Question

How to solve this problem?

A rhombic dodecahedron (a polyhedron made of 12 rhombi, each shaped such that one diagonal is $\sqrt{2}$ times the length of the other one) of side length $n$ is tiled by rhombohedrons (a parallelepiped made of 6 rhombi, each similar to the rhombi in the rhombic dodecahedron) of side length $1.$

Prove that it is possible to label each rhombohedron an integer from $0$ to $n$ inclusive such that:
1) If two rhombohedrons share a face and have the same orientation, then they have the same label.
2) If two rhombohedrons share a face and have different orientation, then the difference of their labels is equal to the distance from some vertex of the rhombic dodecahedron to a point where the plane of their common face intersects an edge of the rhombic dodecahedron.
3) The sum of the labels for the rhombohedrons of a particular orientation is the same across all orientations.

Answer 1

Um I'll get help

Answer 2

@k1ngofpadlet

Answer 3

Following diagram of a rhombic dodecahedron:

[asy]
import three;

size(4cm);
currentprojection = perspective(-4/5,-7/10,2);


triple A = (-1,-1,-1);
triple B = (-1,-1,1);
triple C = (-1,1,-1);
triple D = (-1,1,1);
triple E = (1,-1,-1);
triple F = (1,-1,1);
triple G = (1,1,-1);
triple H = (1,1,1);

draw(A--B--F--E--cycle);
draw(C--D--H--G--cycle);
draw(A--C--G--E--cycle);
draw(B--D--H--F--cycle);
draw(A--E,dashed);
draw(B--F,dashed);
draw(C--G,dashed);
draw(D--H,dashed);

label("$n$", (A+B)/2, W);
label("$n-1$", (A+C)/2, NW,fontsize(8));
label("$n-1$", (A+E)/2, SW,fontsize(8));
label("$n-1$", (B+F)/2, SE,fontsize(8));
label("$n-1$", (B+D)/2, NE,fontsize(8));
label("$n-1$", (C+G)/2, NW,fontsize(8));
label("$n-1$", (D+H)/2, NE,fontsize(8));
label("$n-1$", (F+H)/2, SE,fontsize(8));
label("$n-2$", (C+E)/2, NW,fontsize(8));
label("$n-2$", (D+F)/2, NE,fontsize(8));
label("$n-2$", (G+E)/2, SW,fontsize(8));
label("$n-2$", (H+F)/2, SE,fontsize(8));
[/asy]

Start by labeling the rhombohedron at vertex $A$ with $n.$ We will then label the rhombohedrons adjacent to it, and so on, until we have labeled all the rhombohedrons.

Note that there are two types of rhombi in the rhombic dodecahedron: those that are in the same plane as face $ABFE,$ and those that are in the same plane as face $CDHG.$ For simplicity, you will refer to these as type $X$ and type $Y$ rhombi, respectively.

Since the rhombohedrons are all congruent to each other, you only need to cull the labels of the rhombohedrons adjacent to each other. You only need to set the labels of the rhombohedrons adjacent to vertex $A,$ since you can then elongate this labeling to the entire polyhedron.

The rhombohedron adjacent to vertex $A$ sharing a face with the rhombohedron labeled $n$ must be labeled $n-1,$ since they apportion a type $X$ rhombus. Similarly, the rhombohedron adjacent to vertex $A$ sharing a face with the rhombohedron labeled $n-1$ must be labeled $n-2,$ since they apportion a type $Y$ rhombus. The remaining two rhombohedrons adjacent to vertex $A$ additionally share a type $Y$ rhombus with the rhombohedron labeled $n,$ so they must withal be labeled $n-1.$

Utilizing this pattern, you can label all the rhombohedrons adjacent to vertex $A.$ you then move on to vertex $B,$ which is adjacent to three of the rhombohedrons adjacent to $A.$ We perpetuate in this manner until you have labeled all the rhombohedrons.
You will have constructed a valid labeling, but you require to prove that it satiates the three conditions verbally expressed in the quandary.

Here we go:

Condition 1 is satiated because if two rhombohedrons share a face and have the same orientation, then they are adjacent to the same rhombohedron and hence have the same label.

Condition 2 is gratified because if two rhombohedrons share a face and have different orientation, then they are adjacent to different rhombohedrons and hence have labels that differ by the distance from some vertex of the rhombic dodecahedron to a point where the plane of their prevalent face intersects an edge of the rhombic dodecahedron. To optically discern why this is veridical, note that any two rhombohedrons that share a face have a mundane rhombus, and this rhombus lies in a plane that intersects an edge of the rhombic dodecahedron. The distance from a vertex of the rhombic dodecahedron to this plane is identically tantamount to the label of the rhombohedron adjacent to that vertex.

Condition 3 is satiated because the labeling is consummately symmetric with deference to the orientation of the rhombohedrons. That is, if we perform a symmetry of the rhombic dodecahedron (such as a reflection or a rotation), then the labeling is unchanged.

Ergo, we have shown that it is always possible to label the rhombohedrons as described in the quandary.

You're welcome. Hope it helps.

Answer 4

This solution is completely wrong and looks like it was generated by ChatGPT.

Answer 5

A rhombic dodecahedron is a corner projection of a hypercube, and the tiling becomes some unit hypercubes placed inside the hypercube. Label each unit hypercube with the number of hypercubes below it.