A regular tetrahedron with one black and three white faces is positioned with its black face at the bottom of a plane. The tetrahedron is tipped several times over an edge and finally reaches the same original position.

Is it possible, that the tetrahedron ended up with a white face at the bottom of the plane at its final position?

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    $\begingroup$ While I can see how one arrives at the answer, I'm nowhere near capable of answering this beautifully. +1 $\endgroup$ Commented Oct 19, 2021 at 15:48

3 Answers 3


It is ...

not Possible.


Imagine that the 4 sides have been unrolled to form a Parallelogram.
Let the sides be numbered 1, 2, 3 & 4 which is shown in the Image.
Now all movements in the track along the X axis will repeat in order and this is fixed.
If a movement moves it in the Y axis, it will move up or down into a new track.
Here, again it can move in this track in the X axis which will again be in same order but shifted & reversed.
Moving back to original track will again shift back & reversed back to normal.
Eventually, the Plane will be tiled with the triangles with numbers 1, 2, 3 & 4.


Triangle Tiling


No matter which number or which colour is selected to be the black side, it will come back to the Starting Position without change.

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    $\begingroup$ Is there anything preventing me from doing a diagonal flip? If not then starting where you numbered $1$; $1$, $2$, $3$, $4$, $2$ (below), $4$ (back towards $1$), finally landing on $3$ where we started (assuming that $1$ is the black face), unless I confused myself somehow. $\endgroup$ Commented Oct 19, 2021 at 16:51
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    $\begingroup$ I'm pretty sure there is something wrong with your picture. If you keep one corner fixed to the ground and revolve around it you should get six triangles two of each colour except the face opposite the fixed corner. $\endgroup$
    – loopy walt
    Commented Oct 19, 2021 at 16:53
  • $\begingroup$ @loopywalt , good catch , will update shortly .... $\endgroup$
    – Prem
    Commented Oct 19, 2021 at 17:35
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    $\begingroup$ @Tacoタコス , My earlier Image was wrong ; I hope it makes sense now. Movements are over edges , not on corners. $\endgroup$
    – Prem
    Commented Oct 19, 2021 at 17:44

No, it is not possible.

This is because

all loops have lengths that are a multiple of six, which is a multiple of three, which traps the tetrahedron "T" to only ever have its original position with the black face underneath.

The tetrahedron is composed of four equilateral triangles and "rolling" it fixes one vertex. Continuing to roll it around that vertex will result in a sequence of equilateral triangles all sharing one vertex on the plane. Our basic geometry tells us that six rolls will form a regular hexagon. But any vertex on T is the intersection of exactly three faces, which will cycle in the same order during that roll: * The first and fourth triangles will be made by the first "inside" the white face, * the second and fifth by the second "inside" white face, * the third and sixth (being the last, the same as the original position) by the black face. The "outside" white face never touches the plane during the rolling, because it is not coincident to the vertex which stays on the plane. If we stop and roll backwards on the loop, we are simply returning to the same position, which is pointless. So for T to reach the same position in one loop, it must finish on the same fave it started from. What about multiple loops? Well, they must eventually connect back to one of the three loops that share the original position, and then that loop must be completed or reversed too. Thus, we can never change the face that must land on the original position.

  • $\begingroup$ +1 , Nice Alternative Thought-Process !! While the Conclusion is valid & the reasoning is Partially valid , I think the reasoning is Incomplete : Yes , all loops have length 6 & Paths which go through these Complete loops will come back to Initial Position, .... But there are ways to go around in Partial loops, hop on to other Partial loops, and go in "Arbitrary Directions" , including "Straight lines" and then come back via other routes, all without having Complete loops. $\endgroup$
    – Prem
    Commented Oct 21, 2021 at 11:11
  • $\begingroup$ All loops have a length that is a multiple of six. You can never return to a position without making a multiple of six rolls, or retreating back over the path used (which means returning to the original state as well as position). Returning to the original position is the definition of a loop, whether it is formed of straight lines or crazy curves. $\endgroup$
    – Nij
    Commented Oct 21, 2021 at 20:05
  • $\begingroup$ You wrote "What about multiple loops? Well, they must connect to one of three loops that share the original position, then that loop must be completed or reversed too." :: (1) With Definition = route is one Big loop, there are no "multiple loops", right ? (2) OK, with the Definition = route is one Big loop, having multiple Small loops, my comment was about Partial Smaller loops being "not completed or reversed". In Starting Small loop, I goto sides 1&2, goto other Partial Small loops, goto Starting Small loop on sides 5&6, goto Starting Position 1 skiping sides 3&4 of Starting Small loop. $\endgroup$
    – Prem
    Commented Oct 22, 2021 at 5:39

Imagine coloring the vertex opposite the black face red. As we roll the tetrahedron around, we keep track of the projection of the red vertex onto the plane. It is easy to see that the red vertex will always be constrained to the set of red points below, which form the vertices of a hexagonal grid. In particular, the red vertex will never end up at any of the blue vertices, which is necessary for the tetrahedron to return to the shaded triangle with a different face on bottom.

enter image description here

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    $\begingroup$ I know this is an old answer, but I think the image is incomplete? Not in a way that makes the conclusion wrong, at least, but I'm fairly certain that for each given vertex projected into the center of a triangle, the triangle directly above it should also have a vertex projected into its center. Take any tetrahedron in one of the presented positions with an upwards red vertex and roll it right, up, and left, essentially moving halfway around a hexagon, and it will have its red vertex facing upwards again. Thus the vertices actually end up forming a triangular grid, not a hexagonal one. $\endgroup$
    – Idran
    Commented Oct 3, 2023 at 18:41
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    $\begingroup$ @JustinHilyard Thank you for the comment! I fixed the picture. $\endgroup$ Commented Oct 30, 2023 at 16:14

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