This represents a circular balance.

                       
         o           /|\           o
                      |
                      |					  
                      |       
 

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.]

One does not need to weigh three coins in each of three pans, one can easily weigh two coins in each of three pans.
Similar to the "official solution", if the coin of variance is not among the first six weighed, the pan locations would form the vertices of an equilateral triangle and a second weighing of the remaining three coins will find the coin of variance at the apex of the isosceles triangle formed. If the coin of variance is among the first six weighed, the disparate coin would be in the pan at the apex of the isosceles. The two coins and any one of the others may be reweighed in like fashion, one in each pan, with the coin of variance being at the apex.



Taking a step backward and readdressing a question posed by Gamer: Minimum?

The minimum number of weighings would be one. The scalene scenario in Paul's commentary can also be applied to a nonagon formed with nine pans; or, more minimally, eight pans. By deduction, we can accomplish the task with an octagon. If the polygon is regular, the coin of variance would be the one that wasn't placed on the scale. If irregular, the technique as described in Paul's post would be used.

As it wasn't actually defined, I offer this definition of minimum. Minimum is to be defined as the count of pans, with each pan counted again each time it is used in a weighing. In addition, the minimum is meant to be the count required in which the determination can be absolutely assured. With this definition, the method of three pans and two weighings gives a  count of six as opposed to the single weighing with a count of eight. Thus three pans and two weighings (six) would be the minimum.



As to the case where we have two coins of variance, one lighter and one heavier, using three pans would take three weighings to identify the two coins of variance and to which was the heavier and which was lighter. The total pan count of the three weighings, with the given definition would be nine. Using eight pans and one weighing, in this case, would then be considered minimal -- (equally, it may be done with four pans and two weighings).

With the single weighing, determining the heavier and/or lighter coin(s) on the scale would follow the techniques as described in Paul's post. If only one of the coins is on the scale, the same technique would be used to determine which coin of variance was on the scale and with the second coin of variance, obviously, the one not weighed.

In answer to the question, if there is a known disparate difference in the lighter than normal and heavier than normal coins, the two weighings with three pans can be performed as the minimum.  Only if the two coins of variance to the normal are equal or unknown to be equal in absolute difference will the procedure change the number of weighings for the minimum.

As to the method described in Charlie's first post for this puzzle, if the disk could support each of the coins spaced around the disk's edge, the minimum would indeed require no pans and could be performed in a single weighing.  Of course finding the equilibrium in this manner may be more difficult than using the pans.

Edited on March 28, 2008, 11:28 pm

Posted by Dej Mar on 2008-03-28 12:47:55