# 1980 USAMO Problems/Problem 1

## Problem

A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight $A$, when placed in the left pan and against a weight $a$, when placed in the right pan. The corresponding weights for the second object are $B$ and $b$. The third object balances against a weight $C$, when placed in the left pan. What is its true weight?

## Solution

A balance scale will balance when the torques exerted on both sides cancel out. On each of the two sides, the total torque will be $\text{[some constant amount] (due to the weight, and distribution of the weight, of the arm itself) } + \text{ [the length of the arm] } \times \text{ [the weight of what is sitting in the pan]}$. Thus, the information we have tells us that, for some constants $x, y, z, u$:

$$x + yA = z + ua$$ $$x + yB = z + ub$$ $$x + yC = z + uc$$

In fact, we don't exactly care what $x,y,z,u$ are. By subtracting $x$ from all equations and dividing by $y$, we get:

$$A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)$$ $$B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)$$ $$C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)$$

We can just give the names $X$ and $Y$ to the quantities $\frac{z-x}{y}$ and $\frac{u}{y}$.

$$A = X + Ya$$ $$B = X + Yb$$ $$C = X + Yc$$

Our task is to compute $c$ in terms of $A$, $a$, $B$, $b$, and $C$. This can be done by solving for $X$ and $Y$ in terms of $A$,$a$,$B$,$b$ and eliminating them from the implicit expression for $c$ in the last equation. Perhaps there is a shortcut, but this will work:

$$A = X + Ya\implies \boxed{X = A - Ya}$$ $$B = X + Yb\implies B = A - Ya + Yb\implies Y(b-a) = B-A\implies Y = \frac{B-A}{b-a}\implies X = \boxed{A - a\left(\frac{B-A}{b-a}\right)}$$ $$C = X + Yc\implies Yc = C - X\implies c = \frac{C-X}{Y}\implies c = \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}} \implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}} \implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A} \implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A} \implies c = \frac{Cb - Ca - Ab + Ba}{B-A}$$

So the answer is: $$\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}$$.