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

m (Created page with "== 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 <math>A</math>, when placed in ...")
 
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== Problem ==
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== Solution ==
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A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight <math>A</math>, when placed in the left pan and against a weight <math>a</math>, when placed in the right pan. The corresponding weights for the second object are <math>B</math> and <math>b</math>. The third object balances against a weight <math>C</math>, when placed in the left pan. What is its true weight?
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The effect of the unequal arms and pans is that if an object of weight <math> x</math> in the left pan balances an object of weight <math>y </math> in the right pan, then <math> x = hy + k</math> for some constants <math>h</math> and <math>k</math>. Thus if the first object has true weight x, then <math> x = hA + k, a = hx +  k</math>.
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So <math>a = h^2A + (h+1)k</math>.  
  
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== Solution ==
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Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = \frac{a-b}{A-B} </math> and so
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{{solution}}
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<cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>.
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The true weight of the third object is thus:
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<cmath>
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hC + k = \
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\boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{(A - B)(\sqrt{ \frac {a-b}{A-B}}+ 1)}}
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</cmath>.
  
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== See Also ==
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More readably:
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{{USAMO box|year=1980|before=First Question|num-a=2}}
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<cmath>
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\boxed{ h=\sqrt{\frac{a-b}{A-B}} ;
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\
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\text{weight} = hC + \frac{bA - aB}{(A - B)(h + 1)}}
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</cmath>
  
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[[Category:Olympiad Algebra Problems]]
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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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