Your picture is surely partly correct, in that any planes may simply be adjusted so that they meet the given sphere and still form a trough outside some tetrahedron. However given a regular tetrahedron, that trough turns out to be the wrong shape to accommodate any sphere! I initially thought I had a rather slick argument supporting (B), but further contemplation has revealed the (inevitable) flaw. So (B) is probably false in general, and a little more spadework is indicated --- customary apologies! WFL On 4/6/16, Allan Wechsler <acwacw@gmail.com> wrote:

Does (B) have a typo? I have a pretty good visualization of a sphere tangent to the inside of two faces and the outside of two others. It hovers just outside the common edge of the two faces it's exteriorly tangent to, and the other two faces cut off the ends of the trough it rests in. I do see that m=3 is impossible for the tetrahedron.

On Wed, Apr 6, 2016 at 12:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

I finally wrote up properly all the stuff I've been doing about inspheres and exspheres of a simplex. Finished. Done & dusted. Off my plate.

But as I sat down again to start writing up P*nc*l*t's perishin' p*r*sm, an obnoxious inspiration intruded. Why shouldn't a simplex in Euclidean space also sport an ` m-fold' exsphere, tangent to m facets on their exterior side, and to n-m+1 on their interior side, where perhaps m > 1 ?

In particular, show

(A) That m = 2 is impossible when n = 2 (for a triangle);

(B) That m = 2 is impossible when n = 3 (for a tetrahedron);

(C) What happens for m = 2 and n = 4 (for a pentatope);

(D) What happens for general 0 <= m <= n+1 ?

Fred Lunnon

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By symmetry, the centre of your sphere lies on the "bi-altitude" line joining the mid-points of near and far tetrahedron edges: choose this line as our vertical. All four planes lie at the same angle to the it, therefore their points of tangency to the sphere all lie on the same horizontal plane. By symmetry these points are the vertices of a plane rhombus; and since they lie on the sphere, the rhombus is cyclic. Therefore it is a square; the near and far edges of the tetrahedron meet on the bi-altitude; and your regular tetrahedron degenerates to a single point. However, I now realise that here is a sense in which the `sphere' in this case has merely retreated to (inversive) infinity, making it a most unsuitable candidate for generalisation! Should perhaps have read (B) That m = 2 is not always possible when n = 3 (for a tetrahedron) (now that it's been comprehensively cooked and spoiled); (C) What happens for m = 3 and n = 4 (for a pentatope); WFL On 4/7/16, Allan Wechsler <acwacw@gmail.com> wrote:

Forgive me for still not seeing this; it still looks feasible to me in the regular case. Suppose I put a tiny marble exactly halfway along an edge, exteriorly tangent to the two faces incident to that edge. Now, keeping the center of the sphere on the perpendicular bisector plane of that edge, and keeping it tangent to those two planes, let it inflate slowly. It has to move away from the tetrahedron in order to maintain tangency. At some point in this process, surely the sphere will kiss the two planes that form the ends of the trough. What am I missing? If this goes on I will have to calculate coordinates!

On Wed, Apr 6, 2016 at 6:00 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

Your picture is surely partly correct, in that any planes may simply be adjusted so that they meet the given sphere and still form a trough outside some tetrahedron.

However given a regular tetrahedron, that trough turns out to be the wrong shape to accommodate any sphere!

I initially thought I had a rather slick argument supporting (B), but further contemplation has revealed the (inevitable) flaw.

So (B) is probably false in general, and a little more spadework is indicated --- customary apologies!

WFL

On 4/6/16, Allan Wechsler <acwacw@gmail.com> wrote:

...

Does (B) have a typo? I have a pretty good visualization of a sphere tangent to the inside of two faces and the outside of two others. It hovers just outside the common edge of the two faces it's exteriorly tangent to, and the other two faces cut off the ends of the trough it rests in. I do see that m=3 is impossible for the tetrahedron.

On Wed, Apr 6, 2016 at 12:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

I finally wrote up properly all the stuff I've been doing about inspheres and exspheres of a simplex. Finished. Done & dusted. Off my plate.

But as I sat down again to start writing up P*nc*l*t's perishin' p*r*sm, an obnoxious inspiration intruded. Why shouldn't a simplex in Euclidean space also sport an ` m-fold' exsphere, tangent to m facets on their exterior side, and to n-m+1 on their interior side, where perhaps m

...

