Oops --- in terms of Cartesian vectors, the formulae require scaling weights --- read << Instead, we must compute "pseudo-centroid" P = (A_1 + ... + A_n - A_0) /(n-1) ; set C_i = P (n-1)/(n-2) - A_i /(n-2) ; then reflect C_0 in the new base plane H_0 as before. >> WFL On 2/15/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< WRS uses much the same transformations as I do: a centroid inscription (in one dimension lower), two homotheties and a translation (rather than one homothety and a reflection), which from an animation perspective is more complicated whilst avoiding discontinuity. >> That critique can't be quite right: in even dimension, somewhere in the chain there has to be an orientation-reversal (possibly achieved via homothety).
Now notice centroidally inscribed B is homothetic to original simplex A , via centre the centroid; once that fact has registered, it is a short step to try constructing C via the same method as B , having first simply changed the sign of distinguished vertex A_0 !
Sadly, that doesn't quite work: although each new vertex C_i (including C_0 ) lies on original face plane F_i , and the base edges match up, the slant edges are incorrect. Instead, we must compute "pseudo-centroid" P = A_1 + ... + A_n - A_0 ; set C_i = P - A_i ; then reflect C_0 in the new base plane H_0 as before (dammit!).
But though this is algorithmically neater than previous attempts, it seems harder to motivate ...
WFL
On 2/14/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
WRS uses much the same transformations as I do: a centroid inscription (in one dimension lower), two homotheties and a translation (rather than one homothety and a reflection), which from an animation perspective is more complicated whilst avoiding discontinuity.
There's something to be said for his distinguishing 3-space separately; [Kla79] do the same with 2-space --- a recourse unavailable here!
My possibly oppressive formality is a consequence of having developed computer programs in advance of writing up. The upside is that Maple code for my diagrams could surely be modified easily to deliver an animation, if that served a useful didactic purpose. ___________________________
Now I never did get around to explaining just why this construction is of interest; in pursuit of which I pose
Problem (4) : Denote by R' the circumradius of any simplex D , inscribed to an original A with inradius r . Show that r <= R' . If instead D is exscribed to A with exradius r , deduce |r| <= R' . [ Of course it's "obvious" --- but can you prove it's obvious? ]
Once that's out of the way, apply it immediately to the (similar 'scribed) B,C constructed earlier, to prove two theorems: denoting by R,r circumradius and in/exradius of a simplex in n-space,
*** in-radius r <= R/n ; *** *** ex-radius |r| <= R/(n-2) ; ***
the first due originally perhaps to Fejes-Tóth, the second a shameless rip-off from [Kla79] claimed provisionally by myself.
Problem (5) : In 2-space both the construction of C and this last theorem gallop away to infinity. What should take their place?
[Kla79] Murray S. Klamkin, George A. Tsintsifas (1979) "The Circumradius-Inradius Inequality for a Simplex" Mathematics Magazine Vol. 52, No. 1 (Jan., 1979), pp. 20-22 http://www.jstor.org/stable/2689968?seq=1#page_scan_tab_contents
WFL
On 2/13/16, William R. Somsky <wrsomsky@gmail.com> wrote:
Here is a less formal approach, which is less rigorous, but perhaps easier to visualize (well, at least it was to me before I started to put it in words -- can anyone do an animation?):
Take your tetrahedron [[ n-space simplex ]] and label the vertex A0 as the apex, and the face F0 as the base. Now place it w/ the base on the table and the apex A0 above it. Pick units such that the height of the apex above the table is one.
Within the triangular base [[ n-1 simplex ]], inscribe a similar triangle using the midpoints of the base's sides as the vertices of the similar triangle. This similar triangle [[ simplex ]] will be scaled 1:2 [[ 1:n-1 ]] in relation to the original base.
On top of this similar base, construct a tetrahedron [[ n simplex ]] similar to the original tetrahedron with vertices B. This similar tetrahedron will be in 1/2 [[ 1/(n-1) ]] scale of the original, and it's apex, B0, will be at a height 1/2 [[ 1/(n-1) ]] above the table.
