Kvant Math Problem 1559
Let the given plane be $\Pi$.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 9m11s
Source on kvant.digital
Problem
Does there exist a cube in space such that the distances from its vertices to a given plane are equal to 0, 1, 2, 3, 4, 5, 6, and 7?
V. V. Proizvolov
59th Moscow Mathematical Olympiad (1996), Tournament of Towns (Spring 1996)
Exploration
Let the given plane be $\Pi$. For each vertex of the cube, consider its signed distance to $\Pi$. The ordinary distances are the absolute values of these signed distances.
A cube has three edge directions. If $u,v,w$ are the edge vectors from one vertex, then every vertex has the form
$$A+\varepsilon_1u+\varepsilon_2v+\varepsilon_3w, \qquad \varepsilon_i\in{0,1}.$$
Signed distance to a plane is an affine function of the point. Hence the signed distances at the eight vertices are
$$d_0+\varepsilon_1a+\varepsilon_2b+\varepsilon_3c,$$
where $a,b,c$ are the changes of signed distance along the three edge directions.
Thus the eight signed distances form the vertices of a parallelepiped in one dimension; explicitly they are
$$d_0,\ d_0+a,\ d_0+b,\ d_0+c,\ d_0+a+b,\ d_0+a+c,\ d_0+b+c,\ d_0+a+b+c.$$
The problem asks whether the absolute values of these eight numbers can be exactly
$$0,1,2,3,4,5,6,7.$$
A first thought is that perhaps one can choose the plane so that all vertices lie on the same side. Then the signed distances themselves would be $0,1,\dots,7$. But among the eight numbers above, opposite corners satisfy
$$x+(x+a+b+c)=(x+a)+(x+b+c),$$
and analogous relations. For the set ${0,1,\dots,7}$, the only pairings with constant sum $7$ are
$$0+7,\ 1+6,\ 2+5,\ 3+4.$$
This looks promising.
Suppose the signed distances are exactly $0,1,\dots,7$. Then the smallest value must be $d_0=0$, and the three adjacent vertices have distances $a,b,c$, three distinct numbers from ${1,\dots,7}$. The largest value is $a+b+c=7$. Since the remaining vertex values are $a+b,a+c,b+c$, they must be ${4,5,6}$. The only possibility is ${a,b,c}={1,2,4}$, which indeed produces all numbers $0,\dots,7$.
Now the crucial question is whether such $a,b,c$ can arise from a cube. If the edge length is $s$ and $n$ is a unit normal to the plane, then
$$a=s(u\cdot n),\quad b=s(v\cdot n),\quad c=s(w\cdot n),$$
where $u,v,w$ are mutually orthogonal unit vectors. Hence
$$a^2+b^2+c^2=s^2.$$
For $a,b,c=1,2,4$ this gives
$$1^2+2^2+4^2=21=s^2.$$
There is no contradiction. In fact any triple satisfying $a^2+b^2+c^2=s^2$ can be realized by choosing $n$ with coordinates $a/s,b/s,c/s$ in the orthonormal basis $u,v,w$.
This suggests that the answer is yes.
The delicate point is showing rigorously that the unsigned distances must actually coincide with the signed distances. Since one of the distances is $0$, the set of signed distances already contains $0$. If some signed distances were negative, then because the absolute values are $0,1,\dots,7$, the sum of all signed distances would have magnitude strictly less than $0+1+\cdots+7=28$. On the other hand, for the configuration ${0,1,\dots,7}$ the affine structure works perfectly. A simpler route is to construct an example directly. Existence alone is required.
Problem Understanding
We are given a plane in space and ask whether there exists a cube whose eight vertex-to-plane distances are exactly the numbers
$$0,1,2,3,4,5,6,7.$$
This is a Type D problem, an existence problem.
The core difficulty is to understand the restrictions imposed on distances from cube vertices to a plane. Signed distances vary affinely, so the eight values must be generated from three increments corresponding to the three edge directions.
The answer is yes. The reason is that signed distances at the vertices of a cube can be made equal to
$$0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7$$
by choosing the three edge-direction increments to be $1,2,4$.
Proof Architecture
The first lemma states that signed distances from the vertices of a cube to a plane have the form
$$d_0+\varepsilon_1a+\varepsilon_2b+\varepsilon_3c,$$
with $\varepsilon_i\in{0,1}$; this follows from the affinity of signed distance.
The second lemma states that for a cube of edge length $s$, the increments $a,b,c$ satisfy
$$a^2+b^2+c^2=s^2;$$
this comes from expressing the plane normal in the orthonormal basis of edge directions.
The third lemma states that any numbers $a,b,c$ satisfying
$$a^2+b^2+c^2=s^2$$
can actually occur; one chooses the plane normal with coordinates $a/s,b/s,c/s$.
The hardest point is the realization lemma, because one must show that the geometric data can indeed be chosen so that the prescribed increments occur simultaneously.
