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

(→‎Solution: Submitted a solution. I believe it to be correct.)
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== Problem ==
 
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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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== Solution ==
 
== Solution ==
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{{solution}}
 
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A balance scale will balance when the torques exerted on both sides cancel out.  On each of the two sides, the total torque will be [some constant amount] (due to the weight, and distribution of the weight, of the arm itself) plus [the length of the arm] times [the weight of what is sitting in the pan].  Thus, the information we have tells us that, for some constants x, y, z, u:
 
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x + yA = z + ua
 
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x + yB = z + ub
 
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x + yC = z + uc
 
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In fact, we don't exactly care what x,y,z,u are.  By subtracting x from all equations and dividing by y, we get:
 
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A = (z-x)/y + (u/y)a
 
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B = (z-x)/y + (u/y)b
 
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C = (z-x)/y + (u/y)c
 
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We can just give the names X and Y to the quantities (z-x)/y and (u/y).
 
  
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A = X + Ya
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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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B = X + Yb
 
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C = X + Yc
 
  
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Our task is to compute c in terms of A, a, B, b, and C. This can be done by solving for X and Y in terms of A,a,B,b and eliminating them from the implicit expression for c in the last equation.  Perhaps there is a shortcut, but this will work:
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So <math>a = h^2A + (h+1)k</math>.  
  
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A = X + Ya
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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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=> X = A - Ya
 
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B = X + Yb
 
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=> B = A - Ya + Yb
 
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=> Y(b-a) = B-A
 
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=> Y = (B-A)/(b-a)
 
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=> X = A - (B-A)/(b-a) * a
 
  
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C = X + Yc
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<cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>.
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=> Yc = C - X
 
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=> c = (C-X)/Y
 
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=> c = (C - [A - (B-A)/(b-a) * a]) / [(B-A)/(b-a)]
 
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=> [simplify numerator]
 
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c = (C - A + a(B-A)/(b-a)) / [(B-A)/(b-a)]
 
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=> [multiply numerator and denominator by (b-a)]
 
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c = (C(b-a) - A(b-a) + a(B-A)) / (B-A)
 
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=> [distribute numerator]
 
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c = (Cb - Ca - Ab + Aa + Ba - Aa) / (B-A)
 
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=> [cancel Aa's]
 
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c = (Cb - Ca - Ab + Ba) / (B-A)
 
  
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So the answer is: (Cb - Ca - Ab + Ba) / (B-A).
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The true weight of the third object is thus:
  
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[Someone else feel free to clean up the formatting here.]
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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 13:46, 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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