A Few Fun-"Packed" Links

I just wanted to share a couple of geometry-related links that I ran across this evening.

This article starts with the penny-packing problem and builds up to discussing the unsolved problem of how to best back tetrahedra in space. (It's the most recent Feature Column at the American Mathematical Society.)

Is this the best way to pack pentagons? Nobody knows.

On a related topic, James and I got to talking about putting together polyhedra in nice ways--like to build another one. A cuboctahedron, in particular. That made me think of this paper from Bridges over the summer about making loops out of copies a polyhedron.

And finally, here are some four-dimensional symmetry puzzles. Awesome.

Cut and Paste Transformations

Inspired by Mr. Lanier's challenge sheet (particularly challenge #2--which involved transforming an irregular pentagon into a rectangle), I decided to try to create some of my own versions of cut and paste transformations. I wasn't sure how to go about this at first, so I tried a couple of different methods.
First, I tried this by drawing a rectangle on a sheet of graph paper that was say, 40 units in area. Then, I would try to draw a different shape that was 40 units in area. This was too difficult though, because with basically all other shapes besides squares/rectangles, not all of the individual units would be completely filled, so it was hard to calculate exactly how many units the new shape took up.
I found that it was easiest to take a square, for example, and then cut it up into smaller triangles, squares, etc., and then use those smaller shapes to construct a new shape with the same area as the previous one.

I started with a rectangle (50 units in area):

Then, I cut it up into lots of smaller triangles, and one one by ten unit long rectangle. I put them back together, and ended up with something like this:

I did the same for an octagon, but this time, I am posing it as a challenge for you guys. Here is the shape I got upon dividing an octagon, and then re-constructing it (using the method above). How could you use this to make an octagon? what would the area be? Let me know how you approached this problem.

Hurricanes make a somewhat geometrical shape...

Investigation #4 Found Shapes

I found this on the corner of Bergen&Smith:

 

It's a pretty ordinary fire escape (although the "Z" kind is much more common). It led me to think about the ratio between the V shape and the outer corner, like in this diagram:

So, what is the ratio between the gray and the white?

Icosahedrons and the Platonic Solids

As some of you might have seen in class, Natassia and I have been working on constructing an origami icosahedron (it is a regular polygon with 20 identical equilateral triangular faces, 12 vertices, and 30 edges, for those of you who were wondering).

Now that it is finished, I started doing some research on icosahedrons. I found out that it is one of the five Platonic solids. A Platonic solid is a convex polyhedron, where the same number of congruent, regular polygon faces meet at each vertex. Only five of these polyhedrons exist: 

They were admired for their aesthetic beauty and symmetry, and were extensively studied by the ancient Greeks. The philosopher Plato theorized that each regular polygon was associated with one of the classical elements-- the tetrahedron with fire, the icosahedron with water, the octahedron with air, and the cube with earth (the dodecahedron represented the universe itself). Euclid called these solids the atoms of the Universe. Basically, they believed that all physical matter was composed of the atoms of the Platonic solids.

What I found really fascinating in my research was discovering where these Platonic solids could be found in nature. The tetrahedron, cube, and octahedron all occur in crystal structures. For example, here's a Franklinite octahedral crystal:

In addition, many viruses, such as herpes, have icosahedral shells. The viral structure is made from identical protein subunits, and it is easiest to assemble them as an icosahedron. 

Pretty cool, huh?

3 Points Make a Circle

One of the problems presented in investigation 3 was this: "given three points, [find] the center of a circle that would pass through those three points." This question was posed in four different contexts, including compass and straightedge, ruler and protractor, paper folding and--the one that we chose--an algebraic approach.

So what are the things that we already do know? Well the problem states that we are already given the x and y coordinates of the three points that define the circle.  The definition of a circle is "A curve made up of the of all the points on a two dimensional plane that are a fixed distance away from a central point" according to our glossary of objects. So that means that the length of the lines drawn from each of these points to the unknown center are all the same length.

Point A is unknown. Coordinates of points B, C, and D are all known. All lines between points are equidistant from the center.

So, now we have to figure out the lengths of the lines in terms of the points they are connected to. We can do this by using the pythagorean theorem (I'm just gonna take it for granted this time; feel free to write a blog post proving the theorem if you like!). This example is for any of three points where its 2 coordinates are A and B, the points of the unknown are x and y, and the length of the line is l.

|A-x|^2 + |B-y|^2 = l^2

The absolute value signs are there to account for negative values, because in this case we don't care if line is pointing to the left; we just need to know total heftiness of where it is going (its length): its absolute value. Now all we have to do is plug this in for the other two points. Keep in mind that this does not correspond to the diagram in terms of letters; the capital letters are just the x and y coordinates of the 3 points that we already know. Also remember that throughout these 3 equations x and y remain the same.

|A-x|^2 + |B-y|^2 = l^2

|C-x|^2 + |D-y|^2 = l^2

|E-x|^2 + |F-y|^2 = l^2

This system of equations really shows the relationship between these four points. Now it is purely a matter of only Algebra. But we wanted a little more satisfaction (maybe get rid of the absolute values?), so we plugged this system of equations into Wolfram Mathematica. Well, let me just say that the output was less than satisfactory. There is even a couple of is in there(i is the square root of -1: a scary number). Unfortunately, I don't have the Wolfram Mathematica output with me now, so I will upload it later.

Ending this problem with a slightly bitter taste in my mouth, I wonder if taking a compass and straightedge (or other approach) to this problem will give more elegant results. What if you try the equation with only 2 points? My guess is that the graph of the equations plots a line where the center could be, instead of a single point. But that's just a hunch. It is up to you, dear reader to investigate!

--Solution by Harry LR and Akash V Mehta
--Write up by HLR

Here is a link to a Google Doc where you can fill out that table Justin gave us, ohh and don't screw with.

http://bit.ly/Wts4vx

Our Musings on Transformations

What we came up with during our brainstorm
Courtesy of Paul

1. turn/rotate
2. mirror/reflect/flip/invert
3. scale/project
4. warp/continuous map
5. break/explode/cut & paste/dent
6. stretch
7. translate/move/slide/shift
8. erase/delete/teleport to the void
9. extrude/thicken/solidify
10. twist
11. shadowfy/flatten/detrudue
12. dilate
13. glide reflect

Paper Cube

This is a cube,it is cool, it is paper.

Lucas,Vikram, Josh, Jack and Jack

A poly-polygon

(Sorry for the resolution, the file size was too big for my more detailed scan)

The other night I made this shape. As you can see, it is 4 regular polygons inside of each other all sharing the same side length.

I constructed everything with only compass and straightedge, except for the pentagon and hexagon sides coming out of from the base. For the other 2 (or 3) sides of the pentagon/hexagon I copied the angle of the first two lines coming from the base.

Methods used in this construction.

• Copying lengths and angles
• Making an equilateral triangle
• Erecting a perpendicular (for the square, because it has 90 degree interior angles)

To do list

• Explain why pentagons and hexagons have interior angles of 108 and 120 degrees respectively, as well as give proof for a general case for any regular polygon
• Describe how to execute the 3 methods listed above

P.S. Those two purple crayon smudges are driving me crazy.