I encountered this picture of a construction of the golden section at https://r-knott.surrey.ac.uk/Fibonacci/phi2DGeomTrig.html :
The points P and Q are the midpoints of the edges AE and AC. The point Q divides line segment PR in the golden section: PR/PQ = τ = ½ + ½√5 . Alternatively, PQ : QR = τ-1 = τ – 1 = -½ + ½√5. Then I made a picture myself with geogebra to get more feeling for the geometry of this fact. The first picture shows some geometry:
Points D and E are the midpoints of edges AB and AC. Line segment DE has length ½√3. Line segment DH = ½√3 (½ + ½√5).
Proof: The median AF of triangle ABC is divided in the ratio 2:1 by the centroid O. Hence OA = 1 and OF = ½. Line segment DE divides the median with length 3/2 in two halves of 3/4. The distance of O to line segment DE is 3/4 – 1/2 =1/4. The length of OH is 1 as a radius of the unit circle. With Pythagoras we see that the distance from H to the centre of DE is √(1-1/16) = ¼√15. From the centre of DE to D is ¼√3. Thus DH = ¼√3 + ¼√15 = ½√3 (½ + ½√5) = τ ∙ ½√3.
Then EL = EH = DH – DE = ½√3 ( – ½ + ½√5) = τ-1 ∙ ½√3. Drawing some circles shows two points of intersection I and J. And EK = τ-2 ∙ ½√3 and LK = τ-3 ∙ ½√3 . Then the miracle appears:
The points I, J and M complete D and E to the five vertices of a pentagram.
Proof: EL/EJ = τ. Hence angle EDJ = 36o = π/10 rad. Because EJ = EM = DM = DI, DMEJI is a regular pentagon. Hence DEIMJ is a pentagram.
Point Q is the centre of the pentagram.
Then i put some more equidistant points on the two small circles in the next picture, where I left out the grid and the axes:
The edges MJ and MI are also edges of two decagrams:
The triangular symmetry invites us to draw a third circle with centre F:
Even if we mark the points of intersection of drawn circles, we miss some points to draw a pentagram or decagram in circle F. Adding 5 new points on circle F, one can draw a decagram:
But this doesn’t look very symmetric. Adding another 5 points, we get two decagrams in circle F:
The orientation of the two green decagrams is still different from the orientation of the yellow decagrams in circles D and E. You can add a decagram orientated like the yellow decagrams in circle F. The 30 vertices of the three decagrams are at equidistant points, at 12o on the circle. A decagram has the property that the intersection points of the edges lie on a circle with a radius that is smaller by a factor of τ-1 with respect to the circle of vertices. This fact was the inspiration for the next picture:
I have drawn line segments DF and EF. The smaller point on these edges is such a intersection point of edges of the (green) decagrams. By triangular symmetry it is obvious the distance of the two points on edge EF have a golden ratio, identical to the ratio EL/EK = τ of the pentagram. For clarity I left out the decagrams. By using points on the circle E, I can draw two circles with centre F. The larger circle E with radius ½√3 gives the opportunity to draw the smallest circle in the picture with centre F. Then another miracle happens: the 4 x 30 = 120 points on the 4 circles centered in F are the vertices of the two-dimensional orthogonal projection of a 4-dimensional 600-cell.
The 2 small points then led me to draw a second (smaller projection of a) 600-cell to get a projection of the Witting polytope in 4 complex dimensions or the E8 polytope in 8 real dimensions:
After a while however I realised that the overlap between the circles (marked in blue) is too small.
The nearest neighbours of point F are the 240 points on the 8 circles, if interpreted as projection of the 8-dimensional polytope. For D and E to be nearest neighbour of F they must be one of the 240 points on the 8 circles with centre F. So, in the next picture I consider the initial 600-cell as the smallest in projection and so I added 120 points on 4 circles, larger by a factor τ .
By now we are getting in known territory if you already have looked on this page, where a (in 2D-projection) different pair of nearest neighbours is chosen. Leave the pentagram out to see more clearly the next steps.
The region DEF, where the 3 sets of 8 circles overlap, shows 9 triangles that are concentric with triangle ABC. The vertices are found as the intersection points of three circles. To see the central region in more detail we zoom in to the centre of the picture.
The 27 vertices happen to be the vertices of one of the 240 cells (3-dimensional complex hyperplanes), with symbol 3{3}3{3}3 and known as Hessian polytope, of the complex Witting polytope with symbol 3{3]3{3}3{3}3. The points B and C have a direction perpendicular to the hyperplane with the nine triangular diagonals (they are called diagonals of the cells because their centroids are also the centre of the cell). A reflection of a triangle in the line BC (in the plane of projection) gives a second triangular edge of a complex polytope 3{3}3 with centre F. One example of nine:
Just draw six lines from B and six lines from C to the vertices of the two triangles and connect the non-opposite vertices of the two opposite triangles. The 3{3}3 has eight edges with the shape of a equilateral triangle. There are 240 vertices and consequently 240 cells (Hessian polytopes). They come in 120 opposite pairs, just like the vertices. If you think about it for a while, you see that there are (120 x 9 )/4 = 270 3{3}3 with centre F.
One 3[3]3 for the money, two for the show:
And another two to get the idea:
Now we are going to investigate the nearest neighbours shared by two nearest heigbours, say E and F. We search for the intersection points of the 8 circles E and F. Ofcourse we already have 27 points that are also nearest neighbours of D. It is easily seen there are another 27 points if we reflect in the line EF. Point D and C complete the number of 56 nearest neighbours shared by E and F. It is not just numerology that the number of vertices of a dodecahedron and of three icosahedra is 20 + 12×3 = 56 alltogether. They are really there as the next picture shows:
A red dodecahedron and 3 icosahedra in blue, cyan and black. Compare with this page. More apparent numerology says 7 x 8 = 56 for the vertices of 7 cubes. And yes, there 7 cubes in the same 56 vertices:
The red cube is the one that can be found in Coxeters book Regular Polytopes. The edges of the cubes have the same lenght in 8-D space. This can be understood if we draw (a limited set of green edges of ) a well chosen 8-D orthoplex (a hyper-octahedron):
There are 8 mutually orthogonal directions from F to the vertices of the orthoplex. Except for largest one, their direction and length are reproduced as edges of the cubes. Each
There are 8 mutually orthogonal directions from F to the vertices of the orthoplex. Except for largest one, their direction and length are reproduced as edges of the cubes. Each direction and length occurs as 12 edges, in 3 cubes, 4 in one cube. In one drawing:
An alternative way of choosing 7 cubes is shown in the next picture:
Now the 8 directions are by four in the direction of FE and four are perpendicular to FE in the plane.
Finally, I show the construction of the 3{3]3{3]3 that is identical to the one on this page. Therefore we choose nearest neighbours to F on the smallest circle with blue points. Here are the two sets of 8 circles around two of these points.
Then the 9 triangular diagonals of the 3{3]3[3]3:
And some typical examples of 3{3]3 that have center F.
You can read on this page about the dodechedron and icosahedra whose 56 vertices fit in the 54 vertices of two 3{3]3{3]3 + two points on line BC, and the 7 cubes in the same 56 vertices.
For comparison some more pictures. The projection of the dodecahedron and the icosahedra is more attractive than the previous projection in my opinion:
The third icosahedron has the same size and shape is this larger one. The 7 cubes now have the largest projection direction (FE) in 3 of them as edges:
An icosidodecahedron has its 30 vertices only on circles with centre F and not on circles with centre D or E.
We conclude with our original pentagram and two descendants of equal size.
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Frans Marcelis 