# Tetrahedron Centers

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 Ed Pegg 3 Votes I don't think we've mentioned going over eleven thousand Wolfram Demonstrations. A few of the recent demos include Electric Dance, Mondrian Art Problem, Ponting Square Packing, Rep-tiles of Order Five, Tiling Dragons, Epitrochogons, Canonical Polyhedra, Generalized Gosper Curves, and Integer Tetrahedra. One I just put up is Tetrahedron Centers. Go and get it. It explores a 3D version of Triangle Center. The Encyclopedia of Triangle Centers currently lists 11791 triangle centers.I'm planning to add to this one, but I can use some help. Here are various items I could use help on. Mathematica-readable Triangle Center database. The KimberlingCenters.m file in the collection MathWorldPackages.zip needs an update. In addition to 7000+ new centers, it would be helpful to have the lines a point is on, barycentrics / trilinears, and hammer-level steps (cevian, isogonal conjugate, extouch triangle). An updated list of 2D / 3D code for hammer-level geometry steps. While writing the Tetrahedron Centers demo, I started to see the utility of useful utilities. For example, given a reference triangle (or tetrahedron), find the orthic / extentral / contact / extouch / incentral / medial / antimedial triangle (or tetrahedron). First step is to narrow down the list of tools that are needed. 3D barycentric / trilinear formulas for tetrahedron centers. Many are based on angles, but solid angles don't seem to translate from 2D triangle center formulas. I'm thinking they might apply in spherical trigonometry, with spherical tetrahedra on a unit sphere. There are many constellations of points, lines, and cubics in triangle centers. Translate some of that constellation into the tetrahedron, and figure out which points are most important. Redefined points. The orthocenter isn't considered a valid tetrahedron center, since the altitudes are often skew. They do concur sometime (what are the conditions?) and in those cases the orthocenter concurs with the Monge point. I'm going to redefine the tetrahedron orthocenter as the Monge point. The Nagel and Gergonne points can be constellationally calculated for the tetrahedron, but they don't seem to pick up their normal properties (do they ever?). There are papers that create new Gergonne and Nagel points, but they don't fit the constellation, and I haven't been able to get them to work. Exact values. The Fermat point minimizes the total distance to the vertices. I can approximate it well with NMinimize, but I would like to find an exact value. For the equal parallelians point, I can approximate a point for equal area parallel triangles, but an exact answer would be better. I need to constellationally add the Gergonne, Nagel, Mittenpunkt, Feuerbach, Clawson, Longchamps, Schiffler, Bevan, and Isoperimetric points, then figure out how valid they are. The constellation-based points so far seem confided to a plane, what are good points for getting off-plane? If the tetrahedra centers are planar, that suggests there is a particular triangle in that plane which would have the same centers. A Soddy tetrahedron is one with vertices at the centers of 4 tangent spheres. A smaller fifth sphere can be added tangent to all four. Does the center of that sphere concur with any points in the constellation? What other tetrahedra-with-rare-properties exist? At the bottom of excentral triangle is a list of centers from one triangle that match up with centers from another. What constructed tetrahedra should I consider? Here's a new construction, the Equal Legs tetrahedron. Reference tetrahedron in red. From the circumcenters of each triangle, draw the line through the tetrahedron circumcenter and find the intersections with the circumsphere. (Actually 8 points, not 4). Segments from a vertex to the reference triangle all have equal length. (The two possible equal lengths seem related) Not sure if this is useful. A tetrahedron has 3 pairs of opposing skew edges. Calculate lines perpendicular to each, the points marking the distance between skew lines. it is possible for these lines to concur (what point is that?). (What are the conditions for this special property?) Most of the time they don't concur. When the three distance-between-skew-lines lines don't concur, another set of three distance-between-skew-lines lines can be calculated. Those can be used to calculate a third set. The recursion stops there! The segments in the 2nd and 3rd steps make a non-planar hexagon with all right angles. Most tetrahedra have a skewed hexagon. Can this be used for anything?What are good papers, books, and sources of tetrahedron centers that I should look at? All comments, wish lists, suggestions, or possible help with some of these is welcome.
 Given an irregular tetrahedron, with distinct non-collinear incenter, circumcenter, and centroid. Is there a unique triangle with the same incenter, circumcenter, and centroid? Seems so. Here is code for that image.  ColoredLabeledPoint[color_RGBColor, text_String] := Graphics[{EdgeForm[{Gray}], color, Disk[{-.2, .1}, 1], Black, Text[Style[text, Small], {0, 0}, FormatType -> StandardForm]}, ImageSize -> {15, 15}]; Graphics3D[{Text[ColoredLabeledPoint[Pink, "I"], {4, 5, 0}], Text[ColoredLabeledPoint[Cyan, "O"], {0, 0, 0}], Text[ColoredLabeledPoint[Green, "G"], {3, 3, 0}], Black, Tube[#, .5] & /@ Subsets[ {{1/4 (49 + 31 Sqrt[3]), 1/4 (-31 + 49 Sqrt[3]), 0}, {1/4 (49 - 31 Sqrt[3]), 1/4 (-31 - 49 Sqrt[3]), 0}, {-(31/2), 49/2, 0}}, 2], Red, Tube[#, .5] & /@ Subsets[{{0, -15, 0}, {0, 9, -12}, {12, 9, 0}, {0, 9, 12}}, 2]}, Boxed -> False] Is there an easy way to find this triangle for a given tetrahedron? The same question on math.stackexchange. The method I used to find this example was very messy.