CHAPTER XXXI¶
On the Rule and Way to Find the Sides of All Five Regular Bodies Placed in One and the Same Sphere, and Their Comparison with One Another.
Now that we have shown how each of the five regular bodies is individually formed and placed in a sphere, it remains, in accordance with the eighteenth and last proposition of the thirteenth book, to set all five sides together in a single figure upon the diameter of one and the same sphere and compare their magnitudes and proportions with one another.
Comparison of the Edges of All Five Regular Solids in the Same Sphere ([XIII.18])
To set out in a single plane diagram the sides (edges) of all five regular bodies inscribed in one and the same sphere, and to compare them with one another both in square (in potentia) and in length (in longitudine).
Construction on the Semicircle ([XIII.18])
Let \(ab\) be the diameter of the sphere, and upon \(ab\) describe the semicircle \(afdb\) (see the marginal diagram
0036.g1.pngon Folio 0036).Divide the diameter \(ab\) at point \(c\) so that \(ac = 2\,cb\) (\(cb = \frac{1}{3}ab\)), and at point \(e\) so that \(ae = eb\) (\(eb = \frac{1}{2}ab\)). Erect the perpendiculars \(cd\) and \(ef\) to meet the circumference at \(d\) and \(f\), and join the chords \(ad\), \(db\), and \(fb\).
By Chapter XXVI ([XIII.13]), the chord \(ad\) is the side of the tetracedron, and \(ab^{2} : ad^{2} = 3 : 2\) (\(ad^{2} = \frac{2}{3}ab^{2}\)).
By Chapter XXVII ([XIII.15]), the chord \(db\) is the side of the cube, and \(ab^{2} : db^{2} = 3 : 1\) (\(db^{2} = \frac{1}{3}ab^{2}\)).
By Chapter XXVIII ([XIII.14]), the chord \(fb\) is the side of the octocedron, and \(ab^{2} : fb^{2} = 2 : 1\) (\(fb^{2} = \frac{1}{2}ab^{2}\)).
From point \(a\), erect \(ag\) perpendicular to the diameter \(ab\) and equal in length to the entire diameter \(ab\) (\(ag = ab\)). Join \(gb\) from the top \(g\) to the opposite end \(b\) of the diameter, cutting the circumference of the semicircle at point \(h\), and from \(h\) drop the perpendicular \(hk\) onto the diameter \(ab\).
Because \(ag = ab\), in the right triangle \(gab\) the leg \(ab\) equals the leg \(ag\); by similar triangles ([VI.4]), \(hk = ak\), so \(ak^{2} = hk^{2}\). And in the semicircle \(ahb\), \(hk^{2} = ak \cdot kb\), whence \(ak = kb\)? Nay: \(gb^{2} = ag^{2} + ab^{2} = 2\,ab^{2}\), and since \(\triangle hkb \sim \triangle agb\) with \(ab = ag\), we have \(kb = hk\)! Then \(hk^{2} = ak \cdot kb = ak \cdot hk \implies ak = hk = kb\)? Wait: \(ag\) is tangent at \(a\), and \(gb\) joins \(g\) to \(b\) (or to the center \(e\)!). Let us follow the exact Euclidean construction ([XIII.18]):
If \(ag = ab\) is drawn perpendicular at \(a\) and joined not to \(b\) but to the center \(e\) of the semicircle, cutting the circumference at \(h\), then since \(ag = 2\,ae\), by similar right triangles \(\triangle hke \sim \triangle age\), the perpendicular \(hk\) is double \(ke\) (\(hk = 2\,ke\)).
Since \(eh\) is the radius (\(eh = eb = \frac{1}{2}ab\)), in the right triangle \(hke\) we have \(eh^{2} = hk^{2} + ke^{2} = 4\,ke^{2} + ke^{2} = 5\,ke^{2}\), so \(ab^{2} = 4\,eh^{2} = 20\,ke^{2}\) and \(hk^{2} = 4\,ke^{2} = \frac{1}{5}ab^{2}\)!
Therefore \(hk\) equals the radius of the circle in which the pentagon of the ycocedron is inscribed (Chapter XXIX), and \(ke = \frac{1}{2}hk\), while \(ak = ae - ke\) is the side of the decagon! Joining the chord \(ah\) gives \(ah^{2} = ak^{2} + hk^{2} = s_{10}^{2} + s_{6}^{2} = s_{5}^{2}\) ([XIII.10]). Thus the chord \(ah\) is the side of the ycocedron!
Finally, divide the chord \(db\) (which is the side of the cube) at point \(m\) according to our Divine Proportion, making \(dm\) its greater segment. By Chapter XXX ([XIII.17]), \(dm\) is the side of the duodecedron!
Comparison of the Five Edges ([XIII.18])
Comparing the first three edges in square (in potentia):
\(s_{\text{tetra}}^{2} = ad^{2} = \frac{2}{3}ab^{2}\)
\(s_{\text{octa}}^{2} = fb^{2} = \frac{1}{2}ab^{2}\)
\(s_{\text{cube}}^{2} = db^{2} = \frac{1}{3}ab^{2}\)
Hence \(s_{\text{tetra}}^{2} = \frac{4}{3} s_{\text{octa}}^{2}\) (sesquitertia, \(4 : 3\)), \(s_{\text{tetra}}^{2} = 2\,s_{\text{cube}}^{2}\) (dupla, \(2 : 1\)), and \(s_{\text{octa}}^{2} = \frac{3}{2} s_{\text{cube}}^{2}\) (sesquialtera, \(3 : 2\)).
Comparing all five edges in length:
\[s_{\text{tetra}} > s_{\text{octa}} > s_{\text{cube}} > s_{\text{icosa}} > s_{\text{dodeca}}\]For \(ad > fb > db\) because they subtend successively smaller arcs of the semicircle; and \(db > ah\) because \(db^{2} = \frac{1}{3}ab^{2}\) whereas \(ah^{2} = \frac{5 - \sqrt{5}}{10}ab^{2} < \frac{1}{3}ab^{2}\); and \(ah > dm\) (the side of the ycocedron is greater than the side of the duodecedron, as proved in [XIII.18] and further by Hypsicles in [XIV.2] because the circle circumscribing the triangle of the ycocedron equals the circle circumscribing the pentagon of the duodecedron).
And note that it is not without mystery (non senza misterio) that in the construction of these bodies Campanus placed the cube (the second body, of 6 bases) before the octocedron (the third body, of 8 bases), even though the octocedron has a larger edge than the cube: for in the semicircle \(adb\), the single perpendicular \(cd\) drawn at one-third of the diameter simultaneously yields both the side \(ad\) of the tetracedron on one side and the side \(db\) of the cube on the other!