Mathematical analysis of truncated icosahedron & identical football by applying HCR's formula for platonic solids

Mathematical analysis of truncated icosahedron & football
Application of HCR’s formula for regular polyhedrons (all five platonic solids)
Applications of “HCR’s Theory of Polygon” proposed by Mr H.C. Rajpoot (year-2014)
©All rights reserved
Mr Harish Chandra Rajpoot
M.M.M. University of Technology, Gorakhpur-273010 (UP), India Dec, 2014
Introduction: A truncated icosahedron is a solid which has 12 congruent regular pentagonal & 20 congruent
regular hexagonal faces each having equal edge length. It is obtained by truncating a regular icosahedron
(having 20 congruent faces each as an equilateral triangle) at the vertices to generate 12 regular pentagonal &
20 regular hexagonal faces of equal edge length. For calculating all the parameters of a truncated icosahedron,
we would use the equations of right pyramid & regular icosahedron. When a regular icosahedron is truncated
at the vertex, a right pyramid with base as a regular pentagon & certain normal height is obtained. Since, a
regular icosahedron has 12 vertices hence we obtain 12 truncated off congruent right pyramids each with a
regular pentagonal base.
Truncation of a regular icosahedron: For ease of calculations, let there be a regular icosahedron with edge
length & its centre at the point C. Now it is truncated at all 12 vertices to obtain a truncated icosahedron.
Thus each of the congruent equilateral triangular faces with edge length is changed into a regular
hexagonal face with edge length (see figure 2) & we obtain 12 truncated off congruent right pyramids with
base as a regular pentagon corresponding to 12 vertices of the parent solid. (See figure 1 showing a right
pyramid with regular pentagonal base & normal height being truncated from the regular icosahedron).
Figure 1: A right pyramid with base as a regular pentagon
with side length 𝒂 & normal height h is truncated off from
a regular icosahedron with edge length 𝟑𝒂
Figure 2: Each of the congruent equilateral
triangular faces with edge length 𝟑𝒂 of a regular
icosahedron is changed into a regular hexagonal
face with edge length 𝒂 by truncation of vertices.

Analysis of Truncated icosahedron by using equations of right pyramid & regular icosahedron
Now consider any of 12 truncated off congruent right pyramids having base as a regular pentagon ABDEF with
side length , normal height & an angle between any two consecutive lateral edges (see figure 3 below)
Normal height ( ) of truncated off right pyramid: We know that the normal height of any right pyramid
is given as
√
√ √(√ ) (
√
√ √
)
Figure 3: Normal distance ( ) of regular pentagonal faces is always greater than the normal distance ( ) of regular
hexagonal faces measured from the centre C of any truncated icosahedron.
⇒ √
√ √
√
√
( √ )
√
√
(√ )
√
(√ )
√ √
(√ )
√ √
(√ )
( )

Volume ( ) of truncated off right pyramid: We know that the volume of a right pyramid is given as
( ) ( )
( ) ( )
(√ )
√ √
(
√
√ √
)
(√ )
√ √
(
√
)
( √ )
( √ )( √ )
( √ ) ( √ )
( )
Normal distance ( ) of regular pentagonal faces from the centre of truncated icosahedron: The
normal distance ( ) of each of the regular pentagonal faces from the centre C of truncated icosahedron is
given as
( )
⇒ ( ( ) )
⇒ ( )
(√ )
√ √
( )
√ √ (√ )
√ √
(
√( √ )( √ ) (√ )
√ √
)
(
√ √
√ √
)
⇒ (
√
√ √
)
( √ )
√ √
( √ )
( )
It’s clear that all 12 congruent regular pentagonal faces are at an equal normal distance from the centre
of any truncated icosahedron.
Solid angle ( ) subtended by each of the regular pentagonal faces at the centre of truncated
icosahedron: we know that the solid angle ( ) subtended by any regular polygon with each side of length
at any point lying at a distance H on the vertical axis passing through the centre of plane is given by “HCR’s
Theory of Polygon” as follows
(
√
)
Hence, by substituting the corresponding values in the above expression, we get

