...and you have to determine which coin it is, plus whether it is too heavy or too light.  
You can have three turns at a simple balance scale, using only the coins.  They all look identical, and the only difference between them is the one fake coin's density.  

What is your method to find the fake coin with only 3 uses of the scale?

No tricks, no gimmicks.  Seriously.

12 coins, divide into two groups of 6 coins each and weigh, retain the heaver side.
6 coins, divide into two groups of 3 coins each and weigh, retain the heaver side.
Put one coin on each side of the scale.  If either of the 2 coins on the scale is heavier than the other, retain the one that is heavier; If the 2 coins on the scale are equal, the one that was not weighed is the heavier of the original 12.

I would verify the results - if the last three coins are all the same weight, then repeat the method described but retain the lighter side from the scale.

offhand, I think that not knowing if the density of the fake coin is lighter or heavier would require two passes (as described above).

this is a good riddle, and  I will patiently await the resolution.

I've heard a similar version of this puzzle before.
I know the idea is to divide your possibilities into 3 groups, so you'd begin with 4 on each side of the scale and 4 off the scale.
From there, you hope that the scale balances as that would make things much easier! : )

Yes, if it balances twice in a row, you end up with one lone coin that you know is counterfeit, and all you have to do for the 3rd turn is check it against a known good coin to see if the fake is heavier or lighter.

The trick is how to arrange coins for the second time at the scale.  There are several possible scenarios.

ok, divide 12 coins into 3 Groups
Pass1: weigh two groups for 8known and 4unknown result (=4unknown)
Pass2: weigh 2known&2unknown vs 4known with 2unknown on side result (=2unknown)
Pass3: weigh 1known vs 1unknown with 1unknown on side result (=1unknown)

ok, divide 12 coins into 3 Groups
Pass1: weigh two groups for 8known and 4unknown result (=4unknown)
Pass2: weigh 2known&2unknown vs 4known with 2unknown on side result (=2unknown)
Pass3: weigh 1known vs 1unknown with 1unknown on side result (=1unknown)

I'm not certain though.  If pass one is equal I'm good.  If pass one is not equal, I only know the third group weighs the same as one of the other two groups, but I really don't know which group.  fun.  good riddle.

If the scale balances first time, you have 8 good coins, which is nice enough on the surface, but you also have 4 set aside about which you know only one thing, namely, that the counterfeit (CF) is among them, however, regarding whether the CF is heavy or light, you have no idea.

This lack of information is logically equivalent to the different kind of information deficiency you get when the first pass does not balance.

The reason this is logically equivalent is that when the first pass does not balance, you then know you have only 4 good coins (the ones set aside) but you also know that the 4 coins that were light potentially contain the CF and if they do, it's LIGHT. The same (but opposite) is true of the other 4 coins: potentially CF but HEAVY. By arranging the coins with foresight on the second pass after the first pass didn't balance, you can give yourself the same advantage over the problem that you would have had with a first pass that balanced.

Therefore, you most certainly know SOMETHING about all 8 that did not balance, as you say ("I really don't know which group" of 4 is balanced), and SOMETHING is better than NOTHING. Focus on what you DO know instead of what you DON'T know.

The key is to make use of all this data to your advantage, instead of wasting it -- by not making a plan first and being haphazard.  Accuracy makes all the difference.

So on pass 1, you have:

A) Balances, then 8 good coins and 4 potentially CF and HEAVY OR LIGHT.

B) Not Balanced, then 4 good coins and 4 potentially HEAVY and 4 potentially LIGHT.

Your second pass must always be directed at setting yourself up for the third pass under tolerable conditions, since you only have one more pass to go.

Let's take A first

If it's A, as you say ("If pass one is equal I'm good") your second pass must cover all the bases. You will be lacking any HEAVY/LIGHT data about the 4 set aside. Assuming you check 3 of them on the second pass and they balance too, then you have only one set aside which can be completely evaluated alone on the 3rd pass against one known good coin.

However, if the second pass does not balance, then for the third pass you are left with only one additional good coin (the one set aside), making a total of 9 good coins, and 3 among which is the CF. Since your third pass can have 3 possible outcomes, your challenge is how to arrange the coins so that there is a one-to-one correspondence between the final outcomes of A) Balances, B) Light on the left, and C) Heavy on the left. That means, arrange them so you will know that if it's A then it can only be THAT coin, if it's B then it can only be THAT coin, and if it's C then it can only be THAT coin (for 3 different coins).

Summary:  3 possible outcomes, 3 coins in question, with only one possibility for each coin.

The important thing then is to be sure that you arrange the second pass in such a way that you are without any doubt about the individual potential H/L status of those 3.

So how can you plan on arranging those 3 coins in such a way that you will be able to conclude with certainty whether each one of them (if CF) are HEAVY or LIGHT?

If the first pass balances and the second pass does not balance, how many real coins are you left with now?

pass 1: 4 vs. 4 --> balances ............... 8 good coins
pass 2: 2 vs. 2 --> not balanced ......... 9 good coins  (How do you know have 9 good coins?)

For pass 2, which coins are on the left side of the scale and which are on the right?
Remember, you have to set aside one of the 4 that were set aside during pass 1.
So you're only weighing 3 of those 4, therefore, where do you get 2 vs. 2 with only 3 coins?