At first, I will simply count the matchsticks to determine the number in each of Gemma's patterns. I want to find a way to accurately establish the number of matchsticks in a set of patterns without having to physically count them, especially for large diagrams. I think this method will most likely result in a formula.
To start with, I will answer the first task which instructs me to determine how many matchsticks are in each of Gemma's pictures. My method will be to simply count the number of matchsticks in each diagram, so that I can have a basis for comparing the number of matchsticks in each pattern and then investigate how they relate to the number of matchsticks in patterns with different widths. I can show my initial results as follows:
I notice that there is a relationship between the width of the pattern and the number of matchsticks used. Obviously, as the picture gets wider, more matchsticks are used; but they are not in a direct relationship.
In a direct relationship, if a diagram that is one matchstick wide has 6 matchsticks, then it could be expected that one that is 2 matchsticks wide would have 12, and one that is 3 matchsticks wide would have 18. Mathematically, this could be expressed as the number of matchsticks (n) is equal to 6 times the number of widths (w), or: n=6(w). This formula works for the first picture, but is not accurate for the other two. Clearly, there is a relationship of increasing linear proportions, but it is made more complicated by the fact that each pattern of matchsticks shares a common side. This explains why the sequence is not 6, 12, 18, 24, 30, and 36. Any mathematical or formula representation will have to account for the fact that after the first unit, each additional unit of width lacks the two matchsticks it has in common with its neighbor. I think I should use algebra to try and explain this relationship because it is useful in understanding quantitative relationships, and I think a simple linear function will work.
The first unit of 1 width has six matchsticks. A second unit (or 2 widths) would share two of those matchsticks already
in place and add four more. It would look like this:
|_| 6 Matchsticks |_|_| 10 Matchsticks
If the formula can account for the initial width having 6 matchsticks and all other additions having four, it would be a reliable expression. This could be accomplished by representing the total number of matchsticks as a function of the first width having six and all others having four. By simply adding the number of the first width