You have been given a transparent glass cube, which has a small hole at the center of one of the faces. How can you fill the cube with water occupying exactly one third of its volume?

Edit: The source doesn't offer any other clarification, but presumably the only tool we have are the cube and some mechanism for pouring water in and out (not necessarily at a constant rate). Also, you may assume idealized conditions.

1. This puzzle is taken from this webpage, where it is attributed to Abhishek Tiwari.

2. I do not know the solution to this.

  • $\begingroup$ does the cube have 6 faces ,or one of them is open $\endgroup$
    – Amruth A
    Commented Dec 23, 2016 at 6:34
  • $\begingroup$ @AmruthA Most likely, it has all 6 faces; unfortunately, the source I mentioned above doesn't provide any additional clarification. $\endgroup$
    – Ankoganit
    Commented Dec 23, 2016 at 6:38
  • $\begingroup$ What tools are available for this question? I get that it's a logic problem, but.... just a source of water? Because the obvious solution is "measure it," and I don't see a clear reason why that shouldn't be the answer. $\endgroup$
    – user20
    Commented Dec 23, 2016 at 7:16
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    $\begingroup$ I am looking for a geometric solution - one that only relies on the state of the water in the cube, and not on any measurements over time. Here is what I know: 1) A cube can be divided into 3 equal pyramids, 2) The hole is cut in the center of a face - therefore it also shows both midlines for that face, 3) Water has enough surface tension to supply its own plane, and therefore would always occupy a 3-dimensional geometric space, and 4) The cube is glass for a reason - it needs to be transparent to make geometric observations within. $\endgroup$ Commented Dec 23, 2016 at 7:33
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    $\begingroup$ Love this puzzle, @Ankoganit, and still wonder if it has an overlooked solution without any external aids (other than gravity). Am also euphemistically puzzled by the lack of votes of approval by viewers who spared their time to write answers (for which they expect votes of approval?) and by the votes of disapproval by viewers who didn't bother to indicate what seemed wrong with this beguilingly difficult puzzle. $\endgroup$
    – humn
    Commented Dec 23, 2016 at 19:37

6 Answers 6


As the other answers show, it is easy to fill 1/6th of the cube:

Tilt the cube so that it is balanced on one corner, and fill it until the water level reaches the three corners adjacent to the corner it is standing on. If you consider one of the outer faces as its base, the base has an area of 1/2 and height 1 (assuming a unit cube) and hence volume h*A/3 = 1/6.

You can now use this to work out how far much further to fill the cube:

Place the cube flat on one face. The water reaches 1/6th the height of the cube, and you want the water level to be twice as high. Tilt the cube so that it still rests on one edge, and tilt it until the water level goes through another horizontal edge. The water forms a triangular prism - one side of the water surface has doubled in height relative to the (formerly) vertical edge. Mark the height the point the water level has reached, simply by holding a finger there. Place the cube back flat on a face, and your finger marks the height you need the water to reach when you top it up.

  • 1
    $\begingroup$ Glad to illustrate such a neat (no spills, even) solution. Wish I had thought of it too. $\endgroup$
    – humn
    Commented Dec 30, 2016 at 23:12

Tried looking at volumes obtainable by lining up a fill level with at least 3 well-defined points (vertices and/or the hole) but only found ways to fill 1 6, 1 4, 1 2 or more of the cube’s volume.

So with a drop of and...

...more than a drop of oil...

a. Pour in oil so the cube is 1 6 full — level with 3 corners while the cube is standing on one corner.

b. Add water to 1 2 fill the cube — level with 4 corners while the cube is resting on an edge. The water occupies 1 2 − 1 6 = one third of the cube and settles beneath the oil, which is lighter and forms a separate top layer, as they do not mix.

Now the puzzle statement’s “fill the cube with water occupying exactly one third” could be interpreted in different ways:

c1. Two thirds of the cube should be empty and contain only air.

Carefully pour out that top layer of oil.

c2. The cube should be completely full and water should be a third of the contents.

Top off the cube with anything that does not mix with water, such as more oil.

  • 1
    $\begingroup$ Love this idea. +1 :-) $\endgroup$
    – Ankoganit
    Commented Dec 23, 2016 at 13:33
  • $\begingroup$ That's awesome :) $\endgroup$
    – Rubio
    Commented Dec 23, 2016 at 19:34
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    $\begingroup$ Or $c_3$: One third of the cube is filled water and the rest of the cube doesn't matter. Just leave the oil and water in place. $\endgroup$
    – Lawrence
    Commented Dec 24, 2016 at 1:46

EDIT: Answer below is only valid for a 'hole' which is an entire side of the cube. Upon closer inspection of the question, we need a small hole in the center of the side. Leaving answer in case it's any inspiration.

