7
$\begingroup$

A piece of ribbon 1 metre long is folded in half so that the two ends are on top of each other. This doubled piece of ribbon is then folded in half again. The folded piece of ribbon is then cut right through its midway point (see my drawing below). enter image description here This produces 5 pieces of ribbon of lengths: $\frac{1}{8},\frac{1}{4},\frac{1}{4}, \frac{1}{4}, \frac{1}{8}$.

My question is what happens after $n$ folds?

Trying a few more examples has led me to guess that after $n$ folds: there are $2^n + 1$ pieces of ribbon, with two strands of length $\frac{1}{2^{n+1}}$ and the rest of length $\frac{1}{2^{n}}$. I am having difficulty creating a convincing proof of this.

I came across this problem on an Oxford University Thinking Skills Admissions test ( Question 38, page 23 ). If this is a well known problem, I would also be interested in a source to read about the question in more detail.

$\endgroup$
1
  • $\begingroup$ +1 : Interesting query, nicely presented. With the query showing $n=2$ folds, perhaps, you could edit your query, showing the math (no more pictures needed) re $n=3$, and $n=4$. Then, perhaps you could (in your query) discuss the pattern specifically re $n=3$, $n=4$. Then, you could also (in your query) consider one of two approaches: a direct approach, based on $n$ folds, or an induction approach, based on $N \to (N+1).$ In short, it would be nice if you could show further work, directly in your query. $\endgroup$ Commented Mar 10, 2021 at 0:20

2 Answers 2

2
$\begingroup$

HINT: I think you can convince yourself with this argument:

  • After $N$ folds, how many layers "thick" is the ribbon?

  • The cut will create $2$ open ends per layer. So how many total open ends are there after the cut? Don't forget the $2$ original open ends (the ends of the original ribbon).

  • No. of pieces is related to total no. of open ends is a very simple way.

$\endgroup$
1
$\begingroup$

Imagine the ends of the ribbon are tied together after the first fold, making it into a loop. As antkam hinted, each fold doubles the number of layers, so we end up making $2^n$ cuts, dividing the loop into $2^n$ pieces. Every piece is the same length (as it goes from the cut to the edge and back), and exactly one of these pieces consists of two equal-length parts that we tied together.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .