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"

m (→‎Solution)
m (→‎Solution)
Line 1: Line 1:
−
== 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 <math>A</math>, when placed in the left pan and against a weight <math>a</math>, when placed in the right pan. The corresponding weights for the second object are <math>B</math> and <math>b</math>. The third object balances against a weight <math>C</math>, when placed in the left pan. What is its true weight?
 
−
 
 
== Solution ==
 
== Solution ==
  
Line 8: Line 5:
 
So <math>a = h^2A + (h+1)k</math>.  
 
So <math>a = h^2A + (h+1)k</math>.  
  
−
Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = (a - b)/(A - B)</math> and so
+
Similarly, <math>b = h^2B + (h+1)k</math>. Subtracting gives <math>h^2 = \frac{a-b}{b-a} </math> and so
  
−
<cmath>(h+1)k = a - h^2A = (bA - aB)/(A - B)</cmath>.
+
<cmath>(h+1)k = a - h^2A = \frac{bA - aB}{A - B}</cmath>.
  
 
The true weight of the third object is thus:
 
The true weight of the third object is thus:
Line 16: Line 13:
 
<cmath>
 
<cmath>
 
hC + k = \\
 
hC + k = \\
−
\boxed{\sqrt{ (a-b)/(A-B)} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}}
+
\boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}}
 
</cmath>.
 
</cmath>.
  
Line 22: Line 19:
  
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html
 
Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html
−
 
−
== See Also ==
 
−
{{USAMO box|year=1980|before=First Question|num-a=2}}
 
−
{{MAA Notice}}
 
−
 
−
[[Category:Olympiad Algebra Problems]]
 

Revision as of 14:40, 26 March 2023

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 $h^2 = \frac{a-b}{b-a}$ and so

\[(h+1)k = a - h^2A = \frac{bA - aB}{A - B}\].

The true weight of the third object is thus:

\[hC + k = \\ \boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}}\].

More readably: $\boxed{ h=\sqrt{\frac{a-b}{b-a}} ; \text{weight} = hC + \frac{bA - aB}{A - B} \frac{1}{h + 1}$ (Error compiling LaTeX. Unknown error_msg)

Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_USAMO_Problems/Problem_1&oldid=191385"