How can you fill the cube to one-sixth its volume?

[Loren Booda]

Given only water and a cubic container with one open face, how can you fill the cube to one-sixth its volume?

 

[Redbelly98]

Without thinking too much about this, I imagine the solution has to do with tilting the cube so that the volume between the lowest point of the cube, and the lowest point on the open face, contains 1/6 of the total volume. Also, there should be some simple visual check by which one can tell that the cube is in the desired orientation.

I'll mull this over some more, but probably somebody else will come up with the actual solution before I do.

 

[Mark44]

The seemingly obvious answer is that you fill it to a depth equal to 1/6 the length of a side. Is there something I'm missing here?

 

[Xitami]

 

[berkeman]

Sweet.

 

[The Chaz]

I'm team captain and I pick Xitami first.

 

[Redbelly98]

The seemingly obvious answer is that you fill it to a depth equal to 1/6 the length of a side. Is there something I'm missing here?

How do you tell it's 1/6 full? You have no ruler, so you could only get "eyeball" accuracy.

 

[Mark44]

Fill it all the way full, then divide into two equal parts, which we can do by balancing them.
Then divide one of the halves into three equal parts by distributing into three smaller buckets, and balancing them two at a time.
Pour one of these three buckets into the cube and it will be one-sixth full, and the height of the layer will be 1/6 of the length of a side of the cube.

Of course, that sort of violates the constraints of the problem...

 

[Jerbearrrrrr]

Tilt it until the surface of the water is a triangle, whose sides are diagonals of the faces of the cube.
[edit] oh someone else did it already. doh.
Are there even any other 'perfect' fractional volumes you can measure?
Except 1/2.

 

[Redbelly98]

It is not obvious to everyone, but Xitami seems to have solved the problem. Look at the image in his post #4, and imagine that the red+ blue face is the open one.

Fill the cube more than 1/6 full of water, then tilt the cube so that the yellow corner is straight down and water is running out. Adjust the cube's tilt so that the level water surface just touches the corners indicated by the yellow volume, so that the amount of water is equal to the yellow volume shown.

I am actually having difficulty visualizing why this is equal to 1/6 of the total, but there is probably a simple way to do this. (Too lazy to pick up a pen and paper and sketch this out right now.)

 

[Loren Booda]

Can you not conjure a more mathematically rigorous answer? Some of you may have come close, but there is a simple geometric formula you seem to be missing.

 

[Jerbearrrrrr]

Volume of a pyramid?
1/3*1*1/2*1*1 in this case.

(edit: the 1's are the height and perpendicular dimensions of the base triangle)
(another edit: is there an elementary proof of that formula? ie, the fact that it doesn't matter where you put the "top" point of the pyramid, as long as it's h above the base. not that the calculus is difficult but...)

 

[Loren Booda]

Yea Jerbearrrrrr! The way I put it is:

Fill the cube so the surface of the water intersects one corner on the open face and two corners diagonal to each other on the opposite face. [Remember, the volume of a pyramid, V=(1/3)(base)(height).]

 

[Loren Booda]

Are there even any other 'perfect' fractional volumes you can measure?

Or, is anybody willing to try planes intersecting three or more points on other perfect "solids"?

 

[HallsofIvy]

I don't see why that should be necessary. The volume of a tetrahedron with edges of length 1 has volume 1/6. That shows that Xitami's solution is correct.