# Course:Harris, Fall 08: Diary Week 11

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- Escher Art Project:
- Art project due Friday, this week.
- Written report due Monday, next week
- I pointed out some additional details for the written report, specified on the schedule for this week.
- Written report should normally be a few pages long.

- There are a couple general types of geometry:
- geometry of a surface
- Geodesic--the counterpart for a line--means a curve which is the shortest path between any two of its nearby points.
- Measurements of lengths and angles are fairly obvious.
- Examples:
- plane (Euclidean geometry)
- sphere (sometimes called elliptical geometry)
- any surface you can imagine in 3-space

- abstractly defined geometry
- Measurement must be explained for the particular model.
- Once lengths are defined, a geodesics is, again, a curve which is the shortest path between any two of its nearby points.
- Sometimes geodesics are (for convenience) defined separately from distance measurements, but the result should be the same.

- Example:
- Poincare disk (hyperbolic geometry)
- geodesics: circular arcs intersecting the boundary of the disk at right angles (or diameters of the disk)
- length measurement: hyperbolic units get Euclidean-smaller as one approaches the boundary
- angle measurements: same as Euclidean measurement

- Poincare disk (hyperbolic geometry)

- Measurement must be explained for the particular model.

- geometry of a surface
- Groups took about 25 minutes to do the first Hyperbolic Geometry exploration.
- Left over from spherical geometry: regular spherical tessellations
- This means tessellating a sphere with
- regular polyogons
- all congruent to one another
- with all vertices alike.

- Can one do this on a sphere? Do regular tessellations exist?

- This means tessellating a sphere with

Wed:

- We spent the whole period on regular tessellations of the sphere.
- Recall regular tessellations in general:
- The tessellation is by regular polygons:
- All sides have the same length.
- All angles are congruent.

- All the polygons are congruent to one another.
- All vertices look the same.
- (Maybe that follows automatically? Not sure.)

- The tessellation is by regular polygons:
- On the plane, there are 3 regular tessellations:
- by triangles (6 at a vertex)
- by squares (4 at a vertex)
- by hexagons (3 at a vertex)

- On the sphere:
- Totally unrelated to plane tessellations, we can tessellate a sphere by biangles (2-gons).
- We can have any number <math>k</math> of biangles (all the same size), all fitted around the same pair of antipodal points as vertices:
- Then, since <math>k</math> fit around a vertex, each biangle has an angle of <math>A = 360/k</math> degrees.
- When <math>k = {2}</math>, the biangles are also 1-gons (same as 180-degree 2-gon).
- When <math>k = 1</math>, it's sort of a degenerate situation, with the two sides coinciding and producing a 360-degree 2-gon.

- We can have any number <math>k</math> of biangles (all the same size), all fitted around the same pair of antipodal points as vertices:
- Tessellation by triangles:
- Again, let <math>k</math> = # of triangles at a vertex.
- Let <math>A</math> = the angle in each triangle.
- Since <math>k</math> angles fit around a vertex, the again have <math>A = 360/k</math> degrees, or <math>A = \frac{2\pi}{k}</math> in radians.

- Let <math>{\Sigma}</math> = angle-sum of each triangle.
- Then <math>\Sigma = 3A = \frac{6\pi}{k}</math> radians.

- Let <math>{\delta}</math> = defect.
- Then <math>{\delta} = {\Sigma} - \pi = \frac{6\pi}{k} - \pi = (\frac{6}{k} - 1)\pi</math> radians.

- Let <math>n</math> = total # of triangles in the tessellation. How can we find <math>n</math>?
- We know <math>\delta = \frac{Area(triangle)}{R^2}</math> (so long as we measure angles in radians)
- or <math>Area(triangle)</math> = <math>{\delta}R^2</math>

- We also known <math>Area(sphere) = 4\pi{R^2}</math>.
- We also know <math>n</math> x <math>Area(triangle)</math> = <math>Area(sphere)</math> (because there are <math>n</math> triangles covering the sphere).
- Putting those together: <math>n</math> = <math>Area(sphere)/Area(triangle)</math> = <math>\frac{4{\pi}R^2}{\delta R^2}</math> = <math>\frac{4\pi}{\delta}</math>

- We know <math>\delta = \frac{Area(triangle)}{R^2}</math> (so long as we measure angles in radians)
- Then putting together formulas for <math>{\delta}</math> and <math>n</math>, we get
- <math>n = \frac{4}{\frac{6}{k} - 1}</math>

- For 2 triangles at a vertex (i.e., <math>k = {2}</math>), this gives a total of <math>n = \frac{4}{3-1} = {2}</math> triangles.
- This is really just a tessellation by 2 biangles.

