Commentary on Proposition XIII.14¶
Back to Proposition XIII.14.
I think the accompanying figure will perhaps be clearer than that in Euclid’s text.
EFGH being a square with side equal to BD, it follows that KE, KF, KG, KH are all equal to CB.
So are KL, KM, by construction; hence LE, LF, LG, LH and ME, MF, MG, MH are all equal to EF or BD.
Thus (1) the figure is made up of eight equilateral triangles and is therefore a regular octahedron.
Since KE $= $ KL $= $ KM, the semicircle on LM in the plane LKE passes through E.
Similarly F, G, H lie on semicircles on LM as diameter.
Thus all the vertices of the tetrahedron lie on the sphere of which LM is a diameter.
LE $= $ EM $= $ BD; therefore LM $^2 = 2$ EL $^2 = 2$ BD $^2$ $= $ AB $^2$, or LM $= $ AB.
$$\begin{aligned} \text{\it AB}^2 &= 2\text{\it BD}^2 \\ &= 2\text{\it EF}^2. \end{aligned}$$
If $r$ be the radius of the circumscribed sphere,
$$\text{(edge of octahedron)} = \sqrt{2}\cdot r.$$
Pappus’ method.¶
Pappus (III. pp. 148—150) finds the two equal and parallel sections of the sphere which circumscribe two opposite faces of the octahedron thus.
Analysis.
Suppose the octahedron inscribed, A, B, C; D, E, F being the vertices.
Through ABC, DEF describe planes making the circular sections ABC, DEF.
Since the straight lines DA, DB, DE, DF are equal, the points A, E, F, B lie on a circle of which D is the pole.
Again, since AB, BF, FE, EA are equal, ABFE is a square inscribed in the said circle, and AB, EF are parallel.
Similarly DE is parallel to BC, and DF to AC.
Therefore the circles through D, E, F and A, B, C are parallel; and they are also equal because the equilateral triangles inscribed in them are equal.
Now, ABC, DEF being equal and parallel circular sections, and AB, EF equal and parallel chords “not on the same side of the centres,” AF is a diameter of the sphere. [Pappus has a lemma for this, pp. 136—138].
And AE $=$ EF, so that $\text{\it AF}^2 = 2\text{\it FE}^2$.
But, if $d'$ be the diameter of the circle DEF,
$$d'^2 = \frac{4}{3}\text{\it EF}^2. \qquad \text{[cf. XIII. 12]}$$
Therefore, if $d$ be the diameter of the sphere,
$$d^2 : d'^2 = 3 : 2.$$
Now $d$ is given, and therefore $d'$ is given; hence the circles DEF, ABC are given.
Synthesis.
Draw two equal and parallel circular sections with diameter $d'$, such that
$$d^2 = \frac{3}{2}d'^2,$$
where $d$ is the diameter of the sphere.
Inscribe an equilateral triangle ABC in either circle (ABC).
In the other circle draw EF equal and parallel to AB but on the opposite side of the centre, and complete the inscribed equilateral triangle DEF.
ABCDEF is the octahedron required.
It will be observed that, whereas in the problem of XIII. 13 Euclid first finds the circle circumscribing a face and Pappus first finds an edge, in this problem Euclid finds the edge first and Pappus the circle circumscribing a face.