Knowing that a pyramid volume is computed as 1/3 of base multiplied by the height, how many pyramids may be constructed by connecting the corners of a cube to create pyramids with volume 1/3 of the cube volume?

Using a computer an exhaustive search of 4 out of 8 and 5 out of 8 could produce the result, but than it is not interesting, so explanation is expected.


3 Answers 3


I think,

26 pyramids


Consider a cube with side length $1$. Suppose that one of the sides of a pyramid coincides with a side of the cube. Fix that side as the base. Then picking the other vertex remaining as the apex (if one is remaining) would give volume zero, and picking one on opposite side would give the volume $\frac{1}{3}$ times the area of the base as the height is $1$. For this to be $\frac{1}{3}$ the base area must be $1$ which is only possible when we pick the whole side.
There are $24$ pyramids satisfying the above as there are $6$ possible bases and $4$ possible apices for each base.
Now consider all pyramids with an edge of length $1$. Note that for such a pyramid to not fall in the already considered category, the only possible vertices for that pyramid is the diagonally opposite pair to the already chosen pair. But choosing that gives a rectangle with volume zero, so there are no possibilities here.
Now consider all pyramids with an edge of length $\sqrt{3}$. But whatever vertex we pick next, the pyramid would have an edge of length $1$, so no possibilities here either.
So the only possibility remaining is the tetrahedron with all edges $\sqrt{2}$. Using the tetrahedron volume formula we see that this gives the required $\frac{1}{3}$ volume. Note that $2$ of those exists on the cube, as picking one edge uniquely determines all the other edges and there are $6$ edges to the tetrahedron compared to $12$ possible $\sqrt{2}$ edges, giving a total of $26$ pyramids.

  • $\begingroup$ You select a diagonal. How many per edge? $\endgroup$
    – Moti
    Sep 1, 2019 at 14:53
  • $\begingroup$ @Moti Can you please clarify more? :) Are you asking for each edge how many pyramids contain that edge? $\endgroup$
    – user47134
    Sep 1, 2019 at 15:07
  • $\begingroup$ The answer 2 for the tetrahedron is not sure. A corner has 3 diagonals connected to it - each creates one, and all different since they have at least one diagonal that is not participating in other tetrahedron. Given 12 diagonals, how many selections of 2 out of 12 possible? How many 6 out of 12? $\endgroup$
    – Moti
    Sep 1, 2019 at 15:21
  • $\begingroup$ I'm not trying to consider the tetrahedrons as a group in the solution. I'm only considering the regular tetrahedron: all other tetrahedrons fall under one of the categories I considered possibilities with. Also even though your argument is right, volume 1/3 condition eliminates most of them. I'm sorry if I'm misunderstanding though. $\endgroup$
    – user47134
    Sep 1, 2019 at 15:32
  • $\begingroup$ If you are referring to the regular tetrahedron, a corner does not create 3 different of those, only one, because a tetrahedron on one diagonal includes the other two too. (Try drawing a cube and checking if this is not clear) $\endgroup$
    – user47134
    Sep 1, 2019 at 15:40

There are plenty

First you can take any of the faces as a base and take one of the remaining corners as the apex. The base area is 1 and the height is 1, giving a volume of 1/3.

There are 6 faces and for each 4 corners to choose from. This gives 6 x 4 = 24 pyramids.

And then

You can join any 2 opposite edges and form a rectangle of size 1 by $\sqrt{2}$ cutting the cube diagonally in two. Take any of the 4 remaining corners as the apex.

The base area is $\sqrt{2}$, the height of the pyramid is always $1\over\sqrt{2}$. This again gives a volume of 1/3.

There are 6 ways to choose opposite edges and each time 4 corners to choose from for the apex. Again it gives 6 x 4 = 24 pyramids.

And finally

The two tetrahedra that you can inscribe in the cube happen to also have volume 1/3.

It might look complicated to compute the volume. Luckily the space around the tetrahedron is composed of 4 pyramids. The bases are half a face and the height is 1. This gives a volume of 1/6 for each surrounding pyramid. Since there are 4 of them, they have a total volume of 2/3. Since these and the central tetrahedron add to 1, the volume of the tetrahedron is exactly 1/3.

There are 2 possible ways to inscribe a tetrahedron in a cube.

So the total is:

24 + 24 + 2 = 50 pyramids.


5 points:

Pyramids with 5 vertices are square pyramids. The base of the pyramid must obviously be one of the faces of the cube, and the apex of the pyramid can be any one of the 4 vertices in the opposite face of the cube. This means there are $6*4=24$ such pyramids.

4 points:

Pyramids with 4 vertices are triangular pyramids, i.e. tetrahedra. If three of the vertices lie in the same face of cube, then the pyramid's triangular face is half the cube's square face. For its volume to be $1/3$rd of the cube, its height would then have to be twice the height of the cube, so it is not possible for the apex to be another vertex of the cube.
So no three vertices of the pyramid can be in the same face. This leaves only two ways to choose four cube vertices:
a) The endpoints of two opposite edges of the cube. These lie in the same plane so do not form a tetrahedron.
b) A set of four non-adjacent vertices. There are two such sets. These form a regular tetrahedron with edge length $\sqrt{2}$ assuming the cube is a unit cube. The volume of this tetrahedron is $\frac{a^3}{6\sqrt{2}}=\frac{2\sqrt{2}}{6\sqrt{2}}=\frac{1}{3}$.

The total number of such pyramids is therefore:


  • $\begingroup$ You select a diagonal. How many per edge? $\endgroup$
    – Moti
    Sep 1, 2019 at 14:54
  • $\begingroup$ @Moti: I have no idea what you mean. $\endgroup$ Sep 1, 2019 at 15:19
  • $\begingroup$ My mistake. I mean per corner, not edge. $\endgroup$
    – Moti
    Sep 1, 2019 at 15:22

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