# How to fill 1/3 of the cylinder?

You are given 3 containers - pictured in this order below:

• a box with side 2, height 1, and a cone with base radius 1 and height 1 in the middle.
• a box with side 2, height 1, and a half sphere with radius 1 in the middle
• a cylinder with base radius 1 and height 1

The cylinder is full with water. By moving the water between the containers have only 1/3 of the cylinder filled.

Then have the cylinder filled to 2/3. Not very challenging but might be of interest.

• Which side of each box is open? Is the first box the same as a solid box with the cone shape removed (same for the second box)? If so, isn't it a kind of math-trivia question about the volumes of a cone, a sphere, and a cylinder? – Bubbler Jan 20 at 7:24
• Both boxes are open up. To one a half sphere added and into the other a cone. The volumes of the containers are the volume of a box (4) - the volume of the added object (half sphere or cone). You need to have some basic knowledge or use google to finf the volumes of the three (including the box). – Moti Jan 20 at 7:57
• The volumes of the sphere/cone containers is not a nice ratio of the sphere - thus some work is required. – Moti Jan 20 at 7:58
• Which side is "up"? (I guess it's the large side of the box, opposite of the base of a cone or a half sphere, but optical illusion makes it hard to tell in the first glance.) – Bubbler Jan 20 at 8:06
• You are looking into the containers - you need to subtract from the box the added object into it. – Moti Jan 21 at 6:41

The volumes of the boxes from left to right are

$$4-\pi/3$$, $$4-2\pi/3$$, $$\pi$$
$$2.95$$, $$1.91$$, $$3.14$$

I found it easier to do the second task first.

The first two boxes have volumes that are not rational multiples of $$\pi$$, but their difference is. So we need to fill the smallest of the two from the largest to get that difference of $$\pi/3$$. The rest of the liquid is then $$2\pi/3$$, and it is then just moving those two amounts around to get each of them into the cylinder in turn. Note that the middle box is smaller than $$2\pi/3$$, so this moving around takes more moves than you might expect. $$\begin{array}{c|c|c}box 1 & box 2 & box 3\\ \hline 0 & 0 & \pi \\4-\pi/3 & 0 & 4\pi/3-4 \\ \pi/3 & 4-2\pi/3 & 4\pi/3-4 \\\pi/3 & 0 & 2\pi/3 \\ 0 & \pi/3 & 2\pi/3 \\ 2\pi/3 & \pi/3 & 0 \\ 2\pi/3 & 0 & \pi/3\end{array}$$

• pouring 3->1, 3->2, 1->3, 2->1, 3->2 seems to work with fewer pours. – Bass Jan 20 at 11:18
• @Bass Yes, that is a quicker way to get one third directly, but my solution has the advantage of solving both tasks. – Jaap Scherphuis Jan 20 at 12:16
• Actually solving the 1/3 means that the remaining is 2/3. – Moti Jan 21 at 6:44

Let's do the second part first:

Flip the first box over, and push the circular cone into the cylinder. Some water will spill over.

The cylinder will now have exactly 2/3 of water left, because

The base areas and heights of the cone and the cylinder are identical, so the volume of the cone ($$\frac{1}{3}Ah$$) is exactly one third of the cylinder's volume. ($$Ah$$)

Continuing from here, we can get the one third by

catching the overflow from the previous step in the other box, and swapping it back into the cylinder.

or by

repeating the procedure with the other box: pushing the hemisphere, whose volume is $$0.5 \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$$ into the cylinder, whose volume is $$Ah = \pi r^2 \times r = \pi r^3$$ leaves room for exactly one third of the original water in the cylinder.

• Interesting solution, but you might lose some drops in the process. If you fill the cone container and with it the sphere container you are left with 1/3. You move the water around and you are having the 1/3 in the cylinder and 2/3 in the cone container. I will accept Japp solution, since it met the intent of the question. – Moti Jan 21 at 6:49