# Partition a cube into 6 congruent tetrahedra

A tetrahedron is a solid shape with four triangular faces, not necessarily regular or identical.

Show how to partition a solid cube into 6 tetrahedra that are congruent, meaning identical up to translation, rotation, and possibly reflection.

(It's easy to find solutions by Googling, but please solve it on your own.)

• Must the splitting be done purely with cuts that go all the way through the cube? Just checking. – Dennis Meng Apr 28 '15 at 6:11
• @DennisMeng No, you can cut the piece however you want. – xnor Apr 28 '15 at 6:12
• I am too dumb for this: I only get four congruent ones with length 1,1,sqrt(2) and one with sqrt(2),sqrt(2),sqrt(2). – Alexander Apr 28 '15 at 10:37
• Is wasting some material allowed...or do we need all material of cube to be used equally for all tetrahedron...? – user2408578 Apr 28 '15 at 13:33
• @user2408578 You need to use all of the cube; it can't be "6 congruent tetrahedra and a little left over," it has to be "6 congruent tetrahedra" – Dennis Meng Apr 28 '15 at 16:09

This is an extension of this post about trisecting a cube. A cube is shown to be divided into 3 right-angle pyramids. Then, 6 right-angle tetrahedra can be made by cutting each pyramid diagonally through its square face. This dissection results in 3 left-hand and 3 right-hand tetrahedra, so they are not quite identical.

A better explanation uses a diagram from this post about points on a cube. As shown below, the cube can be divided into 6 tetrahedra by making 3 planar cuts. Each planar cut must follow the long diagonal of the cube (shown in red). Again, this dissection results in 3 left-hand and 3 right-hand tetrahedra.

To obtain 6 identical tetrahedra, the cutting planes need to be rotated by 30 degrees around the longest diagonal. The planar cuts are then bisecting the vertices of the included hexagon which is shown below. The 6 resulting tetrahedra are identical.

For most people, visualizing this dissection is not easy. Perhaps the best way is to imagine standing the cube on a corner (in this case the far corner) so that the longest diagonal is now a vertical axis. By looking straight down this axis, the cube's 6 peripheral corners are seen and the 3 planar cuts can be made to either bisect the corners of the cube or to bisect the vertices of the included hexagon.  Follow-up on Len’s answer. Think of the regular hexagon (2D) as the diagonal view of a cube (3D) as shown below. Divide the hexagon by three diagonals. The corresponding operation in 3D cuts the cube into 6 identical pieces. These identical pieces have 6 edges and thus tetrahedra. I don't agree that the solution above, rotating the cutting planes by 30 degrees around the diagonal, will give identical tetrahedra. Instead, the planes will cut through six of the twelve edges of the cube in the middle of those edges, i.e., at the vertices of the hexagon. This will add six new vertices to the cube, bringing the total number of vertices to fourteen.

Thus, the cuts described will produce six identical (but irregular) pentahedra. Each will have five vertices, consisting of three of the cube's original vertices and two of the new ones. Six pentahedra meet at the two cube vertices at the ends of the long diagonal. Each pentahedron doesn't share its third cube vertex. Two pentahedra meet at each of the six new vertices.

The total count of vertices can be written as 14 = 6*2/6 + 6 + 6*2/2, i.e. six times two cube vertices divided by six-way sharing plus six unshared cube vertices plus six times two new vertices divided by two-way sharing.

• Welcome to SE Puzzling! You're right that the bit in the other solution about rotating the planes by 30 degrees does not turn the (valid) solution with 3 mirror image tetrahedron pairs into a solution with 6 identical tetrahedra. When you have a bit more reputation (50 I think), you will be able to add this to the other post as a comment, which is a more appropriate place for it. – Jaap Scherphuis Apr 8 '17 at 2:37

If ABCDEFGH is a cube measured as being 1 unit as volume, then corresponding line segments measure:

1 = AB = BC = CD = DA = AF = BG = CH = DE = EF = FG = GH = HE,

2^.5 = ca. 1.414 = AC = BF = DF = DH = HF, and

3^.5 = ca. 1.732 = CF, so,

Pairs:

Smallest 2 volumes each with faces of 3 right isoceles triangles [BGF, BGH and FGH, and DEF, DEH and EFH] and 1 equilateral [BFH and DFH]; 3 mutually perpendicular unit 1 leg measures and 3 hypotenuses of ca. 1.414, BFGH and DFHE are congruent;

Mid-sized 2 volumes each with faces of 2 right isoceles triangles [ABC and ABF, and ADC and ADF] and 2 triangles of unit, square and cube diagonal sides [BCF and ACF, and ACF and DCF]; ADCF and ABCF are reflective; and

Largest 2 volumes each with faces of 1 right isoceles triangle [BCH and DCH], 1 equilateral triangle [BFH and DFH] and 2 of unit, square diagonal and cube diagonal sides [BCF and CFH, and CFH and DCF]; DCHF and BCHF are reflective.

Triangles ACF and HCF measure sides 1 unit, 1 square diagonal ca. 1.414, and one cube diagonal ca. 1.732.

Concerning congruencies:

Is this a dissection of a cube into six congruent tetrahedra, or a pair of congruent tetrahedra triplets? https://m.youtube.com/watch?v=ffnVCEAcOns

Tetrahedron: (Lengths) 1; 2^.5; 3^.5 FCBA: AB, AF, DE; AC, BF; CF FCAD: AD, DC, DE; AC, DF; CF FCDE: CD, DE, EF; CE, DF; CF FCEH: CH, EF, EH; CE, FH; CF FCHG: CH, FG, GH; CG, FH; CF FCGB: BC, BG, FG; BF, CG; CF

As EF, HF, GF, BF, AF and DF sequentially alternate about CF, respectively, each corresponding tetrahedron; through a common non-symetrical triangle; reflects its complementary tetrahedron: .. FCBA; ACF; FCAD; DCF; FCDE; ECF; FCEH; HCF; FCHG; GCF; FCGB; BCF; ..*

So, because the observed triangular non-symetries noted above preclude mirror image congruence, alternate tetrahedra, FCBA, FCDE and FCHG are congruent triplets, as FCAD, FCEH and FCGB are, also.

*.. an "ellipset" pair connects a loop. Pending further review, see

https://www.ellipset.info Rectangular prism ABCDEFGH cut across rectangle ACHF yields 2 triangular prisms, ABCHGF and ADCHEF, so, these, cut across triangles BFH and DHF net 2 tetrahedra, BFGH and DFHE, and 2 residual pyramids, ABCHF and ADCHF. These pyramids cut across triangles BCF and DCF net 4 tetrahedra, ABCF, BCHF, ADCF and DCHF, for total net 6 tetrahedra.  No one of the 6 tetrahedra is regular in this partition paradigm.Indeed, the only two equilateral triangles are BFH and DFH. Alternatively, BD is 1 edge of regular tetrahedron BDFH, with 4 right tetrahedra DBCH, DBAF, BFGH and DEFH also capable of perhaps more symmetrically partitioned cube ABCDEFGH into 4 congruent right tetrahedra packed around 1 regular tetrahedron, BDFH. 