Is there a size of rectangle that retains its ratio when it's folded in half?

Question

A hypothetical (and maybe practical) question has been nagging at me.

If you had a piece of paper with dimensions 4 and 3 (4:3), folding it in half along the long side (once) would result in 2 inches and 3 inches (2:3), which wouldn't retain its ratio. For example, here is a piece of paper that doesn't retain its ratio when folded:

Is retaining the ratio technically possible? If so, what is the side length and ratio that fulfills this requirement? Any help would be appreciated.

Update:

I added "once" because I got an answer saying that any recectangle would work, as any rectangle folded twice has the original ratio. Nice answer, but not quite what I was looking for. As for the other answers, I got 3x as much information as I needed! Thanks!

Answer 1

The $1:\sqrt{2}$ ratio ensures exactly that. That is the idea behind the ISO 216 standard for paper sizes, which was adopted from the German DIN 476 standard.

Its most common usage is the A series which especially in Europe is a collection of very common paper sizes. The base size, A0, has an area of a square meter, and every next smaller paper size is constructed by folding it in half.

Answer 2

A ratio of $1:\sqrt{2}$ will do the trick!

The original rectangle will have a ratio of $x:y$, where $y$ is the longer side and $x$ the shorter side. Then the folded rectangle will have a ratio of $\frac{1}{2}y:x$ and we want

$$\begin{align} \frac{x}{y} & =\frac{\frac{1}{2}y}{x} \\ x^2 & = \tfrac{1}{2}y^2 \\ x & = \tfrac{1}{\sqrt{2}} \cdot y \\ y & = \sqrt{2}\cdot x \end{align}$$

Answer 3

$H:W = W:(H/2)$ resolves to $H:W = \sqrt 2$

Answer 4

Well, suppose you have a rectangle of sides with $A$ and $B$ of length, $A$ being the bigger side.

When we fold it along the longest side we end up with a rectangle with sides of length $B$ and $A/2$.

So what you want to know is if there are any values of A and B that satisfy the following condition:

$$\frac{A}{B} = \frac{B}{A/2}$$

From solving the equation we get that $A = B \cdot \sqrt2$. So if the paper's height is $\sqrt2$ times its width, then we can make a rectangle with the properties that you wanted. And this is the only ratio that will work.

Answer 5

Standard Bond paper sizes A4, A3, A2 and A1 have been designed so that areas double up and sides increase by scale $ \sqrt 2 $.

$$ \frac LB = \frac 12 \cdot \frac bl $$

Answer 6

Well, if you have paper that has a ratio equivalent to the square root of two with the length divided by the width, if you fold it in half hamburger style (from width to width), you will get another ratio that's equivalent to the square root of two of the same property, likewise if you fold it more in half that way.

How It Works:

$$\sqrt{2}/1=\sqrt{2}$$$$\sqrt{2}/2=1/\sqrt{2}=0.7071...$$

That's half the square root of two, and when one is divided by it, we get the square root of two again: $1/0.7071...=\sqrt{2}$ and it just keeps going on and on from number to number being divided by the square root of two to get that result. Just divide the dividend by the quotient in those things.

Answer 7

A degenerate rectangle with an aspect ratio of $0$, also known as a line segment, can be folded in half either along its length or across its middle, yielding either the same segment or a segment having half the original length. Either way, the aspect ratio remains $0$. Algebraically, the solutions to the equations $$\frac{a}{b}=\frac{a/2}{b}$$ and $$\frac{a}{b}=\frac{a}{b/2}$$ are both $a=0$. (In order to speak of the ratio being a real number, I'm ruling out $b=0$.)

As the other answers observe, the non-degenerate cases are $$\frac{a}{b}=\frac{b}{a/2},$$ which has the solution $a=b\sqrt2$, and $$\frac{a}{b}=\frac{b/2}{a},$$ which has the solution $a=b/\sqrt2$.

Answer 8

When you fold any rectangular sheet of paper through the middle, clearly the original area is the double of resulting (folded) area.

At the same time, if you have a little rectangle with sides $a$ and $b$ (so area $ab$) (let us say $a>b$), and you want a similar rectangle with area $2ab$, then clearly that larger rectangle must have sides $\sqrt{2}\cdot a$ and $\sqrt{2}\cdot b$, the factor of magnification being $\sqrt{2}$.

Combining these two simple observations, and looking at your illustrations, we see that the long side $a$ of the "half" rectangle is equal to the short side $\sqrt{2}\cdot b$ of the "unfolded" (full) rectangle, so $a = \sqrt{2}\cdot b \Rightarrow \frac{a}{b} = \sqrt{2}$.