Two cubes in unit cube

Question

A cube of side one contains two cubes of sides $a$ and $b$ having non-overlapping interiors. How to prove the inequality $a+b \le 1?$ The same question in higher dimensions.

Answer 1

Here's a proof of the (much weaker) fact that $a+b\leq\sqrt{n}$. Maybe someone can find a way to elaborate upon it to get a solution of the problem.

Let $n$ be the dimension we are considering. Let $C_n\subset\mathbb{R}^n$ be the cube of side $1$. Let $a$ be the side of the first cube (we will call it $A$), $b$ the side of the second cube ($B$). By the classical theorems of convex separation, there is an $(n-1)$-plane $P_{n-1}$ separating the two cubes (since they are disjoint). By translating everything if necessary, assume that $0$ is contained in $P_{n-1}$. Let $L$ denote the line through $0$ that is orthogonal to $P_{n-1}$ and denote by $\pi:\mathbb{R}^n\rightarrow L\cong\mathbb{R}$ be the projection on $L$. It is easy to prove that the projection of a cube of side $x$ on a line is an interval of maximal length $x\sqrt{n}$. Let $\ell$ be the function associating to an interval in $\mathbb{R}$ its length. Then: $$\ell(\pi(C_n))=:z\leq\sqrt{n}$$ $$\ell(\pi(A))=:x\geq a$$ $$\ell(\pi(B))=:y\geq b$$ Thus: $$a+b\leq x+y\leq z\leq \sqrt{n}$$ where the second inequality is given by the fact that $P_{n-1}$ separates $A$ and $B$.

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A possible elaboration of the above would be treating higher dimensional planes instead of the $1$-plane (= straight line) $L$. For example we could obtain an $(n-1)$-plane $Q$ by taking some vector $v\in P_{n-1}$ and defining $Q$ to be its orthogonal complement. Then the projections of $A$ and $B$ on $Q$ would be disjoint, since the $(n-2)$-plane $P_{n-1}\cap Q$ would separate them in $Q$. Maybe some work in this sense could give us better estimates or some way to prove the initial statement by induction.

Answer 2

The following proof basically says the following: (1) you can assume the cubes are in opposite corners, and (2) you can then assume they are aligned with the unit cube, in which case the theorem follows easily.

Preliminary Fact: If we have an arrangement of two n-cubes, A and B, in $\mathbb{R}^n$, then for each standard basis direction, $\vec{e}_i$, we can translate A in either the $+\vec{e}_i$ or the $-\vec{e}_i$ direction forever without intersecting B (this is seen by the convexity of cubes). If A can move in one direction forever, then B can move in the opposite direction forever.

By this fact, if A and B are within the unit cube, we can go through each direction and slide each of them in opposite directions so that they end up in antipodal corners of the unit cube. WLOG let A be in the corner $(0,\ldots,0)$ and B be in the corner $(1,\ldots,1)$. So A cannot slide in the negative basis vector directions without leaving the box, and B cannot slide in any of the positive basis vector directions.

This sliding doesn't change the size of A or B, so we can assume that $a+b$ is the largest possible sum of side lengths ($a$ is side length of A and $b$ is side length of B).

Now for the main technical lemma.

Lemma: Take $x_C = \sup \{y | \{y,y,\ldots,y\}\in C\}$ where $C$ is some cube inside the unit cube which intersects the main diagonal. Then $x_C \geq c$, where $c$ is the side length of $C$.

What this says is that if we take the point $\{x_C,\ldots,x_C\}$ on the intersection of $C$ and the main diagonal which is furthest from the origin, then the cube $[0,x_C]^n$ has longer side length than $C$. We use this to show that A and B must be aligned with the unit cube if they are to have the largest side length.

Proof sketch of lemma: We will sketch the proof since it's an ugly calculation, which can be done if interested. So we can reduce the problem to saying that if an affine unitary transformation T takes the unit cube to the first quadrant, $[0,\infty)^n$, and can't be slid any further in the negative basis directions without leaving the first quadrant, then $x_{T([0,1]^n)} \geq 1$. We can get rid of the affine part by looking at only unitary transformation of $[-1/2,1/2]^n$ and defining $x_{T[-1/2,1/2]}$ w.r.t. $(\min(T_1[-1/2,1/2],\ldots,\min(T_n[-1/2,1/2])$ instead of the origin. Parameterizing the unitary transformations, we can set this up as an optimization problem, which you have to solve for general n.

So all the calculation is encoded in that lemma. The rest is pretty simple. We have A and B at opposite parts of the unit cube, and they are largest possible total side length sum. Now if A is not aligned with the unit cube, we can replace it by a cube at least as big that is aligned by the lemma. It's easy to see this new cube doesn't intersect B. So assume A is aligned. Now, we can use the same argument to replace B by an aligned cube as well (except the lemma is used w.r.t. the antipodal point of the origin this time). Thus we have two aligned cubes with total side length at maximum. From here it's not so bad to see the sum of side lengths must be at most 1.