
To prove that
$$(\text{edge of icosahedron}) > (\text{edge of dodecahedron}),$$
i.e. that MB $>$ NB.