# Relation of different edges in rectangles

Is there any rectangle, that if you divide it into another rectangle (or any other quadrilateral), the relation between the two different edges is the same in both rectangles?

For example - Orginal rectangle:

Divided Rectangle:

Note that these illustrations aren't really exact, it's just to clear the point.

Thanks.

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Do you require the longer side to be divided in half, or do you want the longer side divided so that you get a square and "the rest"? – Arturo Magidin Sep 20 '10 at 18:26
devided in half. – Alon Gubkin Sep 20 '10 at 18:28
it's spelled "divided". It comes from "divide". – Arturo Magidin Sep 20 '10 at 18:34
Sorry for my English. Thanks for the correction :) – Alon Gubkin Sep 20 '10 at 18:35

I presume you want $b/a=a/(b/2)$, that is $(b/a)^2=2$ so the ratio of sides is $\sqrt 2$. The A series of paper sizes is based on this ratio. To the nearest millimetre the ratio of the long to the short size is $\sqrt2$, so tearing a sheet of A4 along a line through its centre parallel to its short sides gives two sheets of A5 etc.
This amounts to solving an equation. For the case you illustrate (where we divide the longer side in half), you fall into one of two cases: if $b\gt 2a$, so that in the second rectangle the horizontal sides are still the longer sides, you want $\frac{b/2}{a}=\frac{b}{a}$. For this to occur, you would need $ba = 2ba$, which is impossible since both $a$ and $b$ are assumed positive.
If $b\leq 2a$, then in the second rectangle it is the vertical side that are longest, so you want $\frac{a}{b/2} = \frac{b}{a}$. This gives you $2a^2 = b^2$, or $\sqrt{2}a=b$ (since $a$ and $b$ are lengths, they will be positive). So, for example, if $a$ is $1$ in the first figure, you need $b=\sqrt{2}$.
Of course, it will all depend on what proportions you want. For example, one can consider the case in which you "cut off" a square of side length $a$ from your original rectangle, leaving you with a rectangle with sides $a$ and $b-a$. Then you would want this latter rectangle to have the same proportions as the original one, which means you want $\frac{b}{a}=\frac{a}{b-a}$ (turns out you cannot do it if $b-a\gt a$). This leads to an equation that you can solve for $b$ in terms of $a$ (or solve for $a$ in terms of $b$), giving you the proportions that the original rectangle must have. Etc.