(
(
( √ )
√ √
)
√ (
( √ )
√ √
)
)
(
( √ )
√ √
√ √
√
( √ )
√
(
√
√ √
)
)
(
( √ )√ √
√ √ √
)
(
( √ )√ √
√ √
) (
( √ )√ √
√ √
)
(
( √ )√ √ √ √
√ √ √ √
) (
( √ ) √
( √ )
)
(
( √ )
( √ )
) (
( √ )( √ )
( √ )( √ )
)
(
( √ √ )
( )
) (
( √ )
( )
)
(
√
) ( )
Normal distance ( ) of regular hexagonal faces from the centre of truncated icosahedron: The
normal distance ( ) of each of the regular hexagonal faces from the centre C of truncated icosahedron is
given as
( )
⇒ ( ( ) )
⇒
( √ )( )
√
( )
⇒
√ ( √ )
( )
It’s clear that all 20 congruent regular hexagonal faces are at an equal normal distance from the centre of
any truncated icosahedron.
It’s also clear from eq(III) & (V) i.e. the normal distance ( ) of regular pentagonal faces is greater
than the normal distance ( ) of regular hexagonal faces from the centre of truncated icosahedron i.e.
hexagonal faces are much closer to the centre as compared to the pentagonal faces in any truncated
icosahedron.

Solid angle ( ) subtended by each of the regular hexagonal faces at the centre of truncated
icosahedron: we know that the solid angle ( ) subtended by any regular polygon is given by “HCR’s Theory
of Polygon” as follows
(
√
)
Hence, by substituting the corresponding values in the above expression, we get
(
(
√ ( √ )
)
√ (
√ ( √ )
)
)
(
(
√ ( √ )
)
√ ( √ )
(√ )
)
(
√ ( √ )
√ √
)
(
√ ( √ )
√ ( √ )
)
(
√ √ √
√
) ( √ √
)
( √ √
)
(
√(√ )
√
)
(
√
√
)
(
√
√
) (
√
) ( )
It’s clear that the solid angle subtended by each of the regular hexagonal faces is greater than the solid angle
subtended by each of the regular pentagonal faces at the centre of any truncated icosahedron.
Important parameters or dimensions of a truncated icosahedron:
1. Inner (inscribed) radius( ): It is the radius of the largest sphere inscribed (trapped inside) by a
truncated icosahedron. The largest inscribed sphere always touches all 20 congruent regular
hexagonal faces but does not touch any of 12 regular pentagonal faces at all since all 20 hexagonal
faces are closer to the centre as compared to all 12 pentagonal faces. Thus, inner radius is always
equal to the normal distance ( ) of hexagonal faces from the centre & is given from the eq(V) as
follows
√ ( √ )
Hence, the volume of inscribed sphere is given as

( ) (
√ ( √ )
)
2. Outer (circumscribed) radius( ): It is the radius of the smallest sphere circumscribing a given
truncated icosahedron or it’s the radius of a spherical surface passing through all 60 vertices of a
given truncated icosahedron. It is calculated as follows (See figure 3 above).
In right
⇒
( )
( )
In right
⇒ √( ) ( ) √( ) (
( √ )
) ( )
√ √ √ √
√ √
√
( √ )( √ )
( √ )( √ )
√ √
√ √ √ √
( )
Hence, outer radius of truncated icosahedron is given as
√ √
Hence, the volume of circumscribed sphere is given as
( ) (
√ √
)
3. Surface area( ): We know that a truncated icosahedron has 12 congruent regular pentagonal & 20
regular hexagonal faces each of edge length . Hence, its surface area is given as follows
( ) ( )
We know that area of any regular n-polygon with each side of length is given as
Hence, by substituting all the corresponding values in the above expression, we get
( ) ( ) √
( √ )

4. Volume( ): We know that a truncated icosahedron with edge length is obtained by truncating a
regular icosahedron with edge length at all its 12 vertices. Thus, 12 congruent right pyramids with
regular pentagonal base are truncated off from a parent regular icosahedron. Hence, the volume (V)
of the truncated icosahedron is given as follows
( ) ( )
( ( √ )( ) ) ( ) ( )
( ( √ ) ) (
( √ )
) ( ( ))
( √ )
( √ )
( √ √ ) ( √ )
( √ )
5. Mean radius( ): It is the radius of the sphere having a volume equal to that of a given truncated
icosahedron. It is calculated as follows
( )
( √ )
⇒ ( )
( √ )
(
( √ )
)
It’s clear from above results that
Construction of a solid truncated icosahedron: In order to construct a solid truncated icosahedron with
edge length there are two methods
1. Construction from elementary right pyramids: In this method, first we construct all elementary right
pyramids as follows
Construct 12 congruent right pyramids with regular pentagonal base of side length & normal height ( )
( √ )
Construct 20 congruent right pyramids with regular hexagonal base of side length & normal height ( )
√ ( √ )
Now, paste/bond by joining all these right pyramids by overlapping their lateral surfaces & keeping their apex
points coincident with each other such that all edges of each pentagonal base (face) coincide with the edges of
five hexagonal bases (faces). Thus, a solid truncated icosahedron with 12 congruent regular pentagonal & 20
congruent regular hexagonal faces each of edge length is obtained.

