The MathTwitterBlogosphere (MTB): Professional Development at its Best
I love to go to conferences and take workshops, but it involves packing, shlepping myself and my stuff to the airport, as well as money and time. I will always continue to seek out “live” learning opportunities, but I have a PD source right at my fingertips 24-7 in the form of the MTB. (Full disclosure: I don’t do twitter, but I more than make up with it by following many blogs.) Every once in a while, a blog post lights a fire under me, guiding me into to a line of inquiry that I can’t leave alone until I play around with the idea and write my own post to make sense of the idea.
3.5 gons? What the heck?
This post continues my pursuit of understanding of 3.5-gons and their ilk. It all started with Dan Meyer’s March 27, 2013 post “Discrete Functions Gone Wild!” His post focused on what a regular polygon would look like when when the number of sides was not a whole number. He used his understanding of how regular polygons with whole number sides behaved to determine the angle between sides, using the formula (n-2)180/n. He let n = 3.5 and determined that the measure of each angle in a 3.5-gon is 77.1 degrees. Then he constructed the 3.5-gon. and created a star shape. Then he left his readers with the following BTWs and went to bed, across the pond in Nottingham:
BTW. One of you enterprising programmers should create the animation that runs through continuous values of n and shows the regular polygon with that many sides. That’d blow my mind. I can only do the discrete values.
BTW. Malcolm Swan demonstrated the 3.5-gon on the back of some scratch paper in the middle of a design session here in Nottingham. That kind of throwaway moment (often before tea, of course) has been a lot of fun these last two months.
BTW. But where is the 3.5 in that shape? Maybe you see how the number 3.5 turned into the number 77.1 and how the number 77.1 turned into that star shape. But where is the 3.5 in the star? I’ll hint at it in the comments but I’ll encourage you to think this through. (It may be helpful to see 3.5 as the rational number 7/2.)
When he awoke, he found that several enterprising programmers had taken his challenge can created the goods.
2013 Mar 28. I love you guys. I fall asleep for a few hours and wake to find out it’s Christmas. Some interesting visualizations of rational regular polygons from Marc Garneau, Eric Berger, Stuart Price, and Josh Giesbrecht.
When I stumbled upon the post, I went to work on BTW # 3, trying to make sense to myself about why a 3.5-gon looks like this:
3.5-gon to 7/2-gon: Different representations of the same value open up new possibilities!
Looking at functions via multiple representations (graphical, numeric, algebraic, verbal) has always served me well. Some representations shine a different light on the function. Putting 3.5 into the form 7/2, a hint that Dan gave, opened up a new door for making sense of the shape.
I played with the different applets that were provided overnight by four readers of Dan’s post…another positive aspect of the MTB…great minds just waiting to use their talent for the betterment of mankind! The 7 was pretty easy to see since there were 7 line segments. The 2 took awhile for me to see a pattern. What I finally noticed by trying out different fractions was that the denominator indicated how many “rotations” around the shape it took to “close up” the polygon. These terms are used loosely, since a full rotation would take one back to the “starting point”, which would “close up” the polygon.
3 – gon, 3/1 – gon, or you may know it as a triangle
For example, in a whole number-gon, such as a 3-gon (triangle), which, by the way, could also be called a 3/1 – gon, follow this procedure:
Start at A. Heading counter-clockwise and draw segment AA’. Construct a 60 degree angle degrees at A’. Draw A’B the same length as AA’. Construct a 60 degree angle at B. Draw BB’ and B’ will coincide with A. You have closed the 3/1 -gon after 1 rotation.
m/n – gons and partial sweeps or revolutions
When dealing with a m/n-gon, when n is not 1, it will take more than one sweep to get exactly back to the starting point I am calling each “sweep around” a “partial revolution”, meaning that it gets close to the starting point, but doesn’t quite make it that time around.
Although my explanation was not mathematical precise…it was more like hand-waving and a “kinda” explanation… I posted my comment to Dan’s post, putting that interpretation out there.
Here is part of my comment to Dan’s post:
Wow! I love math! I was blown away by this post and the idea of thinking of a 3.5 gon as a 7/2 “gon” that means that it has seven “sides” and takes two rotations to complete.
