## Pi litres

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RaptorInsurance
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### Pi litres

This is a problem I heard on the radio a few years ago, and this isn't one that's particularly well known. There are probably many correct solutions, but there is one that is particularly clever.

You have an infinite source of water, a bowl that can hold an indefinite amount of water, but at least 6 litres, and two unmarked cylinders, which can hold three and four litres respectively. The cylinders are the same height, and have different radii. Can you use these objects and nothing else to measure out an amount of liquid approximately equal to pi (i.e., within 0.002 litres of pi)?

This puzzle requires you to be a bit more crafty than in the usual 'get x units of water with jugs that hold y and z units'. If you are having trouble, the following facts are particularly significant:
Spoiler:
The shape of the cylinders

and
Spoiler:
The cylinders are the same height.

The answer isn't particularly long, so I won't create a solution thread unless people want it.
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genewitch
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### Re: Pi litres

Spoiler:
I'll take a stab at it and show off why i'm not a math major. fill the bowl. Fill the three liter cylinder. Put four litre cylinder under bowl. Place three litre cylinder into bowl. the displacement should be roughly equal to Pi (and leaving pi litres in the 4 litre container). If not, then it's probably some repetition of putting the 3 liter into the four liter container. and catching the spillage into the bowl. it has to be displacement, doesn't it?
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RaptorInsurance
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### Re: Pi litres

genewitch wrote:
Spoiler:
I'll take a stab at it and show off why i'm not a math major. fill the bowl. Fill the three liter cylinder. Put four litre cylinder under bowl. Place three litre cylinder into bowl. the displacement should be roughly equal to Pi (and leaving pi litres in the 4 litre container). If not, then it's probably some repetition of putting the 3 liter into the four liter container. and catching the spillage into the bowl. it has to be displacement, doesn't it?

That's not it, but nice out of the box thinking (a bit too far out, actually). You can assume that the walls of the containers are thin enough to be negligible.
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debuggingRL
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### Re: Pi litres

Spoiler:
Fill the 4 liter cylinder with water, then pour its water into the 3 liter cylinder until its full, leaving 1 liter in the 4 and dumping out the 3. Pour this water into the bowl and repeat once, leaving 2 liters, or 14/7 liters of water in the bowl. Now do the same to get 1 liter in the 4 but instead of pouring it into the bowl, pour some into the 3 liter, set them both beside each other, and repeat until the same amount of water is in both cylinders. This will leave 4/7 of a liter in the 4 liter cylinder and 3/7 in the 3 liter. Pour the 4/7 liter into the bowl, leaving 18/7 liters. Repeat once more leaving 22/7 liters or 3.14286 liters in the bowl, which is off from pi by .0013.
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RaptorInsurance
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### Re: Pi litres

Close enough (you should probably put it in spoiler tags). The answer I heard was to:
Spoiler:
Fill the four litre cylinder, pour it into the bowl. Fill the three litre cylinder, and pour half of it into the bowl. This is normally not allowed, but due to the symmetry of the cylinders, you can tilt it until the water level touches opposite corners, thus leaving half. this leaves 5.5 litres, or 22/4. Using the method you used (i.e., pouring water into the cylinders until they fill to the same height) you can then take 4/7 of this, leaving 22/7, the approximation for pi. I suppose the other method is nicer as it doesn't rely on half-empting the cylinders.
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genewitch
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### Re: Pi litres

Spoiler:
you do realize that you said this wasn't a "get two jugs and fill with x and y amount of water" sort of thing. Which is basically what it is
that's why i was trying to come up with another way of doing it that had to do with the radius and shape. which you said were important (but really aren't, cylinders are circles extruded at a right angle to the radius... and therefore i could only imagine it being some obscure method of knowing something special about cylinders that i didn't know) hehe.

