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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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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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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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