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

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

(→‎Solution: Oh god, the formatting was absolutely terrible. At least save the line breaks...)
Line 3: Line 3:
  
 
== Solution ==
 
== 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 [some constant amount] (due to the weight, and distribution of the weight, of the arm itself) plus [the length of the arm] times [the weight of what is sitting in the pan].  Thus, the information we have tells us that, for some constants x, y, z, u:
+
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>:
  
−
  x + yA = z + ua
+
<cmath>x + yA = z + ua</cmath>
−
  x + yB = z + ub
+
<cmath>x + yB = z + ub</cmath>
−
  x + yC = z + uc
+
<cmath>x + yC = z + uc</cmath>
  
−
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:
+
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:
  
−
  A = (z-x)/y + (u/y)a
+
<cmath>A = \frac{z-x}{y} + a\left(\frac{u}{y}\right)</cmath>
−
  B = (z-x)/y + (u/y)b
+
<cmath>B = \frac{z-x}{y} + b\left(\frac{u}{y}\right)</cmath>
−
  C = (z-x)/y + (u/y)c
+
<cmath>C = \frac{z-x}{y} + c\left(\frac{u}{y}\right)</cmath>
  
−
We can just give the names X and Y to the quantities (z-x)/y and (u/y).
+
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>.
  
−
  A = X + Ya
+
<cmath>A = X + Ya</cmath>
−
  B = X + Yb
+
<cmath>B = X + Yb</cmath>
−
  C = X + Yc
+
<cmath>C = X + Yc</cmath>
  
−
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:
+
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:
  
−
  A = X + Ya
+
<cmath>A = X + Ya\implies \boxed{X = A - Ya}</cmath>
−
  => X = A - Ya
+
<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>
−
  B = X + Yb
+
<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}}
−
  => B = A - Ya + Yb
+
\implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}}
−
  => Y(b-a) = B-A
+
\implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A}
−
  => Y = (B-A)/(b-a)
+
\implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A}
−
  => X = A - (B-A)/(b-a) * a
+
\implies   c = \frac{Cb - Ca - Ab + Ba}{B-A}</cmath>
−
 
 
−
  C = X + Yc
 
−
  => Yc = C - X
 
−
  => c = (C-X)/Y
 
−
  => c = (C - [A - (B-A)/(b-a) * a]) / [(B-A)/(b-a)]
 
−
  => [simplify numerator]
 
−
  c = (C - A + a(B-A)/(b-a)) / [(B-A)/(b-a)]
 
−
  => [multiply numerator and denominator by (b-a)]
 
−
  c = (C(b-a) - A(b-a) + a(B-A)) / (B-A)
 
−
  => [distribute numerator]
 
−
  c = (Cb - Ca - Ab + Aa + Ba - Aa) / (B-A)
 
−
  => [cancel Aa's]
 
−
   c = (Cb - Ca - Ab + Ba) / (B-A)
 
  
−
So the answer is: (Cb - Ca - Ab + Ba) / (B-A).
+
So the answer is: <cmath>\boxed{\frac{Cb - Ca - Ab + Ba}{B-A}}</cmath>.
−
 
 
−
[Someone else feel free to clean up the formatting here.]
 
  
 
== See Also ==
 
== See Also ==

Revision as of 18:55, 10 April 2013

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 $\text{[some constant amount] (due to the weight, and distribution of the weight, of the arm itself) } + \text{ [the length of the arm] } \times \text{ [the weight of what is sitting in the pan]}$. 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:

\[A = X + Ya\implies \boxed{X = A - Ya}\] \[B = X + Yb\implies B = A - Ya + Yb\implies Y(b-a) = B-A\implies Y = \frac{B-A}{b-a}\implies X = \boxed{A - a\left(\frac{B-A}{b-a}\right)}\] \[C = X + Yc\implies Yc = C - X\implies c = \frac{C-X}{Y}\implies c = \frac{C - [A - \frac{B-A}{b-a} \cdot a]}{\frac{B-A}{b-a}} \implies c = \frac{C - A + a\left(\frac{B-A}{b-a}\right)}{\frac{B-A}{b-a}} \implies c = \frac{C(b-a) - A(b-a) + a(B-A)}{B-A} \implies c = \frac{Cb - Ca - Ab + Aa + Ba - Aa}{B-A} \implies c = \frac{Cb - Ca - Ab + Ba}{B-A}\]

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_USAMO_Problems/Problem_1&oldid=52431"