# How to find half volume of tetrahedron?

Let's say you have a few juice packs that are shaped as regular tetrahedra.

Question. Is it possible to measure half of the juice there is in one pack?

Edit. You do not have any measuring tools (rules, cups...). But you can use some juice packs.

Edit 2

The original situation is: you have a few handmade paper's piramids (not regular, but identical each other) with water and need to estimate their quality. The measure of quality is a time until the water level is exactly at the middle of piramids.

• What does 'half of juice the one pack' mean? Aug 25 '20 at 9:39
• For example, the full pack is 1L, the half pack is 0.5 L.
– Nick
Aug 25 '20 at 9:44
• Are you asking for the juice level when the pack is stood on one face and is half full? Aug 25 '20 at 10:01
• @weathervane, you are right, one can open the pack and drink an juice, then the opened pack should stay on one face.
– Nick
Aug 25 '20 at 10:06
• So you want a method to identify when the pack is half full? Aug 25 '20 at 10:08

Since we were able to drink half of the juice, there must be a hole in the box somewhere.

If the hole is at one of the vertices, and small, then we can

choose an edge between two vertices that don't have a hole, and hold the box so that the chosen edge is vertical.

This will cause juice to pour out until the liquid level is exactly at the middle of the box.

• This is my favourite answer. Although based on the OP's comments, it looks like the total capacity of the tetrahedron is known, and therefore a simple solution is to simply pour into a measuring glass which is known to hold exactly half of this volume.
– Stef
Aug 25 '20 at 11:40
• Thank you for answer, but how to measure the right angle and middle point?
– Nick
Aug 25 '20 at 12:25
• @Nick hmm? There's no need to measure any middle points, the vertex with the hole will automatically be at middle height. Also, if you want some practical examples like "the corner between any two walls in your house will provide an easy to use vertical alignment tool for the juice box edge", it's good to say so in the question, or at least add the real tag.
– Bass
Aug 25 '20 at 21:09
• @Bass, thank you for the comments, I have added the real tag.
– Nick
Aug 25 '20 at 23:40
• Wouldn't that position the tetrahedron in a way similar to Weather Vane's answer? But then nothing would come out of the hole. I probably misunderstand something. Could you add a visual? Aug 26 '20 at 14:15

My solution is:

Hold the pack so one edge is vertically against a wall.
Insert a thin transparent straw through the hole to touch the bottom.
Place your finger over the end of the straw and withdraw it.
Eye up or measure the level of juice trapped in the straw.

Half the edge length means half the volume is present.
Being slightly off the vertical won't matter much. Possible method

Consider the tetrahedron $$ABCD$$ in the following image Orient the tetrahedron so that the vertex at $$A$$ is pointing down and the waterline is level and coincides with $$BC$$.
Mark the point $$G$$ where the waterline touches $$AD$$.
Now turn the figure upside down so that $$D$$ is pointing downward and the waterline is again level with $$BC$$.
Mark the point $$G'$$ where the waterline touches $$AD$$.
The tetrahedron is half full if and only if $$G=G'$$.

• But I can't see the juice. The packaging is in the way. :-( Aug 25 '20 at 10:06
• You don't even need to measure twice - if the waterline doesn't touch B and C exactly on the first measurement, it's not half full, otherwise, it is. Aug 25 '20 at 18:43
• @NuclearWang I'm assuming here that getting $AD$ exactly vertical here is not a trivial task. Aug 25 '20 at 19:45
• @hexomino If AD isn't vertical, there's an added requirement of needing some way to perfectly reproduce and mirror whatever non-vertical angle the object is held at. If you don't, the overlap of G and G' can be due to differences in how the juicebox is held, and not due to it being exactly half full. The two-measurement method does allow you to verify hall-fullness even if you don't have a vertical wall to align against, but you do need something to align against. Aug 25 '20 at 20:00
• @hexomino Got it, good description. BC must be horizontal or else you can't align the liquid level, so you can't tilt left or right, and there will be only one angle on the front-to-back tilt that actually has the waterline touch BC. On my first read, I interpreted "level with BC" as including a water level that's parallel to BC, but not actually touching it, but we do require that BC is the water line on that side. Elegant solution using symmetry. Aug 25 '20 at 20:48

The usual method:

Pour the juice into a measuring cup with gradations. See how much it measures. Pour from the cup until half the previously measured amount remains.

