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This question is related to my bath towel, which I hang on a rope, so let's have fun (you can use your own towel to do this experiment in bath-o).

There is this rectangle with sides $a. The rectangle is bent along a line that passes through the center of the rectangle. At which angle $\alpha$ (i. e., angle $\angle RAU$ – see the picture) should we bend the rectangle in order to get the minimum area of crossing intersection?

It is obvious that if $|a-b| \gg 0$, then $\alpha\approx\pi/4$. Also, we can look at the pentagon NOPQR. In addition to that, for a certain $a/b$ we get the triangle ($a/b<0.8150237, \alpha =\pi/4$).

All in all, I am looking for a graph: $a/b$ as in terms of $\alpha$ and the area of the crossing intersection, which is $\displaystyle S = [2ab - (a^2+b^2)t + 2abt^2 - (a^2+b^2)t^3]/[4(1-t^2)]$, where $t=\tan \alpha$.

Any help is highly welcome.

MAth towel

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    I do not understand the problem at all. You talk about towels and ratios and areas, but you do not mention the exact facts, and it's confusing. If you want an answer, you should explain the problem better.2011-05-26

1 Answers 1

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The minimum area occurs when $\alpha=\frac{\pi}{4}$ and it is equal to $S_\rm{min}=a^2/2$

Why? Well I solved this problem using a parametric CAD system and then using geometry. If the height of the towel is $a$ and the width is $b$ with $a then the critical aspect ratio is $\eta_c=\tan^{-1}\left(\frac{a}{b}\right)$. When the fold angle $\alpha\leq\eta_c$ then the common area is

$ S_1 = \dfrac{2 a b\cos\alpha-(a^2+b^2)\cos\alpha}{4\cos\alpha (2\cos^2\alpha-1))} $

but when the fold angle is $\alpha > \eta_c$ then the area is

$ S_2 = \dfrac{a^2}{4\cos\alpha\sin\alpha} $

which has a minimum at $\alpha=\pi/4$. The key to the solution is finding the coordinates of the point where the two long sides of the towel intersect after the fold (point $P$ above). This calculates to: $P=(\frac{a}{2}\tan\alpha,\,-\frac{a}{2})$ in a coordinate system in the middle of the towel.

In the end I take the area of half a tower and subtract the triangular parts that stick out. The math I used are based on homogeneous coordinates for the lines, such that a line $L=[A,B,C]$ has an equation equal to $[A,B,C]\cdot[x,y,1]=0$ or $Ax+By+C=0$.

Two points $P$ and $Q$ join with the line $\rm{JOIN}(P,Q)=\left[P_2-Q_2,Q_1-P_1,P_1 Q_2-P_2 Q_1\right]$ where $P=(P_1,P_2)=(x_P,y_P)$ and the similarly for $Q$.

Then two lines $L$, $K$ intersect at $\rm{MEET}(L,K)=\left[\dfrac{L_2 K_3-K_2 L_3}{L_1 K_2 - L_2 K_1}, \dfrac{K_1 L_3 - L_1 K_3}{L_1 K_2 - L_2 K_1}\right]$ where $L_1$, $L_2$ and $L_3$ are the components of line $L$ and similarly for $K$.

The area of a triangle with vertexes $A$, $B$ and $C$ is

$ \rm{AREA}(A,B,C) = \frac{1}{2}\left( (B_1-A_1)(C_2-A_2)-(C_1-A_1)(B_2-A_2) \right)$ that comes from the derivation of the cross product of the vectors $B-A$ and $C-A$.

In addition, the following is true:

  1. The folding line has coordinates $L=[-\sin\alpha,\cos\alpha,0]$.

  2. Point $E$ has coordinates $E=(-\frac{b}{2}\cos2\alpha+\frac{a}{2}\sin2\alpha,\,-\frac{b}{2}\sin2\alpha-\frac{a}{2}\cos2\alpha)$

  3. Point $T$ has coordinates $T=(\frac{b}{2}\cos2\alpha+\frac{a}{2}\sin2\alpha,\,\frac{b}{2}\sin2\alpha-\frac{a}{2}\cos2\alpha)$

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    You could go to wolframalpha.com and plot it yourself. The equations for $S_1$ and $S_2$ are given above.2012-05-04