A two-pan balance is innacurate since its balance arms are of different lengths and its pans are of different weights. Three objects of different weights A, B, and C are each weighed separately. When placed on the left-hand pan, they are balanced by weights A1, B1, and C1, respectively. When A and B are placed on the right-hand pan, they are balanced by A2 and B2, respectively. Determine the true weight of C in terms of A1,B1,C1,A2, and B2.
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Tags: AMC, USA(J)MO, USAMO, algebra unsolved, algebra
infiniteturtle
03.12.2013 03:30
Hmm no solution was ever given. Am I right in thinking this is really just basic algebra? Also is this really a combinatorics problem? Moderator says: indeed, not combinatorial at all. Moved.
P_Groudon
15.11.2020 23:45
Let L1 be the length of the left arm of the balance scale and L2 be the length of the right arm. Let p1 be the mass of the pan on the left and p2 be the mass of the pan on the right. Finally, let m1 be the combined mass of the weight(s) we put on the left side and m2 be the combined mass of the weight(s) we put on the right side.
To balance the scale, we must have (p1+m1)⋅L1=(p2+m2)⋅L2, since pk+mk represents all the mass on one side of the scale.
Distributing and simplifying, we have p1⋅L1−p2⋅L2=L2⋅m2−L1⋅m1. The LHS is constant, since we use the same balance every time, so we just let it equal k.
So k=L2⋅m2−L1⋅m1 for any choice of (m1,m2) that balances the scale. For brevity, let L1=u and L2=v. We have k=vm2−um1. Since k,u, and v are constants, we divide through by u to obtain a new equation with constants k′ and v′: k′=v′m2−m1.
Now, we plug in (m1,m2)∈{(A,A1),(B,B1),(C,C1),(A2,A),(B2,B)} and proceed to bash.
Should come out to C=BC1−AC1B1−A1+BA2−AB2B−A