Pick up some wrapping paper.

Size: px

Start display at page:

Download "Pick up some wrapping paper."

Juniper Warren
5 years ago
Views:

1 Pick up some wrapping paper. What is the area of the following Christmas Tree? There is a nice theorem that allows one to compute the area of any simply-connected (i.e. no holes) grid polygon quickly. It is known as Pick s theorem. It states: A = I + 1 B 1, where A is the area of the region, I is the number of grid points on the inside of the region, and B is the number of grid points on the boundary of the region. Check this out on a square, and a 3 rectangle. Now compute the area of the tree. OK. Why does this work? We will see this in several steps. Step 1: Triangulate the region i.e. cut it into triangles. Abstractly any such region is an n-gon. Notice that connecting two non-adjacent vertices of an n-gon with a line cuts it into a k-gon and an l-gon with k, l < n. (Try this with a pentagon.) To do so draw a (ground) line beneath the region that does not pass through any grid points. Now pick the point in the region closest to the ground. Find the two boundary points closest to it, and connect them with a line. If there are no vertices of the region in this triangle, cut it off to get a triangle and a smaller region. If there are vertices in the resulting triangle, connect the lowest vertex with the second lowest vertex in the triangle. See the figure below. 1

2 Remark: The above argument shows that one can triangulate any polygon. Combined with the dissection of a triangle into a parallelogram and the dissection of a parallelogram into a rectangle, then the dissection of a rectangle into a 1 x rectangle, we see that any polygon may be dissected into a 1 x rectangle. This is the basis of the axiomatic definition of plane area. The analogous dissection procedure will not generalize to three dimensional polyhedra! Step : Grid Triangulate the region i.e. cut it into triangles. Pick the lowest triangle with non-vertex interior grid points. Connect the lowest interior grid point with each vertex. Repeat until no triangle has interior grid points. Now pick the lowest triangle with non-vertex boundary grid points. Connect the lowest non-vertex boundary grid point with the vertex that is not on the same edge. Repeat until there are no triangles with non-vertex grid points. See the figure below. Step 3: Show that the area of a grid triangle with no non-vertex grid points is 1 Use the deck of cards skew trick to show that any such triangle has the same area as one with corners at (0, 0), (1, 0). Note that the third vertex must have y-coordinate equal to ±1 or else the triangle would have with non-vertex grid points. Use the deck of cards skew trick to show that any such triangle has the same area as one with corners at (0, 0), (1, 0) and (0, 1), i.e. 1/. Step 4: Notice that the area of the region is 1 the number of triangles, and compute the number of triangles. To compute the number of triangles, we tape two copies of the grid region together to make a triangulated pillow case. The number of triangles in the region will be half the number of triangles in the pillow case. We now use Euler s formula: V E + F =. Putting dots on each side of each edge, we see that there are E dots ( for each edge) but there are also three dots for each face (triangle) so E = 3F and E = 3 F. Thus

3 V 1 F = giving 1 F = V Now, A = Area of Region = 1 (Number of Triangles in Region) = 1 ( ) 1 (Number of faces in pillow) = 1 (Number of vertices in pillow ) = 1 ( (Number of interior grid points of region) + Number of boundary grid points of region ) = I + 1 B 1. Think about everything that went into this result: similar triangles, cut a square into triangles, deck of cards for area of triangle, phone book for circle, cut a cube for volume of pyramid, volume of cone, tilted squares and pythagorean s theorem, a turtle plus b turtle equals c turtle, volume of a sphere, drill to surface area of sphere, slice the sphere into two triangles and six wedges to get Euler s theorem... All of that work is why this is called a theorem. Wrapping paper Euler s theorem also leads to the classification of wallpaper patterns (we ll call them wrapping paper patterns for the season). First recall where the in Euler s theorem came from. We had α + β + γ = 1 + A πr, where α, β and γ were the angles of a triangle measured in rotations. The expression A measures the excess angle in a curved triangle. We call it the total curvature of the πr triangle. Remark: Angles are usually measured in radians in technical papers. The same is true with curvature. We will continue to measure angles and curvature in rotations for this exposition. When we combined the angles in all of the triangles of a triangulation of the sphere the total curvature terms totaled to the area of the sphere over πr. Since the area of a sphere of radius R is 4πR, we see that the total curvature of a sphere is. This is the 3

