Solution to Problem 7: "Balancing Act"

(I guess the assumption is that the physics in the problem is well-known.  If this isn't the case, then each of the horizontal lines represents a balanced system of the weights underneath it.  Each weight contributes a torsion force equal to its weight times its distance from the pivot.  For instance, the top bar has three forces on it: the first of distance d and weight (A+B+...+H), the second of distance -d and weight (G+I), and the third of distance 4 and weight J.  Since the bar is balanced, the sum of all these forces is 0, so we have A + B + ... + H = (G + I) + 4J.)

The sum of all ten weights is 55. 3G = 2I, so the sum of those two weights (which we'll call X) is a multiple of 5. Also, since the top pivot balances, we have (55-X-J) = X + 4J, or:

55 = 2X + 5J.

There are several integral solutions to this equation, but only one where 55-X-J is an even number (which it must be, since the pivot right above E also balances). This is J=7, X=10 (so that G=6 and I=4).

The remaining weights are 1, 2, 3, 5, 8, 9, 10: a total of 38. EFH must weigh 19 and be such that H=F+2E. The only solution for that is E=1, F=8, H=10. This leaves 2, 3, 5, and 9 such that D=B+2A, so D=9, B=5, A=2, and C=3. So, the value of all of the weights from left to right is 2 5 3 9 1 8 6 10 4 7.

Comments: I didn't try this one during the contest, and it's one of the ones that has me kicking myself now.  It took me less than five minutes to solve it once I settled down Saturday evening to take a second look at the puzzles!  Any time you can pick up 3 points per minute, you know you're on the right track.