Difference between revisions of "1980 USAMO Problems/Problem 1"
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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 | A balance scale will balance when the torques exerted on both sides cancel out. On each side, the total torque will be | ||
| − | <cmath>\text{[arm+pan torque] | + | <cmath>\text{[arm+pan torque]} + \text{[arm length]} \times \text{[object weight]}</cmath> |
Thus, for some constants <math>x, y, z, u</math>: | Thus, for some constants <math>x, y, z, u</math>: | ||
Revision as of 13:16, 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 
, when placed in the left pan and against a weight 
, when placed in the right pan. The corresponding weights for the second object are 
and 
. The third object balances against a weight 
, 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
![\[\text{[arm+pan torque]} + \text{[arm length]} \times \text{[object weight]}\]](http://latex.artofproblemsolving.com/9/2/2/922594d04856e71385a4cbc60f4e7f8b93f98faa.png)
Thus, for some constants 
:
![\[x + yA = z + ua\]](http://latex.artofproblemsolving.com/1/a/f/1af6a9374021ffb043d18b0abb8d771b5d1311e4.png)
![\[x + yB = z + ub\]](http://latex.artofproblemsolving.com/1/5/3/15340a88bc18303142f0b1334c83e8f26bd71435.png)
![\[x + yC = z + uc\]](http://latex.artofproblemsolving.com/6/4/8/648687d88e8a986a35afc243b31e23b3c9930ef7.png)
In fact, we don't exactly care what 
are. By subtracting 
from all equations and dividing by 
, we get:
![\[A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)\]](http://latex.artofproblemsolving.com/e/5/8/e58c096b55eb98091bfb59be454222144a07f502.png)
![\[B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)\]](http://latex.artofproblemsolving.com/e/a/5/ea55121cf120f2d49450369f185cabe63e2dc3e6.png)
![\[C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)\]](http://latex.artofproblemsolving.com/a/0/f/a0f8407f526aa4fdf9444931b8af5b0acc19ec13.png)
We can just give the names 
and 
to the quantities 
and 
.
![\[A = X + Ya\]](http://latex.artofproblemsolving.com/e/7/3/e730933b73293d555310bf98793c8a3ee4e98942.png)
![\[B = X + Yb\]](http://latex.artofproblemsolving.com/2/c/e/2ce988e326ebcaae2e621d37689a905f54c83aa2.png)
![\[C = X + Yc\]](http://latex.artofproblemsolving.com/5/0/1/5019b1fed38e6dc38657a32152a78020f43fcf9f.png)
Our task is to compute 
in terms of , , , , and . This can be done by solving for and in terms of ,,, and eliminating them from the implicit expression for in the last equation. Perhaps there is a shortcut, but this will work:
![\[A = X + Ya\implies \boxed{X = A - Ya}\]](http://latex.artofproblemsolving.com/f/5/f/f5f37f1ea2a7ae57b451147c500de3061b9492a5.png)
![\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*}](http://latex.artofproblemsolving.com/9/8/3/983d725b273665f2b618d20e7dd7238abd6d3349.png)
So the answer is: ![\[\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}\]](http://latex.artofproblemsolving.com/8/0/8/8084ecfa518d7b7a3a3abcab389b205fc2fe3ba5.png)
.
See Also
| 1980 USAMO (Problems • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
