Commentary on Proposition XIII.15

Back to Proposition XIII.15.

AB is divided so that AC $= 2$ CB; CD is drawn at right angles to AB, and BD is joined.

KG is, by construction, a cube of side equal to BD.

To prove (1) that it is inscribable in a sphere.

Since KE is perpendicular to EH, EF, KE is perpendicular to EG.

Thus, KEG being a right angle, E lies on a semicircle with diameter KG. The same thing is proved in the same way of the other vertices F, H, L, M, N.

Thus the cube is inscribed in the sphere of which KG is a diameter.

  1. KG $^2 = $ KE $^2 + $ EG $^2$ $= $ KE $^2 + 2$ EF $^2$ $= 3$ EK $^2$. Also AB $= 3$ BC, while AB $:$ BC $= $ AB $^2 : $ AB $\cdot$ BC $= $ AB $^2 : $ BD $^2$;

therefore AB $^2 = 3$ BD $^2$.

But BD $= $ EK;

therefore KG $= $ AB.

  1. AB $^2 = 3$ BD $^2$ $= 3$ KE $^2$.

If $r$ be the radius of the circumscribed sphere,

$$\text{(edge of cube)} = \frac{2}{\sqrt{3}} \cdot r = \frac{2}{3}\sqrt{3} \cdot r.$$

Pappus’ solution.

In this case too Pappus (III. pp. 144—148) gives the full analysis and synthesis.

Analysis.

Suppose the problem solved, and let the vertices of the cube be A, B, C, D, E, F, G, H. Draw planes through A, B, C, D and E, F, G, H respectively; these will produce parallel circular sections, which are also equal since the inscribed squares are equal. And CE will be a diameter of the sphere. Join EG. Now, since EG $^2 = 2$ EH $^2 = 2$ GC $^2$, and the angle CGE is right,

CE $^2 = $ GC $^2 + $ EG $^2 = \frac{3}{2}$ EG $^2$.
But CE $^2$ is given;
therefore EG $^2$ is given, so that the circles EFGH, ABCD, and the squares inscribed in them, are given.
H G E F D C A B

Synthesis.

Draw two parallel circular sections with equal diameters $d'$, such that

$$d^2 = \frac{3}{2}d'^2,$$

where $d$ is the diameter of the given sphere. Inscribe a square in one of the circles, as ABCD. In the other circle draw FG equal and parallel to BC, and complete the square on FG inscribed in the circle EFGH.

The eight vertices of the required cube are thus determined.