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 $A_1$, $B_1$, and $C_1$, respectively. When $A$ and $B$ are placed on the right-hand pan, they are balanced by $A_2$ and $B_2$, respectively. Determine the true weight of $C$ in terms of $A_1, B_1, C_1, A_2$, and $B_2$.
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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 $L_1$ be the length of the left arm of the balance scale and $L_2$ be the length of the right arm. Let $p_1$ be the mass of the pan on the left and $p_2$ be the mass of the pan on the right. Finally, let $m_1$ be the combined mass of the weight(s) we put on the left side and $m_2$ be the combined mass of the weight(s) we put on the right side.
To balance the scale, we must have $(p_1 + m_1) \cdot L_1 = (p_2 + m_2) \cdot L_2$, since $p_k + m_k$ represents all the mass on one side of the scale.
Distributing and simplifying, we have $p_1 \cdot L_1 - p_2 \cdot L_2 = L_2 \cdot m_2 - L_1 \cdot m_1$. The LHS is constant, since we use the same balance every time, so we just let it equal $k$.
So $k = L_2 \cdot m_2 - L_1 \cdot m_1$ for any choice of $(m_1, m_2)$ that balances the scale. For brevity, let $L_1 = u$ and $L_2 = v$. We have $k = vm_2 - um_1$. Since $k, u,$ and $v$ are constants, we divide through by $u$ to obtain a new equation with constants $k'$ and $v'$: $k' = v'm_2 - m_1$.
Now, we plug in $(m_1, m_2) \in \{(A, A_1), (B, B_1), (C, C_1), (A_2, A), (B_2, B)\}$ and proceed to bash.
Should come out to $C = \frac{BC_1 - AC_1}{B_1 - A_1} + \frac{BA_2 - AB_2}{B - A}$