Project each of the vertices B of the similar tetrahedron in a line from A0 by a factor of S, giving a new tetrahedron [[ simplex ]] with vertices C, similar to the original w/ scale S:2 [[ S:n-1 ]].
Note that each vertex C1,...,Cn projected from the inscribed base is on an extended face of the original tetrahedron. And the new apex, C0? A0 was at height 1 above the tabletop; B0 was at height 1/2 [[ 1/[n-1] ]]; so projecting to C0 places it at height 1 - S ( 1 - 1/2 ) [[ 1 - S ( 1 - 1/[n-1]) ]].
If we adjust S such that the height above the table of C0 is zero, it will lie on the original face F0, giving us the desired excribed tetrahedron [[ n simplex ]]. This occurs when S = 2 [[ S = (n-1)/(n-2) ]], and the excribed tetrahedron is in scale S:2 = 1:1 [[ S:n-1 = 1:n-2 ]] to the original.
On 2016-02-13 11:23, Fred Lunnon wrote:
On 2/13/16, William R. Somsky <wrsomsky@gmail.com> wrote:
Yeah, I see how to do it. In n>=3 space, you end up w/ a 1/(n-2) copy.
WRS wins the chocolate frog!
Below is my solution, extracted from a rather terse text-based summary of results about simplex exradii which I'll post separately when polished within an inch of its (and my) life.
WFL
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In Euclidean n-space, let A be an arbitrary given simplex, with vertices A_0,...,A_n and (n-1)-dimensional facets F_0,...,F_n opposite the respective vertices. Similarly B,C have vertices B_i,C_i and facets G_i,H_i respectively. Distinguished vertex A_0 acts as `apex' of A , and facet F_0 as its `base'.
DEFINITION: B is `inscribed' to A when vertex B_i meets facet F_i ; C is `exscribed' to A when vertex C_i meets the hyperplane extending facet F_i , and C lies entirely on the side of F_0 opposite to A_0 . [The analogy is with insphere and exsphere of A .]
Suppose now that B is inscribed to A , with B_i the centroid of F_i for all i . It is a familiar fact that B is then similar to A , but scaled in the ratio B : A = (-1)^n : n . Similarity follows from noting that G_i is parallel to F_i for all i ; the absolute ratio 1 : n via induction on n ; similarity is direct for n even but `reverses orientation' for n odd, again via induction on n .
LEMMA: For any simplex A in Euclidean n-space, there exists C exscribed to A and similar to A , scaled in ratio C : A = (-1)^n : (2-n) . The similarity bijection is `natural': if vertex C_i opposite facet H_i meets facet F_i opposite vertex A_i , then H_i , F_i are similar, as are the vertex figures truncating C_i , A_i .
Proof: Let B be inscribed in the facet centroids of A as above. The first stage of construction dilates B from centre A_0 yielding C --- [Kla79] calls this a `homothety' --- by a scale factor s , chosen so that the distance between the new base H_0 and the original base F_0 equals the altitude of C .
Letting the altitude of A be unity, the distances of F_0, G_0, H_0 from A_0 equal 1, (n-1)/n, s(n-1)/n ; the altitudes of A, B, C equal 1, 1/n, s/n respectively. So s(n-1)/n - 1 = s/n , whence s = n/(n-2) , and the required transformation in Cartesian vector notation becomes C_i = n/(n-2) B_i - 2/(n-2) A_0 .
The base vertices C_1,...,C_n of C remain on hyperplanes extending F_1,...,F_n ; however the apex C_0 points downwards. This is remedied by reflecting C in its own base H_0 , so that C_0 now also meets F_0 or its extension. It is similar to B , and hence to A . Reflection in the base reverses the orientation of B , so the final scale factor with respect to A equals (-1)^n / (2-n) . QED
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