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 r³, 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 (r²). 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 (=2¹, one dimensional) and four (=2², two dimensional).
What would its dimension be?
 
 

Exercise: Describe a standard 4-simplex.