# Can six “$1\times1\times1$ pyramids” fit inside a $1\times1\times1$ cube? [duplicate]

Consider the pyramid $\left \{ (x, y, z): x,y,z\geq 0, x+y+z\leq 1 \right \}$.

This is a pyramid that has $3$ pairwise-orthogonal edges, all of length $1$ (just like a corner of a room): From now on I'll call it a "unit pyramid" (UP). One can easily check that the volume of a UP is exactly $\frac{1}{6}$. So that raises the question:

Can we take six UPs and fit them all perfectly inside a unit cube?

When I thought about it, my initial answer was an obvious yes. So I tried to imagine such arranging, but I couldn't. I was quite surprised since I usually have a very good spatial vision. Then I tried drawing a solution and failed.

Since I'm not tech-savvy enough to use a 3D graphics software, I decided to take this problem into the real world.

So here's a unit pyramid: It's composed of 3 edges of length 1 (which I set to be $5$ cm) and 3 edges of length $\sqrt2$ ($\approx 7$ cm).

Now, there's only one reasonable way to attach another one to it, if our goal is a unit cube: (since now we have our square base). Now I added all the edges necessary to form a cube: And as your 3D mind can see, now we have four unit pyramids: It's pretty hard to see the region that has stayed uncovered by our 4 UPs, so here it is, emphasized: And what do you know - that's a regular tetrahedron! A quick reality check: a regular tetrahedron with edges of length $\sqrt 2$ has a volume of $\sqrt 2 (\sqrt 2)^3/12=4/12=1/3$, which suits the fact that we entered 4 UPs inside.

But now, I'm pretty certain that you can't fit another 2 of them inside. To be sure, I constructed them seperately (last picture, I promise!): On the left there's a UP, and on the right - our regular tetrahedron (would you believe that one is twice the volume of the other?) Now you'll just have to trust me that it won't fit.

Well, that convinced me that the mission of fitting 6 unit pyramids inside a unit cube is impossible, but of course I wouldn't call it a mathematical proof. Does anyone have a more mathematically-convincing argument? I'd be curious to hear your ideas and thoughts about this.

## marked as duplicate by Rahul, Community♦Apr 9 '17 at 17:22

• Nice art, and camera too. – Jaideep Khare Apr 9 '17 at 16:59
• There is a dissection of the cube into six pyramids, but of a different shape than your pyramids: papercraftetc.blogspot.com/2013/09/… – Michael Lugo Apr 9 '17 at 16:59
• Michael, I'm actually aware of that, found that when I googled this problem. But thanks! And thank you too, Jaideep – 35T41 Apr 9 '17 at 17:00
• I think the answer is "no", as seen here: math.stackexchange.com/questions/26052/… – Michael Lugo Apr 9 '17 at 17:05
• The angles are 90 (three per pyramid), 60 (3), and 45 (6). The cube faces must be composed of pyramid faces. The only way to the get the 24 right angles of the cube is there the right and 45 degree angles of the pyramids. Bit of fussing and the only way to do that is the way you did. So it's a matter of identifying to remaining "hole" solid inside the cube. Although the hole solid has the volume of two pyramids (I believe) it'll be easy (once you actually see the danged thing) to show it isn't possible. (I think). – fleablood Apr 9 '17 at 17:16

However, each of the pyramids contributes either three complete edges of the cube, or none at all (not even partially). As the cube has twelve edges, there must be exactly four pyramids of the first kind. But such pyramids also contribute $3/2$ to the surface, so together they contribute all the surface. This is only possible if each surface is cut into two triangles and hence the apices of these pyramids are on diagonally opposite vertices of faces. This enforces that the apices form a tetrahedron (formed from four of the eight vertices of the cube).