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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 of the two sides, the total torque will be <math>\text{[a constant] (due to the weight, and distribution of the weight, of the arm itself)} + \text{[the length of the arm]} \times \text{[the weight of the object in the pan]}</math>.  Thus, the information we have tells us that, for some constants <math>x, y, z, u</math>:
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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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<cmath>\text{[arm+pan torque])} + \text{[arm length]} \times \text{[object weight]}</cmath>
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Thus, for some constants <math>x, y, z, u</math>:
  
 
<cmath>x + yA = z + ua</cmath>
 
<cmath>x + yA = z + ua</cmath>

Revision as of 13:15, 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

A balance scale will balance when the torques exerted on both sides cancel out. On each side, the total torque will be

\[\text{[arm+pan torque])} + \text{[arm length]} \times \text{[object weight]}\]

Thus, for some constants $x, y, z, u$:

\[x + yA = z + ua\] \[x + yB = z + ub\] \[x + yC = z + uc\]

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:

\[A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)\] \[B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)\] \[C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)\]

We can just give the names $X$ and $Y$ to the quantities $\frac{z-x}{y}$ and $\frac{u}{y}$.

\[A = X + Ya\] \[B = X + Yb\] \[C = X + Yc\]

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:

\[A = X + Ya\implies \boxed{X = A - Ya}\]

\begin{align*} B &= X + Yb\\ \implies B &= A - Ya + Yb\\ \implies Y(b-a) &= B-A\\ \implies Y &= \frac{B-A}{b-a}\\ \implies X &= \boxed{A - a\left(\frac{B-A}{b-a}\right)} \end{align*} \begin{align*} C &= X + Yc\\ \implies Yc &= C - X\\ \implies c &= \frac{C-X}{Y}\\ \implies c &= \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}}\\ \implies c &= \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}\\ \implies c &= \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}\\ \implies c &= \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}\\ \implies c &= \frac{Cb - Ca - Ab + Ba}{B-A} \end{align*} So the answer is: \[\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}\].

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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