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1980 USAMO Problems/Problem 1: Difference between revisions

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== Solution ==
== Solution ==


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


<cmath>\text{[arm+pan torque]} + \text{[arm length]} \times \text{[object weight]}</cmath>
So <math>a = h^2A + (h+1)k</math>.


Thus, for some constants <math>x, y, z, u</math>:
Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = (a - b)/(A - B)</math> and so


<cmath>x + yA = z + ua</cmath>
<cmath>(h+1)k = a - h^2A = (bA - aB)/(A - B)</cmath>.
<cmath>x + yB = z + ub</cmath>
<cmath>x + yC = z + uc</cmath>


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:
 
<cmath>A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)</cmath>
<cmath>B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)</cmath>
<cmath>C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)</cmath>
 
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>.
 
<cmath>A = X + Ya</cmath>
<cmath>B = X + Yb</cmath>
<cmath>C = X + Yc</cmath>
 
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:
 
<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>


<cmath>
<cmath>
\begin{align*}
hC + k = \\
B &= X + Yb\\
\sqrt{ (a-b)/(A-B)} C
\implies B &= A - Ya + Yb\\
+ (bA - aB)/(A - B) \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}
\implies Y(b-a) &= B-A\\
</cmath>.
\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*}
</cmath>
So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.


== See Also ==
== See Also ==

Revision as of 13:26, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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