There are a few methods that rely on weight:

Using a balance scale: Empty two packs of juice to as close as half full as you can get, without going below. Check that they contain an identical amount of juice by confirming that they balance. Now put both half-full packs on one side of the balance, and on the other side, put one full and one empty pack. Both sides should have the exact same weight of 1 pack's worth of juice and 2 empty packs. If they don't weigh the same, empty some more out of the half-full packs and try again.

If you can pour juice into a pack, just empty one entirely, and pour juice from a full pack to an empty one until they balance.

Using an absolute scale: Weigh a full pack and an empty pack separately, and take the difference to determine the weight of the juice. Now just empty the full pack until it weighs as much as the empty pack plus half the weight of the juice.

And here's one based on volume:

If you only have a ruler: Freeze a full pack, unwrap it, and measure the edge length of the frozen juice tetrahedron. From this, calculate the volume. For any other pack that may or may not be half-full, freeze it (with the bottom face laying horizontally), unwrap it, and measure it its height. The volume of the truncated tetrahedron is straightforward to calculate, and the volume of a half-full pack will be exactly one-half the volume of the full pack.

• One solution needs two half-packs. Rot13(V unq gubhtug gb cbfg n fbyhgvba hfvat n onynapr jvgu bar nez gjvpr gur yratgu bs gur bgure. Unat n shyy grgencnx va n yvtugjrvtug arg sebz gur fubeg nez, naq unat gur grgencnx ba grfg sebz gur ybat nez. Nffhzr gur cnpx jrvtug vf artyvtnoyr gbb). Aug 25 '20 at 20:07

Here is a method using

other packs to measure a right angle which is only used to mark the midpoint of one of the edges. We then pierce that edge in the middle and suspend the box on the tips of two other boxes. The pierced box will lose juice until half full.

No, it is not possible to measure the volume in litres of half a pack if you are not allowed to use any measuring tools. Measuring requires using a reference measure and a tool for comparing. Such a tool is used in measuring and should be classed as a measuring tool.

Note
This is an overliteralist answer which might cause annoyance, but it does answer the question. The question could be amended to say "pour out" instead of "measure".

• Thank you for the answer and comment. The original situation is: you have a few handmade paper's piramids (not regular, but identical each other) with water and need to estimate thier quality. The measure of quality is a time until the water level is exactly at the middle of piramids.
– Nick
Aug 27 '20 at 23:58

A tetrahedron that is resting on one of its faces, on a flat, horizontal surface, is half-full when it is filled to about 0.2063* of its height, which is close to 1/5 (the error is a bit over 3%)

I would pick an edge of the tetrahedron and mark off five equal segments on it. Then we can make a small hole at the 1/5th height. If more than just a little liquid drains out, it was more than half full. If it has to be significantly tilted to get anything to pour out, it is less than half full.

There are ways to divide the edge into five equal segments. If we can somehow obtain a sequence of equally spaced parallel lines, it's obtainable that way. Without special equipment, we can iterate on a process whereby we tear out a piece of paper, a bit wider than 1/5th of the edge length, from one of the tetra packs. Then if five widths of that piece of paper exceed the length of the container, we can trim it a little bit and try again.

1. This figure is obtained as 1 - (0.5)1/3. The reason is that the volume of a cone/pyramid is proportional to the cube of its height. If we take a section of length h of a cone, from its cusp to a plane parallel with its base, that section has a volume proportional to h3. Therefore h is proportional to the cube root of the volume. The cube root of half the volume, 0.5, is about 0.7937. So that is to say, if we are filling the cone/pyramid from cusp toward base, then it becomes half-full at 0.7937 of the height. Thus the half-empty part is the remaining 0.2063 of the height. 