The A4 paper puzzle

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Introduction

I first heard about this problem from the Stand-up Maths video on the topic and I found it utterly fascinating. I'm going to present here the problem, the solution, and finally discuss a more general version.

The problem

The problem is quite simple. Find the length of the perimeter of the following plane figure:

A kite-shaped figure where the side corners are marked as being right angles — The shape for which we want to find the perimeter

The shape is obtained by folding an A4 piece of paper (whose sides have lengths in a ratio of $1 : \sqrt{2}$ ) two times: on one corner until the short side reaches the adjacent long side, and then again on the corner of the leftover rectangle, until the leftover on the long side touches the folded short side. Check Matt's video if you're confused, or consider the following picture: from the initial rectangle, fold diagonally to bring $U$ to $H$ and then $V$ to $K$ :

Same shape as above, but with auxiliary lines showing how it can be obtained from a rectangle with 1:√2 — How to obtain the shape from an A4 sheet of paper, with reference points for clarity in the solution

The solution

The solution to the problem is quite straightforward: we simply need to add up the lengths of the segments composing the perimeter of our figure.

We'll take as reference for the length the short side of the sheet of A4 paper, so the bottom segment $A D$ , which is the long side of the sheet, has length $\bar{A D} = \sqrt{2}$ .

The long diagonal segment $A B$ is the diagonal of a square $A U B H$ , whose side is the short side of the sheet: the side has length $\bar{A U} = \bar{U B} = \bar{B H} = \bar{A H} = 1$ , and therefore the diagonal length is $\bar{A B} = 1 \cdot \sqrt{2} = \sqrt{2}$ .

The short diagonal $B C$ was obtained folding the $V$ angle of the leftover rectangle $H B V D$ , so let's focus for a moment on this rectangle. The long side of $H B V D$ is the short side of the original A4 paper sheet, and is thus of length $\bar{B H} = \bar{V D} = 1$ ; the length of the short side of this rectangle, on the other hand, is equal to the difference in length between the long and short sides of the A4 paper sheet, and is thus $\bar{B V} = \bar{H D} = \bar{A D} - \bar{A H} = \sqrt{2} - 1$ .

By folding the corner of this rectangle, the resulting side is the diagonal of the $B V C K$ square, whose side is the short side of $H B V D$ , and its length is thus

$\bar{B C} = (\sqrt{2} - 1) \cdot \sqrt{2} = 2 - \sqrt{2} .$

The last segment $C D$ is the long side of the rectangle $K C D H$ , which is the “leftover of the leftover” after the second fold: its length is thus the difference between the long and short sides of the leftover rectangle $H B V D$ :

$\bar{C D} = \bar{V D} - \bar{V C} = \bar{V D} - \bar{B V} = 1 - (\sqrt{2} - 1) = 2 - \sqrt{2} .$

(Notice how, curiously, this last segment is equal in length to the previous, diagonal segment! We have $\bar{BC} = \bar{CD}, which will come in handy next.)

We now have the lengths of all the segments:

The kite shape, with the side lengths written in — Side lengths of our figure

and the length of our perimeter can then be computed as:

$\bar{A B} + \bar{B C} + \bar{C D} + \bar{A D} = \sqrt{2} + \sqrt{2} + (2 - \sqrt{2}) + (2 - \sqrt{2}) = 2 + 2 = 4 .$

Generalization

The beauty of the problem obviously comes from the fact that the cancelling square roots give us a nice and clean integer despite the original piece of paper having an irrational aspect ratio, thus resulting in irrational lengths for all sides. But what happens if we don't start from an A4 (or more in general, from an A-series) sheet? What if the original rectangle has a different aspect ratio?

So let's assume now that we're starting from a rectangle with aspect ratio $1 : a$ with $a \geq 1$ .

The bottom segment now has length $\bar{A D} = a$ .

The long diagonal segment still has length $\bar{A B} = \sqrt{2}$ , since the short side of the original rectangle still has length $1$ .

