Given that:
a+b+c= 2022
1/(a2022) + 1/(b2022) + 1/(c2022) = 1/2022
Find the value of:
1/(a2023) + 1/(b2023) + 1/(c2023)
**** Adapted from a problem appearing in Vietnamese IMO, 2022
Note: the given equations are symmetrical in a,b,c.
Note2: we are asked to find 'the' sum, implying that there are not many solutions.
Call the two given equations eq1 and eq2.
Starting with eq2, take the simpler case of 1/(a2022) = 1/2022, then a^x=x, so a is the 2022th root of 2022, a number rather near 1.
Then let b=c= (2022-a)/2, a number around 1000, whose 2022nd power will be immense, making the terms 1/(b2022) + 1/(c2022) negligible, and 1/(b2023) + 1/(c2023) even smaller.
Hence we have shown that there is a solution to eq2 and that solution occurs very near
a =2022^(1/2022)
But in fact we don't need to know a or b or c, just the aggregate value of the third expression.
If 1/a^x=1/x, then 1/a^(x+1) =1/(x*(x)^(1/x)) is a real solution.
Since the terms in b and c are negligible, we conclude that the required solution is 1/(2022*2022^(1/2022))
Edited on May 30, 2024, 2:31 am
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Posted by broll
on 2024-05-30 01:27:12 |
some thoughts
