To find ratio of Length and Breadth of a Rectangle

Question

Given a rectangular paper sheet. The diagonal vertices of the sheet are brought together and folded so that a line (mark) is formed on the sheet. If this mark length is same as the length of the sheet, what is the ratio of length to breadth of the sheet?

Answer 1

Let the length be $l$, and the width be $w$. When the sheet is folded so that the diagonals meet, the fold crosses the sheet from one of the long sides to the other, and meets the long sides the same distance from the opposite corners. Let the distance from the intersection of the fold and the long side to the nearest corner be $d$, and the length of the fold be $f$.

Then, from the geometry of the fold, we can write: $$w^2+(l-2d)^2=f^2$$ $$w^2+d^2=(l-d)^2$$

Solving for $d$ from the second equation results in: $$d=\frac {l^2-w^2}{2l}$$

Substituting it into the second equation, and using your stipulation that $f=l$: $$w^2-2(l^2-w^2)+\frac{(l^2-w^2)^2}{l^2}=0$$

Finally if we define the ratio $r = \frac l w$ and rearrange the above equation, we arrive at: $$r^4-r^2-1=0$$

Thus $r^2$ is the golden ratio, so $r$ is the square root of the golden ratio.

Interesting problem!

Answer 2

It is just square-root of Golden Ratio.

Let $ 2a, 2b $ be length and breadth. Then by folding the slope of diagonal= b/a ;

From your condition we have

$$ 2 b = 2 a \cos \theta $$

$$ 2 \sin^2\theta + \sin \theta -1 =0 \rightarrow \sin \theta = \frac{\sqrt5 -1}{2} = \frac{1}{\phi } $$ $$ \frac{a}{b} = \sec \theta = \sqrt \phi. $$

Nice and simple problem.

Answer 3

$x$ is the length of the fold, and $l$ and $w$ are the respective length and width of the rectangle.

By using Pythagoras' theorem repeatedly, we can state:

$w^2 + a^2 = (l-a)^2$

$w^2 + (l-b)^2 = b^2$

$w^2 + (b-a)^2 = x^2$

From the first two equations, you can derive the values of $a$ and $b$:

$a = \frac 12(l - \frac{w^2}{l})$

$b = \frac 12(l + \frac{w^2}{l})$

Put those into the third equation and equate $x = l$ (given condition) to get:

$\frac{w^4}{l^2} + w^2 = l^2$

Now let $k = \frac lw$, the ratio between length and width.

You can now simplify to:

$k^4 - k^2 - 1 = 0$

which is a biquadratic, allowing you to find $k^2 = \frac{1 + \sqrt 5}{2}$, the golden ratio.

so the required ratio is the square root of the golden ratio.

EDIT: I solved this independently, but it really looks like a dupe of Matt's solution. Apologies for that, it was completely unintentional. But I had already done the bitmap art, so I didn't want to waste it. Hopefully, it will come in handy in visualisation.