## Monday, October 29, 2012

### 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.

## Sunday, October 28, 2012

Hurricanes make a somewhat geometrical shape...

## Wednesday, October 24, 2012

### 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?

## Tuesday, October 23, 2012

### 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?

## Monday, October 22, 2012

### 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

## Thursday, October 18, 2012

### 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
12. dilate
13. glide reflect

## Tuesday, October 16, 2012

### Paper Cube

This is a cube,it is cool, it is paper.
Lucas,Vikram, Josh, Jack and Jack

## Saturday, October 13, 2012

### 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.

## Friday, October 12, 2012

how to make a square with only a compass and straight edge

james h and will s

## Thursday, October 11, 2012

Sangaku (one example above, carved into a wooden tablet) are japanese geometrical problems originating in Japanese temples of the 17th century. The problems usually consist of a group of shapes that share one surrounding tangent shape. Sangaku were presented in temples in honor of the gods, and this is the most fascinating thing about Sangaku; the fact that geometric contemplations could have religious properties. Any combination of art math and faith is an enigma, but also seemingly natural, why sacrifice an animal or fast for the purpose of appeasing a higher power, when one could create mathematical conjecture and theorize about perfect shapes in honor of any ethereal forces or traditions you observe. Doesn’t geometrical theory seem so much purer and theological then a hymn or diet? My surprise and confusion at the utility of geometry as a sacred persuit compelled me to share this idea with any who don’t know about these types of problems and to open one of the problems I found as open discussion for the class.
After researching some existing sangaku problems and their solutions, I found a problem I would like to explore and dissect (pictured above) but I haven’t been able to piece together the construction of the problem after assigning the value of diameter of the largest circle (tangent) as one, so obviously there is more progress to be made. Anyone who is interested in solving or analyzing this problem, please contribute, I’m posing this as a problem and a challenge.

### Investigation #2: Attempting 'Dimension'

The idea of 'Dimension' is to design coordinate systems for various shapes, ranging from cubes to barbells. As a point of reference, we can use the sphere, which happens to be the shape of our very own Earth.

Beginning with the cube, it is fairly simple to design a coordinate system reminiscent of our own. All you need is two perpendicular lines along the cube, each one going through the exact centers of four faces. To visualize it, picture a present wrapped up in a ribbon. Or, alternatively, refer to the unfolded cube below. The lines have been drawn in such a way that when the cube is folded, they will be in the right place.

These two lines will be the 0s of latitude and longitude, equivalent to the equator and the prime meridian on Earth. We now have the power to specify the location of any point of the surface of the cube with relation our two lines. Just say how many units 'north' it is from the equator, and how many units 'east' it is from the prime meridian, and that's all we need. In terms of specifying points, the only difference between the cube and Earth is that on the cube, we need to turn a corner every now and again. This distinction is purely superficial, and makes no difference to the actual difficulty of the problem beyond our not being used to it (as a matter of fact, the sphere is actually much more unintuitive, since the units get smaller as you near the poles).

So, after the relatively simple problem of the square, we move on to the donut (or 'torus', if you want to be fancy about it). With the donut, the two lines are going to need to be placed a little more creatively.
One will run along the inside of the donut, and the other will run perpendicular to the first. See the image below.

In this illustration, we see the donut divided into sections. Any two perpendicular lines you see above you could be chosen to be the 'equator' and the 'prime meridian'; what matters is that you can describe any other point on the donut in terms of its relation to those two lines.

Now, that's about as far as I've gotten so far, but there is certainly more to think about the topic. For example, what if we tried, instead of a donut, a croissant:

Well, as you can tell from the lines already drawn in, it's obviously possible. Don't let the fact that it comes to a point deceive you: there is little theoretical difference, I think, between the croissant and Earth (with the two points corresponding to the two poles of Earth). How about this:

We are now entering a realm which I am really not prepared to discuss, so I'll leave it open to you guys. How would one even go about trying this? Is it possible to do with the methods we've been able to use so far? What do all those lines scribbled on it mean? (I actually have no idea, I just grabbed it off google images).

## Wednesday, October 10, 2012

### Hudson Cooke Is Smarter Than Us All

In the paper "Explorations into Euclidian and Taxi Cab Minimum Distances" by Hudson Cooke he explains our size problem and gives proof and reason for why distances between two points are the shortest. He explains that you can simply find the distance between two points on an open plain using the Pythagorean theorem. He then says that in a Taxi Cab plain, (where you can't draw lines diagonally, but instead have to follow lines all with 90 degree angles) if both point A and B are at opposite corners of a plain then there is only one distance that can take you from one point to another. It doesn't matter which specific route you take it will end up being the same distance as any other route.

Then I realized that if you draw your route and make unnecessary turns along the way then it will obviously take much longer and be a longer line drawn from point A to point B.

I'm probably just completely missing the point that Hudson is making but if anyone else has something to say about this please do, because if we figure this out then I think we will have a definite answer to the size problem.

la mosquita en la biblioteca con coolio

## We,(Paul and Mars) came up with a sort of solution to the camera problem. Our solution works with shapes that do not have walls. Here is an example of a walled shape;

What we found was that there was two distinct sub groups of shapes; Concave and Convex;
What we realized was that all convex shapes need only to have one camera, while the concave ones split into two categories; regular and irregular concave polygons. Regular Concave Polygons, or RCPs are concave polygons that from any point, and it does not need to be every point, in fact it can be one single point, a line can be drawn from that point to any point on the perimeter of the shape. The reverse is the definition for a ICP, or Irregular Concave Polygon. An example of a Irregular Concave Polygon is; ^^^^^^^^

## Thursday, October 4, 2012

### Readings Related to Investigation #2

I've updated the Readings page with several readings related to each of the five problems from Investigation #2 and the geometric properties they focused on. You might enjoy browsing through them--particularly those related to the problem you investigated. Writing a reflection on one of these readings--or a recommendation/review--could make a great blog post.

Enjoy!

We (Lucy and Natassia) tried tackling the 'how many 1 inch circles on a piece of paper that is 8.5"x11"' problem and discovered:

The sheet of paper is 8.5"x11" inches but half circles are not allowed. So the easiest way to organize the circles is simply in a grid like format of 8 circles x 11 circles, which gives a grand total of 88 circles in a 8.5" x 11" rectangle. However there is a way to pack more circles in this space...

...which is by staggering the rows of circles (see picture above). Instead of lining them up as if they were in a 8"x11" grid, if you move the first row up against one side, and the next row on top up against the other side and so on, therefore it is then possible to fit 12 rows instead of only 11 and a total of 96 circles in the rectangle.

We were wondering how this might work in a 3D version, which would be a 8.5"x8.5"x11" rectangular cube filled with 1 inch diameter spheres. Possibly each plane/sheet of spheres (96 in total according to above proof) stacked in a staggered fashion on top of each other would allow for a new "sheet" of spheres, like the staggering in 2D allowed for another row? Would the total amount of spheres be 9x8x11 or just 8x8x11? Please comment with ideas!

Here is a video showing our thought process: