For those who already know
what Platonic Solids are, one might wonder as to whether there are any
other 5 Platonic Solids than the ones which we already know?
For those who do not know
what Platonic Solids are, they are those solid objects whose sides are
made of a unique regular polygon i.e. they are regular polyhedra.
The 5 platonic solids are:
No.
Platonic Solids
No. of sides
of the polyhedra
base polygon
1
Tetrahedron
4
Equilateral Triangle
2
Cube
6
Square
3
Octahedron
8
Equilateral Triangle
4
Dodecahedron
12
Regular Pentagon
5
Icosahedron
20
Equilateral Triangle
Now the question arises.
Why do we have only 5 Platonic Solids? Are there any other regular
polyhedra which haven't been discovered yet?
In mathematics we always
make a logical analysis to find an equation to solve a particular type of
problem with available data or to find a solution to a puzzle.
In the above case the
puzzle is the existence of only 5 regular polyhedra. To solve this puzzle,
first we shall look at the available data and build a formula to verify
the existence of a given regular polyhedra with a given number of
sides with a given regular polygon.
The data available is that
for a given polyhedra there is only one regular polygon which makes up its
sides. The other data which is required to form any solid object regular
or irregular (i.e. with irregular and/or different polygons ) is
that the sum of the angles between the sides of the polygons at any
given vertex of the polyhedra should always be less than 360 degrees else
the polyhedra will flatten out at that particular vertex and the formation
of a closed solid object would be impossible.
So we have two sets of data:
Sum of the angles of
all the sides at any given vertex of the regular polyhedra has
to be less than 360o
The number of sides
which meet at a vertex is same for all vertices (for a regular
polyhedra) and the angles between all sides and the total angle at any
vertex is the same at all vertices (as all the sides are made up of
regular polygons).
Thus if 'n' is the total
number of the sides which meet at a vertex of the polyhedra and 'Ø' is
the angle between any two sides of the regular polygon, then the sum
of the angles between all the sides at any vertex of the polyhedra
is given by
Ø+Ø+Ø+Ø+Ø+Ø.. n
times = nØ
We know that for the polyhedra to
exist, nØ hs to be less than 360
i.e. nØ <
360 - (1)
Now we shall see how many
regular polyhedra are possible which satisfy the above condition.
We know that a minimum of
3 sides are required to meet at a vertex to form a polyhedra. Thus n>2
is also an essential condition.
Therefore, let us
start with n=3, i.e. let 3 regular polygons meet at each vertex. Let
the first regular polygon we start with be the one with the smallest
number of sides i.e. a equilateral triangle. So we have Ø = 60. Thus
nØ = 3 x 60 = 180 < 360. Hence a regular polyhedra with the above
combination is possible as this satisfies our condition (1) and we
know that this a tetrahedron.
Now let n=4 and we
continue with an equilateral triangle as the regular polygon. We
have
nØ = 4 x 60 = 240 < 360. Since our condition is satisfied, we have
another regular polygon i.e. an Octahedron.
Let n=5 and we still
continue with an equilateral triangle as our regular polygon and we
have nØ = 5 x 60 = 300 which still satisfies our condition and here
we have a Icosahedron.
Let n= 6 and we
continue with the equilateral triangle i.e. Ø=60. Thus
nØ = 6 x 60 = 360 = 360. Here our condition (1) is not satisfied and
hence a regular polyhedra with this combination cannot exist in
nature. With Ø=60 no other regular polyhedra is possible for n =6 or
greater.
So now we switch to
the next smallest regular polygon i.e. a square. Hence we have Ø=90.
As usual, we shall start with n=3. We have Ø=90 for a square. Thus
nØ = 3 x 90 = 270 < 360. Thus it is possible to have a regular
polyhedra made up of squares with 3 squares meeting at each vertex.
This is our most familiar Platonic Solid i.e. a cube.
Now let n=4 and Ø=90
i.e. let us continue with the square. Now nØ = 4 x 90 = 360 = 360.
Our condition (1) is not satisfied. Hence a regular polyhedra with
this combination is not possible nor is it possible for higher values
of n with Ø = 90.
Let us now switch to
the next polygon i.e. a pentagon which has 5 equal sides and 5
interior angles with each angle being 108 degrees i.e. Ø=108.
Starting with n=3 we have
nØ = 3 x 108 = 324 < 360. Hence such a regular polyhedra can exist
and it is a Dodecagon.
Now when n=4 and
Ø=108 we have nØ = 4 x 108 = 432 > 360. Hence this regular
polyhedra cannot exist as our condition (1) is not satisfied.
Similarly it cannot exist for higher values of n with Ø=108.
Now we move on to
hexagon as our regular polygon i.e. Ø=120. starting with n=3 we see
that
nØ = 3 x 120 = 360 = 360 . Our condition (1) is not satisfied and
this regular polyhedra cannot exist nor can it exist for any higher
values of n or Ø as all such combinations will be greater than 360.
Thus we have only 5
Platonic Solids whose existence we have proved in the above reasoning and
we also have proved that no other regular polyhedra can exist in nature.
If you have any queries feel free to contact
me.
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