Good puzzle none-the-less
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RaptorInsurance
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### Re: Pi litres

genewitch wrote:
Spoiler:
you do realize that you said this wasn't a "get two jugs and fill with x and y amount of water" sort of thing. Which is basically what it is
that's why i was trying to come up with another way of doing it that had to do with the radius and shape. which you said were important (but really aren't, cylinders are circles extruded at a right angle to the radius... and therefore i could only imagine it being some obscure method of knowing something special about cylinders that i didn't know) hehe.

Good puzzle none-the-less

I didn't say that it wasn't a 'get two jugs and fill with x and y amounts of water', I just said that it wasn't a usual one and required a bit more ingenuity. It's true the puzzle would work with other shapes, but I think that a more regular shape like a cuboid would probably give more 'false' solutions due to the difference in properties between these shapes and the cylinders.
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Mo' Money
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### Re: Pi litres

I have a way to get an arbitrarily precise approximation of pi.

Spoiler:
This uses the base-3 expansion, and the assumption that the containers are very thin. Suppose we want to get 3-n liters of water. First, take 3-n+1 liters of water and pour it into the 3-liter container. Then place the 3-liter container into the four-liter container. Pour water into the four-liter container until it reaches the level in the 3-liter container. (Bonus: you will know when this happens because the 3-liter container will float, assuming the walls are massless.) Since the space between the containers has a total capacity of 1 liter and has vertical sides, the amount of water now in the 4-liter container but outside the 3-liter container is 3-n liters.

Xanthir
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### Re: Pi litres

Heh, cool. That can be used to get an arbitrary approximation of *any* number, assuming you know its trinary expansion.
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MatrixFrog
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### Re: Pi litres

I find the answer rather unsatisfying as it relies on the fact that pi is about 22/7, not any of the other properties that make pi such an important number. I guess the "within 0.002" should have been a clue, but I still think there must be answer that relies on the fact that the containers are circles. In that spirit, I present the following problem, which is, I guess you could say, a sequel to the original.

You have just completed RaptorInsurance's problem, and as a reward, he or she has provided you with a shiny new protractor. However, while you were admiring its beautiful design, an evil demon came along and stole the 3L container from the original problem, as well as kidnapping RaptorInsurance. Thus you are now left with the same bowl and 4L container as before, and a protractor. The task now is to end up with exactly pi liters of water. Your solution may not depend on the fact that pi is very close to some particular fraction, and the only reason for anything less than perfect precision is minor errors in measurement, pouring, etc. not in the method.

I have an outline of how I would do this, but not a complete solution. By the way, as has been pointed out, you could get arbitrarily close by using the ternary expansion of pi, but I'm not granting you infinite time. After all, you already have infinite water, so don't be so greedy! Besides, I think that would count as repeated applications of the already-forbidden close-to-a-certain-fraction rule.

Now that I think about it, the physics is beyond me, but I think you could also get the answer if you put the 4L cylinder into a centrifuge, and you calculated the exact RPM necessary to get 4-pi L to spill out over the edges. Of course, this assumes you have access to information such as the viscosity of water, and so on. Anyway, that's just silly.

Lord Aurora
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### Re: Pi litres

MatrixFrog wrote:The task now is to end up with exactly pi liters of water.

MatrixFrog wrote:I'm not granting you infinite time.

Pi is an irrational number, therefore, without infinite time (to measure out the infinite digits of pi) it's impossible to measure out exactly pi liters of water. Even given infinite time, there is really no true way to do it, as there is no value that represents exactly pi liters without approximating at some point in the process. I would even question the assumption that there is a specific "value" of pi, given that it goes on forever and ever.
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MatrixFrog
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### Re: Pi litres

Okay, fair enough. But if I have a perfect circle which is known to have a radius of 0.5, and I measure a string to be the same length as the circumference of the circle, we would say the string has a length of pi. This is dependent on a few assumptions, of course, such as the fact that there is such a thing as a perfectly circular object in the physical world, and probably a massless string, and so on. But we always allow such assumptions in this type of problem, don't we?

The answer I'm thinking of would be exact in the same way this string example is exact. If you actually tried it out in real life, it would probably be way off, but only due to the imperfections of the physical world, not due to flaws in the method.