Solution
Let the cube have edge length $s$. Choose one vertex $A$, and let $u,v,w$ be the three edge vectors issuing from $A$. Then $u,v,w$ are mutually orthogonal and satisfy
$$|u|=|v|=|w|=s.$$
Let $\Pi$ be the given plane, and let $n$ be a unit normal vector to $\Pi$. The signed distance from a point $X$ to $\Pi$ is an affine function of $X$; hence there exists a constant $d_0$ such that the signed distance of
$$A+\varepsilon_1u+\varepsilon_2v+\varepsilon_3w$$
from $\Pi$ equals
$$d_0+\varepsilon_1a+\varepsilon_2b+\varepsilon_3c,$$
where
$$a=u\cdot n,\qquad b=v\cdot n,\qquad c=w\cdot n.$$
Consequently the eight signed distances at the vertices are
$$d_0,\ d_0+a,\ d_0+b,\ d_0+c,\ d_0+a+b,\ d_0+a+c,\ d_0+b+c,\ d_0+a+b+c.$$
Take
$$a=1,\qquad b=2,\qquad c=4.$$
Then these eight numbers become
$$d_0,\ d_0+1,\ d_0+2,\ d_0+4,\ d_0+3,\ d_0+5,\ d_0+6,\ d_0+7.$$
Choose $d_0=0$. The signed distances are then exactly
$$0,1,2,3,4,5,6,7.$$
It remains to show that such values of $a,b,c$ can be realized by a cube and a plane.
Let
$$s=\sqrt{1^2+2^2+4^2}=\sqrt{21}.$$
Choose an orthonormal basis $e_1,e_2,e_3$ and set
$$u=se_1,\qquad v=se_2,\qquad w=se_3.$$
Define
$$n=\frac1s(1,2,4).$$
Since
$$|n|^2=\frac{1^2+2^2+4^2}{s^2} =\frac{21}{21}=1,$$
the vector $n$ is a unit vector and may serve as a normal to a plane.
Then
$$u\cdot n=1,\qquad v\cdot n=2,\qquad w\cdot n=4.$$
Hence the corresponding increments of signed distance are precisely $1,2,4$.
Finally, choose the plane to pass through the vertex $A$. Then the signed distance of $A$ is $0$, and the eight signed distances of the cube vertices are
$$0,1,2,3,4,5,6,7.$$
All of them are nonnegative, so the ordinary distances from the vertices to the plane are the same numbers.
Thus there exists a cube whose vertex-to-plane distances are
$$0,1,2,3,4,5,6,7.$$
The required cube is obtained by taking edge length $\sqrt{21}$ and a plane whose unit normal has coordinates proportional to $(1,2,4)$ in the orthonormal system of edge directions.
$$\boxed{\text{Yes, such a cube exists.}}$$
Verification of Key Steps
The first delicate step is the description of the vertex distances. Signed distance to a plane with unit normal $n$ has the form
$$X\mapsto X\cdot n+c.$$
For a vertex
$$A+\varepsilon_1u+\varepsilon_2v+\varepsilon_3w,$$
the value becomes
$$(A\cdot n+c)+\varepsilon_1(u\cdot n)+\varepsilon_2(v\cdot n)+\varepsilon_3(w\cdot n),$$
which is exactly
$$d_0+\varepsilon_1a+\varepsilon_2b+\varepsilon_3c.$$
No additional geometric assumptions are used.
The second delicate step is the realization of $a=1$, $b=2$, $c=4$. For a cube of edge length $s$, writing the unit normal in the orthonormal edge basis as
$$n=\alpha e_1+\beta e_2+\gamma e_3,$$
gives
$$a=s\alpha,\qquad b=s\beta,\qquad c=s\gamma.$$
The condition $|n|=1$ becomes
$$a^2+b^2+c^2=s^2.$$
With $a,b,c=1,2,4$, choosing $s=\sqrt{21}$ satisfies this exactly.
The third delicate step is passing from signed distances to ordinary distances. Because the plane is chosen through the vertex $A$, the smallest signed distance is $0$. The other seven signed distances are
$$1,2,3,4,5,6,7,$$
all positive. Hence taking absolute values does not change any number.
Alternative Approaches
One may begin with the combinatorial structure of the eight vertex values. Any affine function on a cube is determined by its increments along the three edge directions, so the vertex values must form
$$d_0+{;0,a,b,c,a+b,a+c,b+c,a+b+c;}.$$
Requiring these values to be exactly $0,1,\dots,7$ forces
$${a,b,c}={1,2,4}.$$
After identifying this pattern, one checks that a cube can realize these increments because
$$1^2+2^2+4^2=21.$$
This yields the same construction.
The presented proof is preferable because it constructs the cube and plane directly. The affine description and the geometric realization appear naturally in the same argument, and no separate combinatorial reconstruction is required.