7
$\begingroup$

If we assume the classic balance puzzle (Twelve balls and a scale) is phrased like this:

You have 12 coins that are identical except for one counterfeit that is heavier or lighter. Can you use a balance scale three times to determine the counterfeit coin and say if it was heavier or lighter?

Can you design an object that could properly be called a "balance scale" that could solve the above problem with a single weighing instead of three?

Obviously any such object is a far cry from the intention of the original problem, and may be hard to draw or describe. I just think us Engineers deserve a little fun at the expense of of you math-riddle folk sometimes.

Part B and Part C will contain something like spoilers as to the two ways I thought of answering this, so if you want to take an unbiased stab at this, I would avoid those for now.

And before anybody mentions it, coins are always better than "balls" because they don't roll off the table and get lost under your desk. It is befuddling to me that so many people do these puzzles with balls.

$\endgroup$
  • 1
    $\begingroup$ Some of us math-riddle folks are engineers ... but I think the engineering answer is to challenge the requirements! You'll need a better story than that! :-) :-) $\endgroup$ – deep thought Jan 26 at 19:03
7
$\begingroup$

I have a fun solution:

We can use the idea of water pressure to solve this question. Imagine a mechanism with 12 pistons each with water inside. The 12 pistons are connected to each other. Now, place a coin on each piston. Due to a difference in weight, one of the pistons will be at a different height from the others. This allows you to identify the counterfeit coin, and also whether it is heavier or lighter. Technically it's a 12 pan hydraulic balance. :)

$\endgroup$
  • $\begingroup$ I love it! You might have to get clever to make the small weight difference noticeable for an operator but as a basic approach that's perfectly sound. $\endgroup$ – Dark Thunder Jan 26 at 16:07
5
$\begingroup$

This might be not what an engineer likes to hear, but

one could go to outer space (no air friction, no graviational potential) and form a equally spaced ring out of the balls/coins.

Then,

the counterfeit one will break the symmetry of the center of mass of the circle.

If the counterfeit is lighter than the rest

the circle will contract away from it

else

it will contract towards it.

$\endgroup$
  • $\begingroup$ Amazing! Looking past the nit-pick that you don't have a "balance scale" object, can you actually differentiate between those two outcomes? I'm really asking, I can't tell. $\endgroup$ – Dark Thunder Jan 26 at 15:59
  • $\begingroup$ @DarkThunder I replaced the drawings by simulations, so that it's easier to see the difference. $\endgroup$ – A. P. Jan 26 at 17:52
  • $\begingroup$ Would low earth orbit be enough, or would tidal forces mess that up? $\endgroup$ – deep thought Jan 26 at 18:32
  • $\begingroup$ @deepthought Of course the exact answer would depend on how small the mass differences are, but I would say that tidal forces don't have much influence on the result. As you can see in the figures the outcome of the test is rather local, while tidal forces would have a more long-range effect, which would probably make the circle more elliptical. And judging from the fact that with a similar setup it is possible to detect gravitational waves without being disturbed by tidal forces, it must be possible to measure very small mass differences of the coins. $\endgroup$ – A. P. Jan 26 at 19:14
5
$\begingroup$

Another answer, which is probably closer to what the OP intended:

One could think of just extending a 3-pan balance to a 12-pan balance by just adding 3 equally spaced arms into each gap, so that we end up with arms spaced by 30°.

The problem with this is that

you can't distinguish whether one coin is lighter or whether the coin on the other side is heavier. This results from the symmetry of the balance. Hence we have to create a balance without any pair of coins lying on opposing sides. This sounds like the balance could never be balanced, but the 3-pan balance is an example which fulfills this criterion and is still balanced (if all coins are equal).

To make a

12-pan balance fulfilling the criterion out of it, we can just add the additional arms in other angle combinations than (30°, 30°, 30°). To make sure that at the same time the balance is intrinsically balanced we can group triples of arms to regular 3-pan balances. As long as they share their center of mass the whole construction is balanced. In the following drawing the triangles representing the 3-pan balances are rotated by 15° relative to each other, having the coins sit on the corners of an imaginary 24-gon.

$\endgroup$
  • 2
    $\begingroup$ That is the exact thing I came up with, so you win a cookie (sorry, I don't know how this forum works, I'm assuming it's a cookie). The YouTube channel Numberphile had a video on balancing a centrifuge not that long ago that dealt with surprisingly similar issues, which I thought was neat. $\endgroup$ – Dark Thunder Jan 27 at 0:54
  • $\begingroup$ @DarkThunder If it's the right answer, you can click the tick near the top of the answer to accept it as correct. $\endgroup$ – ZanyG Jan 27 at 3:52
  • $\begingroup$ @DarkThunder I just saw that video, and then I come here and see your problem. Neat! $\endgroup$ – LinuxBlanket Jan 27 at 13:37
2
$\begingroup$

You could have some kind of

Metal tube with 12 springs attached to it and a plate at the end of each spring. The heaviest/lightest coin would be the one that's lowest/highest

$\endgroup$
  • $\begingroup$ Haha, yeah I guess you could. Each coin is balanced against its own spring and doesn't interact with each other. Not what I had in mind but that is literally the point of asking this sort of question. $\endgroup$ – Dark Thunder Jan 26 at 21:25
2
$\begingroup$

What about

putting smaller balances on larger balances

for example you can obtain

one large 3-pan balance, three medium 2-pan balances, six small 2-pan balances, and stack them in the obvious way.

Or even

four different sizes of 2-pan balances, with four extra known-genuine coins.

$\endgroup$
  • 1
    $\begingroup$ One could weld 4 3-pan balances together and say this is one device. :-) $\endgroup$ – A. P. Jan 26 at 19:17
  • 1
    $\begingroup$ @A.P. But I'm cheap and I was planning on borrowing the equipment, so I need to return it in good condition! :-) :-) $\endgroup$ – deep thought Jan 26 at 19:23
  • 2
    $\begingroup$ Definitely cheaper than flying to outer space. $\endgroup$ – A. P. Jan 26 at 19:25
  • $\begingroup$ Nesting balances was one of my ideas too. Can you do it with only 2-pan balances but without additional genuine coins? $\endgroup$ – Dark Thunder Jan 26 at 20:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.