- What is the number of diagonals in a regular 50-gon?
- What is a formula to calculate the number of diagonals in a regular n-gon?
With that, I told the class I was awaiting their answers.
Of course, this led to a lot of grumbling, staring, and not a lot of activity. I used this as an opportunity to tell the class that every one of them had the knowledge and ability to answer these questions. They needed to think about what was going on and use what they already knew to answer the questions. I also told them I wouldn't be asking them these questions if I didn't believe they could answer them.
The class settled down and started working. A few students still were sitting there looking distracted. I talked to each one about what they were doing. I told them that avoiding the problem was not going to help find a solution and re-directed them to start writing down what they knew and drawing figures to help them consider the dynamics that were occurring as we moved from a square to a pentagon to a hexagon.
After a while, students started observing and making notes of patterns. Most saw that the difference in the number of diagonals increased by one each time. A few students actually wrote out the sequence all the way to the 50-gon in order to determine the number of diagonals. One student noticed that the number of diagonals connected to a single vertex increased by one each time, i.e. in a square a single diagonal is connected to each vertex, in a pentagon two diagonals are connected to each vertex, etc.
It took a while for students to get to this point. I then tried to push them to answer the second problem. Here, students really struggled. A couple of groups were able to write out an informal recursive formula. No one was able to create the closed formula, although a couple of students did look up the formula online, saying D = n (n - 3) / 2.
I focused on the recursive formula first. Students don't get to see recursive formulas very often and they aren't presented in a more formal setting. I wanted to contrast the recursive formula with a closed formula in the hope of contrasting that the closed formula is a function solely based on the number of vertices versus a function that is based on the preceding value.
I took what students had and re-wrote it using subscripts: Dn = Dn-1 + n-2. We discussed what this showed, how to use it, and how you could work through each step to get to an answer, say for D50. I then said that, theoretically, every recursive formula could be re-written as a closed formula. Sometimes the conversion could be complex, but it could be done.
We turned our attention to the closed formula. I told everyone that if they couldn't explain the formula they couldn't use the formula. I then pushed them to consider what the n represented, why was three being subtracted from n, why were these two values being multiplied together and then divided by 2?
Students immediately said that the n in the formula represented the number of sides. I drew a pentagon and its diagonals on the board. I asked if n is the number of sides, that does n - 3 represent? I erased three of the sides. Students were baffled. I told the class the formula should connect back to reality.
After thinking about it some more, most students still believed that n still represented the number of sides. I asked, if this was the case, why are you multiplying the number of sides by three less than the number of sides? Again the class had no response.
I asked what else n could possibly represent. A few students said it could be the number of vertices rather than the number of sides. I told the class that when faced with evidence that contradicts or doesn't support their assumption of n representing the number of sides, that, perhaps, they should rethink their assumptions rather than ignore the evidence.
We focused on a single vertex. How many diagonals could not be drawn from this vertex? Well, you can't draw a diagonal to the vertex itself. You also cannot draw a vertex to the two adjacent vertices. This means that there are n - 3 vertices where diagonals can be drawn. For how many vertices total can this happen? Well, there are n vertices in total. So, how many total diagonals can be drawn? Well, you have to add up n - 3 each time for every vertex. What's a shorthand way to do repeated addition? Multiplication! So n (n - 3) is counting the number of diagonals that can be drawn.
But why do we divide this number by 2?
At this point time was running out on the class. I told students to focus on the physical process that was occurring here. Some started to see that by the time you went around all the vertices that you actually had drawn to diagonals between each pair of non-adjacent vertices.
I told the class we would have a quiz on Friday. The quiz is one question: explain how the formula n (n - 3) / 2 models the number of diagonals in a regular polygon.
I know that students can look up the result online and that is fine with me. I want them to see that formulas aren't just these arbitrary results. I want them to see and understand that mathematics is simply a way of describing reality. And that by describing reality, we can make use of the math to answer questions that would be difficult or impossible to answer otherwise, such as how many diagonals does a regular 500-gon contain?
Students immediately said that the n in the formula represented the number of sides. I drew a pentagon and its diagonals on the board. I asked if n is the number of sides, that does n - 3 represent? I erased three of the sides. Students were baffled. I told the class the formula should connect back to reality.
After thinking about it some more, most students still believed that n still represented the number of sides. I asked, if this was the case, why are you multiplying the number of sides by three less than the number of sides? Again the class had no response.
I asked what else n could possibly represent. A few students said it could be the number of vertices rather than the number of sides. I told the class that when faced with evidence that contradicts or doesn't support their assumption of n representing the number of sides, that, perhaps, they should rethink their assumptions rather than ignore the evidence.
We focused on a single vertex. How many diagonals could not be drawn from this vertex? Well, you can't draw a diagonal to the vertex itself. You also cannot draw a vertex to the two adjacent vertices. This means that there are n - 3 vertices where diagonals can be drawn. For how many vertices total can this happen? Well, there are n vertices in total. So, how many total diagonals can be drawn? Well, you have to add up n - 3 each time for every vertex. What's a shorthand way to do repeated addition? Multiplication! So n (n - 3) is counting the number of diagonals that can be drawn.
But why do we divide this number by 2?
At this point time was running out on the class. I told students to focus on the physical process that was occurring here. Some started to see that by the time you went around all the vertices that you actually had drawn to diagonals between each pair of non-adjacent vertices.
I told the class we would have a quiz on Friday. The quiz is one question: explain how the formula n (n - 3) / 2 models the number of diagonals in a regular polygon.
I know that students can look up the result online and that is fine with me. I want them to see that formulas aren't just these arbitrary results. I want them to see and understand that mathematics is simply a way of describing reality. And that by describing reality, we can make use of the math to answer questions that would be difficult or impossible to answer otherwise, such as how many diagonals does a regular 500-gon contain?
No comments:
Post a Comment