4 in Euler s theorem. A generalization of this result known as the Gauss-Bonnet theorem reads V E + F = total curvature measured in rotations, for any closed surface, where V is the number of vertices, E is the number of edges, and F is the number of faces in a cellular decomposition of the surface. It is worth computing the total curvature of a cone. One can make a cone by putting a slit in a piece of paper. One then widens the slit by reducing the angle down from one full rotation (one could also decrease the slit by increasing the angle past one rotation by adding extra paper) then tapes the edges of the slit together. See the following figure: Now consider making a triangle. We start drawing part of the first edge, turn a corner, draw a bit, then turn a second corner. After deciding the angle of the third corner we can change the length of the third edge to make it match up with the start. (See the following figure.) If we then cut a slit in the paper with some angle, we will need to add the same angle to the third corner of the triangle (perhaps adjusting the length of the third edge) to make it close up. 4

5 Thus, we see that the curvature of a cone is just the angle of the slit used to make the cone. Let s apply this idea to compute the total curvature of a torus. Consider the torus in the figure below. We see that each outside corner has a 1/4 rotation slit. Similarly, each inside corner has an extra 1/4 rotation (negative curvature here). Since there is the same number of outside corners as inside corners, the total curvature of the torus is zero. To check this we could cut the top and bottom square-with-hole into four trapezoids, and then count the number of vertices, edges and faces to get V E + F = 0. Extra Credit: Cut the two, three,... g holed tori into cells or look at the curvature to compute V E + F for these shapes. Use this to derive a version of Pick s theorem for a region with h holes. What does this have to do with wrapping paper? Nice wrapping paper will have two independent symmetries. If we roll up the paper to match the images under one translation, then roll it in the second direction to match up the images of the second translation, we will get a torus. We will now outline the steps in the classification of symmetries of wallpaper (I mean wrapping) paper patterns. Step 1: The total curvature of the rolled up torus is zero, as we just saw. Step : If we fold something up into several say l layers (think 3 if you wish). The the total curvature of the folded thing will be 1/l of the curvature of the original thing. For example, the total curvature of a sphere is, so the total curvature of a hemisphere is 1. A sphere can be folded in two layers to make a hemisphere, so the principle works in this case as 1 = (1/) 1. If the wrapping paper has extra symmetry, folding up the torus with this symmetry will produce some number of layers. The total curvature of the 5

6 folded up item will be 1/l of the curvature of the torus, which is zero. Thus, the total curvature of the pillow case for any wallpaper pattern is zero. Step 3: Figure the cost of various symmetry elements. Now consider making a pillow case by adding symmetry elements to a sphere. If you draw a pattern on a hemisphere with a 1/7th rotational symmetry, you could slit the hemisphere open and roll it up into a conehead shaped region with 1/7th of the total curvature of the hemisphere. Since a hemisphere has total curvature equal to one, the total curvature of the result would be 1/7. To add such a symmetry element to a sphere we would remove a hemisphere (removing 1 unit of curvature) and then glue back the conehead (adding back 1/7 unit of curvature). Thus we see that the cost of this is 1 1 = 6/7. A 1/nth 7 rotational symmetry is also called an order n cyclic symmetry. This could be denoted by C n and we see that such would cost 1 1 = (n 1)/n. n We already saw that folding the sphere in half cost 1 unit of curvature. This leaves a boundary, so we could denote it by B. It is what happens when one adds a reflection symmetry. A dihedral symmetry is the combination of a rotational symmetry of order n with a reflection symmetry having the center of rotation as a fixed point. We could denote it by D n. This just folds the cyclic symmetry in half, so it costs half as much, i.e. (n 1)/n. Of course to have a dihedral symmetry one must already have a reflection (boundary) symmetry B. We will use T to denote the situation with only translational symmetries. This means we get rid of the sphere and replace it with a torus. This costs us units of curvature. We can add two boundary components to a sphere by considering a wallpaper pattern with two reflection symmetries. This costs two units of curvature. It is denoted by BB or in Conway-Thurston notation. It is just a cylinder. It has zero curvature. If we have 6

7 a reflection symmetry and a glide reflection, the pillowcase will just be a Mo bius band. Thus we see that the total cost of one reflection (boundary = B) and one glide reflection (X) [so BX in total] is the same as the cost of two reflections (BB) and conclude that the cost of a glide reflection is the same as the cost of a reflection, i.e. X costs 1 unit. Step 4: Go shopping. Find all ways to spend exactly units of curvature on symmetry elements. This is exactly what we did in the last wrangle. We discovered that there were 17 ways to do so. Those 17 ways correspond to the 17 wallpaper patterns classified by symmetry elements. Here is a nice version of the picture behind the Euler theorem. Figure 1: Spherical Triangle We ll wrap things up with a few wall paper patterns. Try to find all of the inequivalent symmetry elements in each pattern. Hint: The total cost will be units of curvature. See the wallpaper file on the class webpage. 7

Geometry 10 and 11 Notes

Geometry 10 and 11 Notes Area and Volume Name Per Date 10.1 Area is the amount of space inside of a two dimensional object. When working with irregular shapes, we can find its area by breaking it up into