1 ?

In particular, show

(A) That m = 2 is impossible when n = 2 (for a triangle);

(B) That m = 2 is impossible when n = 3 (for a tetrahedron);

(C) What happens for m = 2 and n = 4 (for a pentatope);

(D) What happens for general 0 <= m <= n+1 ?

Fred Lunnon

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On Wed, Apr 6, 2016 at 10:21 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

By symmetry, the centre of your sphere lies on the "bi-altitude" line joining the mid-points of near and far tetrahedron edges: choose this line as our vertical. All four planes lie at the same angle to the it, therefore their points of tangency to the sphere all lie on the same horizontal plane.

I don't see why this follows. If you translated one of the planes "upwards", it would stil remain at the same angle to the vertical line, but the point of tangency would change. Or to put it another way, two of the planes intersect our vertical line at one, lower, point, and two intersect it at a different, higher, point. So there is no symmetry that says that the points of tangency should have the same z-coordinate. Or to put it a third way, the same proof shows there is no sphere inside the tetrahedron tangent to all four planes, and the fallacy is in the assertion that all four points of tangency lie in the same horizontal plane. I think your claim is true, but your proof seems faulty, and I'm having trouble patching it up. Andy

I posted those "simplex extra-sphere" puzzles under the impression that I actually knew the answers, a delusion quickly exposed. Now that I finally do know (most of) them, I shall try to make amends. ________________________________________________________________________________ Understanding what's going on here requires adopting notions from contact geometry. A general "cycle" in inversive n-space comprises a sphere equipped with an extrinsic orientation, naturally represented by the corresponding open ball (positive radius), or the closed ball's complement (negative radius). Special cases of cycles include points (unoriented), primes (hyperplanes), and inversive infinity. Two cycles are "tangent" when their spheres touch and their half-spaces intersect. "Eversion" of a cycle changes just its orientation. The generalised Apollonian problem (GAP) demands construction of a cycle bearing given relations (in terms of tangent length or intersect angle) to n+1 given cycles. It is a theorem that this problem has in general just two solutions, which --- in traditional fashion --- may be real, double or complex; they may also be congruent (equal except for orientation), or one or both at infinity. [ The only construction known to me involves the Clifford algebra representation of the Lie-sphere group of transformations preserving tangency. ] A particular extra-sphere (extended exsphere) of a simplex occupies some convex region bounded by facet hyperplanes, exterior to a given subset, and interior to the remainder. In terms of contact-geometry, it can (almost) be identified with some cycle tangent to every facet prime, interior facets oriented positively, exterior negatively. Of the two GAP solutions, one lies at infinity (to which all primes are tangent) and may be discarded. However, notice that the Euclidean configuration might equally well be represented by everting every cycle. Since the unoriented solutions of both problems are congruent, regions fall into complementary pairs, such that just one of each pair contains an extra-sphere! ____________________________________________________________ Previously I discussed the construction of a simplex similar to the original, inscribed to all but one of its facet primes and exscribed to the other. By induction on m an analogous construction applies in the more general situation of m negative facets, the original instead scaled by a factor (-1)^(n+m) / (n-2m) . Using the circumradius of this "extra-medial" simplex as an upper bound immediately yields constraint 0 < |r(m)| <= R/(n - 2m) on extra-radius r(m) in terms of circumradius R , at any rate for 2m < n . I conjecture that when 2m = n , the upper bound is not attained. Which now implies that for 2m <> n+1 , the nearer one ( 2m < n+1 ) of a complementary pair of regions is always occupied by an extra-sphere, while the further one ( 2m > n+1 ) is always empty! For 2m = n+1 , things are more complicated. There are binomial(n+1, m)/2 Euclidean spheres, to be allocated to twice as many regions: so in general, just half those regions may be occupied, while their complements lie empty! Worse still, symmetry may cause a pair of complementary regions to be congruent, in which case the only possible resolution involves both spheres retreating to infinity: this occurs for example when the simplex is regular. In fact, expressing simplex content two ways in terms of facet contents and scribed radii leads immediately to relation between extra-radius and inradius (n+1-2m) |r(m)| = (n+1) r(0) for a regular simplex: eg. r(2) = oo for a tetrahedron, as painfully deduced earlier. _____________________________________________________________ Fred Lunnon [14/04/16]

