Poncelet-style systems of tetrahedra in 3-space are rather different from the classical 2-space flavour, in that they arise not from an Chapple-style equality constraint, but more directly from imbalance of available freedoms. We attempt to investigate how the freedom of a tetrahedron in a 3-space porism behaves for sphere-pairs with given R,r,d at various extremes of feasibility. Given some sphere-pair, a compact method of parameterising accommodated tetrahedra is in terms of a pair of vectors, the first from circumcentre to (some arbitrarily chosen) "apex" vertex, the second from incentre to tangency of the opposite "base" plane. To each apex vector corresponds a region of feasible base vectors, and vice-versa: an effective algorithm for computing basis region given apex etc would be illuminating, but at present this remains uncharted territory. For lack of such analytical aids, along with most of the algebraic results which might facilitate the investigation, we endeavour to progress utilising symmetry. There is a helpful diagram of the various flavours of symmetrical tetrahedra encountered at https://en.wikipedia.org/wiki/Disphenoid#/media/File:Tetrahedron_symmetry_tr... so we quote their terminology. Modulo isometry, tetrahedral shape is specified by 6 independent edge-lengths, constrained only by inequalities ensuring volume real and positive: denote them [u,v,w,x,y,z] , ordered so that u,v,w form a triangle and oppose z,y,x resp. A sphere-pair is specified by just 3 quantities: circumradius, inradius, centre displacement R,r,d . In general therefore, any given sphere-pair accommodates a system of shapes with freedom 6-3 = 3 . [ An alternative way to look at this is instructive: the tetrahedron vertices are specified in space by 3*4 = 12 Cartesian components, subject to 4+4 = 8 constraints confining both vertices to lie on the given circumsphere, and faces to touch the given insphere. Hence remaining tetrahedron freedom is 12-8 = 4 --- hang on, that's too much --- ah yes, and subtract 1 for rotation symmetry about the axis line joining sphere centres: leaving freedom 3 modulo isometry, as before. ] Easy special case: d = f_3 = 0 . Then R = 3 r , and by the theorem in Klamkin 1979 the tetrahedron is "regular" with edges [u,u,u,u,u,u] . Its freedom is 0 , while the concentric sphere-pair symmetry clearly has maximal freedom 3 . Next consider d = 0 , f_3 > 0 . Via symmetry there is essentially only one choice for the apex, then freedom-1 for the base, hence freedom 1 in total: the tetrahedra are "rhombic" isohedral [u,v,w,w,v,u] . Sphere-pair symmetry again has freedom 3 . Elementary algebra suffices to express R,r in terms of u,v,w --- see https://en.wikipedia.org/wiki/Disphenoid http://mathworld.wolfram.com/IsoscelesTetrahedron.html [ Hey, this doesn't look too hard, so far ... ] Finally suppose f_3 = 0 , d > 0 . Uh-huh. Anyway, factor out freedom-1 sphere-pair symmetries by fixing the apex to lie on some chosen diametral plane. For a lower bound, consider only shapes with symmetry about that plane: "mirrored" with two faces isosceles and two faces congruent, [u,v,v,y,y,z] . It is not difficult to visualise a freedom-1 system of these, starting from tall "trigonal" [u,u,u,z,z,z] (pyramid), progressing via "digonal" [u,v,v,v,v,z] to squat "trigonal" at half-way, before returning via another congruent "digonal" to the start. Hence at f_3 = 0 , shape freedom >= 1 . [ Cases "digonal" and "trigonal" deserve special mention because they are cases for which we have succeeded in expressing R,r,d,f_3 in terms of edge-lengths. This happy outcome does not currently extend to "mirrored", discouraging implementation of a simulation of the symmetric porism described above. ] However, a numerical experiment which iteratively relaxes a random initial tetrahedron into one for which f_3 ~ 0 uncovers approximate examples which plainly have no symmetry: for example, R, r, d, f = 1.000000000, 0.1037995262, 0.8718244802, -1.8E-9 ; u, v, w, x, y, z = 0.7479954183, 0.3745446035, 0.6545259316, 