# Practice Paper 4 Question 5

A thin square piece of paper of side \(1\) is folded recursively. The first fold is along a diagonal of length \(\sqrt{2},\) thus obtaining an isosceles triangle. All successive folds are along a segment of length \(l_i,\) with \(i=1,2,\ldots,\) that results in a new triangle that is similar (geometrically) to the previous one. Find the sum of all \(l_i\) as this process continues indefinitely.

## Related topics

## Hints

- Hint 1After the first fold, where would the fold be so that the resulting triangle is geometrically similar to the previous one? A diagram may be useful.
- Hint 2Find the relationship between the value of \(l_i\) and the adjacent side of the corresponding triangle.
- Hint 3How does that help you find the relationship between the value of \(l_i\) and \(l_{i+1}?\)
- Hint 4Do you notice anything familiar about the sum of all \(l_i\) given the above relationship?

## Solution

The first fold is of length \(\sqrt{2}\). This is a special case though as it turns the square into an isosceles right triangle.

All future triangle-to-triangle folds should be along the altitude of the larger triangle. The fold halves the right triangle, so the adjacent side of the larger triangle becomes the hypotenuse of the smaller one. This gives us the ratio of \(l_i\) to \(l_{i+1}\) to be \(\frac{1}{\sqrt 2},\) and that would also be the ratio for our geometric series.

Hence after \(n\) folds the total length is:\[ \begin{align} L(n) &= \sum_{i=1}^{n}{\frac{1}{\sqrt 2^i}} \\ \lim_{n \rightarrow \infty} L(n) &= \sqrt{2} + 1\\ \textrm{or equivalent: }&=\frac{\sqrt{2}}{2-\sqrt{2}}=\frac{1}{\sqrt{2}-1}. \end{align} \]

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