A damn' sight more irritating than I thought ... mea culpa! Three weeks ago I had apparently finished typesetting all my material concerning inequality |r/R| <= 1/(n-2) for circum-, ex-radius R, r of a simplex in n-dimensional Euclidean space --- spent forever typesetting diagrams, mildly high on self-congratulation, on the point of publication to all and sundry (well, maybe half-a-dozen correspondents who might be mildly interested) --- when (as mentioned here earlier) I realised that the method could be generalised to spheres tangent externally to m facets for m > 1 . So I went ahead and generalised it: an additional section of the paper would be called for, but no big deal. Now there's a quotation somewhere in Knuth about having proved a theorem, but not yet written a program to verify it. When eventually I got around to verifying my belated inspiration computationally, it turns out to be cobblers: for m > 1 , there is actually _no_ finite upper bound on |r/R| at all. << Picture the Viscount's great surprise! He scarcely could believe his eyes! >> Which implied that my proof of the original (and successfully verified) result had to be incorrect; and once I'd accustomed myself to the inevitability of that, the (grossly elementary) mistake was not hard to track down. But now I've no idea how repair it, so my precious theorem remains currently in limbo. Ignorance was bliss. And vice-versa. Sigh! Fred Lunnon On 4/14/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:

I posted those "simplex extra-sphere" puzzles under the impression that I actually knew the answers, a delusion quickly exposed. Now that I finally do know (most of) them, I shall try to make amends.

________________________________________________________________________________

Understanding what's going on here requires adopting notions from contact geometry. A general "cycle" in inversive n-space comprises a sphere equipped with an extrinsic orientation, naturally represented by the corresponding open ball (positive radius), or the closed ball's complement (negative radius). Special cases of cycles include points (unoriented), primes (hyperplanes), and inversive infinity. Two cycles are "tangent" when their spheres touch and their half-spaces intersect. "Eversion" of a cycle changes just its orientation.

The generalised Apollonian problem (GAP) demands construction of a cycle bearing given relations (in terms of tangent length or intersect angle) to n+1 given cycles. It is a theorem that this problem has in general just two solutions, which --- in traditional fashion --- may be real, double or complex; they may also be congruent (equal except for orientation), or one or both at infinity. [ The only construction known to me involves the Clifford algebra representation of the Lie-sphere group of transformations preserving tangency. ]

A particular extra-sphere (extended exsphere) of a simplex occupies some convex region bounded by facet hyperplanes, exterior to a given subset, and interior to the remainder. In terms of contact-geometry, it can (almost) be identified with some cycle tangent to every facet prime, interior facets oriented positively, exterior negatively. Of the two GAP solutions, one lies at infinity (to which all primes are tangent) and may be discarded.

However, notice that the Euclidean configuration might equally well be represented by everting every cycle. Since the unoriented solutions of both problems are congruent, regions fall into complementary pairs, such that just one of each pair contains an extra-sphere!

____________________________________________________________

Previously I discussed the construction of a simplex similar to the original, inscribed to all but one of its facet primes and exscribed to the other. By induction on m an analogous construction applies in the more general situation of m negative facets, the original instead scaled by a factor (-1)^(n+m) / (n-2m) . Using the circumradius of this "extra-medial" simplex as an upper bound immediately yields constraint

0 < |r(m)| <= R/(n - 2m)

on extra-radius r(m) in terms of circumradius R , at any rate for 2m < n . I conjecture that when 2m = n , the upper bound is not attained.

Which now implies that for 2m <> n+1 , the nearer one ( 2m < n+1 ) of a complementary pair of regions is always occupied by an extra-sphere, while the further one ( 2m > n+1 ) is always empty!

For 2m = n+1 , things are more complicated. There are binomial(n+1, m)/2 Euclidean spheres, to be allocated to twice as many regions: so in general, just half those regions may be occupied, while their complements lie empty! Worse still, symmetry may cause a pair of complementary regions to be congruent, in which case the only possible resolution involves both spheres retreating to infinity: this occurs for example when the simplex is regular.

In fact, expressing simplex content two ways in terms of facet contents and scribed radii leads immediately to relation between extra-radius and inradius

(n+1-2m) |r(m)| = (n+1) r(0)

for a regular simplex: eg. r(2) = oo for a tetrahedron, as painfully deduced earlier.

_____________________________________________________________

Fred Lunnon [14/04/16]