It is possible.
Label the coins A-L, with L being the known genuine coin.
Weigh ABCD vs EFGH vs IJKL.
If no pan rises, then both heavy coins are on the same pan.
There are ${4\choose 2} = 6$ combinations where both heavy coins are on Pan 1, and 6 combinations where both are on Pan 2, but only ${3\choose 2} = 3$ combinations were both are on Pan 3, since we know coin L is genuine. That adds up to 15 possibilities.
If Pan 3 rises, then one heavy coin is on Pan 1 and another on Pan 2, for a total of $4\times 4 = 16$ combinations. If Pan 2 rises, then there are only $12$ combinations, again because we know coin L is genuine. Likewise for Pan 1.
If no pan rose on the first weighing, then the second weighing should be ABEF vs CGIJ vs DHKL. The results of this would be:
- None: AB, EF, IJ
- Pan 1: CD, GH, IK, JK
- Pan 2: AD, BD, EH, FH
- Pan 3: AC, BC, EG, FG
In the first case, for the third weighing, use AE vs BI vs EJ. The two heavy coins are now split and one of the three pans will rise.
In the other three cases, there are four patterns of which at least two intersect by one coin. Put the intersecting coin on Pan 1, and each of the others from the pairings on Pans 2 & 3. Pick one other pattern and split its coins between Pans 2 & 3. Put other coins aside and fill in with known genuine coins (of which you have several).
For example, if Pan 1 rose, you would weigh KA vs IC vs JD. If the pattern was GH, then no Pan would rise. CD would raise Pan 1, IK would raise Pan 3, and JK would raise Pan 2.
If Pan 3 rose on the first weighing, then one of the heavy coins is in ABCD, and the other in EFGH, for a total of $16$ possibilities. Coins I,J,K, & L are now all known to be genuine.
Weigh ABGH vs CDIJ vs EFKL.
- No Pan: AG, AH, BG, BH
- Pan 1: CE, CF, DE, DF
- Pan 2: AE, AF, BE, BF
- Pan 3: CG, CH, DG, DH
In all four of these cases, you have four patterns where one coin is from one pair and the other coin from the other pair. The weighing pattern for this is similar to above, but requiring 2 genuine coins.
No Pan: weigh AI vs BG vs HJ.
AG would raise Pan 3; AH would raise Pan 2; BG would raise no pan, and BH would raise Pan 1.
A similar pattern would work for any of the results of the second weighing.
If Pan 1 or 2 rose on the first weighing, follow the same logic for Pan 3 rising, with the different sets of coins. Any of the possibilities where coin L came out heavy would obviously not happen, but otherwise the logic would be the same.
QED
