Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 – Housefly 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
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.

share|cite|improve this answer
Thanks a lot man – Housefly 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 $t = \sqrt \phi$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.