1taking a-b=x , b-c=y , c-a=z
we get
(y-z)^2 + (z-x)^2 + (x-y)^2 = x^2 + y^2 + z^2
but we want
\\ (x-y)^2=x^2\\ (y-z)^2=y^2\\ (z-x)^2=z^2
so only possibility is x=y=z=0
or a=b=c
so ratio is 1:1:1
a,b, c are real numbers such that
(a+b-2c)^2 + (b+c-2a)^2 + (c+a-2b)^2 = (a-b)^2 + (b-c)^2 + (c-a)^2
Find a:b:c
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6 Answers
21one possible ratio is 1:1:1
as the variables are symmetric!!
and also lhs = rhs = 0
1
9no sir no more possibilities exist
opening up squares and solving u wud end up with
Σ(a-b)2 = 0
implying a=b=c