I got a puzzle for christmas.
12 congruent wooden sticks.
When assembled, they appear to interpenetrate.
Cross section of stick is 60° diamond.
Ends are mitered—miter axis along long diagonal of diamond.
The two mitered ends of a stick are parallel.
Sticks come in sets of 3s.
A set is of one sort of wood.
4 kinds of wood.
The three sticks of a set meet at a shared edge which is a body diagonal of a cube.
By “edge” I mean edge joining the two mitered ends of a stick.
All edges are the same length.
⊦ Sticks of a set are parallel to each other.
Symmetry of Puzzle is that of oriented cube, except for wood types and hidden voids and cleavage planes.
Note that an end of a body diagonal comes to three faces of a cube.
An end of a stick touches one of the three faces connected to the stick’s body diagonal’s end.
It is the stick’s vertex at the end of the miter axis that touches the cube’s face.
The touching vertex of the same stick at the stick’s other end belongs to the opposite edge of that stick.
A rotation of 120° about a body diagonal permutes the three sticks in that diagonal’s set.
The view of the puzzle projected on a cube face is a bunch of squares rotated by 45°.
We adopt coordinates so that the cube’s corners are at <±4, ±4, ±4>.
A stick edge touches a face at <x, y, 4> where x ≅ 2 and y ≅ 0.
(Other touchings follow from symmetry.)
The other end of the edge to <x, y, 4> is <−x, 4, y>.
The opposite edge of that stick runs from <x', −x', x'> to <−x', x', −x'>.
Note that this edge is along the body diagonal of the set.
I think that x = x'.
Assuming that x = x' = 2 and y = 0 we have that out favorite stick is the convex hull of {
<2, 0, 4>, <3, −1, 3>, <2, −2, 2>, <1, −1, 3>,
<−2, 4, 0>, <−1, 3, −1>, <−2, 2, −2>, <−3, 3, −1>}.
The first four and last four in this vertex list each describe a mitered end of the stick.
This fits all of the observations.
The group of an oriented cube is generated by
<x, y, z> → <y, z, x> and
<x, y, z> → <y, −x, z>.
I just realized that each edge of a stick including those at the ends,
is parallel to some body diagonal.
This is true of the model above.
Can we find bricks in the above?
The convex hull of
{<2, 0, 4>, <3, −1, 3>, <2, −2, 2>, <1, −1, 3>,
<1, 1, 3>, <2, 0, 2>, <1, −1, 1>, <0, 0, 2>}
and three more each translated from the original by <−1, 1, −1> partition the stick into 4 congruent affine cubes.
Alas the bricks of other sticks, even of the same set, are not parallel—thus they do not partition the puzzle.
Just one of the sticks has a cleavage plane and part of that stick is thus the key piece.
I orient the puzzle so that piece touches <2, 0, 4> and remove the key piece.
That frees the sticks at <0, 2, 4> and <0, 4, 2> to move down (-y) but not all of the way out.
If I displace <0, 4, 2> by <2, −2, −2> then stick at <3, 4, 1> can move by <2, −2, −2>.