Smallest possible dimensions of a piece of paper after one fold.

Question

So I've been thinking, can someone explain mathematically if I started with a square of paper dimensions 1x1, what the smallest single side dimension could be reached after one fold. My gut instinct is that after only one fold there's no way to make the paper less than half the height/width. When I say single side dimension, imagine drawing a bounding box around the folded paper and taking the smaller of the two side dimensions.

Answer 1

We fold the $1\times 1$ paper once and cover it with a rectangle of sides $a,b$.

We want $x=\min\{a,b\}$ such that the smaller side is minimized.

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Showing $x\le 1/2$ is easy. Fold the paper in half and align the opposite edges. We get exactly a $1/2$ by $1$ rectangle. The smallest rectangle that covers a rectangle, is that rectangle.

Now we need to also prove $x\ge1/2$ and we are done. Consider that the paper is cut along the folding line. We get two pieces. It is clear that the rectangle covering the folded paper must at least be able to cover each of these pieces individually.

If the cut goes over two neighbouring sides, it is clear that we need at least a $1$ by $1$ covering rectangle which is not minimal as we already know $1\not \le 1/2$. In the limiting case where cuts are along the opposite corners, we have $x\ge \frac{\sqrt2}{2}\approx 0.7071\not \le 0.5$ by rotating to align the height of the triangular piece to the bounding rectangle, which still isn't the true lowest lower bound. Hence, assume otherwise: the cut goes over two opposing sides of the paper.

Both cut pieces $i=1,2$ will be right angle trapezoids with sides $1,a_i,b_i$ where $a_i||b_i$, and $a_i,b_i$ are perpendicular to side $1$, and such that $1=a_1+a_2=b_1+b_2=1$. (The area of larger of the two trapezoids is then at least $A\ge 1/2$.)

WLOG $a_1\ge b_1$. This implies that $b_2\ge a_2=1-a_1$.

We need to find the minimal covering rectangles for these two right angled trapezoids. WLOG consider that the covering rectangle is parallel to the coordinate axes, and that the trapezoid is rotated by some $\phi_i\in[0,\pi/2]$. We want to find $\phi_i$ such that the rectangles $r_i$ of sides $p_i,q_i$ covering the right angle trapezoids $i\in\{1,2\}$ have minimal smaller sides $\min_{\phi_i}\{p_i,q_i\}$.

If the rotation is $0$ or $\pi/2$ then sides $1,a_i,b_i$ are parallel to the sides of the covering rectangle $r_i$. Therefore, the larger side of the covering rectangle is $p_i=1$ and the smaller is $q_i=\max\{a_i,b_i\}$. We want to be able to cover at least the larger of the two trapezoid pieces:

$$x\ge\min\{\max\{a_1,b_2\},1\}=\min\{a_1,1-a_1\}=1/2$$

This is a true lower bound if for $\phi_i\in(0,\pi/2)$, we can't do lower than $1/2$.

Assume we can. WLOG $q_i\ge 1$ is larger and $p_i\le 1/2$ is a better (or equal) smaller side. We can imagine extending $q_i$ infinitely since we only care about minimizing $p_i$, the smaller dimension. Now it is clear that $p_i$ is minimized by minimizing the height of the highest vertex of the trapezoid, which is precisely when the fourth side of the trapezoid is parallel to $q_i$.

The larger of the minimal dimensions $p^*_i$ (of the two $i=1,2$) needed to cover both is the one for some $i$ such that $a^*_i=\max\{a_i,b_i\}$. The $p^*_i$ is the height from point where $a^*_i,1$ sides meet, to the fourth side. The $1$ side needs to be extended beyond where $b^*_i,1$ meet to form a right angled triangle. The extension is by $\frac{a^*_i}{\sqrt{{c^{*}_i}^2-1}}$ (See the similar triangles). Similarly, the extension of the fourth side in the same direction is $\frac{a^*_i}{\sqrt{1-{c^{*}_i}^{-2}}}$. The $c^*_i$ value is the length of the fourth side. $$c^*_i=\sqrt{(a^*_i-b^*_i)^2+1}$$

The area of the right angled triangle obtained by the extension is:

$$ \frac12 p^*_i \left(c^{*}_i+\frac{a^*_i}{\sqrt{1-{c^{*}_i}^{-2}}}\right)=\frac12 a^*_i \left(1+\frac{a^*_i}{\sqrt{{c^{*}_i}^2-1}}\right) $$

Where we have now $p^*_i=f(a^*_i,b^*_i)$ which we want to minimize to get $a$. Notice we have $a^*_i\ge 1-a^*_i,a^*_i\ge 1-b^*_i$ and $a^*_i \ge b^*_i\ge 0$ because these are the maximal dimensions among the two pieces.

The global minimum is precisely when $a^*_i=b^*_i=0.5$ giving a $1/2$ by $1$ rectangle, implying the lower bound $x\ge 1/2$. This is precisely our initial $1/2$ by $1$ covering rectangle case we had as an upper bound at the beginning. (You can use wolfram to double check.)

That is, lower bound = upper bound = $1/2 = x$, regardless of rotation.

This finishes the proof.