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

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 $y$ in the right pan, then $x = hy + k$ for some constants $h$ and $k$ . Thus if the first object has true weight x, then $x = hA + k, a = hx + k$ .

So $a = h^2A + (h+1)k$ .

Similarly, $b = h^2B + (h+1)k$ . Subtracting gives $h^2 = (a - b)/(A - B)$ and so

$(h+1)k = a - h^2A = (bA - aB)/(A - B)$ .

The true weight of the third object is thus:

$hC + k = \\ \sqrt{ (a-b)/(A-B)} C + (bA - aB)/(A - B) \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}$ .

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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