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 side, the total torque will be

$\text{[arm+pan torque])} + \text{[arm length]} \times \text{[object weight]}$

Thus, 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:

$A = X + Ya\implies \boxed{X = A - Ya}$

$\begin{align*} B &= X + Yb\\ \implies B &= A - Ya + Yb\\ \implies Y(b-a) &= B-A\\ \implies Y &= \frac{B-A}{b-a}\\ \implies X &= \boxed{A - a\left(\frac{B-A}{b-a}\right)} \end{align*} \begin{align*} C &= X + Yc\\ \implies Yc &= C - X\\ \implies c &= \frac{C-X}{Y}\\ \implies c &= \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}}\\ \implies c &= \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}\\ \implies c &= \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}\\ \implies c &= \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}\\ \implies c &= \frac{Cb - Ca - Ab + Ba}{B-A} \end{align*}$ So the answer is: $\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}$ .

1980 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
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1980 USAMO Problems/Problem 1

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