1.217042748, 1.569973616, 1.107308554 . For r close to R/3 , most examples found are quite close to "mirrored"; but as r decreases towards 0 , the asymmetry increases. We conclude that the shape freedom >= 2 ; now the last great problem remains to decide wild *** Conjecture: Given R,r,d with f_3 = 0 , d > 0 *** *** accommodate tetrahedral shapes with freedom = 2 . *** Finally, exact test cases. Heronian tetrahedra have rational coordinates, edge-lengths, face areas and volume: the smallest has edges [117, 84, 51, 80, 53, 52] and (Cartesian) vertices at [0, 0, 0], [0, 0, 84], [0, 48, 64], [27, 36, 108] . They also have rational circumcentres, in/excentres, and in/exradii. But a survey finds none up to diameter 20,000 having also rational circumradius or displacement, excepting fairly mundane instances including "rhombic" (d = 0) , and shards with circumcentre along the longest edge (R = u) ; and none in the critical region f_3 = 0 . Fred Lunnon On 1/13/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
In pursuit of my declared ambition to extend the Euler-Chapple relation between inradius r , circumradius R , and centre displacement d to tetrahedra in 3-space (never mind old Poncelet), I can't resist setting off along an ever-so-slightly inappropriate tangent in pursuit of excircles and exspheres. My initial excuse is that they come pretty much for free, simply by allowing the inradius to assume negative values --- legerdemain unfortunately justiable only via a substantial digression into Lie geometry of oriented spheres.
In 2-space the classical Poncelet triangular porism arises because (modulo isometry) triangles have freedom 3, whereas circum/incircle pairs accommodating them have freedom only 2 via Chapple's constraint; hence each pair must host a freedom 1 system of triangles. But when r < 0 there follows a partner porism for excircles:
*** Circles accommodating some triangle as circum- and excircle *** *** must necessarily accommodate a continuum of triangles! ***
In 3-space tetrahedral circum- and inradius r, R satisfy inequality
*** 3 r <= R for r > 0 ! ***
(incidentally with equality only for a regular tetrahedron): instructive proofs are to be found in
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 ; free to view at http://www.jstor.org/stable/2689968?seq=1#page_scan_tab_contents
The result is also an immediate corollary of my conjecture proposed earlier
*** f_3 == (R - 3 r)(R + r) - d^2 >= 0 ? ***
--- and extending to r < 0 yields its (conjectural) exradius partner
*** |r| <= R for r < 0 ? ***
But it does not seem straightforward to modify either Klamkin et al proof suitably for exradii --- would anyone care to try?
The updated scatter diagram at https://www.dropbox.com/s/fo59jy5b83v4bt3/f_3_plot.png shows R,r,d for 10K random tetrahedra scaled to R = 1 , where r denotes in- or exradius according to sign and colour (green or blue); the ellipse f_3 = 0 (brown) can be seen to hug their convex hull tightly, as expected. But take a gander at an additional, impressively emphatic, exradius boundary
*** d + |r| >= R for r < 0 ? ; ***
apparently a partner of the inradius trivium
*** d + r <= R for r > 0 ! , ***
despite which it at present lacks plausible justification --- any suggestions?
Fred Lunnon
On 1/7/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
[ Errata: In the diagram, each green spot represents a random tetrahedron scaled to R = 1 with coordinates (d, r) ; the brown curve is f_3(1, r, d) = 0 . The approximate boundary is extracted via convex hull. ] WFL
On 1/7/16, Fred Lunnon <fred.lunnon@gmail.com> wrote:
No direct generalisation of the Chapple equation exists for tetrahedra: for example, if r << R then d may take any value between 0 and R - r , via contemplation of needle disphenoids with one pair of short edges versus tall pyramids with narrow base. On the other hand, if R = 3 r then only d = 0 is possible, and the tetrahedron is regular; so some nontrivial inequality f(R, r, d) >= 0 must operate (besides the weak, trivial R >= r + d ).
