1980 USAMO Problems/Problem 1

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 $h^2 = \frac{a-b}{A-B}$ and so

$(h+1)k = a - h^2A = \frac{bA - aB}{A - B}$ .

The true weight of the third object is thus:

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

More readably: $\boxed{ h=\sqrt{\frac{a-b}{B-A}} ; \text{weight} = hC + \frac{bA - aB}{A - B} \frac{1}{h + 1}$ (Error compiling LaTeX. Unknown error_msg)

Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html

Page

Toolbox

Search

1980 USAMO Problems/Problem 1

Solution