Theorem: Five Platonic Solids

There exist exactly five platonic solids:
{{begin-itemize}}
{{item|(1):|the regular tetrahedron}}
{{item|(2):|the cube}}
{{item|(3):|the regular octahedron}}
{{item|(4):|the regular dodecahedron}}
{{item|(5):|the regular icosahedron.}}
{{end-itemize}}
{{:Euclid:Proposition/XIII/18/Endnote}}

Proof:
A solid angle cannot be constructed from only two planes.
Therefore at least three faces need to come together to form a vertex.
Let $P$ be a platonic solid.
Let the polygon which forms each face of $P$ be a equilateral triangles.
We have that:
:each vertex of a regular tetrahedron is composed of $3$ equilateral triangles
:each vertex of a regular octahedron is composed of $4$ equilateral triangles
:each vertex of a regular icosahedron is composed of $5$ equilateral triangles.
$6$ equilateral triangles, placed together at a vertex, form $4$ right angles.
From {{EuclidPropLink|book = XI|prop = 21|title = Solid Angle contained by Plane Angles is Less than Four Right Angles}}:
:a solid angle is contained by plane angles which total less than $4$ right angles.
Thus it is not possible to form $P$ such that its vertices are formed by $6$ equilateral triangles.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $6$ equilateral triangles.
Hence there are only $3$ possible platonic solids whose faces are equilateral triangles.
We have that each vertex of a cube is composed of $3$ squares.
$4$ squares, placed together at a vertex, form $4$ right angles.
Thus, again from {{EuclidPropLink|book = XI|prop = 21|title = Solid Angle contained by Plane Angles is Less than Four Right Angles}}:
:it is not possible to form $P$ such that its vertices are formed by $4$ squares.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ squares.
Hence there is only $1$ possible platonic solid whose faces are squares.
We have that each vertex of a regular dodecahedron is composed of $3$ regular pentagons.
From {{EuclidPropLink|book = XIII|prop = 18|qual=Lemma|title = Comparison of Sides of Five Platonic Figures}}:
:the vertices of a regular pentagon equal $1 \dfrac 1 5$ right angles.
$4$ regular pentagons, placed together at a vertex, form $4 \dfrac 4 5$ right angles.
Thus, again from {{EuclidPropLink|book = XI|prop = 21|title = Solid Angle contained by Plane Angles is Less than Four Right Angles}}:
:it is not possible to form $P$ such that its vertices are formed by $4$ regular pentagons.
For the same reason, it is not possible to form $P$ such that its vertices are formed by more than $4$ regular pentagons.
Hence there is only $1$ possible platonic solid whose faces are regular pentagons.
$3$ regular hexagons, placed together at a vertex, form $4$ right angles.
Thus, again from {{EuclidPropLink|book = XI|prop = 21|title = Solid Angle contained by Plane Angles is Less than Four Right Angles}}:
:it is not possible to form $P$ such that its vertices are formed by $3$ or more regular hexagons.
Regular polygons with more than $6$ sides have vertices which are greater than those of a regular hexagon.
Therefore $3$ such regular polygons, placed together at a vertex, form more than $4$ right angles.
Thus, again from {{EuclidPropLink|book = XI|prop = 21|title = Solid Angle contained by Plane Angles is Less than Four Right Angles}}:
:it is not possible to form $P$ such that its vertices are formed by $3$ or more regular polygons with more than $6$ sides.
Hence the $5$ possible platonic solids have been enumerated and described.
{{qed}}