What's the smallest convex shape that can wrap a cube?

Question

As shown in this question, a square with area $8$ can wrap a cube. What is the smallest convex shape by area that can wrap a cube in a similar way?

This question was motivated by the linked question.

Answer 1

I don't know whether area can be $6$ or not. However, by using very slim rectangles, one can make the area as close to $6$ as one wish.

First some conventions:

• Let $\mathcal{S}$ be the set of alphabets $\{ S, L, R \}$.

• For any $N \in \mathbb{Z}_{+}$, the cartesian product $\mathcal{S}^N$ can be viewed as the set of strings of length $N$ formed from alphabets $\mathcal{S}$.

• For any $M, N \in \mathbb{Z}_{+}$, $u \in \mathcal{S}^M$, $v \in \mathcal{S}^N$, we will use $uv$ to denote the string in $\mathcal{S}^{M+N}$ obtained by concatenating $u$, $v$ together and $u^N$ for the string in $\mathcal{S}^{MN}$ obtained by concatenating $N$ copies of $u$.

Consider rectangle $[0,1] \times [0,N]$ in the plane.

For any $X = x_1\cdots x_N \in \mathcal{S}^N$ create a folding of above rectangle by the rules:

• If $x_k$ is $R$, make a valley fold along the line $y = k-1$ and a mountain fold along the line $y - x = k-1$.
• If $x_k$ is $L$, make a valley fold along the line $y = k-1$ and a mountain fold along the line $y + x = k$.
• Otherwise, $x_k$ is $S$ and do nothing to $k^{th}$ square of rectangle.

For any integer $n > 2$, let $N = 2n+4$. Consider folding corresponds to the string $$X_{n} \stackrel{def}{=} SRS^{n-1}LSLS^{n-1}R$$ Since first character of $X_n$ is $S$, the $1^{st}$ square $[0,1]^2$ of rectangle isn't moved. It is not hard to see the resulting figure cover the $n \times 2$ rectangle $[0,n] \times [0,2]$ with the original top edge of rectangle moved from $(0,N)\to (1,N)$ to $(0,2) \to (1,2)$.

When $n = 2k$ is even, we can

1. stack $k$ copies of $X_{2k}$ on top of each other and folding $Y_k \stackrel{def}{=} X_{2k}^k$ gives us a square of side $n$.
2. stack $k$ copies of $X_{8k}$ on top of each other and folding $Z_k \stackrel{def}{=} X_{8k}^k$ gives us a rectangle of dimension $4n \times n$.

Now take $2$ copies of $Y_k$ and one copy of $Z_k$ and glue them in the order $Y_kZ_kY_k$, the resulting figure is a $T$-hexomino consists of $6$ squares of side $n$.

Scaling the figure down by a factor $\frac1n$ and notice $T$-hexomino is one of the unfolding of the cube. We find for even $n$,

one can fold a $\frac1n \times 6(n+1)$ rectangle of area $6 + \frac6{n}$ to wrap/cover the surface of an unit cube.

Since $n$ can be as large as one want, we can make the area as close to $6$ as one like.

Answer 2

This is an extended comment I posted to the question, showing the solution for area 7 times the cube face area:

While I have no proof that this is the minimum, examining the properties of three faces sharing a corner tells me it would be logical for it to be.

This was created by taking one of the cube cube nets with convex hull having area 8 (the two in the bottom right corner; all others have area 8.5 (the two center ones in the bottom row) or 9 times the cube face area), and moving polygonal face parts (here, triangles with edge ratio 1:3) to minimize the area of the convex hull for the resulting wrapping.