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

Revision as of 14:37, 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:

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

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

1980 USAMO (Problems • Resources)
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Difference between revisions of "1980 USAMO Problems/Problem 1"

Revision as of 14:37, 26 March 2023

Problem

Solution

See Also

@@ Line 15: / Line 15: @@
 <cmath>
- hC + k = \\
+hC + k = \\
- \sqrt{ (a-b)/(A-B)} C
+\boxed{\sqrt{ (a-b)/(A-B)} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}}
- + (bA - aB)/(A - B) \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}
 </cmath>.
+More readably: <math>\boxed{ h=\sqrt{\frac{a-b}{b-a}} ; \text{weight} = hC + \frac{bA - aB}{A - B} \frac{1}{h + 1}</math>
 Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html