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

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So <math>a = h^2A + (h+1)k</math>.  
 
So <math>a = h^2A + (h+1)k</math>.  
  
−
Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = \frac{a-b}{b-a} </math> and so
+
Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = \frac{a-b}{A-B} </math> and so
  
 
<cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>.
 
<cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>.
Line 13: Line 13:
 
<cmath>
 
<cmath>
 
hC + k = \\
 
hC + k = \\
−
\boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}}
+
\boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{(A - B)(\sqrt{ \frac {a-b}{A-B}}+ 1)}}
 
</cmath>.
 
</cmath>.
  
−
More readably: <math>\boxed{ h=\sqrt{\frac{a-b}{b-a}} ; \text{weight} = hC + \frac{bA - aB}{A - B} \frac{1}{h + 1}</math>
+
More readably:  
 +
<cmath>
 +
\boxed{ h=\sqrt{\frac{a-b}{A-B}} ;
 +
\\
 +
\text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}
 +
</cmath>
  
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html

Latest revision as of 14:46, 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

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