## Problem 1

A balance has unequal arms and pans of unequal weight. It is used to weigh two objects of unequal weight. 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 $\mathrm{B}$ and $\mathrm{b}$. A third object balances against a weight C, when placed in the left pan. What is its true weight?


## Solution

The effect of the unequal arms and pans is that if an object of weight $x$ in the left pan balances an object of weight $\mathrm{y}$ in the right pan, then $\mathrm{x}=\mathrm{hy}+\mathrm{k}$ for some constants $\mathrm{h}$ and $\mathrm{k}$. Thus if the first object has true weight
$\mathrm{x}$, then $\mathrm{x}=\mathrm{hA}+\mathrm{k}, \mathrm{a}=\mathrm{h} x+\mathrm{k}$. So $\mathrm{a}=\mathrm{h}^{2} \mathrm{~A}+(\mathrm{h}+1) \mathrm{k}$. Similarly, $\mathrm{b}=\mathrm{h}^{2} \mathrm{~B}+(\mathrm{h}+1) \mathrm{k}$. Subtracting gives $\mathrm{h}^{2}=(a-$ b) /(A - B). and hence $(h+1) k=a-h^{2} A=(b A-a B) /(A-B)$.

The true weight of the third object is thus $h C+k=\sqrt{ }((a-b) /(A-B)) C+(b A-a B) /(A-B) 1 /(\sqrt{ }(a-b) /(A-B))$ $+1)$.

