Commentary on Proposition XIII.17

Back to Proposition XIII.17.

In this proposition we find Euclid using two propositions which precede but are used nowhere else, notably VI. 32, which some authors, in consequence of their having overlooked its use here, have been hard put to it to explain.

Euclid’s construction in this case is really identical with that given by Mr H. M. Taylor, and also referred to by Henrici and Treutlein under “crystal-formation.”

Euclid starts from the cube inscribed in a sphere, as in XIII. 15, and then finds the side of the regular pentagon in which the side of the cube is a diagonal.

Mr Taylor takes $l$ to be the diagonal of a regular pentagon of side $a$, so that, by Ptolemy’s theorem,

$$l^2 = al + a^2,$$

constructs a cube of which $l$ is the edge, and gets the side of the pentagon by drawing ZX from Z, the centre of the cube, perpendicular to the face BF and equal to $\frac{1}{2}(l + a)$, then drawing UV through X parallel to BC, and making UX, XV both equal to $\frac{1}{2}a$.

Euclid finds UV thus.

Draw NO, MH bisecting pairs of opposite sides in the square BF and meeting in P.

Draw GK, HL bisecting pairs of opposite sides in the square BD and meeting in Q.

Divide PN, PO, QH respectively in extreme and mean ratio at R, S, T (PR, PS, TQ being the greater segments); draw RU, SV, TW outwards perpendicular to the respective faces of the cube, and all equal in length to PR, PS, TQ.

Join BU, UV, VC, CW, WB.

Then BUVCW is one of the pentagonal faces of the dodecahedron; and the others can be constructed in the same way.

Euclid now proves

  1. that the pentagon BUVCW is equilateral,

  2. that it is in one plane,

  3. that it is equiangular,

  1. that the vertex U is on the sphere which circumscribes the cube, and hence

  2. that all the other vertices lie on the same sphere, and (6) that the side of the dodecahedron is an apotome.

  1. To prove that the pentagon BUVCW is equilateral. We have BU $^2 = $ BR $^2 + $ RU $^2$ $= ($ BN $^2 + $ NR $^2) + $ RP $^2$ $= ($ PN $^2 + $ NR $^2) + $ RP $^2$ $= 3$ RP $^2 + $ RP $^2$ [XIII. 4] $= 4$ RP $^2$ $= $ UV $^2$. Therefore BU $= $ UV.

U X V E M F y N R P S O B H C T β W Z α' α w β' G Q K Y A L D v u

Similarly it may be proved that BW, WC, CV are all equal to UV or BU.

[Mr Taylor proceeds in this way. With his notation, the projections of BU on BA, BC, BE are respectively $\frac{1}{2}a$, $\frac{1}{2}(l - a)$, $\frac{1}{2}l$. Therefore BU $^2 = \frac{1}{4}a^2 + \frac{1}{4}(l - a)^2 + \frac{1}{4}l^2$ $= \frac{1}{2}(l^2 - al + a^2)$ $= a^2$. Similarly for BW, WC etc.]

  1. To prove that the pentagon BUVCW is in one plane. Draw PX parallel to RU or SV meeting UV in X. Join XH, HW.

Then we have to prove that XH, HW are in one straight line. Now HP, WT, being both perpendicular to the face BD, are parallel. For the same reason XP, HT are parallel. Also, since QH is divided at T in extreme and mean ratio, QH $:$ QT $= $ QT $:$ TH. And QH $= $ HP $,\quad$ QT $= $ WT $= $ PX. Therefore HP $:$ PX $= $ WT $:$ TH. Consequently the triangles HPX, WTH satisfy the conditions of VI. 32 ; hence XHW is a straight line.

[Mr Taylor proves this as follows : The projections of WH, WX on BE are $\frac{1}{2}a$ and $\frac{1}{2}(a + l)$, and the projections of WH, WX on BA are $\frac{1}{2}(l - a)$ and $\frac{1}{2}l$ ; and

$$a : (a + l) = (l - a) : l,$$

since

$$al = l^2 - a^2.$$

Therefore WHX is a straight line.]

  1. To prove that the pentagon BUVCW is equiangular. We have BV $^2 = $ BS $^2 + $ SV $^2$ $= ($ BN $^2 + $ NS $^2) + $ SP $^2$ $= $ PN $^2 + ($ NS $^2 + $ SP $^2)$ $= $ PN $^2 + 3$ PN $^2$, since NS is divided in extreme and mean ratio at P [XIII. 5], so that NS $^2 + $ SP $^2 = 3$ PN $^2$. [XIII. 4] Consequently BV $^2 = 4$ PN $^2$ $= $ BC $^2$, or BV $= $ BC. The $\Delta\text{s }$ UBV, WBC are therefore equal in all respects, and $\angle$ BUV $= \angle$ BWC. Similarly $\angle$ CVU $= \angle$ BWC. Therefore the pentagon is equiangular. [XIII. 7]

  2. To prove that the sphere which circumscribes the cube also circumscribes the dodecahedron we have only to prove that, if Z be the centre of the sphere, ZU $= $ ZB, for example. Now, by XI. 38, XP produced meets the diagonal of the cube, and the portion of XP produced which is within the cube and the diagonal bisect one another. And ZU $^2 = $ ZX $^2 + $ XU $^2$ $= $ NS $^2 + $ PS $^2$ $= 3$ PN $^2$, as before. Also (cf. XIII. 15) ZB $^2 = $ ZP $^2 + $ PB $^2$ $= $ ZP $^2 + $ PN $^2 + $ NB $^2$ $= 3$ PN $^2$. Hence ZU $= $ ZB.

