Comparison of Sides of Five Platonic Figures

== Theorem ==
=== Lemma ===
== Proof ==
:
Let $AB$ be the diameter of the given sphere.
Let $AB$ be bisected at $C$.
Let $AB$ be cut at $D$ where $AD = 2 \cdot DB$.
Let $AEB$ be a semicircle on the diameter $AB$.
Let $CE$ be drawn from $C$ perpendicular to $AB$.
Let $DF$ be drawn from $D$ perpendicular to $AB$.
Let $AF, FB, EB$ be joined.
We have that:
:$AD = 2 \cdot DB$
Therefore:
:$AB = 3 \cdot DB$
and so:
:$AB = \dfrac 3 2 \cdot AD$
We have that $\triangle AFB$ is equiangular with $\triangle AFD$.
So from:
:
and:
:
it follows that:
:$BA : AD = BA^2 : AF^2$
Therefore:
:$BA^2 = \dfrac 3 2 \cdot AF^2$
But from :
:the square on the diameter of the given sphere is $1 \frac 1 2$ times the square on the side of the regular tetrahedron inscribed within it.
But $AB$ is the diameter of the given sphere.
Therefore $AF$ is equal to the side of the regular tetrahedron inscribed within.
We have that:
:$AD = 2 \cdot DB$
Therefore:
:$AB = 3 \cdot DB$
So from:
:
and:
:
it follows that:
:$AB : BD = AB^2 : BF^2$
Therefore:
:$AB^2 = 3 \cdot BF^2$
But from :
:the square on the diameter of the given sphere is $3$ times the square on the side of the cube inscribed within it.
But $AB$ is the diameter of the given sphere.
Therefore $BF$ is equal to the side of the cube inscribed within.
We have that:
:$AC = CB$
Therefore:
:$AB = 2 \cdot BC$
But:
:$AB : BC = AB^2 : BE^2$
But from :
:the square on the diameter of the given sphere is $2$ times the square on the side of the regular octahedron inscribed within it.
But $AB$ is the diameter of the given sphere.
Therefore $BE$ is equal to the side of the regular octahedron inscribed within.
Let $AG$ be drawn from $A$ perpendicular to $AB$.
Let $AG = AB$.
Let $GC$ be joined.
Let $H$ be the point where $GC$ cuts the semicircle $AEB$.
Let $HK$ be drawn perpendicular to $AB$ from $H$.
We have that:
:$GA = 2 \cdot AC$
We also have that:
:$GA : AC = HK : KC$
Therefore:
:$HK = 2 \cdot KC$
Therefore:
:$HK^2 = 4 \cdot KC^2$
Therefore:
:$HK^2 + KC^2 = 5 \cdot KC^2$
But by :
:$HK^2 + KC^2 = HC^2$
We also have that:
:$HC = CB$
So:
:$CB^2 = 5 CK^2$
We have that:
:$AB = 2 \cdot BC$
and:
:$AD = 2 \cdot DB$
Therefore:
:$BD = 2 \cdot DC$
Therefore:
:$BC = 3 \cdot CD$
and so:
:$BC^2 = 9 \cdot CD^2$
But:
:$BC^2 = 5 CK^2$
Therefore:
:$CK^2 > CD^2$
and so:
:$CK > CD$
Let $CL = CK$.
Let $LM$ be drawn from $L$ perpendicular to $AB$.
Let $MB$ be joined.
We have that:
:$BC^2 = 5 CK^2$
and:
:$AB = 2 \cdot BC$
and:
:$KL = 2 \cdot CK$
Therefore:
:$AB^2 = 5 KL^2$
But from :
:the square on the diameter of the given sphere is $5$ times the square of the radius of the circle from which the regular icosahedron has been described.
We have that $AB$ is the diameter of the given sphere.
Therefore $KL$ is the radius of the circle from which the regular icosahedron has been described.
Therefore from :
:$KL$ is equal to the side of a regular hexagon in that circle.
From :
:the diameter of the given sphere is composed of the side of the regular hexagon and two of the sides of the regular decagon inscribed in that circle.
We have that $AB$ is the diameter of the given sphere.
Also we have that $KL$ is equal to the side of a regular hexagon in that circle.
Also:
:$AK = LB$
Therefore each of $AK$ and $LB$ is the side of the regular decagon inscribed in that circle.
We have that:
: $LB$ is the side of such a regular decagon
: $ML$ is the side of such a regular hexagon
Then:
:$HK$ and $HK$ are the same distance from $C$, the midpoint of $AB$
Also, each of $HK$ and $KL$ is twice $KC$.
So:
:$ML = KL = HK$
Therefore from :
:$MB$ is the side of the regular pentagon inscribed in that circle.
But from :
:the side of the regular pentagon inscribed in that circle equals the side of the regular icosahedron inscribed within the given sphere.
Therefore $MB$ is equal to the side of the regular icosahedron inscribed within.
We have that $FB$ is equal to the side of the cube inscribed within the given sphere.
Let $FB$ be cut in extreme and mean ratio at $N$ such that the greater segment is $NB$.
From the :
:$NB$ equals the side of the regular dodecahedron inscribed within the given sphere.
We have that the square on the diameter of the sphere is:
:$1 \frac 1 2$ times the square on the side $AF$ of the regular tetrahedron
:$2$ times the square on the side $BE$ of the regular octahedron
:$3$ times the square on the side $FB$ of the cube
inscribed within it.
Therefore the square on:
:the side $AF$ of the regular tetrahedron is $\dfrac 4 6$ 
:the side $BE$ of the regular octahedron is $\dfrac 3 6$ :the side $FB$ of the cube is $\dfrac 2 6$
times the square on the diameter of the sphere.
Therefore:
:the square on the side $AF$ of the regular tetrahedron is $\dfrac 4 3$ times the square on the side $BE$ of the regular octahedron
:the square on the side $AF$ of the regular tetrahedron is $2$ times the square on the side $FB$ of the cube
:the square on the side $BE$ of the regular octahedron is $\dfrac 3 2$ times the square on the side $FB$ of the cube.
That is, the sides of the regular tetrahedron, the regular octahedron and the cube are to one another in rational ratios.
From :
:the side of the regular icosahedron is a minor.
From :
:the side of the regular dodecahedron is an apotome.
Thus the remaining two, that is the sides of the regular icosahedron and the regular dodecahedron, are not in rational ratios either to one another or the sides of the first three:
From :
:$\triangle FDB$ is equiangular with $\triangle FAB$.
So from :
:$DB : BF = BF : BA$
Thus $DB, BF, BA$ are proportional.
So from:
:
and:
:
it follows that:
:$DB : BA = DB^2 : BF^2$
That is:
:$AB : BD = FB^2 = BD^2$
But:
:$AB = 3 \cdot BD$
Therefore:
:$FB^2 = 3 \cdot BD^2$
But we have that:
:$AD = 2 \cdot DB$
and so:
:$AD^2 = 4 \cdot DB^2$
Therefore:
:$AD > FB$
and so:
:$AL > FB$
From :
:$LK$ is the side of the regular hexagon and $KA$ the side of the regular decagon inscribed in that circle.
So:
:when $AL$ is cut in extreme and mean ratio, the greater segment is $KL$
and:
:when $FB$ is cut in extreme and mean ratio, the greater segment is $NB$.
Therefore:
:$KL > NB$
But:
:$KL = LM$
Therefore:
:$LM > NB$
Therefore $MB$, which is the side of the regular icosahedron, is greater than $NB$, the side of the regular dodecahedron.
== Also see ==
* Five Platonic Solids
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Platonic Solids