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Place 3 coins in each of 3 pans.
Adjust the pan locations until an "isosceles" equilibrium is achieved (the pans form the vertices of an isosceles triangle).
The disparate coin is in the pan at the apex of the isosceles.
These three coins are reweighed in a similar fashion with the abnormal coin again being at the apex.
This utilises the Principle of Moments.
If three identical weights were to be placed around the track they would form the vertices of an equilateral triangle. Viewed from the side we would have 2 weights being 1 unit away from the fulcrum while the third weight would be 2 units from the fulcrum.
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Since the radius of the circular track cannot change, then if the load at the apex of the triangle is light the other two weights must be brought closer thus widening the apex angle. In contrast an apex angle less than 60° would signify a heavier weight.
Courtesy of Dej Mar's comment:
One does not need to weigh three coins in each of three pans, one can easily weigh two coins in each of three pans.
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.
At the time of publication I had no solution to that marked
"For additional discussion:" but Paul, in his comment at this link, Paragraph 4 paves the way for rather easy solutions to various scenarios. He details a "scalene equilibrium" which might be more readily understood by reading "An Experiment".
1. If perchance the imperfect coins are both in the same pan, we have an isosceles equilibrium. We can then differentiate among all three by virtue of a second weighing using the "scalene" proposal.
2. If we have a scalene scenario to begin with, then we need to do two isosceles weighings to identify them. We identify the heavy and light piles from the scalene vertices (as demonstrated by Paul and "An Experiment"). Using two further weighings we use the isosceles procedure to determine the faulty coins.
3. If the disparity of weight makes the two faulty coins equal to two regular ones, then we have more work to do.
If the first two weighings produce and "equilateral equilibrium" then we will get our result by scalene on the third.
In a somewhat similar vein is A Three Way Scale. That and FrankM's Unreliable Scale were largely my inspiration although I had been toying with an idea along these lines, but it would never have worked. |