- For 3 triangles at a vertex (i.e., <math>k = {3}</math>), this gives a total of <math>n = \frac{4}{2-1} = 4</math> triangles.
- For 4 triangles at a vertex (i.e., <math>k = 4</math>), this gives a total of <math>n = \frac{4}{\frac{3}{2}-1} = 8</math> triangles.
- For 5 triangles at a vertex (i.e., <math>k = {5}</math>), this gives a total of <math>n = \frac{4}{\frac{6}{5}-1} = 20</math> triangles.
- For 6 triangles at a vertex (i.e., <math>k = 6</math>), this gives a total of <math>n = \frac{4}{1-1}</math> = infinity triangles.
- This is really just a reflection of what happens in the plane, where 6 triangles around a vertex works very well.

- For <math>k = 7</math> or larger, we get a negative number for <math>n</math>, so those are all impossible.
- Thus, the only possible combinations for triangles tessellating the sphere are
- 4 triangles, 3 at a vertex
- 8 triangles, 4 at a vertex
- 20 triangles, 5 at a vertex

- What about tessellating by squares?
- Defect changes in that <math>\delta = \Sigma - 2\pi</math> (since Euclidean angle-sum for quadrilateral is 360 degrees).
- But we again have <math>\delta = \frac{Area(square)}{R^2}</math> (since both defect and area come from two triangles making up the quadrilateral).
- What else changes from triangles?
- What remains the same as with triangles?

- Totally unrelated to plane tessellations, we can tessellate a sphere by biangles (2-gons).

- Recall regular tessellations in general:

Fri:

- Art Projects were turned in.
- Reports are due Monday. Things to look for including:
- the base polygonal tessellation (by parallelograms, for instance) and its symmetry group
- the means by which the base polygonal tessellation was transformed
- Was a changed line on one side simply translated to the next side?
- Was a changed line glide-reflected to the next side?
- Was a changed line rotated about an endpoint?
- Was a changed half of a line rotated about its midpoint?

- How did the artistic vision influence the choice of base polygonal tessellation, method of changing it, symmetries in finished product?
- How did symmetries in polygonal pattern influence the development of the art?

- Reports are due Monday. Things to look for including:
- We spent most of the time on looking at spherical regular tessellations by polygons with 4 or more sides. With <math>k</math> = number of polygons at each vertex and <math>n</math> = total number of polygons on the sphere:
- We did squares on the board.
- We found <math>n = \frac{4}{\frac{8}{k} - {2}}</math>.
- only one possible:
- 6 squares, 3 at a vertex

- 4 at a vertex is the plane.
- 5 or more at vertex is impossible.

- only one possible:

- We found <math>n = \frac{4}{\frac{8}{k} - {2}}</math>.
- Groups worked on pentagons.
- We found <math>n = \frac{4}{{\frac{10}{k}} - 3}</math>.
- Possible:
- 12 pentagons, 3 at a vertex

- 4 or more at a vertex is impossible.

- Possible:

- We found <math>n = \frac{4}{{\frac{10}{k}} - 3}</math>.
- We looked at hexagons on the board.
- We found <math>n = \frac{4}{\frac{12}{k} - 4}</math>.
- No possibles:
- 3 at a vertex is the plane.
- 4 or more at a vertex is impossible.

- No possibles:

- We found <math>n = \frac{4}{\frac{12}{k} - 4}</math>.
- Polygons of 7 or more sides will have nothing possible.

- We did squares on the board.
- No exercises for this weekend. We'll go back to hyperbolic geometry next week.