# Course:Harris, Fall 08: Diary Week 11

Mon:

• 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
• 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?

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.)
• 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 $k$ of biangles (all the same size), all fitted around the same pair of antipodal points as vertices:
• Then, since $k$ fit around a vertex, each biangle has an angle of $A = 360/k$ degrees.
• When $k = {2}$, the biangles are also 1-gons (same as 180-degree 2-gon).
• When $k = 1$, it's sort of a degenerate situation, with the two sides coinciding and producing a 360-degree 2-gon.
• Tessellation by triangles:
• Again, let $k$ = # of triangles at a vertex.
• Let $A$ = the angle in each triangle.
• Since $k$ angles fit around a vertex, the again have $A = 360/k$ degrees, or $A = \frac{2\pi}{k}$ in radians.
• Let ${\Sigma}$ = angle-sum of each triangle.
• Then $\Sigma = 3A = \frac{6\pi}{k}$ radians.
• Let ${\delta}$ = defect.
• Then ${\delta} = {\Sigma} - \pi = \frac{6\pi}{k} - \pi = (\frac{6}{k} - 1)\pi$ radians.
• Let $n$ = total # of triangles in the tessellation. How can we find $n$?
• We know $\delta = \frac{Area(triangle)}{R^2}$ (so long as we measure angles in radians)
• or $Area(triangle)$ = ${\delta}R^2$
• We also known $Area(sphere) = 4\pi{R^2}$.
• We also know $n$ x $Area(triangle)$ = $Area(sphere)$ (because there are $n$ triangles covering the sphere).
• Putting those together: $n$ = $Area(sphere)/Area(triangle)$ = $\frac{4{\pi}R^2}{\delta R^2}$ = $\frac{4\pi}{\delta}$
• Then putting together formulas for ${\delta}$ and $n$, we get
• $n = \frac{4}{\frac{6}{k} - 1}$
• For 2 triangles at a vertex (i.e., $k = {2}$), this gives a total of $n = \frac{4}{3-1} = {2}$ triangles.
• This is really just a tessellation by 2 biangles.
• For 3 triangles at a vertex (i.e., $k = {3}$), this gives a total of $n = \frac{4}{2-1} = 4$ triangles.
• For 4 triangles at a vertex (i.e., $k = 4$), this gives a total of $n = \frac{4}{\frac{3}{2}-1} = 8$ triangles.
• For 5 triangles at a vertex (i.e., $k = {5}$), this gives a total of $n = \frac{4}{\frac{6}{5}-1} = 20$ triangles.
• For 6 triangles at a vertex (i.e., $k = 6$), this gives a total of $n = \frac{4}{1-1}$ = infinity triangles.
• This is really just a reflection of what happens in the plane, where 6 triangles around a vertex works very well.
• For $k = 7$ or larger, we get a negative number for $n$, 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 $\delta = \Sigma - 2\pi$ (since Euclidean angle-sum for quadrilateral is 360 degrees).
• But we again have $\delta = \frac{Area(square)}{R^2}$ (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?

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?
• We spent most of the time on looking at spherical regular tessellations by polygons with 4 or more sides. With $k$ = number of polygons at each vertex and $n$ = total number of polygons on the sphere:
• We did squares on the board.
• We found $n = \frac{4}{\frac{8}{k} - {2}}$.
• only one possible:
• 6 squares, 3 at a vertex
• 4 at a vertex is the plane.
• 5 or more at vertex is impossible.
• Groups worked on pentagons.
• We found $n = \frac{4}{{\frac{10}{k}} - 3}$.
• Possible:
• 12 pentagons, 3 at a vertex
• 4 or more at a vertex is impossible.
• We looked at hexagons on the board.
• We found $n = \frac{4}{\frac{12}{k} - 4}$.
• No possibles:
• 3 at a vertex is the plane.
• 4 or more at a vertex is impossible.
• Polygons of 7 or more sides will have nothing possible.
• No exercises for this weekend. We'll go back to hyperbolic geometry next week.