% book13.tex --- Book XIII of Euclid's Elements: Platonic Solids.
%
% All 18 propositions encoded.  Book XIII is the climax of the Elements:
% lemmas on the golden section (XIII.1-XIII.6), the construction of the
% five regular solids inscribed in a sphere (XIII.13-XIII.17), and the
% proof that there are exactly five (XIII.18).
%
% Wording follows Heath (1908).

\section{Book XIII --- Platonic Solids}
\label{sec:book-XIII}

\begin{claim}[Proposition XIII.1: Square on the whole plus square on half segment]
\label{prop:XIII.1}
If a straight line be cut in extreme and mean ratio, the square on
the greater segment added to the half of the whole is five times the
square on the half.
\end{claim}
\begin{evidence}[Proof of XIII.1]
\label{ev:XIII.1}
Let $AB$ be cut at $C$ in extreme and mean ratio with $AC > CB$.
Let $D$ be the midpoint of $AB$.  Apply II.6: $(AB/2 + AC)^2 =
(AB/2)^2 + AB \cdot AC + AC^2$.  By the defining relation $AC^2 =
AB \cdot CB$, simplification gives $(AB/2 + AC)^2 = 5 (AB/2)^2$.
\dependson{XIII.1}{II.6}
\dependson{XIII.1}{II.11}
\dependson{XIII.1}{def:XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.2: Square on segment plus square on smaller part]
\label{prop:XIII.2}
If the square on a straight line be five times the square on a
segment of it, then, when the double of the said segment is cut in
extreme and mean ratio, the greater segment is the remaining part of
the original straight line.
\end{claim}
\begin{evidence}[Proof of XIII.2]
\label{ev:XIII.2}
Converse of XIII.1: assume the squared relation and deduce the
extreme-and-mean cut using II.6 / II.11.
\dependson{XIII.2}{II.6}
\dependson{XIII.2}{II.11}
\dependson{XIII.2}{XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.3: Squares on extreme-and-mean parts]
\label{prop:XIII.3}
If a straight line be cut in extreme and mean ratio, the square on
the lesser segment added to the half of the greater segment is five
times the square on the half of the greater segment.
\end{claim}
\begin{evidence}[Proof of XIII.3]
\label{ev:XIII.3}
Apply II.6 / II.11 to the lesser-segment configuration; algebraic
analogue of XIII.1.
\dependson{XIII.3}{II.6}
\dependson{XIII.3}{XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.4: Squares on whole, greater, and lesser are commensurable]
\label{prop:XIII.4}
If a straight line be cut in extreme and mean ratio, the square on
the whole and the square on the lesser segment together are triple
of the square on the greater segment.
\end{claim}
\begin{evidence}[Proof of XIII.4]
\label{ev:XIII.4}
$AB^2 + CB^2 = 3 \cdot AC^2$ where $C$ cuts $AB$ in extreme-and-mean
ratio (greater $AC$).  Use $AC^2 = AB \cdot CB$ and II.4 to verify
the identity.
\dependson{XIII.4}{II.4}
\dependson{XIII.4}{II.11}
\dependson{XIII.4}{XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.5: Extension preserves extreme-and-mean property]
\label{prop:XIII.5}
If a straight line be cut in extreme and mean ratio, and there be
added to it a straight line equal to the greater segment, the whole
straight line is cut in extreme and mean ratio, and the original
straight line is the greater segment.
\end{claim}
\begin{evidence}[Proof of XIII.5]
\label{ev:XIII.5}
Extending by the greater segment $AC$ to $A'$ (so $A'A = AC$, $AB$
the original), check that $A'A : AB = AB : (A'A + AB - AC)$, which
reduces via the original extreme-and-mean relation to the same form.
\dependson{XIII.5}{II.11}
\dependson{XIII.5}{XIII.1}
\dependson{XIII.5}{def:XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.6: Greater segment of a rational divided in extreme-and-mean is an apotome]
\label{prop:XIII.6}
If a rational straight line be cut in extreme and mean ratio, each of
the segments is the irrational straight line called apotome.
\end{claim}
\begin{evidence}[Proof of XIII.6]
\label{ev:XIII.6}
Solve $x^2 + ax = a^2$ for the greater segment $x = a(\sqrt{5} -
1)/2$; this is the form of an apotome relative to the rational $a$
(Book X classification).
\dependson{XIII.6}{X.73}
\dependson{XIII.6}{II.11}
\dependson{XIII.6}{XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.7: Three angles of equilateral pentagon equal implies all equal]
\label{prop:XIII.7}
If three angles of an equilateral pentagon, taken either in order or
not in order, be equal, the pentagon will be equiangular.
\end{claim}
\begin{evidence}[Proof of XIII.7]
\label{ev:XIII.7}
The five interior angles sum to $3 \cdot 180^\circ$ (I.32 extended);
combined with three equal angles, the constraint forces all five to
be equal.
\dependson{XIII.7}{I.32}
\dependson{XIII.7}{I.5}
\end{evidence}

