This represents a circular balance.

                       
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A light circular disk is needle point mounted. It has a graduated scaled track "o" from which identical measuring pans may be suspended.

Suggest the minimum number of pans as well as the minimum weighings needed, and the strategy required to determine which one of 9 coins is in weight variance with the other 8 in a visually identical collection. You should also be able to determine if the variant coin is lighter or heavier than each of the others.

At the end of each weighing, the coins being weighed will be on the balance, and the balance will be in equilibrium.

For additional discussion:
The set of nine has 7 regular coins with one weighing lighter, and the other weighing heavier than the others.
How might the circular balance assist?
Does a disparate difference in weight of the lighter to the normal than the heavier to normal influence the procedure? [This thought was raised at review.]

I've also been struggling to understand exactly what "equilibrium" means in the context of this puzzle and how this circular balance operates. I imagined that the balance operated like a normal "one-dimensional" two pan balance, where the degree of tilt of the arm of the balance is measured (or, equivalently, the relative heights of the two pans.) I though this balance worked the same way except now the entire plane of the circle is able to tilt in two dimensions, reducing to the one-dimensional case when there are exactly two pans at opposite ends of a diameter of the circle. Based on the comments below, it seems this is an incorrect assumption.

There are two other possibilities that I haven't seen considered here that could allow for a true horizontal equilibrium and yet still offer a solution to the problem. Let me describe them in turn. I don't believe my approach to weighing would change, nor the number of weighings; only how I interpreted the balance.

The first possibility is that the balance, while "needle point mounted" as described, need not have the needle coincident with the center of the circular disc of the balance. In this case, then if we use three equally spaced pans and load them up, then to "read" the balance we must move the needle point until the disk remains horizontal. Clearly, if all three pans have the same weight, the needle will not move, and equally clearly the center of mass of the entire system of balance plus weights will lie within the circle of the balance itself, so this is always possible. It's easier to calculate that point if we assume the balance itself has no weight, but that's not necessary to answer the questions posed here. In part one, we need to identify when one pan has a higher or lower weight than the other two. In this case, the center of mass (the point at which the needle must be mounted for "equilibrium" will move along the diameter that passes through one pan and bisects the chord connecting the other two pans (by symmetry). If the pan moves towards the lone pan, then that pan is heavier; if it moves toward the pair of pans, then the lone pair is lighter. Since we are only measuring relative differences and not absolute amounts, that's all we need to solve part one. Everywhere in my original proposed solution I mention finding the pan that was "lower" than the other two, instead find the pan that's in line with the needle point and the center of the circle and is closer to the needle. Finding the pan "higher" than the other two means finding the point. The second part of the problem introduces a new possibility that all thre pans have different weights. That's fine too -- the pan with the light coin is farthest from the needle point and the pan with the heavy coin is nearest the needle point.  So with this "movable mount point" balance we can always have horizontal equilibrium and still solve both problems as described in my original proposal.

There is another possible approach, which is that the needle mount is fixed, but that the pans themselves must be moved around the balance until they balance, and then the balance is "read" by examining the positions of the pans. This approach fails for the two-pan case (the pans are a fixed distance from the center and so the sum of two pans' moments can't be zero unless they're the same weight), but is plausible for three or more pans. We're left with the same task of understanding how to read this new balance variant. Here, if one pan is heavier than the other two (and the other two are the same) then the two similar pans will shift towards one another  and the chord that connects them will bisect the diameter that contains the odd pan, again by symmetry. Now, we read that the odd pan is heavier than the other two when the other two are closer together than 120 degrees and it's lighter then the other two then they're farther away than 120 degrees. In principle the magnitude of the deviation from equidistant could be graded to measure the exact amount of the weight difference but that's not important to solving this problem. So, with THIS balance, everywhere in my original proposed solution where I refer to one pan being "lower" or "higher" than the other two, replace that statement with one in which a chord connecting two pans bisects the diameter containing the third pan (which is the odd one). "Lower" corresponds to the two pans being closer together than 120 degrees and "higher" corresponds to them being farther apart. The first part can be solved using the same strategy but reading the balance in this new manner. For the second part we again must consider the case where all three pans are of different weights. In this case the triangle formed by the three pans will be scalene when the balance is in equilibrium instead of isoceles when one pan is odd or equilaterall when all pans match.  In this case, a bit of calculation will reveal that the heaviest pan is opposite the longest side of the triangle and the lightest pan is opposite the shortest side of the triangle. (Actually, this has been true all along for this balance, but it's only with three unique weights that this fact becomes particularly useful.) So in this case, we can easily determine what in my first proposal I called the "highest" and "lowest" pans by substituting instead "opposite the shortest side" and "opposite the longest side". Once again, the strategy is unchanged, only the physical mechanism of using and reading the balance.

I have no idea if either of these variations is what the author intended, but they do both have the virtue of having the circle of the balance be perfectly horizontal when in equilibrium and also allowing the problem to be solved.  Thoughts?

Posted by Paul on 2008-03-27 02:35:45