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

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== Solution ==
 
== Solution ==
  
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A balance scale will balance when the torques exerted on both sides cancel out.  On each side, the total torque will be
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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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<cmath>\text{[arm+pan torque]} + \text{[arm length]} \times \text{[object weight]}</cmath>
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So <math>a = h^2A + (h+1)k</math>.
  
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Thus, for some constants <math>x, y, z, u</math>:
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Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = (a - b)/(A - B)</math> and so
  
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<cmath>x + yA = z + ua</cmath>
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<cmath>(h+1)k = a - h^2A = (bA - aB)/(A - B)</cmath>.
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<cmath>x + yB = z + ub</cmath>
 
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<cmath>x + yC = z + uc</cmath>
 
  
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In fact, we don't exactly care what <math>x,y,z,u</math> are.  By subtracting <math>x</math> from all equations and dividing by <math>y</math>, we get:
+
The true weight of the third object is thus:
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<cmath>A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)</cmath>
 
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<cmath>B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)</cmath>
 
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<cmath>C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)</cmath>
 
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We can just give the names <math>X</math> and <math>Y</math> to the quantities <math>\frac{z-x}{y}</math> and <math>\frac{u}{y}</math>.
 
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<cmath>A = X + Ya</cmath>
 
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<cmath>B = X + Yb</cmath>
 
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<cmath>C = X + Yc</cmath>
 
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Our task is to compute <math>c</math> in terms of <math>A</math>, <math>a</math>, <math>B</math>, <math>b</math>, and <math>C</math>.  This can be done by solving for <math>X</math> and <math>Y</math> in terms of <math>A</math>,<math>a</math>,<math>B</math>,<math>b</math> and eliminating them from the implicit expression for <math>c</math> in the last equation.  Perhaps there is a shortcut, but this will work:
 
−
 
 
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<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>
 
  
 
<cmath>
 
<cmath>
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\begin{align*}
+
hC + k = \
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B &= X + Yb\
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\sqrt{ (a-b)/(A-B)} C
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\implies B &= A - Ya + Yb\
+
+ (bA - aB)/(A - B) \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}
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\implies Y(b-a) &= B-A\
+
</cmath>.
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\implies Y &= \frac{B-A}{b-a}\
 
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\implies X &= \boxed{A - a\left(\frac{B-A}{b-a}\right)}
 
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\end{align*}
 
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\begin{align*}
 
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C &= X + Yc\
 
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\implies Yc &= C - X\
 
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\implies c &= \frac{C-X}{Y}\
 
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\implies c &= \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}}\
 
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\implies c &= \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}\\
 
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\implies c &= \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}\
 
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\implies c &= \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}\
 
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\implies c &= \frac{Cb - Ca - Ab + Ba}{B-A}
 
−
\end{align*}
 
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</cmath>
 
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So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 14:26, 26 March 2023

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 $A$, when placed in the left pan and against a weight $a$, when placed in the right pan. The corresponding weights for the second object are $B$ and $b$. The third object balances against a weight $C$, when placed in the left pan. What is its true weight?

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 $h^2 = (a - b)/(A - B)$ and so

\[(h+1)k = a - h^2A = (bA - aB)/(A - B)\].

The true weight of the third object is thus:

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

See Also

1980 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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