# Construction of Regular Dodecahedron within Given Sphere

## Theorem

In the words of Euclid:

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.

### Porism

In the words of Euclid:

From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron.

## Proof

Consider a cube inscribed in the given sphere.

Let $ABCD$ and $CBEF$ be two faces of this cube which are perpendicular to each other.

Let the edges $AB, BC, CD, DA, EF, EB, FC$ of this cube be bisected at $G, H, K, L, M, N, O$ respectively.

Let $GK, HL, MN, NO$ be joined.

Let the straight lines $NP, PO, HQ$ be cut in extreme and mean ratio at the points $R, S, T$ respectively.

Let the greater segments be $RP, PS, TQ$.

From the points $R, S, T$ let $RU, SV, TW$ be constructed perpendicular to the planes $ABCD$ and $CBEF$ appropriately, towards the outside of the cube.

Let $RU = SV = TW = RP$ and so equal also to $PS$ and $TQ$.

Let $UB, BW, WC, CV, VU$ be joined.

It is to be demonstrated that the pentagon $UBWCV$ is regular and lies all in one plane.

Let $RB, SB, VB$ be joined.

We have that:

$NP$ has been cut in extreme and mean ratio at $R$
$RP$ is the greater segment
$PN^2 + NR^2 = 3 \cdot RP^2$

But:

$PN = NB$

and:

$PR = RU$

Therefore:

$BN^2 + NR^2 = 3 \cdot RU^2$
$BR^2 = BN^2 + NR^2$

Therefore:

$BR^2 = 3 \cdot RU^2$

So:

$BR^2 + RU^2 = 4 \cdot RU^2$
$BU^2 = BR^2 + RU^2$

Therefore:

$BU^2 = 4 \cdot RU^2$

Therefore:

$BU = 2 \cdot RU$

But also:

$SR = 2 \cdot PR = 2 \cdot RU$

and so:

$VU = 2 \cdot UR$

Therefore:

$BU = UV$

Similarly it can be proved that:

$BW = WC = CV = BU = UV$

Therefore the pentagon $BUVCW$ is equilateral.

Next it is to be demonstrated that $BUVCW$ lies in one plane.

Let $PX$ be drawn from $P$ parallel to each of $RU$ and $SV$ and towards the outside of the cube.

Let $XH$ and $HW$ be drawn.

We have that:

$HQ$ has been cut in extreme and mean ratio at $T$
$QT$ is the greater segment.

Therefore:

$HQ : QT = QT : TH$

But:

$HQ = HP$

and:

$QT = TW = PX$

Therefore:

$PH : PX = WT : TH$

Also, each of $HP$ and $TW$ is perpendicular to the plane $BD$.

$HP \parallel TW$

Also, $TH$ and $PX$ are both perpendicular to the plane $BF$.

$TH \parallel PX$

But we have that $\triangle XPH$ and $\triangle HTW$ have two sides which are proportional to two sides placed together at one angle so that their corresponding sides are parallel.

the remaining straight lines $XH$ and $HW$ are in a straight line.

Therefore $XHW$ is a straight line.

$XHW$ lies completely in one plane.

Therefore the pentagon $UBWCV$ lies all in one plane.

It is next to be demonstrated that the pentagon $UBWCV$ is equiangular.

We have that:

$NP$ has been cut in extreme and mean ratio at $R$
$RP$ is the greater segment

Also:

$PR = PS$
$NS$ has been cut in extreme and mean ratio at $P$
$NP$ is the greater segment.
$NS^2 + SP^2 = 3 \cdot NP^2$

But:

$NP = NB$

and:

$PS = SV$

Therefore:

$NS^2 + SV^2 = 3 \cdot NB^2$

So:

$SN^2 + VS^2 + NB^2 = 4 \cdot NB^2$

But:

$SB^2 = SN^2 + NB^2$

Therefore:

$BS^2 + SV^2 = 4 \cdot NB^2$

But $\angle VSB$ is a right angle.

$BS^2 + SV^2 = BV^2$

and so:

$BV^2 = 4 \cdot NB^2$

Therefore:

$VB = 2 \cdot BN$

Therefore:

$BV = BC$

Also we have that:

$BU = BW = UV = WC$
$\angle BUV = \angle BWC$

Similarly it can be demonstrated that:

$\angle UVC = \angle BWC$

Therefore:

$\angle BWC = \angle BUV = \angle UVC$
the pentagon $UBWCV$ is equiangular.

So we have that the pentagon $UBWCV$ is equiangular and equilateral.

Therefore the pentagon $UBWCV$ is regular.

We also have that $UBWCV$ is constructed so that $BC$ is one edge of the cube.

The same construction can be used to place a regular pentagon on each of the $12$ edges of the cube.

Hence a polyhedron will have been constructed which is contained by $12$ regular pentagons.

This is then, by definition, a regular dodecahedron.

$\Box$

It is next to be demonstrated that this regular dodecahedron can be inscribed in the given sphere.

Let $XP$ be produced, and let the produced straight line be $XZ$.

Thereby $PZ$ meets the diameter of the cube.

$PZ$ and the diameter of the cube bisect one another.

Let them meet at $Z$.

Therefore $Z$ is the center of the sphere in which the cube is inscribed.

Also, $ZP$ equals half the side of the cube.

Let $UZ$ be joined.

We have that:

$NS$ has been cut in extreme and mean ratio at $P$
$NP$ is the greater segment.
$NS^2 + SP^2 = 3 \cdot NP^2$

But:

$NP = PZ$

and:

$XP = PS$

and so:

$NS = XZ$

But as $PS = RP$:

$PS = XU$

Therefore:

$ZX^2 + XU^2 = 3 \cdot NP^2$

But:

$UZ^2 = ZX^2 + XU^2$

Therefore:

$UZ^2 = 3 \cdot NP^2$
the square on the diameter of the given sphere is three times the square on the side of the cube.

Therefore the square on the radius of the given sphere is three times the square on half the side of the cube.

Therefore $UZ$ equals the radius of the given sphere.

We have that $Z$ is the center of the sphere.

Therefore $U$ is on the surface of the sphere.

Similarly we can show that each of the remaining vertices of the dodecahedron is also on the surface of the sphere.

Therefore the dodecahedron has been inscribed within the sphere.

$\Box$

It remains to be demonstrated that the side of the dodecahedron is the irrational straight line called apotome.

We have that:

$NP$ has been cut in extreme and mean ratio at $R$
$RP$ is the greater segment

and that:

$PO$ has been cut in extreme and mean ratio at $S$
$PS$ is the greater segment

We have that:

$N : PR = PR : RN$
the same is true of the doubles also.

Therefore:

$NO : RS = RS : NR + SO$

But:

$NO > RS$

Therefore:

$RS > NR + SO$

Therefore $NO$ is cut in extreme and mean ratio, and $RS$ is the greater segment.

But:

$RS = UV$

Therefore when $NO$ is cut in extreme and mean ratio, $UV$ equals the greater segment.

We have that the diameter of the given sphere is rational.

We also have that the square on the diameter equals three times the square on side of the cube.

Therefore $NO$, which equals the side of the cube, is rational.

$UV$ is therefore an apotome.

But $UV$ is the side of the dodecahedron.

Therefore the side of the dodecahedron is an apotome.

$\blacksquare$

## Historical Note

This proof is Proposition $17$ of Book $\text{XIII}$ of Euclid's The Elements.
Some sources credit this construction to Hippasus of Metapontum. Others credit it to Theaetetus of Athens.