Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

1980 USAMO Problems/Problem 1: Difference between revisions

← Older editNewer edit →
Etude (talk | contribs)
181 edits
No edit summary
Integralarefun (talk | contribs)
editor
149 edits
→Solution: ; the wiki doesn't know align* exists for some reason
Line 4: Line 4:
== Solution ==
== Solution ==


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{[some constant amount] (due to the weight, and distribution of the weight, of the arm itself) } + \text{ [the length of the arm] } \times \text{ [the weight of what is sitting in the pan]}</math>.  Thus, the information we have tells us that, for some constants <math>x, y, z, u</math>:
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>:


<cmath>x + yA = z + ua</cmath>
<cmath>x + yA = z + ua</cmath>
Line 25: Line 25:


<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>
<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>
<cmath>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)}</cmath>
 
<cmath>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}}
\begin{align*}
\implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}
B &= X + Yb\\
\implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}
\implies B &= A - Ya + Yb\\
\implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}
\implies Y(b-a) &= B-A\\
\implies   c = \frac{Cb - Ca - Ab + Ba}{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*}


So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.
So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.

Revision as of 18:05, 20 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 of the two sides, the total torque will be $\text{[a constant] (due to the weight, and distribution of the weight, of the arm itself)} + \text{[the length of the arm]} \times \text{[the weight of the object in the pan]}$. Thus, the information we have tells us that, 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_USAMO_Problems/Problem_1&oldid=191124"