Alright, trying to build a structured approach:

  • Do we know of any way to easily create a shape that holds 1/3 of the cube?
  • Can a said shape be created by using a plane (= water surface) and the sides of the glass cube in an 'easy to verify' way?

As for the first part:

Yes it can. We know that for pyramids, the volume is given as Sxh/3 (with S the surface of the bottom, and h the height. Put simply, use the bottom square as base surface, and put the top on one of the top corners.

As for the second part:

Can we find a position in which we can pour water from the 'open end', which gives us such a pyramidal content in the cube? Once again, we can. There's a (fairly well known) way to split one cube into three pyramids. As for how that would look, I suggest a look here: https://www.youtube.com/watch?v=65t5cPmt7qU . To mimic this shape, all one needs to do is hold the cube so that two opposing corners are aligned at the same height, and one of the other 4 corners has to be 'as far up as possible'. Just for completion, take into account that the 'holey-side' is on the top half, not the bottom half.

  • $\begingroup$ You can't fill the cube with water in the shape of one of those pyramids, because those pyramids have two unbounded faces each, and only one of those faces can be the water level. You'd need some kind of forcefield to hold the other face in. $\endgroup$ Commented Dec 24, 2016 at 0:12

With a big enough bucket of no precise size, I could do this.
Let's start by calling the length of the cube's sides $\large\bf{s}$.

Tilt the cube so that three corners are the same height above the ground, and a fourth points straight down. (The other 4 corners mirror this; one will point straight up, the last 3 will be at a single height above the ground.)

The three equal height corners and the downward-pointing corner form an inverted triangular pyramid. Its base is an equilateral triangle whose sides are the diagonals of the three downward-facing cube sides. As diagonals of a cube face, the sides of the pyramid's base each have length $\bf{s\sqrt2}$. The (non-base) sides of the pyramid are the cube sides between the three equal height corners and the bottom-most corner; since they are just cube sides, their lengths are each $\bf{s}$.

Now partially fill the cube with water

until the water is level with the base of your inverted pyramid. How much water does this take?
The volume of a triangular pyramid is $\bf{V=\frac13 Bh}$, where $\bf{B}$ is the area of the pyramid's base and $\bf{h}$ is the pyramid's height.

The base of our pyramid is an equilateral triangle. We'll find many useful equations at Wikipedia: for example, we know that the area is found by using $\bf{A=\frac14a^2\sqrt3}$ where $\bf{a}$ is the side length. In our case, $\bf{a=s\sqrt2}$, so $\bf{A=\frac14(s\sqrt2)^2\sqrt3 = \frac14(2s^2)\sqrt3 = \frac{s^2\sqrt3}2}$, giving

The height of our pyramid extends from the midpoint of its base perpendicularly to the apex corner. The midpoint of the base triangle sits thusly, where $h_a$, $h_b$, and $h_c$ intersect:
enter image description here
For ease of reference, the base midpoint (where $h_a$, $h_b$, and $h_c$ meet) we'll call $\bf{M}$, and the midpoint of side $AB$ we'll call $\bf{P}$.

The pyramid's height $\bf{h}$ will be the third side of a triangle whose other two sides are an edge of the pyramid (e.g. between $B$ on the diagram and the pyramid's apex), and the line between that same edge and the midpoint of the base (e.g. between $B$ and $\bf{M}$). This is a right triangle, as the pyramid's height is perpendicular to its base.

Wikipedia tells us $\beta$ is 60°; from this, we see a 30°/60°/90° triangle on the diagram: $\Delta{B\bf{PM}}$, with $\angle{\bf{M}}B\bf{P}=30°$ and $\angle{\bf{PM}}B=60°$. Wikipedia also tells us the height from the center of each side to $\bf{M}$ is $\bf{\frac{h}3}$ where $\bf{h=\frac{\sqrt3}2a}$. Here, $\bf{a}$ is (still) $\bf{s\sqrt2}$, so we find $\bf{\overline{PM}=\frac{h}3=(\frac13)(\frac{\sqrt3}2a)=(\frac13)(\frac{\sqrt3}2s\sqrt2)=\frac{s\sqrt6}6}$, so by the ratio for 30°/60°/90° triangles, we know $\overline{B\bf{M}}\bf{=\frac{s\sqrt6}3}$.

We can now find $\bf{h}$ as we know the other two sides of the right triangle it is in.
- Side 1 is $\bf{h}$.
- Side 2 is $\bf{=\frac{s\sqrt6}3}$.
- Hypotenuse is $\bf{s}$, as it is a pyramid (non-base) edge and thus a cube side.
Pythagorean's now gives us

On the home stretch now ...