\begin{claim}[Proposition XIII.8: Diagonals of a regular pentagon cut each other in extreme-and-mean]
\label{prop:XIII.8}
If in an equilateral and equiangular pentagon straight lines subtend
two adjacent angles, they cut one another in extreme and mean ratio,
and the greater segments are equal to the side of the pentagon.
\end{claim}
\begin{evidence}[Proof of XIII.8]
\label{ev:XIII.8}
Construct the pentagon inscribed in a circle (IV.11).  Two diagonals
form an isosceles triangle with vertex angle $36^\circ$ (I.32 / IV.10);
by similarity (VI.4) the diagonal-segment ratio matches the
extreme-and-mean ratio.
\dependson{XIII.8}{IV.10}
\dependson{XIII.8}{IV.11}
\dependson{XIII.8}{VI.4}
\dependson{XIII.8}{def:XIII.1}
\end{evidence}

\begin{claim}[Proposition XIII.9: Hexagon side plus decagon side in extreme-and-mean]
\label{prop:XIII.9}
If the side of the hexagon and that of the decagon inscribed in the
same circle be added together, the whole straight line has been cut
in extreme and mean ratio, and its greater segment is the side of
the hexagon.
\end{claim}
\begin{evidence}[Proof of XIII.9]
\label{ev:XIII.9}
The hexagon side equals the radius (IV.15); the decagon side
satisfies the 36-72-72 triangle relations (IV.10); their sum is in
the golden ratio to the hexagon side.
\dependson{XIII.9}{IV.10}
\dependson{XIII.9}{IV.15}
\dependson{XIII.9}{XIII.8}
\end{evidence}

\begin{claim}[Proposition XIII.10: Pentagon side squared equals hexagon plus decagon sides squared]
\label{prop:XIII.10}
If an equilateral pentagon be inscribed in a circle, the square on
the side of the pentagon is equal to the squares on the side of the
hexagon and on that of the decagon inscribed in the same circle.
\end{claim}
\begin{evidence}[Proof of XIII.10]
\label{ev:XIII.10}
This is the Pythagorean relation $p^2 = h^2 + d^2$ in the inscribed
polygons of a unit circle.  Proven via I.47 applied to the right
triangle formed by the centre, a pentagon-vertex, and a decagon-vertex.
\dependson{XIII.10}{I.47}
\dependson{XIII.10}{IV.11}
\dependson{XIII.10}{XIII.9}
\end{evidence}

\begin{claim}[Proposition XIII.11: Side of inscribed pentagon in rational circle is minor irrational]
\label{prop:XIII.11}
If in a circle which has its diameter rational an equilateral
pentagon be inscribed, the side of the pentagon is the irrational
straight line called minor.
\end{claim}
\begin{evidence}[Proof of XIII.11]
\label{ev:XIII.11}
By XIII.10 the pentagon side is $\sqrt{h^2 + d^2}$ with $h$ rational
and $d$ an apotome (XIII.6); the resulting form falls in Book X's
minor class (Definition XIII.4 / X.76).
\dependson{XIII.11}{X.76}
\dependson{XIII.11}{XIII.6}
\dependson{XIII.11}{XIII.10}
\dependson{XIII.11}{def:XIII.4}
\end{evidence}

\begin{claim}[Proposition XIII.12: Side of inscribed equilateral triangle squared is three times square on radius]
\label{prop:XIII.12}
If an equilateral triangle be inscribed in a circle, the square on
the side of the triangle is triple of the square on the radius.
\end{claim}
\begin{evidence}[Proof of XIII.12]
\label{ev:XIII.12}
For inscribed equilateral triangle, side $= r \sqrt{3}$.  Proven via
I.47 on the perpendicular bisector configuration.
\dependson{XIII.12}{I.47}
\dependson{XIII.12}{IV.2}
\end{evidence}

\begin{claim}[Proposition XIII.13: Construct a regular tetrahedron in a sphere]
\label{prop:XIII.13}
To construct a pyramid (regular tetrahedron), to comprehend it in a
given sphere, and to prove that the square on the diameter of the
sphere is one and a half times the square on the side of the pyramid.
\end{claim}
\begin{evidence}[Proof of XIII.13]
\label{ev:XIII.13}
Inscribe an equilateral triangle (IV.2); erect an apex above the
centroid at height $r \sqrt{2/3}$ where $r$ is the circumradius.
The four equal edges form the tetrahedron; place the sphere through
its four vertices.  The diameter-squared $/$ side-squared $= 3/2$.
\dependson{XIII.13}{IV.2}
\dependson{XIII.13}{XI.11}
\dependson{XIII.13}{XIII.12}
\end{evidence}

