Waste less time on Facebook — follow Brilliant.
×

Triangle Matchstick Investigations 2

We are investigating the number of matchsticks that we need to form several unit equilateral triangles in 2-D. You should read the first part if you have not already done so.

Let \( f(n) \) denote the minimum number of matchsticks that are needed to form \(n\) equilateral triangles. In the previous investigation, we tried to determine the value of \(f(6) \). From the collective attempts, we arrived at \(f(6) = 12 \) through the arrangement above. In fact, we have the following table of values:

\[ \begin{array} { l | l } n & f(n) \\ \hline 1 & 3\\ 2 & 5 \\ 3 & 7 \\ 4 & 9 \\ 5 & 11\\ 6 & 12\\ \end{array} \]

Strictly speaking, we don't have any proof that these values hold, other than "We tried our best, but couldn't do better". The accurate statement would be that our attempts show us that \( f(6) \leq 12 \). There is the slight chance that \( f(6) = 11 \), because we weren't smart enough to think of something.

Now, let us try and discover more about the function \( f(n) \). We have listed out some initial values, but there isn't an immediately observable pattern.

1) What is the value of \( f(7), f(8), f(9), f(10)\)?

2) Explain why \( f(n) \leq f(n+1) \).

3) Is it possible that \( f(n) = f(n+1) \)? Why, or why not?

4) Show that \( f(n+1) \leq f(n) + 2 \).
This can be tricky to justify.

5) For what values of \(n\) do we have \( f(n+1) = f(n) + 1 \)?

6) Create a table of values of \(f(n) \). Can you find any pattern?


As before, we are working in 2-D. If you are brave, go ahead and try the 3-D version. It is quite challenging.

Note by Chung Kevin
3 years, 3 months ago

No vote yet
1 vote

Comments

There are no comments in this discussion.

×

Problem Loading...

Note Loading...

Set Loading...