Regular Polygons & a Spherical Arrangement for them

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Definition. A regular polygon is an arrangement of equal length line segments joined end to end so that all the corner points lie on a circle. The picture on the right has a 3gon, a 4gon, and so on up to an 8gon.

Theorem. This is a basic property of circles, which seems to be useful for understanding the geometric properties of regular polygons, and which we can get out of the way pronto.

For a circle centered at O, with fixed points A,B on the rim, and a moveable point C also on the rim but restricted to lying on the greater of the two arcs between A and B, then the angle c between AC and CB is always the same, and is in fact half the angle between AO and BO.

There is a rough proof in the picture on the right.

The idea is that if two points close to each other are fixed on a circle, then the angle between those points and a third point on the circle ( not between the first two ), is always the same no matter where the third point is.

Another way of looking at this would be to see that triangles meeting at a point on the circle based on equal length cords of the circle all have the same angle at the meeting point. That is part of the property in the next paragraph.

A Property of a Regular Polygon

The angle between any two adjacent diagonals of a regular n-gon is Π/n. 

This is because the 2 angles between the tangent and the polygon are Π/n, and the fact that due to the theorem just described above, the remaining Π-2Π/n must be divided into n-2 equal angles.

(Π-2Π/n)/(n-2) = (Π(n-2)/n)/(n-2) = Π (n-2)/n(n-2) = Π/n

Arrange n regular n-gons around a point and a side of one of the polygons, so that all polygons meet at one corner and they are all distributed equally around. So each polygon is a copy of the first rotated around the central point by 2Π/n, and the whole thing has n-fold rotational symmetry.

On the left are arrangements for 3,5,7 and 9 sided polygons, and on the right 4,6,7, and 10.

Notice the following properties:

A For the even sided polygons there is only one way of doing this. For odd numbered ones there are two arrangements which mirror eachother.

The reason for this is that any even arrangement will have 2-fold rotational symmetry - which is the same as a reflection, whereas an odd arrangement wont. The two odd arrangements are not only mirror images of eachother but are also the same arrangement rotated by Π/n. 

B In the even arrangements n sides meet at the centre, whereas in the odd arrangement 2n sides meet at the centre. So for example in the arrangement of 5 pentagons you can see 10 lines radiating from the center.

If you think of the polygons as leafs you can also see that the even number polygons cover the area around the center 1 time for the 4-gon, 2 times for the 6-gon, 3 times for the 8-gon, 4 times for the 10-gon, and so on, whereas for the odd numbered polygons the area is covered 1/2 by the 3-gons, 3/2 for the 5-gons, 5/2 by the heptagons, and 7/2 for th nonagons. 

C All the polygons in the flower can be obtained from the others by translation alone ( ie no rotation or reflection is required ).

Consider the odd numbered polygons in the picture on the left above. Consider then the patterns you get if you included the n n-gons as well as the same n-gons rotated by Π/n. This gives a fuller flower pattern as you can see on the left. ( This pattern is exactly the same as that used to create a Zome ).

Consider a Π/n slice of these patterns, as in the picture on the right.

The Π/n slice is an unfolded regular n pointed star, or equivaletly, a matchstick and rulers construction for Π/n.

To see this, note:

• All stick lengths are the same.
• There are n sticks.
• Each stick points a different way.
• Each polygon has sticks pointing in n directions, and in the full flower, the polygons rotated by Π/n actually contain sticks pointing in the same directions.
• Therefore the slice, which is an n stick subset of the full flower with sticks pointing in all different directions, must have sticks pointing in the n directions of the original polygons.
• And the angles between the sticks in the slice are therefore multiples of Π/n.

The photo below shows a folded heptagonal slice. Note the dotted lines in the picture on the right are the other diagonals of the n stars.

Using two rulers and n sticks, you can construct one of these slices by starting with an isosceles triangle with a stick as a base and by varying the length of the rulers until the slice fits nicely. Some websites mention this as a "neusis" construction when n=7.

This is a fascinating ( for some ) link between the geometry of a 7 sided heptagon on one hand, and the geometry of a square on the other, utterly unexpected ( for me ), and definitely worth mentioning here. As we all know, the length of the diagonal of a unit square is root(2), so how can we geometrically link a square with a heptagon ? ...

See this triangle with its diagonals and its line segments marked d. How long are they ? Answer: root(2). Proof is below. 1) The ratios you get from similar triangles: 2) The right and left hand sided right-angle triangles: and 3) Put the equations in 2) together to eliminate h. 4) Expand 3) to get Note from 1) we can get 5) Substitute in 4) to get Note from 1) we can get so  6) Substitute in 5) to get Finally simplify 6) to get so

Here are a some other root(2) lines radiating from the corner of a heptagon with some inner smaller heptagons marked out.

Consider a flower of n regular n-gons, where n is odd. Imagine the polygons are rigidly flat, but loosely coupled at the central point and all lying on the ground. Pull the central point upwards and push the sides on the ground inwards. The sides on the base will merge on their corners and so will the sides emanating from the middle, so that when the corners on the ground merge there will only be n/2 sticks emanating from the center. You end up with a polyhedron consisting of n regular n-gons heaped like a pyramid around another equal n-gon at the base. For n=3 you get a tetrahedron. For n=5 you get most of an icosahedron. For n=7 you get an interesting heptagonal thing. Imagine then "axles" through the centers of the leaning polygons. Because of symmetry these axles must all meet at some point O inside the polyhedron. So all the corners of all the polygons, including the one at the base, are the same distance from this central point O. In other words the corners of the polyhedron lie on a sphere! Things to notice.. ( if you take one of these spherical polyhedrons with an n-gon at the base and n n-gons meeting at the top. Project the sticks outwards from the center so that they form  segments of great circles ). The n lines at the top are at angles of 2Π/n to each other. Each line is/can be part of 2 n-gons - one on either side of the center of the sphere. Angles between lines on junctions of such polygon pairs is 2Π/n. So angles rising from base are also 2Π/n.

If you take a subset consisting of the polygon at the base and two of the leaning ones sharing an edge near the top, a notable pattern can be observed. The picture on the right here illustrates the n=7 case, where you can discern a red triangle that has by logic to be equilateral.

For n=3 you the red triangle diminishes to a dot. For n=5 the triangle is part of the icosahedron. For n=7 you get the shape on the right ( a triangular formation of 3x2 sticks with an embedded equilateral triangle ). For n=9 you get a triangular formation of 3x3 sticks, for n=11 you get a triangular formation of 3x4 sticks and so on.

You can see more about the heptagonal dome here.