Source — Euclid's Elements, encoded as an rrxiv paper

1% book13.tex --- Book XIII of Euclid's Elements: Platonic Solids.
2%
3% All 18 propositions encoded.  Book XIII is the climax of the Elements:
4% lemmas on the golden section (XIII.1-XIII.6), the construction of the
5% five regular solids inscribed in a sphere (XIII.13-XIII.17), and the
6% proof that there are exactly five (XIII.18).
7%
8% Wording follows Heath (1908).
9
10\section{Book XIII --- Platonic Solids}
11\label{sec:book-XIII}
12
13\begin{claim}[Proposition XIII.1: Square on the whole plus square on half segment]
14\label{prop:XIII.1}
15If a straight line be cut in extreme and mean ratio, the square on
16the greater segment added to the half of the whole is five times the
17square on the half.
18\end{claim}
19\begin{evidence}[Proof of XIII.1]
20\label{ev:XIII.1}
21Let $AB$ be cut at $C$ in extreme and mean ratio with $AC > CB$.
22Let $D$ be the midpoint of $AB$.  Apply II.6: $(AB/2 + AC)^2 =
23(AB/2)^2 + AB \cdot AC + AC^2$.  By the defining relation $AC^2 =
24AB \cdot CB$, simplification gives $(AB/2 + AC)^2 = 5 (AB/2)^2$.
25\dependson{XIII.1}{II.6}
26\dependson{XIII.1}{II.11}
27\dependson{XIII.1}{def:XIII.1}
28\end{evidence}
29
30\begin{claim}[Proposition XIII.2: Square on segment plus square on smaller part]
31\label{prop:XIII.2}
32If the square on a straight line be five times the square on a
33segment of it, then, when the double of the said segment is cut in
34extreme and mean ratio, the greater segment is the remaining part of
35the original straight line.
36\end{claim}
37\begin{evidence}[Proof of XIII.2]
38\label{ev:XIII.2}
39Converse of XIII.1: assume the squared relation and deduce the
40extreme-and-mean cut using II.6 / II.11.
41\dependson{XIII.2}{II.6}
42\dependson{XIII.2}{II.11}
43\dependson{XIII.2}{XIII.1}
44\end{evidence}
45
46\begin{claim}[Proposition XIII.3: Squares on extreme-and-mean parts]
47\label{prop:XIII.3}
48If a straight line be cut in extreme and mean ratio, the square on
49the lesser segment added to the half of the greater segment is five
50times the square on the half of the greater segment.
51\end{claim}
52\begin{evidence}[Proof of XIII.3]
53\label{ev:XIII.3}
54Apply II.6 / II.11 to the lesser-segment configuration; algebraic
55analogue of XIII.1.
56\dependson{XIII.3}{II.6}
57\dependson{XIII.3}{XIII.1}
58\end{evidence}
59
60\begin{claim}[Proposition XIII.4: Squares on whole, greater, and lesser are commensurable]
61\label{prop:XIII.4}
62If a straight line be cut in extreme and mean ratio, the square on
63the whole and the square on the lesser segment together are triple
64of the square on the greater segment.
65\end{claim}
66\begin{evidence}[Proof of XIII.4]
67\label{ev:XIII.4}
68$AB^2 + CB^2 = 3 \cdot AC^2$ where $C$ cuts $AB$ in extreme-and-mean
69ratio (greater $AC$).  Use $AC^2 = AB \cdot CB$ and II.4 to verify
70the identity.
71\dependson{XIII.4}{II.4}
72\dependson{XIII.4}{II.11}
73\dependson{XIII.4}{XIII.1}
74\end{evidence}
75
76\begin{claim}[Proposition XIII.5: Extension preserves extreme-and-mean property]
77\label{prop:XIII.5}
78If a straight line be cut in extreme and mean ratio, and there be
79added to it a straight line equal to the greater segment, the whole
80straight line is cut in extreme and mean ratio, and the original
81straight line is the greater segment.
