A New Pi. For GA2 and TPC

I must credit Trevor Kann and his most recent post with this idea.   Click on his blog.  I'll summarize below, but I recommend you see his blog as the "original source." He also gives, in his blog, a lot of the answers to the questions I ask in my blog.   He explored the ratio of the circumference to the diameter of a circle (otherwise known as pi) in "Taxicab Geometry."  We had discussed this as a tangent in class, but being a good thinker and good blogger, he started asking himself, "What if...."

To set this up, in class we talked about a new way to look at  "point," "line," and "plane," after all, they are undefined terms.  What if we defined these things such that a plane was a grid, much like city streets on a NS EW grid. Points would be lattice points, or intersections of the "streets" on the grid, and we could travel only on the grid its self, much like a taxicab would drive.  A circle is still the set of all points equidistant to a single point.

In the grid to the right, the center point of a "circle" is shown with a circle of radius 2 (diameter = 4).  Remembering that you can only travel on the lines and points are only places where the lines intersect, find the circumference. Then take the ratio of the circumference to the diameter.  This ratio is known as pi.  (On a Euclidean plane, like we have been using, the ratio is about 3.14159....). Verify that this calculation of the new "pi" is indeed 4. Make your own grid; identify a circle with a different radius.  Is the ratio still 4? 

The above is our "tangent" in class last week.

So what did he do that was so cool?  Trevor asked, What if we have the same rules about lattice points, traveling only on the lines, but we set up the plane to be equilateral triangles.  What would pi be? (Actually, he said, "This got me thinking....") To the right is a snip of his new grid.  One unit is the length of the side of an equilateral triangle; points are only vertices of the triangle. You can see the center and a circle shown. The radius of this circle is 2.  Find the circumference.  Find the value that is the ratio of the circumference to the diameter ("pi"). It's not 3.14159.....  and it's not 4.


Then he thought some more.  What if the grid was tessellating hexagons.  In the process of thinking, he figured out that the only tessellating regular polygons are squares, triangles, and hexagons.   To the right is his image of the tessellating hexagons. Of course, he used a radius of 2 and found a value for the circumference, diameter and the ratio of the circumference to the diameter.  Go ahead, find that ratio. (Answer is in his blog :) )



So ok.  In this next blog, I know Trevor is still thinking about this (DO keep thinking, Trevor!).

For your blog, you can expand on this idea.  OR better yet, find something like, "Why are there only three regular polygons that tessellate?" or "What is the relationship between regular tessellations and regular polyhedra?" or "What if we put a circle on a soccer ball?" (Trevor's question, lay off that one, I think he may be writing about that one.....) BTW, a soccer ball is also a "truncated icosahedron."  OR even better, think of a tangent we took in class (or maybe your brain just took in class without us being with you on that mental adventure....) As yourself, "What if..." or be like Trevor and just start thinking....