Several people, including Miguel Tenorio and David Stanton, figured out how to solve this problem in four weighings. But it can actually be done in three. Here’s how:

Number the coins 1 through 12 and split them into three groups:

A [1 2 3 4]
B [5 6 7 8]
C [9 10 11 12]

1. Weigh A against B. If they balance, we know that the fake is in C and all the coins in A and B are real. Let’s pursue that outcome for now.
    
2. Take any three coins from A or B (they are all real) and weigh against three coins from C, say [1 2 3] against [9 10 11]. There are two possible outcomes:

   a. They balance, so the fake is 12. A third weighing of 12 against any other coin determines if it is lighter or heavier.

   b. They don’t balance, so the fake is 9, 10, or 11. (And, if [9 10 11] are lighter than [1 2], the fake is lighter than the real coins. If [9 10 11] are heavier, the fake is heavy.)

3. For the third weighing take two coins of [9 10 11] and weigh them against each other. If they balance, the fake is the other coin. If they don’t balance, then you can pick the fake, since you already know whether it is lighter or heavier than the rest.

Now let’s go back to step 1 for the other alternative, which presents a more complicated problem than in the first scenario.

1. We’ve weighed A against B and they don’t balance, so the fake is somewhere in A or B. We also know that the coins in C are all real.*
    
2. Let’s say that A [1 2 3 4] are heavier than B [5 6 7 8]. Let’s split A so that we are weighing [1 2] against [3 4]. But let’s add one coin from B on the left and one from C on the right. So we are weighing [1 2 5] against [3 4 12]. Note that we know 12 is real. There are three possible things that can happen:

   a. [1 2 5] are heavier. We know that 3, 4, and 5 are real because we changed their positions on the scale with changing the balance; 3 and 4 moved from left to right, while 5 moved from right to left. So the fake is 1 or 2. We also now know that the fake is heavier. A final weighing of 1 against 2 reveals the fake, which is the heavier one.

   b. [3 4 12] are heavier. This means that the fake has moved. One possibility is that the fake is 3 or 4 and is heavier. Or, the fake could be 5, which is lighter. A final weighing of 3 against 4 will nail it down. If they balance, the fake is 5. If they don’t, the fake is the heavier of 3 and 4.

   c. [1 2 5] and [3 4 12] balance. This means the fake isn’t on either side of the scale, so it must be in three remaining coins of B, that is, [6 7 8]. We also know that the fake is lighter, since in the first weighing, [1 2 3 4] were heavier than [5 6 7 8]. A third weighing with one coin on each side of the scale reveals the fake. If we weigh 6 against 7 and they balance, 8 is the fake. If they don’t balance, the lighter one is the fake.

This puzzle is described in more detail in:

Michalewicz, Zbigniew, and David B. Fogel. How to Solve It: Modern Heuristics.New York: Springer,2000.