  3. Similarly for ZV, ZW etc.

  1. Since PN is divided in extreme and mean ratio at R, NP $:$ PR $= $ PR $:$ RN. Doubling the terms, we have NO $:$ RS $= $ RS $: ($ NR $+ $ SO $),$ so that, if NO is divided in extreme and mean ratio, the greater segment is equal to RS.

Now, since the diameter of the sphere is rational,

and                     $(\text{diam. of sphere})^2 = 3 (\text{edge of cube})^2$,

the edge of the cube (i.e. NO) is rational. Consequently RS is an apotome.

[This is proved in the spurious XIII. 6 above ; Euclid assumes it, and the words purporting to quote the theorem are probably interpolated, like XIII. 6 itself.]

As a matter of fact, with Mr Taylor’s notation,

$$l^2 = la + a^2,$$

and                             $a = \frac{\sqrt{5} - 1}{2} l$.

Since, if $r$ is the radius of the circumscribing sphere, $r = \sqrt{3} \cdot \frac{l}{2}$,

$$a = \frac{r}{\sqrt{3}}(\sqrt{5} - 1) = \frac{r}{3}(\sqrt{15} - \sqrt{3}).$$

Pappus’ solution.

Here too Pappus (III. pp. 156—162) finds four circular sections of the sphere all parallel to one another and all passing through five of the vertices of the dodecahedron.

Analysis.

Suppose (he says) the problem solved, and let the vertices of the dodecahedron be A, B, C, D, E ; F, G, H, K, L ; M, N, O, P, Q ; R, S, T, U, V.

Then, as before, ED is parallel to FL, and AE to FG ; therefore the planes ABCDE, FGHKL are parallel.

But, since PA is parallel to BH, and BH to OC, PA is parallel to OC ; and they are equal ; therefore PO, AC are parallel, so that ST, ED are also parallel.

Similarly RS, DC are parallel, and likewise the pairs (TU, EA), (UV, AB), (VR, BC).

Therefore the planes ABCDE, RSTUV are parallel ; and the circles ABCDE, RSTUV are equal, since the inscribed pentagons are equal.

Similarly the circles FGHKL, MNOPQ are equal, since the pentagons inscribed in them are equal.

Now CL, OU are parallel because each is parallel to KN ; therefore L, C, O, U are in one plane.

And LC, CO, OU, UL are all equal, since they subtend angles of equal pentagons.

Also L, C, O, U are on a plane section, i.e. a circle ; therefore LCOU is a square.

Therefore                 OL $^2 = 2$ LC $^2 = 2$ LF $^2$

(for LC, LF subtend angles of equal pentagons).

And the angle OLF is right; for PO, LF are equal and parallel chords in two equal and parallel circular sections of a sphere [Pappus’ lemma, p. 138, 20—26].

Therefore                     OF $^2 = $ OL $^2 + $ FL $^2 = 3$ FL $^2$. [from above]

And OF is a diameter of the sphere; for PO, LF are on opposite sides of the centres of the circles in which they are [Pappus’ lemma, pp. 136—8].

D E C A B L F K M N G H Q O P V U R T S x y

Now suppose $p$, $t$, $h$ to be the sides of an equilateral pentagon, triangle and hexagon in the circle FGHKL, $d$ the diameter of the sphere.

Then                             $d : $ FL $= \sqrt{3} : 1$ [from above]

$= t : h$ ; [Eucl. XIII. 12] and it follows alternando (since FL $= p$) that $d : t = p : h$.

Now let $d'$, $p'$, $h'$ be the sides of a regular decagon, pentagon and hexagon respectively inscribed in any one circle.

Since, if FL be divided in extreme and mean ratio, the greater segment is equal to ED, [XIII. 8] FL $:$ ED $= h' : d'$. [VI. Def. 3, XIII. 9]

And FL $:$ ED is the ratio of the sides of the regular pentagons inscribed in the circles FGHKL, ABCDE, and is therefore equal to the ratio of the sides of the equilateral triangles inscribed in the same circles.

Therefore             $t : (\text{side of }\triangle\text{ in }$ ABCDE $) = h' : d'$.

But                                 $d : t = p : h$

$= p' : h'$ ; therefore, ex aequali, $d : (\text{side of }\triangle\text{ in }$ ABCDE $) = p' : d'$.

Now $d$ is given ; therefore the sides of the equilateral triangles inscribed in the circles ABCDE, FGHKL respectively are given, whence the radii of those circles are also given.

Thus the two circles are given, and so accordingly are the equal and parallel circular sections.

Synthesis. Set out two straight lines x, y such that $d$, x, y are in the ratio of the sides of a regular pentagon, hexagon and decagon respectively inscribed in one and the same circle. Find two circular sections of the sphere with radii $r, r'$, where $r^2 = \tfrac{1}{3}$ x $^2,\quad {r'}^2 = \tfrac{1}{3}$ y $^2.$ Let these be the circles FGHKL, ABCDE respectively, and draw the equal and parallel circles on the other side of the centre, namely MNOPQ, RSTUV. In the first two circles inscribe regular pentagons with their sides respectively parallel, ED being parallel to FL. Draw equal and parallel chords (on the other sides of the centres) in the other two circles, namely ST equal and parallel to ED, and PO equal and parallel to FL; and complete the regular pentagons on ST, PO inscribed in the circles. Thus all the vertices of the dodecahedron are determined. The proof of the correctness of the construction is clear from the analysis. Pappus adds that the construction shows that the circles containing five vertices of the dodecahedron are the same respectively as those containing three vertices of the icosahedron, and that the same circle circumscribes the triangle of the icosahedron and the pentagonal face of the dodecahedron in the same sphere.