COMPONENTS DEFINED BY PLATONIC SOLIDS.
© 2009. RAIMOND A. STRUBLE, PhD.
DRAFT COPY ONLY.
9/27/2009.
© Raimond A. Struble.
Send comments and correspondence to: Raimond A. Struble,
P. O. Box 50376, Raleigh, NC 27650-6376
and emails to: George.Moore4@va.gov
Problem: Components defined by planes extended from faces
of a Platonic solid.
There are five Platonic solids (convex regular polyhedra), namely,
tetrahedron, cube, octahedron, dodecahedron, and icosahedron.
Consider a Platonic solid in Euclidean space, with a plane extended from each
face of the solid. How many distinct components, i. e., maximal connected
open sets, are defined by these planar boundaries?
Submitter's solution: 9 components for the regular tetrahedron.
There are four triangular faces of the regular tetrahedron, viewed from the
top, namely, left, right, front, bottom:
4612.
RAS notes. One external component, each face. One external component,
each vertex. One interior component. (Nothing left out.)
From one vertex, if we remove the opposing face-plane, then the remaining
three face-planes through that vertex, create two outward, unbounded
components. When the missing face-plane is restored, then there will remain
two unbounded components and one bounded component, namely, the interior
of the tetrahedron. Therefore, the vertices produce 2 × 4 = 8
unbounded components and one (common) bounded component, for a total
of 9 components.
Bill Moore's narrative. There are four vertices. Passing through the
top vertex, three triangular face-planes extend downward and upward, to form
two cone-like surfaces, resembling a conical party-hat and its reflection:
4611.
The top face-plane forms one unbounded component. The bottom face-plane cuts
the bottom triangular face, to form a bounded component within
the tetrahedron, and an unbounded component extending below the tetrahedron.
A similar construction obtains for each of the four vertices of the
tetrahedron, with the bounded component always the interior of the
tetrahedron. Thus the four vertices produce 2 × 4 = 8 unbounded
components and one (common) bounded component, for a total of 9
components.
Submitter's solution: 27 components for the cube.
There are six square faces of the cube, viewed from the front, namely,
front, back, left, right, top, bottom:
4601.
RAS notes. One external component, each face, 6.
One external component, each vertex, 8. One external component,
each edge, 12. One bounded component, the interior of the cube.
Total, 6+8+12+1=27.
Each face generates a channel with four sides. Five-sided component
(6).
Each edge generates a channel with with two sides which are cut off
by two planes. Four-sided component (12).
Each vertex generates a Three-sided component (8).
(No duplication. Nothing left out.)
One can imagine a package of cookies, with cardboard dividers between
the individual cookies, corresponding to four of the six planar dividers,
as viewed from the top:
4591.
In the central volume between the top and bottom planes, there is one
bounded component and eight unbounded components. Four of these components
are unbounded in one direction, and four components are unbounded
in two directions, for a total of nine.
Above the top plane, there is one component unbounded in one direction,
four components unbounded in two directions, and four components unbounded
in three directions, for a total of nine. Likewise, below the bottom plane,
there is one component unbounded in one direction, four components unbounded
in two directions, and four components unbounded in three directions,
for a total of nine.
Thus there are 9+9+9=27 components all told.
Submitter's solution: xx components for the regular octahedron.
There are eight triangular faces of the regular octahedron,
viewed from the top, namely, ..................
4603.
RAS notes.
Triangular unbounded cones for each face. Four-sided component.
Quadrangular unbounded cones for each vertex. Five-sided component.
Submitter's solution: xx components for the regular dodecahedron.
There are twelve pentagonal faces of the regular dodecahedron,
viewed from the top, namely, ...........
4604.
RAS notes. Each face: Five-sided spike with a base (six sides).
These spikes produce unbounded five-sided cones. Each original vertex
produces an unbounded cones. 47?
Submitter's solution: xx components for the regular icosahedron.
There are twenty triangular faces of the regular icosahedron,
viewed from the top, namely, ................
4605.
RAS notes. Its surface is blistered with twenty tetrahedra.
Each blister produces an unbounded triangular cone; as per tetrahedron.
Each vertex produces a five-sided, unbounded cone. Tet blisters (20);
Tet blister triangular cones (20); Vertex cones (12).
The last three proofs are left to the reader.
Last updated: 9/27/2009, by Raimond A. Struble, PhD.