Difference between revisions of "1980 USAMO Problems/Problem 1"

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More readably: <math>\boxed{ h=\sqrt{\frac{a-b}{B-A}} ; \text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}</math>
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More readably: <math>\boxed{ h=\sqrt{\frac{a-b}{A-B}} \\
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\text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}</math>
  
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html

Revision as of 13:45, 26 March 2023

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:

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

More readably: $\boxed{ h=\sqrt{\frac{a-b}{A-B}} \\ \text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}$

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