Let's now look at the leftover rectangle. The long side still has length $1$ , and the other has length $a - 1$ .

We must now be careful, since the second fold gives us the behavior we have already seen for the A4 sheet only as long as the short side of the leftover rectangle is not longer than the other side, i.e. only if $\bar{B V} \leq \bar{V D}$ , i.e. if $a - 1 \leq 1$ , i.e. only if $a \leq 2$ .

In this case, the second fold thus gives us a short diagonal of length

$\bar{B C} = (a - 1) \sqrt{2},$

and the remaining segment has length

$\bar{B C} = 1 - (a - 1) = 2 - a .$

We have lost the beautiful relationship between the last two segments, but we can still compute the final perimeter, with length

$\bar{A B} + \bar{B C} + \bar{C D} + \bar{A D} = a + \sqrt{2} + (a - 1) \sqrt{2} + 2 - a = 2 + a \sqrt{2} .$

We can thus see that the only case where we can a nice integer value for the length of the perimeter is when $a \sqrt{2}$ is an integer, which can only happen when $a = n \sqrt{2}$ for some integer $n$ , but since it must be $1 \leq a \leq 2$ , the case $n = 1, a = \sqrt{2}$ is the only possible case.

In the extreme case where $a = 1$ , the second fold produces a diagonal of length 0, and we get an isosceles triangle obtained by halving the square, with length $2 + \sqrt{2}$ .

In the extreme case where $a = 2$ , the second fold matches the first one, and we get another isosceles triangle, this time with two sides of length $\sqrt{2}$ , and the other with length 2, and thus a perimeter of length $2 + 2 \sqrt{2} = 2 \cdot (1 + \sqrt{2})$ .

Thin stripe case

We can, in fact, also do the case $a > 2$ , i.e. the case when the original rectangle is a “thin stripe” (long side more than twice longer than short side). In this case, the second fold would stop when the (now short) right side $V D$ of the leftover rectangle $H B V D$ meets the base, giving us a trapezium:

The trapezium created by folding the two short sides of a long stripe — The “thin stripe” case

We still have $\bar{A D} = a$ , $\bar{A B} = \sqrt{2}$ , and this time also $\bar{C D} = \sqrt{2}$ . Finally, the short base $B C$ is obtained subtracting from the long base the short sides of the original rectangle (due to the folds), and thus $\bar{B C} = a - 2$ .

The perimeter of the figure is therefore

$\bar{A D} + \bar{B C} + \bar{C D} + \bar{A D} = a + a - 2 + 2 \cdot \sqrt{2} = 2 \cdot (a - 1 + \sqrt{2}) .$

(Note that for the extreme case $a = 2$ we obtain again $2 \cdot (1 + \sqrt{2})$ .)

If we want the perimeter of the new figure to be an integer $n$ , we need first of all that $a + \sqrt{2} - 1$ to be rational, and thus $a = q - \sqrt{2}$ for some rational $q$ . Then $n = 2 \cdot (a - 1 + \sqrt{2}) = 2 \cdot (q - 1)$ , hence $q = 1 + \frac{n}{2}$ , and thus finally

$a = 1 + \frac{n}{2} - \sqrt{2} .$

Observe that since this was done under the hypothesis $a \geq 2$ , from this we also get $\frac{n}{2} \geq 1 + \sqrt{2}$ whence $n \geq 2 + 2 \sqrt{2}$ , and since $n$ must be integer, $n \geq 2 + 3 = 5$ .

The smallest integer perimeter we can get for the “thin stripe” case is $5$ , achieved with $a = \frac{7}{2} - \sqrt{2}$ (for the curious, this is barely more than $2$ , as $a = 2.0857864376 ...$ —and yes, it's the trapezium in the figure). Subsequent integer perimeters are obtained by incrementing $a$ in steps of $\frac{1}{2}$ .