$\bf{V=\frac13 Bh = (\frac13)(\frac{s^2\sqrt3}2)(\frac{s}{\sqrt3})=\frac13(\frac{s^3}2)=\frac12(\frac{s^3}3)}$

Now, we wanted to fill exactly $\bf{\frac13}$ the cube's volume with water. And we hopefully know that the cube's volume is simply $\bf{s^3}$. So we want an amount of water equal to $\bf{\frac{s^3}3}$.

We just found our pyramid has exactly half the desired amount!

  • We just partially filled the cube with a volume of water $\bf{\frac12(\frac{s^3}3)}$.
  • Pour that water into our arbitrarily sized but big enough bucket.
  • Partially fill the cube in exactly the same way as before.
  • Adjust the cube so the hole is on top.
  • Pour the contents of the bucket carefully into the hole.

You now have the cube filled one third of its own volume with water.

  • $\begingroup$ If there's a way to do this without resorting to a holding bucket, I'd love to see it. $\endgroup$
    – Rubio
    Commented Dec 23, 2016 at 11:38
  • $\begingroup$ No bucket, but a different aid . . . $\endgroup$
    – humn
    Commented Dec 23, 2016 at 12:49
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    $\begingroup$ That's a rather complicated way to show the pyramid is 1/6 of the total volume. Taking one of the right-angled faces as the base, its base is half a square, so has area s^2/2. It's height is a full edge of the cube, so is s. Therefore its folume is h*A/3 = s^3/6. $\endgroup$ Commented Dec 23, 2016 at 13:22
  • $\begingroup$ Where were you when I was working this out? hehe Visualizing the cube was hard for me for some reason, and I didn't think to try the pyramid in a different orientation. $\endgroup$
    – Rubio
    Commented Dec 23, 2016 at 19:36

We can use the volume formula of a pyramid. Start filling it, then point one of the cube's vertices on the holey face downwards, tilting it so that the water level is on the diagonal of the faces around the vertex. When the water forms such a "perfect" right-angled triangle pyramid, its volume is 1/6th of the whole cube's. Then pour the remaining water and do it again to get 1/3rd of the cube's volume's worth of water out of the cube. Then fill the cube with that water, with the holey face pointing upwards.

  • $\begingroup$ Thanks for the explanation. As for whether we can reuse previously measured water, strictly speaking that's not allowed (but I appreciate trying different approaches) , but Rubio has done something similar in his answer with a bucket. $\endgroup$
    – Ankoganit
    Commented Dec 24, 2016 at 4:45

Fill up the cube completely, make the cube downward, and measure the time that the water takes to flow out through the hole completely due to the gravity of earth. Take this time and divide it by 3 and let this be $x$ secs. So, fill again the cube and let the water flow through the hole under gravity for time, $x$ secs which was calculated before, now the volume will be 1/3 rd of it.

  • 2
    $\begingroup$ 1. I'm not very good in physics, but my intuition says the rate of flow under gravity won't be constant; it would vary depending on the height of the liquid remaining at any given instant. 2. Presumably we don't have any other measuring instrument(s) than implies in the statement; otherwise we could just take a graduated cylinder. $\endgroup$
    – Ankoganit
    Commented Dec 23, 2016 at 6:43
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    $\begingroup$ This solution is based on an assumption that water will leave the cube at the exact same rate as it came in. There is no way to know if this is true. Unless a separate hole is cut to regulate air pressure, a stream of air will be flowing against the water stream in both directions. $\endgroup$ Commented Dec 23, 2016 at 6:44
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    $\begingroup$ NOT accounting for air flow, the more water in the cube the faster it will flow out, so the rate of outflow will definitely not be constant. But since air has to replace water, we quickly get into too many unknowns to decide what happens. (For a small enough hole, no water comes out at all.) $\endgroup$
    – Rubio
    Commented Dec 23, 2016 at 6:49
  • $\begingroup$ It's generally not good to rely on assumptions to solve problems involving geometry and/or physics. It's a huge leap to assume that the rate of water flow is exactly the same in both directions. There's simply too many variables that would have to be considered. $\endgroup$ Commented Dec 23, 2016 at 6:56
  • $\begingroup$ Actually, given that this is a mathematics puzzle (and not intended to be a physics puzzle), I think it's fair to assume ideal conditions and ignore random irregularities of air pressure etc. But this particular approach still doesn't work because of Rubio's point above. $\endgroup$
    – Ankoganit
    Commented Dec 23, 2016 at 7:00

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