\begin{claim}[Proposition XIII.14: Construct a regular octahedron in a sphere]
\label{prop:XIII.14}
To construct an octahedron and comprehend it in a sphere, as in the
preceding case; and to prove that the square on the diameter of the
sphere is double of the square on the side of the octahedron.
\end{claim}
\begin{evidence}[Proof of XIII.14]
\label{ev:XIII.14}
Take two perpendicular diameters in a circle; through the centre
erect a perpendicular axis equal in length to the diameter.  The
four endpoints in the circle and two endpoints on the axis form the
six vertices of the octahedron.  Diameter-squared / side-squared $= 2$.
\dependson{XIII.14}{XI.11}
\dependson{XIII.14}{XIII.13}
\end{evidence}

\begin{claim}[Proposition XIII.15: Construct a cube in a sphere]
\label{prop:XIII.15}
To construct a cube and comprehend it in a sphere, as in the
preceding case; and to prove that the square on the diameter of the
sphere is triple of the square on the side of the cube.
\end{claim}
\begin{evidence}[Proof of XIII.15]
\label{ev:XIII.15}
Take a square base (IV.6); erect a parallel square at height equal
to the side.  The eight vertices form the cube; the sphere through
them has diameter $\sqrt{3}$ times the side.
\dependson{XIII.15}{IV.6}
\dependson{XIII.15}{XI.11}
\dependson{XIII.15}{XIII.14}
\end{evidence}

\begin{claim}[Proposition XIII.16: Construct a regular icosahedron in a sphere]
\label{prop:XIII.16}
To construct an icosahedron and comprehend it in a sphere, as in the
case of the aforesaid figures; and to prove that the side of the
icosahedron is the irrational straight line called minor.
\end{claim}
\begin{evidence}[Proof of XIII.16]
\label{ev:XIII.16}
Inscribe a regular pentagon in a circle (IV.11); arrange two parallel
pentagons rotated $36^\circ$ from each other, plus two apex points.
Twelve vertices form the icosahedron.  The side is a minor straight
line by XIII.11.
\dependson{XIII.16}{IV.11}
\dependson{XIII.16}{XIII.11}
\dependson{XIII.16}{XIII.15}
\end{evidence}

\begin{claim}[Proposition XIII.17: Construct a regular dodecahedron in a sphere]
\label{prop:XIII.17}
To construct a dodecahedron and comprehend it in a sphere, like the
aforesaid figures; and to prove that the side of the dodecahedron is
the irrational straight line called apotome.
\end{claim}
\begin{evidence}[Proof of XIII.17]
\label{ev:XIII.17}
The dodecahedron has twelve regular pentagonal faces; the side is
the apotome formed when the cube-edge is cut in extreme and mean
ratio (XIII.6).  Inscribe by placing pentagonal faces on the six
square faces of the inscribed cube (XIII.15).
\dependson{XIII.17}{IV.11}
\dependson{XIII.17}{XIII.6}
\dependson{XIII.17}{XIII.15}
\dependson{XIII.17}{XIII.16}
\end{evidence}

\begin{claim}[Proposition XIII.18: There are exactly five regular solids]
\label{prop:XIII.18}
To set out the sides of the five figures and to compare them with
one another; and that no other figure, besides the said five
figures, can be constructed which is contained by equilateral and
equiangular figures equal to one another.
\end{claim}
\begin{evidence}[Proof of XIII.18]
\label{ev:XIII.18}
Compare the side lengths: tetrahedron $\sqrt{2/3}$, octahedron
$\sqrt{1/2}$, cube $1/\sqrt{3}$, icosahedron (minor irrational),
dodecahedron (apotome).  For the uniqueness clause: at each vertex of
a regular polyhedron, the sum of face-angles must be less than four
right angles (XI.21).  Equilateral triangles ($60^\circ$): 3, 4, or
5 around a vertex --- tetrahedron, octahedron, icosahedron.  Squares
($90^\circ$): only 3 around a vertex --- cube.  Regular pentagons
($108^\circ$): only 3 around a vertex --- dodecahedron.  Hexagons
($120^\circ$): three would tile flat, no vertex --- impossible.
Larger polygons: even three exceed $360^\circ$.  Hence exactly five
regular polyhedra exist.
\dependson{XIII.18}{I.32}
\dependson{XIII.18}{XI.21}
\dependson{XIII.18}{XIII.13}
\dependson{XIII.18}{XIII.14}
\dependson{XIII.18}{XIII.15}
\dependson{XIII.18}{XIII.16}
\dependson{XIII.18}{XIII.17}
\end{evidence}
