Same As Before

Ok, time to solve out the weighing medals problem from last time.

First, its easy to see that the gold medal will never be on the scale, there is nothing to balance it against, so we might as well throw it away and if we can't find the fake in the others then the fake was the gold medal. Next, note that we can't leave all the silvers off the first weighing, because then they must all go on the second weighing (since the gold is never being weighed, we must weigh everything else at least once or we can't find the fake) and there is no way to divide up the 3 silvers into two piles. Similarly, we must put at least one bronze on the first weighing.

It makes the most sense to weigh 2 bronze and 1 silver against 2 bronze and 1 silver for our first one, this way, if one side is heavy we have immediately identified 2 bronze and 1 silver as the "suspect group". We can use our second measurement to weigh the two suspect bronze against eachother and figure our which of the three medals is the fake.

If our first measurement comes out balanced, we have the gold, 1 bronze, and 1 silver as our suspect group. We put the suspect bronze with a genuine silver on the left against the suspect silver and a genuine bronze. Now 'left heavy' 'right heavy' and 'balanced' all point to a unique medal.

This solves the problem, and as stated isn't very difficult. What I found sort of cool was if you reconsider that solution without the different types of medals. It is just the 9 coins and 2 weighings problem and you start by weighing 3 vs 3 to get a suspect group of size 3. Of course, if the solution to the different medals problem is to work for arbitrary values of masses of genuine medals m_bronze, m_silver, m_gold, then it must work in the particular case where all those m's are equal (I suppose this isn't true if your solution divided by m_bronze-m_silver or something, but we aren't doing that).

Anyway, thats all I got.