[this was posted the 1st time in the Google-plus "Mathematics community" in Nov 6, 2015]
Feuerbach's theorem (bis)
The so-called *9-point circle* is, according to *Feuerbach's theorem*, tangent to the *inscribed circle* (*10th point*) & to the *3 escribed circles* of the triangle, so it should be called the 13-point circle, at the very least.. (lol) [cf.http://arxiv.org/pdf/1107.1152.pdf ]
*Feuerbach's theorem*: The middle points of the sides of a triangle, the feet of the altitudes & the points at mid-distance from the orthocenter & the vertices are on a circle whose radius is 1/2 of the one of the circumscribed circle & whose center is the point at mid-distance from the orthocenter & the center of the circumcircle. And this circle is tangent to the inscribed circle & to the 3 escribed circles of the triangle.
The pic today is of a handsome solid by the not easy name of *rhombic diskaiheptacontahedron* (72 faces, all *rhombi* of 2 kinds, 18 narrow ones (dark brown) & 54 wide ones (light brown); the face vector of the 'animal' is [V=74, F=72, E=144] and *symmetry* is *3-fold dihedral* with *vertical reflection* - *D3d* with Rv.
Please be happy you all & enjoy being alive!
To-day I have for my friends in this bunch no less than a "hecaton hexakaidecahedron" (116 faces, all equilateral trigs to the number of 116).
Face vector is [V= 60, F= 116, E= 174} & symmetry is T (tetrahedral chiral); this time vertices are of 5 different kinds:
12 (valence 5) (1-2-2-3-6)
12 (valence 5) (3-5-7-4-6)
12 (valence 6) (5-7-8-9-11-9)
12 (valence 6) (1-2-3-5-9-8)
12 (valence 7) (1-6-4-10-4-7-8)
Please be well & happy, everybody!
[more info]
If one were to allow for 'polyhedra' of non-planar faces, our polyhedron could be described in any of the following 3 ways: (1) a dodecahedron whose 'faces' are 12 equilateral non-planar decagons centered at the 12 verts of valence 10; (2) a icosahedron whose 'faces' are 20 non-planar equilateral hexagons centered at the 20 verts of valence 6; (3) a triacontahedron with 30 non-planar equilateral quadrilaterals centered at the 30 verts of valence 4. [12*10= 20*6= 30*4= 120 faces all]
And here is a view of the exact "sphericon", which is no more a polyhedron but a "surface" like a sphere or a torus.
(pay attention it has "4 vertices", "2 edges" (semicircles) and "1 unique face") Here is the view:
Here is the *ncx-rhtrc* (non-convex rhombic triacontahedron) turned inside-out (fig 1); in fig 2 this was glued over an ordinary cx-rhtrc giving a full cube.
It was: 1/8 cx + 1/8 ncx = 1 cube
it is now: 1 cx + 1 ncx = 8 cubes = 1 cube double the edge
Fig 3 represents a *mixed rhtrc* with 4 convex trios of rhombi & 4 concave ones.
Fig 1 & 2 symmetry is *Th* (*tetrahedral complete*), fig 3 symmetry is T (*tetrahedral chiral*); the solid in fig 3 is a stand alone *space-filler*.
**4-hypercube** or **tesseract** [coordinates: (+-1,+-1,+-1,+-1)]
Take the **convex hull** in R^4 of remaining 15 points.
The figure we would get is shown here as the **net** (onto R^3) of the said 15 vertex hypersolid. The **hypersolid** is made of 8 cells, 4 ordinary cubes (in **blue**) and 4 cubes with a 3-right-angled tetrahedron missing at one vertex [**red**]
4 **red** solids are glued to the 4 faces of a **regular tetrahedron** at the center of the figure. Total **9 cells** (fixed)
This is an *equifaced* (or *isohedral*) polyhedron of 24 congruent faces, which are *irreg. concave pentagons*
Face vector: [V=38, F=24, E= 60]
Symmetry is *O* (*Octahedral chiral*)
faces= 1 type (solid angle subtended at center = 30 deg)
edges= 3 types (two= 6.8, two= 11.6, one= 14.4)
vertices= 3 types (6-v4; 8-v3; 24-v3)
