Post by Jim FerryPost by JEMebiusPost by RobertPost by Chip EasthamPost by William ElliotPost by Patrick D. RockwellIf you have two Tetrahedrons, one smaller than the other,
and just small enough so that if you turned it up side down
and inscribed it into the larger one, it's vertices would touch
the center of each triangular face of the larger, what would
be the ratio of their volumes? Of their sides? Thanks in
1/8, 1/2
See the very nice explanation by "achille"
ahead of you in this thread. Anyway "sides"
probably means faces in this context, not
edges, so we should have answers of the form
r^3 for volumes and r^2 for sides.
regards, chip
which centre? which tetrahedron? for circumcentres the tetrahedron
with vertices 0, (1,0,0), (0,1,0), (0,0,1) will result in a smaller
tetrahedron with zero volume.
I'm unclear what you are asking. In the original
post of this thread, a method for inscribing one
tetrahedron inside another is described: the
smaller tetrahedron will have vertices centered
on the faces of the larger one. As achille showed,
the smaller tetrahedron is similar to the larger
one with (linear) ratio 1/3.
Using your four points as vertices of the larger
tetrahedron, I make the vertices of the smaller
tetrahedron to be (1/3,1/3,1/3), (0,1/3,1/3),
(1/3,0,1/3), and (1/3,1/3,0). As the larger
tetrahedron has volume 1/6, the smaller one
will have volume (1/6)*(1/27) = 1/162.
If the construction were repeated, a sequence
of tetrahedrons would be obtained which tend
to "zero volume" centered on the same point as
the original tetrahedron, (1/4,1/4,1/4).
Is that close to what you were asking?
regards, chip
Nice problem, nice find!
(A) Equidecomposability problem: can the original tetrahedron be cut up into 27 congruent
tetrahedrons of 1/3 of the linear size of the original?
(B) Generalize these problems to N dimensions.
More difficult problem: what remains of all this in infinite-dimensional spaces? In
separable Hilbert space? In inseparable Hilbert space?
Johan E. Mebius
This problem does not generalize to infinite-dimensional space.
Consider the N-d regular simplex, which has N+1 vertices. To write
instead of writing coordinates in R^N, we write them in the subspace
of R^(N+1) in which all coordinates sum to 1. The coordinates of the
regular simplex (with side length sqrt(2)) can then be taken as
(1,0,0,...,0), (0,1,0,...,0), ..., (0,0,0,...,1). The centroid is
(1,1,1,...,1)/(N+1).
The infinite-dimensional regular simplex is well
defined. Including the interior, it is simply the set of points
(x,y,z,...) which (a) sum to 1, and (b) are all non-negative.
Isn't the convex hull of these countably many points
limited to those with only finitely many nonzero
coordinates? (co-finitely zero)
Post by Jim FerryOne might think that the smaller, inscribed simplex would degenerate
to the centroid of the original simplex, but it's worse than that.
There is no centroid to which to degenerate. If there were it would
be (0,0,0,...,0), but this point lies outside the space.
To go back (forth?) to the equidecomposability
problems, we should be looking for a tetrahedron
(resp. N-dimensional simplex) which tiles space
(of the appropriate dimension), since if one of
these is equidecomposable into self-similar parts
of equal size, then turning the construction
around lets us construct a tiling of outwardly
expanding hierarchies of that simplex.
Eike Hertel showed that the tetrahedra which can
be dissected into m^3 congruent pieces by the
"standard construction" (dividing each edge into
m equal segments), each similar to the whole
(reptiles), are exactly the Hill tetrahedra of
the first type:
[Self-similar Simplices (2000)]
http://emis.impa.br/EMIS/journals/BAG/vol.41/no.2/b41h2her.pdf
and conjectured that the only tetrahedra which
are k-reptiles are these Hill tetrahedra.
I don't know if further progress has been made
on this conjecture or extensions to higher
dimensions.
regards, chip