82\end{claim}
83\begin{evidence}[Proof of XIII.5]
84\label{ev:XIII.5}
85Extending by the greater segment $AC$ to $A'$ (so $A'A = AC$, $AB$
86the original), check that $A'A : AB = AB : (A'A + AB - AC)$, which
87reduces via the original extreme-and-mean relation to the same form.
88\dependson{XIII.5}{II.11}
89\dependson{XIII.5}{XIII.1}
90\dependson{XIII.5}{def:XIII.1}
91\end{evidence}
92
93\begin{claim}[Proposition XIII.6: Greater segment of a rational divided in extreme-and-mean is an apotome]
94\label{prop:XIII.6}
95If a rational straight line be cut in extreme and mean ratio, each of
96the segments is the irrational straight line called apotome.
97\end{claim}
98\begin{evidence}[Proof of XIII.6]
99\label{ev:XIII.6}
100Solve $x^2 + ax = a^2$ for the greater segment $x = a(\sqrt{5} -
1011)/2$; this is the form of an apotome relative to the rational $a$
102(Book X classification).
103\dependson{XIII.6}{X.73}
104\dependson{XIII.6}{II.11}
105\dependson{XIII.6}{XIII.1}
106\end{evidence}
107
108\begin{claim}[Proposition XIII.7: Three angles of equilateral pentagon equal implies all equal]
109\label{prop:XIII.7}
110If three angles of an equilateral pentagon, taken either in order or
111not in order, be equal, the pentagon will be equiangular.
112\end{claim}
113\begin{evidence}[Proof of XIII.7]
114\label{ev:XIII.7}
115The five interior angles sum to $3 \cdot 180^\circ$ (I.32 extended);
116combined with three equal angles, the constraint forces all five to
117be equal.
118\dependson{XIII.7}{I.32}
119\dependson{XIII.7}{I.5}
120\end{evidence}
121
122\begin{claim}[Proposition XIII.8: Diagonals of a regular pentagon cut each other in extreme-and-mean]
123\label{prop:XIII.8}
124If in an equilateral and equiangular pentagon straight lines subtend
125two adjacent angles, they cut one another in extreme and mean ratio,
126and the greater segments are equal to the side of the pentagon.
127\end{claim}
128\begin{evidence}[Proof of XIII.8]
129\label{ev:XIII.8}
130Construct the pentagon inscribed in a circle (IV.11).  Two diagonals
131form an isosceles triangle with vertex angle $36^\circ$ (I.32 / IV.10);
132by similarity (VI.4) the diagonal-segment ratio matches the
133extreme-and-mean ratio.
134\dependson{XIII.8}{IV.10}
135\dependson{XIII.8}{IV.11}
136\dependson{XIII.8}{VI.4}
137\dependson{XIII.8}{def:XIII.1}
138\end{evidence}
139
140\begin{claim}[Proposition XIII.9: Hexagon side plus decagon side in extreme-and-mean]
141\label{prop:XIII.9}
142If the side of the hexagon and that of the decagon inscribed in the
143same circle be added together, the whole straight line has been cut
144in extreme and mean ratio, and its greater segment is the side of
145the hexagon.
146\end{claim}
147\begin{evidence}[Proof of XIII.9]
148\label{ev:XIII.9}
149The hexagon side equals the radius (IV.15); the decagon side
150satisfies the 36-72-72 triangle relations (IV.10); their sum is in
151the golden ratio to the hexagon side.
152\dependson{XIII.9}{IV.10}
153\dependson{XIII.9}{IV.15}
154\dependson{XIII.9}{XIII.8}
155\end{evidence}
156
157\begin{claim}[Proposition XIII.10: Pentagon side squared equals hexagon plus decagon sides squared]
158\label{prop:XIII.10}
159If an equilateral pentagon be inscribed in a circle, the square on
160the side of the pentagon is equal to the squares on the side of the
161hexagon and on that of the decagon inscribed in the same circle.
