Determining the volume of the unit
simplex

using Platonic solids and scaling of
volumes (full screen).

What portion of a cube is occupied by a pyramid spanning
four adjacent

corners? In other words, if you stood in a corner of
the floor of

a cubical room facing its center, and divided off a section
with a

piece of plywood which went straight from the corners
of the floor

to your right and left to the corner of the ceiling above
you,

how much of the room would be walled off? This may be
answered

using calculus by finding the volume of the portion of
the

unit cube (0<=x<=1,0<=y<=1,0<=z<=1)
with x+y+z<=1. This latter

inequality describes the `(0,0,0)' (you) side of the
plane containing

the three corner points (1,0,0) (floor to your right),
(0,1,0) (floor

to your left), and (0,0,1) (ceiling above you). Each
of these points

satisfies the equation of that plane x+y+z=1, and together
comprise

three points determining it.

This dissection determines that the volume of that pyramid,
called

a standard 3-simplex by mathematicians (the standard
2-simplex is

the triangle x+y<=1 in the unit square 0<=x<=1,0<=y<=1,
and so on)

to be 1/6 that of the cube without calculus. The key
observation

we need to do this is, is that when similar solids are
scaled so that

corresponding lengths are scaled by a factor r, then
corresponding

volumes are scaled by r^{3}, r cubed. This is
apparent for a cube,

where exactly 8 half-scaled cubes fit in the full cube,
27 third-scaled

cubes fit in the full cube, and so on. If the volumes
of other shaped

solids could be built from or approximated by cubes,
then their

volumes should also scale by the cube of the length scaling.

The first stage of the dissection removes four congruent
corner

pyramids from the cube, two containing the floor, reflections

of each other across a diagonal of the room, and two
containing

the ceiling, reflections of each other across the other
diagonal.

This leaves a regular tetrahedron with four congruent
equilateral

triangle faces in the center. The volume of this tetrahedron
is

equal to that of two of the corner pyramids, which is
demonstrated

in the next two stages.

The second stage divides the tetrahedron into two equal
volumes

by cutting off four similar half-scale corners, separated
at the

midpoints of each edge. Since each half-scale tetrahedron
represents

one eighth the volume of the original tetrahedron, we
have removed

four eighths or one half of the volume.

The third stage exhibits the remaining truncated tetrahedron
is

in fact a regular octahedron, whose six vertices are
the

centers of the original cube. (They are the midpoints
of the

tetrahedron each of whose six edges was a diagonal on
which

the original cube was sliced.) It is broken into eight pieces
each

of which is a half-scale version of the original corner
pyramid.

Thinking again of the octahedron as sitting inside the
original

cube, each of these pieces arises from joining the center
point

to the centers of three adjacent faces, and spanning
those

center points with a plane. This is the same construction
we

used to slice off the corners, with the sides which play
the

roles of the cube edges now half as long (center of cube
to

center of face.) Then the volume of each of these pieces

is one eighth that of the original corner pieces, and
since

eight of them comprise the regular octahedron, its volume

is equal to that of the original corner.

In summary, the volume of the octahedron equals that of
the

corner pyramid, the volume of the tetrahedron is twice
that

of the corner pyramid, and the volume of four corner
pyramids

plus tetrahedron equals four corner pyramids plus two
corner

pyramids equals the volume of the cube. If the volume
of

the unit cube is set to one, then the volume of the

corner pyramids and the octahedron is one-sixth, and
the

volume of the regular tetrahedron is one third.

Exercise: This construction involves three of the five
Platonic

regular solids, the cube and octahedron (duals of each
other-

the vertices of one interchange with the centers of the
faces

of the other) and the tetrahedron (self-dual). Examine
the other

two (icosahedron,dodecahedron).

Exercise: Compute the length of the sides of the equilateral
triangle

face of a corner pyramid, and its area.

Exercise: Compute the length of the sides of the regular

tetrahedron which remains after the corners of the cube
are removed,

the area of its faces, and its altitude. Compute the
face area

and volume of a tetrahedron with unit side length.

Exercise: Determine the length of the sides, the altitude,

and the area of the faces of the octahedron. Scale these

to determine the altitude/diameter, area of sides, and

volume of an octahedron with unit side length.

Exercise: Check that all corresponding lengths of similar
figures

scale by the scale factor of any correspoding lengths,
and corresponding

areas scale by the length scale factor squared (r^{2}).
Can you

construct any objects which scale by powers which are
not whole

numbers? A candidate for such an object, called a fractal,
might

be something which is equal to three half-scaled copies
of itself,

between two (=2^{1}, one dimensional) and four
(=2^{2}, two dimensional).

What would its dimension be?

Exercise: Describe a standard 4-simplex.