Cubes and Boxes have Rupert’s passages in every direction¹¹1This article has been accepted for publication in American Mathematical Monthly, published by Taylor & Francis.

Statement 1 follows from the visual geometric observation that in an infinite wedge bounded by two half planes with an acute angle, no solid cube can reach the edge of the wedge on its convex side. Indeed, if this would be possible, then a continuity argument would imply that the contact is possible also with one face of the cube on one of the bounding half planes, a contradiction.

Statement 2 is a well known property of the cube. Figure 2 becomes a proof without words, once it is noticed that the dot product of a unit edge vector and the vertical unit vector has two geometric meanings. On the one hand, it is equal to the area of the shadow of that face which is perpendicular to the unit edge vector. On the other hand, it is equal to the length of the perpendicular projection of the unit edge vector on the $z$-axis.

Statement 3 must be also an old known property of the cube. Here we present a short proof via matrices. Let $\overrightarrow{P},\overrightarrow{Q}$ and $\overrightarrow{R}$ be the three normal vectors with third coordinates p, q and r. Let $A$ be the matrix whose row vectors are these normal vectors. The linear transformation $A$ is an isometry so for all column vectors $\alpha,\beta\in\mathcal{R}^{3}$, $(A\alpha)^{T}\cdot(A\beta)=\alpha^{T}\cdot\beta$. Equivalently, $(\alpha^{T}A^{T})\cdot(A\beta)=\alpha^{T}\cdot\beta$. Thus, $A^{T}A=I$, which in view of the row-column multiplication means that the column vectors of $A$ form an orthonormal basis. In particular, the third column vector is a unit vector, thus $p^{2}+q^{2}+r^{2}=1$. ∎

Let $A$ be one of the vertices of the hexagonal projection $H$ of the cube. Let $V$ be that vertex of the cube whose vertical projection is $A$. Let us label the three unit edge vectors emanating from $V$ by $\overrightarrow{P},\overrightarrow{Q}$ and $\overrightarrow{R}$. If $p,q$ and $r$ denote the absolute values of the $z$-coordinates of these vectors, then the lengths of the projections of these vectors are $\sqrt{1-p^{2}},\sqrt{1-q^{2}}$ and $\sqrt{1-r^{2}}$. Let us orient the hexagon $H$ so that one pair of opposite sides is horizontal and the lengths of these sides equal to $\sqrt{1-p^{2}}$. Since all angles of $H$ are obtuse (Part 1 of Lemma 1) we may assume the labelling of side lengths and the areas of the sub-parallelograms are as shown on Figure 3.

In order to do a little computation, we introduce an $xy$-coordinate system centered at $A$ so that the horizontal line is the $x$ coordinate axis. Let $B^{\prime}$ and $C^{\prime}$ be the adjacent vertices of $A$ in $H$. Choose points $B$ and $C$ on the horizontal sides of $H$ at distance $1$ from $A$. Let $B$ the point at distance $b$ from $B^{\prime}$ and $B^{\prime}$ at distance $b^{\prime}$ from the $y$-axis. Distances $c$ and $c^{\prime}$ are introduced analogously for points $C$ and $C^{\prime}$. We have

Let $D$ be the vertex which completes $A,B,C$ to a square. Since both $H$ and $ABCD$ are centrally symmetric, $D$ lies on the horizontal line through the vertex of $H$ opposite to $A$. Now, the first and the last equivalent inequalities mean that $D$ is inside of $H$ if $q+r\leq 1+p$. Since all $p,q,r\in(0,1)$, this holds, if $p$ is the largest or the second largest among $p,q,r$, which proves the first two properties listed in Lemma 2. A small shift followed by a small rotation of these unit square ensures that there is a unit square which lies in the interior of the hexagon. ∎

Let $R$ be a rectangle of sides $a$ and $b$, $a\leq b$. If the plane of the cross section is horizontal, then Lemma 3 is obviously true. Otherwise, denote the angle between the plane of the tube’s base $S$ and the plane of the cross section $P$ by $\alpha$ (Figure 4). $P$ is a parallelogram with opposite sides at least at distances $a$ and $b$ respectively. Introduce an $x,y$ coordinate system in the horizontal plane of $R$ so that the $x$ axis belongs also to the plane of $P$. Let us fold the plane of the cross section into the horizontal plane around the $x$-axis. It is clear that there is a congruent image $P^{*}$ of $P$ so that the linear map $(x,y)\rightarrow(x,\frac{1}{\cos\alpha}y)$ takes the rectangle $R$ to parallelogram $P^{*}$. We denote the image of a point $p$ under this mapping by $p^{*}$.

Let us label the vertices of rectangle $R$ by $1,2,3$ and $4$ in the order of their $y$ coordinates (Figure 1). It is easy to see that $\frac{1}{\cos\alpha}>1$ implies that $P^{*}$ has obtuse angles at vertices $2^{*}$ and $3^{*}$. Depending on how the diagonal $2^{*}3^{*}$ splits the obtuse angle at $2^{*}$, we distinguish two cases:

Let $B$ be a rectangular box with edge lengths $a\leq b\leq c$. Shorten the longest edges of $B$ to $b$. It is enough to show that the new box $B^{\prime}$ contains a rectangle of dimensions $a,b$. Let hexagon $H$ be a shadow of $B^{\prime}$. Two of the six sides of $H$ are projections of edges of length $a$, while the remaining four sides are projections of edges of length $b$. To depict this property we label the sides and their lengthes with $pr(a)$ or $pr(b)$ on Figure 5. We have two pairs of opposite vertices which are endpoints of $pr(a)$ sides. In view of Lemma 2 at least one of these two pairs have vertices, say $A$ and $A^{\prime}$, so that moving a cube with edge length $a$ to the corner of the pre-image of $A$, the projection of the cube contain a square whose vertex is $A$. Now it is easy to see that one can extend one pair of parallel sides of this square to $b$ so that the new rectangle is still contained in $H$, what we wanted. A small shift followed by a small rotation of this rectangle ensures that there is a unit square which lies in the interior of the hexagon. ∎

Assume that a homothetic box $\lambda B$ can be passed through a stationary box $B$ with sides $a\leq b\leq c$. Although Theorem 2 is obviously true if $\lambda\leq 1$, we present an argument which works for all $\lambda$. On Figure 6, which illustrates the proof, we choose a $\lambda$ much smaller than $1$ so that the figure becomes less crowded and the proof can be better understood. Let $H_{1}$ be the shadow of the stationary rectangular box $B$ and $H_{\lambda}$ be the shadow of the traveling homothetic rectangular box $\lambda B$. We proved already that the hexagon $H_{\lambda}$ contains a rectangle $\lambda R$ of sides $\lambda a$ and $\lambda b$. Since both $H_{1}$ and $\lambda R$ are central symmetric, the rectangle $\lambda R$ can be shifted in center position while remaining inside of the hexagon $H_{1}$. Now, choose two adjacent vertices, say $U$ and $V$, of this relocated rectangle. Since $U,V$ are inside of the shadow of $B$, we can choose two points $U^{*}$, $V^{*}$ inside of the box $B$ directly over $U$ and $V$. Finally, reflect $U^{*}$, $V^{*}$ through the center of the box $B$ two get a parallelogram $P$ in box $B$. $P$’s shadow is the rectangle of sides $a$ and $b$. According to Lemma 3, parallelogram $P$ contains a rectangle of sides $a$ and $b$. ∎