If anyone is actually interested in solving, the protractor is not intended as a red herring. I really want you to solve by measuring angles and such. I guess I'll allow that the evil demon has also left you with the scientific calculator of your choosing.

Edit: And no, you can't ask for a calculator whose volume is exactly (4-pi) liters, smarty-pants.

Xanthir
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### Re: Pi litres

Lord Aurora wrote:Even given infinite time, there is really no true way to do it, as there is no value that represents exactly pi liters without approximating at some point in the process. I would even question the assumption that there is a specific "value" of pi, given that it goes on forever and ever.

You would be incorrect to question that assumption. Pi has a specific, well-defined value. Calculating it fully does require infinite time (or at least infinite operations, which under current models of physics probably can't be done in finite time), but that's irrelevant for our purposes.

MatrixFrog's example is correct, though. Assuming an ideal circle and string, when you measure the circumference it is indeed exactly pi. This does not contradict what I said above - an ideal circle encodes the full value of pi within itself (assuming Euclidean geometry), eliminating the need for you to calculate it directly.
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Arancaytar
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### Re: Pi litres

And oh look! Our containers are cylindrical, ie. circular, which means that a factor of Pi is inherent in some of its dimensions. You just need to coax it out in such a way that the amount of water contains it too.

This is the approach I'd instinctively attempt in the first problem. I haven't yet been able to solve it though.

The second problem seems easy, assuming that your protractor has negligible volume and can measure length as well as angles, like a slide rule. If you can measure the radius of the cylinder, you know that the volume of a certain height in the cylinder is Pi*r^2*h. Knowing r^2, figure out h such that the volume comes out to Pi. Fill the cylinder to this height. Done.

Thoughts on the first puzzle:
Spoiler:
Call a the 4 liter cylinder, b the 3 liter one. h is the height of the cylinders, which is equal.

Va = 4 = Pi * ra2 * h
Vb = 3 = Pi * rb2 * h

4/3 = ra2 / rb2

2/sqrt(3) = ra / rb.

Congratulations, we have found the proportion between the radii of the two cylinders. Can we use this somehow?
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MatrixFrog
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### Re: Pi litres

Good idea, but not what I had in mind. Kind of along the right lines though.
Spoiler:
To start with, you should actually be able to find the radius and height by pouring out half the water (you'll know when you're halfway because the water's surface intersects with the edge of the cylinder), and then measure the angle the bottom of the cylinder makes with the table.
From there, it's a lot of calculation, but I'm pretty sure it should be possible in theory at least.

Buttons
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### Re: Pi litres

I think the big problem with the original puzzle is that it's possible to get "within 0.002" of anything, because the following operations

• Pour something into one cylinder. Pour water into the other cylinder so that the heights are equal.
• Divide something into two cylinders so that the heights are equal.

allow you to divide or multiply any volume by 4/3 or 4/7. Since the logs of these are not rationally related, any number can be approximated within ε as (4/3)m(4/7)n, where m and n are integers.

parallax
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### Re: Pi litres

If you have a string, you can wrap it around the cylinder to get the length 2πR. Then draw it across the cylinder to get the length 2R. Use those two length to construct a right triangle with base 4R and height πR. Use that triangle and the height of the cylinder to construct the length (π/4)H. Fill the cylinder to that height with water.
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greeniguana00
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### Re: Pi litres

Here is how I would do it:

Spoiler:
First, dump 3 into the bowl (easy enough ). Now, fill the 3 liter cylinder and dump the water into the 4 liter cylinder. It will make it 3/4 of the way up. Then you can fill the 3 liter cylinder to to the same height as the 4 liter w/3 liters of water in it, and you will have 3*(3/4) or 9/4 liters. This amount of water would fill up the 4 liter cylinder 9/16 of the way up, and you could use this using the same method to get 3*(9/16) or 27/16 liters in the 3 liter cylinder. Then you keep going until you find the greatest fraction less than 0.1415... to add to the bowl. Then you continue the process until you have achieved the level of accuracy you desire.
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