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 ___________________________________ 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 ___________________________________________