Difference between revisions of "1980 USAMO Problems/Problem 1"
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== Solution == | == Solution == | ||
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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 = | + | 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 = | + | <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: | ||
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<cmath> | <cmath> | ||
hC + k = \\ | hC + k = \\ | ||
− | \boxed{\sqrt{ | + | \boxed{\sqrt{ \frac{a-b}{A-B}} C + \frac{bA - aB}{A - B} \frac{1}{\sqrt{ (a-b)/(A-B) }+ 1}} |
</cmath>. | </cmath>. | ||
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Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html | Credit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html | ||
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Revision as of 13:40, 26 March 2023
Solution
The effect of the unequal arms and pans is that if an object of weight in the left pan balances an object of weight in the right pan, then for some constants and . Thus if the first object has true weight x, then .
So .
Similarly, . Subtracting gives and so
.
The true weight of the third object is thus:
.
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