Five Platonic Solids/Proof 2

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Theorem

There exist exactly five platonic solids:

$\paren 1: \quad$ the regular tetrahedron
$\paren 2: \quad$ the cube
$\paren 3: \quad$ the regular octahedron
$\paren 4: \quad$ the regular dodecahedron
$\paren 5: \quad$ the regular icosahedron.

In the words of Euclid:

I say next that no other figure, besides the said five figures, can be constructed which is contained by equilateral and equiangular figures equal to one another.

(The Elements: Book $\text{XIII}$: Proposition $18$ : Endnote)


Proof

Consider a convex regular polyhedron $P$.

Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$.

Let $n$ be the number of those polygons which meet at each vertex of $P$.

From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$.


The sum of the internal angles must be less than $360^\circ$.

So:

\(\ds n \paren {180^\circ - \dfrac {360^\circ} m}\) \(<\) \(\ds 360^\circ\)
\(\ds \leadsto \ \ \) \(\ds n \paren {1 - \dfrac 2 m}\) \(<\) \(\ds 2\)
\(\ds \leadsto \ \ \) \(\ds n \paren {m - 2}\) \(<\) \(\ds 2m\)
\(\ds \leadsto \ \ \) \(\ds n \paren {m - 2} - 2 \paren {m - 2}\) \(<\) \(\ds 2m - 2 \paren {m - 2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {m - 2} \paren {n - 2}\) \(<\) \(\ds 4\)


But $m$ and $n$ are both greater than $2$.

So:

if $m = 3$, $n$ can only be $3$, $4$ or $5$
if $m = 4$, $n$ can only be $3$
if $m = 5$, $n$ can only be $3$

and $m$ cannot be greater than $3$.

There are $5$ possibilities in all.

Therefore all platonic solids have been accounted for.

$\blacksquare$


Sources

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