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.]

Gamer's strategy is clearly correct. It omits only making explicit the assumption that the three pans are placed evenly spaced on the balance (at the corners of an inscribed equilateral triangle.)

I can't think of what else might be meant by the comment that the equilibrium requirement had been neglected, and personally I believe that requirement is excessively vague if it's intended to convey any nontrivial element of the problem.

Gamer's strategy can also be extended to the case of two odd coins, one heavy and one light, among the nine, although the number of weighings now depends on whether the weight of heavy+light = normal+normal or not.

If heavy+light != normal + normal, two weighings still suffice. If the two odd coins are initially in different pans, then the three pans will equilibrate at different heights; the heavy is in the lowest pan and the light is in the highest pan. Put one each of the heavy pan coins and one each of the light pan coins into the three pans (in pairs) for the second weighing. Again, if the two odd coins are in different pans the three will have different heights. If the two are in the same pan then the other two will be at the same height, and the two have been identified. If on the first weighing the two are in the same pan, then it will be a different height then the other two at equilibrium; weigh the three coins in that pan on the three scales for the second weighing. They'll necessarily be three different weights and allow you to determine the two odd coins.

 

If heavy+light = normal+normal then when the two odd coins are in the same pan, it balances as if no odd coins were, and it's possible that all three pans will be the same height. If that happens on the second weighing, then rotate each coin from the first weighing's light pan to the clockwise neighboring pan and make a third weighing. Regardless of which pan held the two odd coins, they're now in different pans and so the result will have three different heights. If the first weighing shows all three pans balancing at the same height, then move one coin from each pan to its clockwise neighboring pan and one to its counterclockwise neighboring pan and weigh again. Regardless of which pan had both odd coins, they're now in different pans and this weighing will result in three different heights, which we know from the earlier case requires only one more weighing to resolve.  So in this variant, three weighings are sufficient to identify the two odd coins whose delta weight from the norm is the same but with opposite sign.

 

Posted by Paul on 2008-03-21 19:50:59