# Bath towel on the rope: minimize the area of self-intersection of a folded rectangle

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 $a<b$. 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.

• I assume alpha is the angle that the line through the center makes with one of the sides. Which side? – Matthew Conroy Feb 25 '11 at 19:14
• The rectangle is bent along the line which goes through the diagonal crossing under alpha angle. ( I have a picture, but cannot upload it) – Mikhail G Feb 25 '11 at 19:29
• Am I correct that you have given us the expression $S$ for the area as a function of $\alpha, a ,b$? All what you want to know is which value of $\alpha$ minimizes $S$ for a certain $b/a$? – Fabian Feb 26 '11 at 14:21
• Yes, but please note that if $a<<b$, then you can bent $\pi/4$. If you have a=b, then I got: $\alpha=\pi/8$ What is beetween? – Mikhail G Feb 26 '11 at 19:17
• Agree, $a/b \approx 0.81$ there are two shapes in the crossing intersaction. Have you ever thought about the sizes of paper towels? Why do they have the particular sizes? I searched for eg German paper towels and found that $a/b>0.81$ (very close to $0.9$ or even $1$. So, do you think the manufacture can slightly reduce a/b with no changes in property of the current paper towels but adding a new one (while bending)? – Mikhail G Feb 28 '11 at 17:59

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 $a<b$ 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 $T=(\frac{b}{2}\cos2\alpha+\frac{a}{2}\sin2\alpha,\,\frac{b}{2}\sin2\alpha-\frac{a}{2}\cos2\alpha)$

• Thank you. Could you plot a picture how the area depends of the fold angle and a/b? – Mikhail G May 4 '12 at 19:00
• You could go to wolframalpha.com and plot it yourself. The equations for $S_1$ and $S_2$ are given above. – ja72 May 4 '12 at 21:50