Donning thoroughly inadequate protective gear, I clambered into the cage with Maple for the first of several increasingly protracted bouts of numerical exploration and gruesome algebra, in the course of which it was established to my considerable dissatisfaction that --- Monte-Carlo trials yielded a scatter plot with a well-defined boundary f = 0 ; however (probably due to ill-conditioning) it did not appear feasible to approximate coefficients of putative polynomial f via SVD (least squares). And available CAS's, along with my own geometric skills, fall well short of hacking polynomials to express R,r,d,f in terms of the tetrahedron shape modulo isometry, in the form of its six edge-lengths.
Rescue from the quicksands of laborious computation comes unexpectedly in the ethereal guise of symmetry. Consider the special case of a triangular pyramid with height h , base edge length t , both centres lying on its altitude. Via Pythagoras and similar triangles, easily 12(h r + t^2/12)^2 = (h^2 + t^2/12)t^2 , 2 h R = (h^2 + t^2/3) , R + r = d + h ; eliminating t, h yields a quartic which conveniently factorises into 3 h^2 times the nontrivial polynomial *** f_3 = (R - 3 r)(R + r) - d^2 ; ***
Now consider f_3 qua function of edge lengths. Rotation about the axis leaves both spheres invariant: therefore by symmetry pyramids occupy a stationary curve of f_3 . Also it is intuitively plausible that perturbing a pyramid while fixing R,d will cause r to decrease, whence f_3 = 0 should be a local minimum, at least. And bingo --- when f_3 = 0 is superposed on the scatter data, it shows a perfect fit along the boundary! See https://www.dropbox.com/s/fo59jy5b83v4bt3/f_3_plot.png
However, a proof that f_3 >= 0 everywhere remains elusive --- in fact, I don't 'ave a bleedin' clue, mate). In principle the question is expressible in terms of semi-algebraic sets, so decidable; however as noted above the associated computations, whether involving Tarsky-style resolution, Groebner bases or classical resultants, appear currently intractable.
Notes: For pyramids d may take sign +, - in a solution to f_3 = 0 , corresponding respectively to squat, tall pyramids with different heights h = R + r - d but the same radii R,d . In the (improper) limit as h -> 0 , also r -> 0 and R - d -> 0 ; whence f_3 -> 0 simultaneously, so factor 3 h^2 of the quartic above may be discarded. More generally as the volume vanishes but R = 1 (say), when r, R - d -> 0 we have f_3 -> 0 also, even for asymmetric tetrahedra. Furthermore, it turns out that pyramids are not the only proper tetrahedra with f_3 = 0 : I intend to explore this more fully in a subsequent instalment.
Fred Lunnon
On 12/29/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Recapitulating "Euler's theorem", badly garbled in an earlier post of mine:
Relates circumradius R , inradius r , centre displacement d of triangle by R^2 - 2 R r - d^2 = 0 .
Actually due to Chapple, according to http://mathworld.wolfram.com/EulerTriangleFormula.html .
Synthetic proof at
https://en.wikipedia.org/wiki/Euler's_theorem_in_geometry#/media/File:GeometryEulerTheorem.png .
Provides foundation of Poncelet's porism for triangles in Euclidean 2-space; see http://mathworld.wolfram.com/PonceletsPorism.html .
My Scrooge's Christmas present (to myself, naturally) was to investigate the extent to which Chapple's result generalises to higher dimensions: specifically,
*** What can be established about the relation between circumradius, *** *** inradius and displacement of a tetrahedron in Euclidean 3-space? ***
I'll post more about this problem after a few days, in case other social misfits out there are motivated to tackle it themselves, or alternatively can disabuse me of any delusional claim to originality.
Incidentally, Poncelet's porism now has an entire book devoted to it, Vladimir Dragović, Milena Radnović "Integrable Billiards" Birkhäuser 2011 ; chapter 1 free online at
http://www.springer.com/cda/content/document/cda_downloaddocument/9783034800... .
Fred Lunnon