Given the Sum, Find the Sum (Posted on 2024-05-29)

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

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
1/a+1/b+1/c=1/2022=1/(a+b+c) => 1/(a+b+c) -1/a =1/b+1/c => -(b+c)/(a(a+b+c))= (b+c)/bc) => (b+c)(1/(bc) + 1/a(a+b+c)) = 0 => (b+c) [(a^2 +ab+ac+bc)abc(a+b+c)] =0 => (b+c)(a+b)(c+a)/(abc(a+b+c)) =0 => (b+c)(a+b)(c+a)=0. Case 1: b+c=0 Case 2: a+b=0 Case 3: c+a =0 Considering Case 1: b+c=0=> b=-c Then, a+b+c = 2022=> a = 2022 Hence, we have: 1/a^2023 + 1/b^2023 - 1/b^2024 = 1/2022^2023. Each of the other two cases gives the same result for 1/a^2023 + 1/b^2023 + 1/c^2023. Consequently, the required value of the expression is 1/2022^2023.

	Subject	Author	Date
	Question	Brian Smith	2024-07-25 21:05:35
	some thoughts	broll	2024-05-30 01:27:12