162\end{claim}
163\begin{evidence}[Proof of XIII.10]
164\label{ev:XIII.10}
165This is the Pythagorean relation $p^2 = h^2 + d^2$ in the inscribed
166polygons of a unit circle.  Proven via I.47 applied to the right
167triangle formed by the centre, a pentagon-vertex, and a decagon-vertex.
168\dependson{XIII.10}{I.47}
169\dependson{XIII.10}{IV.11}
170\dependson{XIII.10}{XIII.9}
171\end{evidence}
172
173\begin{claim}[Proposition XIII.11: Side of inscribed pentagon in rational circle is minor irrational]
174\label{prop:XIII.11}
175If in a circle which has its diameter rational an equilateral
176pentagon be inscribed, the side of the pentagon is the irrational
177straight line called minor.
178\end{claim}
179\begin{evidence}[Proof of XIII.11]
180\label{ev:XIII.11}
181By XIII.10 the pentagon side is $\sqrt{h^2 + d^2}$ with $h$ rational
182and $d$ an apotome (XIII.6); the resulting form falls in Book X's
183minor class (Definition XIII.4 / X.76).
184\dependson{XIII.11}{X.76}
185\dependson{XIII.11}{XIII.6}
186\dependson{XIII.11}{XIII.10}
187\dependson{XIII.11}{def:XIII.4}
188\end{evidence}
189
190\begin{claim}[Proposition XIII.12: Side of inscribed equilateral triangle squared is three times square on radius]
191\label{prop:XIII.12}
192If an equilateral triangle be inscribed in a circle, the square on
193the side of the triangle is triple of the square on the radius.
194\end{claim}
195\begin{evidence}[Proof of XIII.12]
196\label{ev:XIII.12}
197For inscribed equilateral triangle, side $= r \sqrt{3}$.  Proven via
198I.47 on the perpendicular bisector configuration.
199\dependson{XIII.12}{I.47}
200\dependson{XIII.12}{IV.2}
201\end{evidence}
202
203\begin{claim}[Proposition XIII.13: Construct a regular tetrahedron in a sphere]
204\label{prop:XIII.13}
205To construct a pyramid (regular tetrahedron), to comprehend it in a
206given sphere, and to prove that the square on the diameter of the
207sphere is one and a half times the square on the side of the pyramid.
208\end{claim}
209\begin{evidence}[Proof of XIII.13]
210\label{ev:XIII.13}
211Inscribe an equilateral triangle (IV.2); erect an apex above the
212centroid at height $r \sqrt{2/3}$ where $r$ is the circumradius.
213The four equal edges form the tetrahedron; place the sphere through
214its four vertices.  The diameter-squared $/$ side-squared $= 3/2$.
215\dependson{XIII.13}{IV.2}
216\dependson{XIII.13}{XI.11}
217\dependson{XIII.13}{XIII.12}
218\end{evidence}
219
220\begin{claim}[Proposition XIII.14: Construct a regular octahedron in a sphere]
221\label{prop:XIII.14}
222To construct an octahedron and comprehend it in a sphere, as in the
223preceding case; and to prove that the square on the diameter of the
224sphere is double of the square on the side of the octahedron.
225\end{claim}
226\begin{evidence}[Proof of XIII.14]
227\label{ev:XIII.14}
228Take two perpendicular diameters in a circle; through the centre
229erect a perpendicular axis equal in length to the diameter.  The
230four endpoints in the circle and two endpoints on the axis form the
231six vertices of the octahedron.  Diameter-squared / side-squared $= 2$.
232\dependson{XIII.14}{XI.11}
233\dependson{XIII.14}{XIII.13}
234\end{evidence}
235
236\begin{claim}[Proposition XIII.15: Construct a cube in a sphere]
237\label{prop:XIII.15}
238To construct a cube and comprehend it in a sphere, as in the
239preceding case; and to prove that the square on the diameter of the
240sphere is triple of the square on the side of the cube.
