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You are 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?

This is my first question on this site, so if this is not a good question please help.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Alex Becker May 23 '12 at 6:40
    
okay thank you....i will follow that from my next post :) –  Archie.bpgc May 23 '12 at 6:45
    
Do you mean that diagonally opposite corners of the sheet are brought together? –  Brian M. Scott May 23 '12 at 6:46
    
yeah...and the sheet is folded(or pressed) so that it forms an oblique line on the sheet –  Archie.bpgc May 23 '12 at 6:51
    
@Archie.bpgc I have changed the title and formatted your question. Hope it is fine. –  user17762 May 23 '12 at 7:09

2 Answers 2

up vote 6 down vote accepted

$\hskip 2.2in$ enter image description here

The above figure was done using grapher on mac osx.

Let $l$ be the length (i.e. the sides $AD$ and $BC$) and $b$ be the breadth (i.e. the sides $AB$ and $CD$). Once you get the diagonal vertices together, i.e. when $D$ coincides with $B$, the length $EB$ is the same as the length $ED = l-a$.

Hence, for the right triangle, we have that $$a^2 + b^2 = (l-a)^2\\ b^2 = l^2 - 2al\\ a = \frac{l^2 - b^2}{2l}$$ The length of $BF$ is $l-a$ and is given by $$l-a = l - \frac{l^2 - b^2}{2l} = \frac{2l^2 - l^2 + b^2}{2l} = \frac{l^2 + b^2}{2l}$$ The distance between the two points is $$EF^2 = r^2 = b^2 + (l-2a)^2 = b^2 + \left(l - \frac{l^2 - b^2}{l} \right)^2 = b^2 + \frac{b^4}{l^2}$$ This is so since the vertical distance between $E$ and $F$ is $l-2a$.

You are given that $r = l$ and hence $$l^2 = b^2 + \frac{b^4}{l^2}$$If we let $$\frac{l}{b} = x,$$ then we get that $$x^2 = 1 + \frac1{x^2}\\ x^4 = x^2 + 1$$ which gives us that $$x = \sqrt{\frac{1}{2} (1+\sqrt{5})} = \sqrt{\phi}$$ where $\phi$ is the golden ratio.

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Thanks a lot man –  Archie.bpgc May 23 '12 at 6:54
    
@Zev Chonoles Thanks. –  user17762 May 23 '12 at 21:19

$\hskip 2in$ enter image description here

We are given that $ZY = BC$, and we want to find $BC/AB$. To keep things simple, choose units so that $AB=1$; and let $s = AC$ and $t = BC$. Then we want to find $t$.

Triangle $XYC$ is similar to triangle $BAC$, so $CX/XY = BC/AB$.
$ZY=2XY$ and $AC=2CX$, so $AC/ZY = BC/AB$.
And we are given that $ZY=BC$, so $AC/BC = BC/AB$.

Thus $s = t^2$. And by Pythagoras, $s^2 = 1 + t^2$. Substituting for $s$, we get $t^4 - t^2 - 1 = 0$, which (as Marvis already showed) gives $t = \sqrt \phi$.

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