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By subtracting 12 from our list of triangle numbers, 3 and 9 must be either side of the 12, and so we will choose to place the 3 on the right. We can keep doing this to get the following:
Clock: going clockwise 12, 3, 7, 8, 2, A, B, C, D, E, F, 9
By the same reasoning A is either 1 or 4, and F is either 1 or 6. But if F=1, then E=4, and so F=6, but 11+6=17 is not a triangle number. So F is not a 1, and so must be a 6.
So E=4, A=1, and so D=11, C=10 and B=5.
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.