241\end{claim}
242\begin{evidence}[Proof of XIII.15]
243\label{ev:XIII.15}
244Take a square base (IV.6); erect a parallel square at height equal
245to the side.  The eight vertices form the cube; the sphere through
246them has diameter $\sqrt{3}$ times the side.
247\dependson{XIII.15}{IV.6}
248\dependson{XIII.15}{XI.11}
249\dependson{XIII.15}{XIII.14}
250\end{evidence}
251
252\begin{claim}[Proposition XIII.16: Construct a regular icosahedron in a sphere]
253\label{prop:XIII.16}
254To construct an icosahedron and comprehend it in a sphere, as in the
255case of the aforesaid figures; and to prove that the side of the
256icosahedron is the irrational straight line called minor.
257\end{claim}
258\begin{evidence}[Proof of XIII.16]
259\label{ev:XIII.16}
260Inscribe a regular pentagon in a circle (IV.11); arrange two parallel
261pentagons rotated $36^\circ$ from each other, plus two apex points.
262Twelve vertices form the icosahedron.  The side is a minor straight
263line by XIII.11.
264\dependson{XIII.16}{IV.11}
265\dependson{XIII.16}{XIII.11}
266\dependson{XIII.16}{XIII.15}
267\end{evidence}
268
269\begin{claim}[Proposition XIII.17: Construct a regular dodecahedron in a sphere]
270\label{prop:XIII.17}
271To construct a dodecahedron and comprehend it in a sphere, like the
272aforesaid figures; and to prove that the side of the dodecahedron is
273the irrational straight line called apotome.
274\end{claim}
275\begin{evidence}[Proof of XIII.17]
276\label{ev:XIII.17}
277The dodecahedron has twelve regular pentagonal faces; the side is
278the apotome formed when the cube-edge is cut in extreme and mean
279ratio (XIII.6).  Inscribe by placing pentagonal faces on the six
280square faces of the inscribed cube (XIII.15).
281\dependson{XIII.17}{IV.11}
282\dependson{XIII.17}{XIII.6}
283\dependson{XIII.17}{XIII.15}
284\dependson{XIII.17}{XIII.16}
285\end{evidence}
286
287\begin{claim}[Proposition XIII.18: There are exactly five regular solids]
288\label{prop:XIII.18}
289To set out the sides of the five figures and to compare them with
290one another; and that no other figure, besides the said five
291figures, can be constructed which is contained by equilateral and
292equiangular figures equal to one another.
293\end{claim}
294\begin{evidence}[Proof of XIII.18]
295\label{ev:XIII.18}
296Compare the side lengths: tetrahedron $\sqrt{2/3}$, octahedron
297$\sqrt{1/2}$, cube $1/\sqrt{3}$, icosahedron (minor irrational),
298dodecahedron (apotome).  For the uniqueness clause: at each vertex of
299a regular polyhedron, the sum of face-angles must be less than four
300right angles (XI.21).  Equilateral triangles ($60^\circ$): 3, 4, or
3015 around a vertex --- tetrahedron, octahedron, icosahedron.  Squares
302($90^\circ$): only 3 around a vertex --- cube.  Regular pentagons
303($108^\circ$): only 3 around a vertex --- dodecahedron.  Hexagons
304($120^\circ$): three would tile flat, no vertex --- impossible.
305Larger polygons: even three exceed $360^\circ$.  Hence exactly five
306regular polyhedra exist.
307\dependson{XIII.18}{I.32}
308\dependson{XIII.18}{XI.21}
309\dependson{XIII.18}{XIII.13}
310\dependson{XIII.18}{XIII.14}
311\dependson{XIII.18}{XIII.15}
312\dependson{XIII.18}{XIII.16}
313\dependson{XIII.18}{XIII.17}
314\end{evidence}
315