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Slovenia — National Math Olympiad

Algebra Difficulty 4.6 AIME Prove it Slovenia

For real numbers aa, bb and cc we have
(2b−a)2+(2b−c)2=2(2b2−ac). (2b - a)^2 + (2b - c)^2 = 2(2b^2 - ac).
Prove that the numbers aa, bb and cc are three consecutive terms in some arithmetic sequence.

Solution

The given equation is equivalent to 8b2−4ab−4bc+a2+c2=4b2−2ac8b^2 - 4ab - 4bc + a^2 + c^2 = 4b^2 - 2ac or
4b2−4ab−4bc+a2+2ac+c2=0. 4b^2 - 4ab - 4bc + a^2 + 2ac + c^2 = 0.
This can be further rewritten as (a+c)2−4b(a+c)+4b2=0(a + c)^2 - 4b(a + c) + 4b^2 = 0 and finally as
(a+c−2b)2=0. (a + c - 2b)^2 = 0.
From here we conclude that a+c=2ba + c = 2b or, equivalently, b−a=c−bb - a = c - b. Hence, aa, bb and cc are consecutive terms of an arithmetic sequence.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.