Out of the denizens of the MTB, a comment appeared from none other than Michael Serra, author of one of my favorite Geometry books, Discovering Geometry: An Investigative Approach. Of course, he has a better and more mathematically succinct way to describe how the “n” behaves. Before I explain his way, which is skipping vertices, and much less wishy-washy than my explanation, I will share the following comment that he made:
BTW: Expressing the n as an improper fraction opens the door to two ways of expression each star polygon. The star polygon 12/5 is equivalent to the star polygon 12/7. The numerator expressing the number of vertex points and the denominator expressing how many points to count from one vertex to the next vertex.
It is cool that Elaine Watson noticed that the denominator is also the number of cycles to complete the star polygon. I hadn’t seen that before.
This comment by Michael Serra served two purposes:
(1) it made me curious to find out more about 12/5-gons and 12/7-gons, and
(2) it fed my ego to have a mathematics educator that I highly respected comment on my post that I had noticed something that he didn’t!
So, in honor of Michael Serra, I am going try to make sense of my explanation of the role of the denominator and define more clearly what I mean by “it takes two rotations to complete.” While doing that, I am also going to refer to Michael’s explanation of the role of the denominator, which expresses “how many points to count from one vertex to the next vertex.”
Closer look at 7/n – gons
7/1 – gon
My explanation of the role of the denominator was influenced by how Dan created the original 3.5-gon. He started with a line segment and the angle measure and added the sides one at a time, until the last segment met up with the starting vertex. On the other hand, Michael Serra’s more precise explanation required that the vertices needed to be placed first, followed by drawing in the sides using the rule for skipping the required number of vertices.
GeoGebra can conveniently create a regular polygon. To do this, construct two points and a line segment connecting the points to represent one side. Using the “polygon” tool, choose “regular polygon” and create a 7-gon. Below is a 7/1-gon. The 7 means there are 7-sides. Using my revolution interpretation, the 1 means that it takes one revolution to complete the polygon. Using Michael’s vertex interpretation, the 1 mean that you go directly to the next vertex, without any skipping over vertices. This forms the regular 7-gon, we all know and love.
In a 7/2-gon, use the vertices of the 7/1-gon as a skeleton. The 7/2-gon will be represented by the black segments. Choose a starting vertex and a direction (I chose counter-clockwise) around the polygon to the 2nd vertex and draw the line segment between the two vertices. Repeat this method until you return to the starting vertex. Since you are skipping vertices, you will have to go around more than once. In this case of a 7/2-gon, you will go around “about” twice to pick up the missing vertices before you get back to the starting one.
There is still one side to draw to close the 7/2-gon. This is not a full sweep. The other two sweeps each created 3 sides. The final sweep will create 1 side, so we’ll call it Sweep 2 1/3.
What about a 7/3-gon? Start with the skeleton of the 7/1-gon. Choose a starting vertex and count counter-clockwise to the 3rd vertex and draw the segment from the starting vertex to the 3rd vertex. Repeat that process until you close up the polygon by returning to the starting vertex. Each “partial revolution” or “sweep” is illustrated below:
As in the 7/2-gon, we’re not quite there yet. There is one more segment to draw. Each sweep filled in 2 sides. This last sweep will fill in 1 side, so we’ll call it sweep 3 1/2. Although the denominator doesn’t exactly define how many revolutions, it hovers around the ballpark.
Here is the final 7/3 – gon.
Both of the final sweeps here have required only one more line to be drawn. Questions to consider:
- Will there always be one leftover side to complete the shape?
- If not, what are the possibilities for leftover sides?
- Explain and illustrate your conjectures.
Comparing and Contrasting 12/7-gons and 12/5-gons
My next project was to figure out what Michael Serra meant when he said “Expressing the n as an improper fractions opens the door to two ways of expression each star polygon. The star polygon 12/5 is equivalent to the star polygon 12/7.” What did he mean by equivalent?
As I worked through my ideas, I simultaneously began creating an investigation for students, Comparing and Contrasting Congruent Star Polygons.
This is a very rough first draft, so I’m not ready to post it. I may have scaffolded it too much for it to be considered an investigation. If anyone has any comments or suggestions, please pass them on.
Conclusion: Fan Letter to the MTB and All of the Math Nerds out there who Keep the Flame Lit
To conclude, all of this started with one post by Dan Meyer that intrigued me. Because Dan has such a following, there were many more contributors who created applets, comments, and their own ideas. I don’t think that professional development gets much better than this. Being actively involved in keeping up with math blogs and then continuing on with my own investigations sparked by the posts, has made me much more reflective about my work and better at what I do. Books and articles have their place in my life; I’ll never be willing to burn my large math library. However, reading is a solitary pursuit. The MTB is a community pursuit. I need both.