A visual solution to the cube cutting problem

Question

A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$. How many pieces are there? (No cheese is moved until all three cuts are made.)

This problem was in the AHSME (American High School Mathematics Exam) and also has a solution here on SE. I'm still having a hard time visualizing the 6 pieces of the cube. I do not have access to any 3D modelling software and so far was able to come up with only this messy GeoGebra visualisation:

Could someone show the shapes of the six individual pieces the cube is cut into, in detail?

Answer 1

Here are my illustrations of the six pieces. The pieces are arranged in a hexagon, and pieces which next to each other in the hexagon will meet along a similarly colored triangular face. The color code is this:

• Red: $x=y$ plane.
• Blue: $y = z$ plane.
• Green: $x = z$ plane.
• Gray: Other.

Exception: The $z<y<x$ piece (bottom row right) has a red face, but I did not illustrate the red face it connects to on the $z<y<x$ piece (middle row right). This is because the red face would block the view of the three other faces.

Answer 2

If the point ${\bf r}=(x,y,z)\in C$ does not lie on one of the cutting planes its three coordinate values are different, and conversely. Consider two points ${\bf r}_0$, ${\bf r}_1$ of the cube having the same order of their coordinate values, say $$x_0<y_0<z_0\qquad\wedge\qquad x_1<y_1<z_1\ .$$ The moving point $${\bf r}(t):=(1-t){\bf r}_0+t{\bf r}_1\qquad(0\leq t\leq 1)$$ then has the same coordinate order for all $t\in[0,1]$; hence the segment $[{\bf r}_0,{\bf r}_1]$ does not intersect one of the cutting planes. It follows that ${\bf r}_0$ and ${\bf r}_1$ belong to the same piece of the partition.

Since there are $3!=6$ possible orders among different numbers $x$, $y$, $z$ we have six different pieces. By symmetry all these pieces look alike up to rotation and reflection. It is therefore sufficient to look at one of these pieces, e.g., to the piece $S$ belonging to the order $x>y>z$.

Given $x\in[0,1]$ we are allowed to have $y\in[0,x]$. Therefore all points of $S$ are lying over the triangle $S'=\{(x,y)\>|\> 0\leq x\leq 1, \ 0\leq y\leq x\}$ in the $(x,y)$-plane. From each point $(x,y)\in S'$ emerges a stalk in the $z$-direction containing the points $(x,y,z)$ with $0\leq z\leq y$. In other words, our $S'$ obtains a skew roof $z=y$ that emerges from the interval $[0,1]$ of the $x$-axis and extends triagonally to the peak $(1,1,1)$. The full set $S$ then is a nonregular tetrahedron, as shown in the last of Mike Earnest's figures.

For the other pieces: Three square faces of $C$ meet at $(0,0,0)$. Draw in each of these squares the diagonal beginning at $(0,0,0)$. The three squares are then partitioned into $6$ triangles $S_i'$, and each of these $S_i'$ is the base of one piece $S_i$, like the $S'$ above is the base of $S$.

Answer 3

I think the best way to understand what's happening is to carefully draw a cube, then carefully draw, using different colors, the lines where each of the cutting planes intersects the surface of the cube. Finally, Oscar Lanzi's comment comes in very handy: you'll find that the three cutting planes have a line of common intersection. At this point you have all the edges of all the pieces, and you should then be able to see the six tetrahedra Furthermore, you can compute all the edge lengths. You will find that each tetrahedron is made by piecing together two isosceles right triangles with edge lengths $1$, $1$, $\sqrt{2}$, and two right triangles with edge lengths $1$, $\sqrt{2}$, $\sqrt{3}$. This is something you can actually build by cutting the pieces from card stock and taping them together. Better, I think, than any two-dimensional visualization.

Note that the line Oscar Lanzi mentioned is a three-fold rotation axis of the cube, and that $120^\circ$ rotations cyclically permute each of two sets of three pieces. All six pieces have an edge in common on that axis. The tetrahedra come in left- and right-handed versions, and the handedness alternates as one makes a circuit around the axis.

Added: If one of the isosceles-triangle faces is used as the base of the tetrahedron, then one of the three edges not on the base is vertical. The other isosceles face and one of the non-isosceles faces both abut this vertical edge and make a dihedral angle of $90^\circ$ with the base.

Another thing you can do with these tetrahedra that I think is rather pretty is to arrange eight of them (four of each handedness) into a pyramid with $2\times2$ square base and faces that are isosceles triangles with sides $2$, $\sqrt{3}$, $\sqrt{3}$. The square base will be made of eight isosceles right triangles and the vertical edges of the eight tetrahedra with these right triangles as bases lie along the common line joining the center of the base of the pyramid to the apex.

Six of these pyramids (or $48$ tetrahedra) may be arranged apex-to-apex along orthogonal axes to form $2\times2\times2$ cube. This cube can also be subdivided into eight $1\times1\times1$ cubes decomposed into tetrahedra the original way and variously rotated.

If two such pyramids are joined back-to-back they form an octahedron. Six such octahedra (or $96$ tetrahedra) sharing a vertex and oriented along orthogonal axes in the same way as the six pyramids above form a rhombic dodecahedron.