2. Machining a solid sphere: It is a method of machining, first we select a blank as a solid sphere of certain
material (i.e. metal, alloy, composite material etc.) & with suitable diameter in order to obtain the maximum
desired edge length of truncated icosahedron. Then, we perform facing operation on the solid sphere to
generate 12 congruent regular pentagonal & 20 congruent regular hexagonal faces each of equal edge length.
Let there be a blank as a solid sphere with a diameter D. Then the edge length , of a truncated icosahedron of
maximum volume to be produced, can be co-related with the diameter D by relation of outer radius ( ) with
edge length ( )of a truncated icosahedron as follows
√ √
Now, substituting ⁄ in the above expression, we have
√ √
√ √
Above relation is very useful for determining the edge length of a truncated icosahedron to be produced
from a solid sphere with known diameter D for manufacturing purposes.
Hence, the maximum volume of truncated icosahedron produced from the solid sphere is given as follows
( √ ) ( √ )
(
√ √
)
( √ )
( √ )√ √
( √ )( √ )
( √ )( √ )√ √
( √ )
√ √
( √ )
√ √
Minimum volume of material removed is given as
( ) ( ) ( )
( √ )
√ √
(
( √ )
√ √
)
( ) (
( √ )
√ √
)
Percentage ( ) of minimum volume of material removed

(
( √ )
√ √
)
(
( √ )
√ √
)
It’s obvious that when a truncated icosahedron of maximum volume is produced from a solid sphere then
about of material is removed as scraps. Thus, we can select optimum diameter of blank as a solid
sphere to produce a truncated icosahedron of maximum volume (or with maximum desired edge length)
Analysis of Football: let there be a football with a diameter D (outside) having 12 congruent regular
spherical pentagons & 20 congruent regular spherical hexagons on its surface. This case of the football is
similar to the case of a truncated icosahedron. The only difference is that a football has spherical faces instead
of flat faces of a truncated icosahedron. But in both the cases, the values of solid angle subtended by each of
regular pentagonal & each of regular hexagonal faces are constant & are independent of the geometrical
dimensions (i.e. diameter & edge length) of both the football & the truncated icosahedron. Hence, we have
the following important parameters of a football with a certain diameter (outside)
1. Solid angle subtended by each of 12 congruent regular spherical pentagons is given as
(
√
)
2. Area of each of 12 congruent regular spherical pentagons is given as
( ) ( )
( ) ( (
√
))
( (
√
))
3. Solid angle subtended by each of 20 congruent regular spherical hexagons is given as
(
√
)
4. Area of each of 20 congruent regular spherical hexagons is given as
( ) ( )
( ) ( (
√
))
( (
√
))
It’s clear from the above results that the solid angle subtended by each of 12 congruent regular spherical
pentagons is smaller than the solid angle subtended by each of 20 congruent regular spherical hexagons at
the centre of the football & hence the area of a regular spherical pentagon is smaller than that of a regular
spherical hexagon. Solid angle subtended by any of the faces of football is independent of the diameter.

Conclusions: let there be any truncated icosahedron having 12 congruent regular pentagonal & 20
congruent regular hexagonal faces each with edge length then all its important parameters are
calculated/determined as tabulated below
Congruent
polygonal faces
No. of
faces
Normal distance of each face from the
centre of the given truncated icosahedron
Solid angle subtended by each face at the
centre of the given truncated icosahedron
Regular
pentagon
12 ( √ )
(
√
)
Regular
hexagon
20 √ ( √ )
(
√
)
Inner (inscribed)
radius ( )
√ ( √ )
Outer (circumscribed)
radius ( )
√ √
Mean radius ( )
(
( √ )
)
Surface area ( ) ( √ )
Volume ( ) ( √ )
Note: Above articles had been developed & illustrated by Mr H.C. Rajpoot (B Tech, Mechanical Engineering)
M.M.M. University of Technology, Gorakhpur-273010 (UP) India Dec, 2014
Email: rajpootharishchandra@gmail.com
Author’s Home Page: https://notionpress.com/author/HarishChandraRajpoot
Courtesy: Advanced Geometry by Harish Chandra Rajpoot

Mathematical analysis of truncated icosahedron & identical football by applying HCR's formula for platonic solids

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