geometryproblem solving
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.
Have you picked up a piece of paper and folded it? The dotted line shows where the triangle came from. The triangles above and below the highest diagonal line are supposed to be congruent. You want to maximize the area of the folded part, which minimizes the area left.
The length is
$$\frac{45}4\,\text{ft}=11.25\,\text{ft}$$
One way to obtain this result is using coordinates. Use $E$ as the center of the coordinate system, then the corners are $A,B,C,D=\left(\pm\frac92,\pm\frac{12}2\right)$. The diagonal $AC$ has the equation $y=\frac{12}{9}x=\frac43x$ so the perpendicular line has $y=-\frac34x$. For $x=-\frac92$ you get $y=\frac{27}{8}$, so you have $F=\left(-\frac92,\frac{27}8\right)$ and likewise $G=\left(\frac92,-\frac{27}8\right)$. The line between these two has a length of
$$\lvert FG\rvert=\sqrt{9^2+\left(\frac{27}4\right)^2}=\sqrt{\frac{1296+729}{16}}=\sqrt{\frac{2025}{16}}=\frac{45}4$$
Best Answer
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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.