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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