c is not equal to a. because in the equation of x :
a(1-x)^2+2bx+c(1+x^2)=2
a-ax^2+2bx+c+cx^2-2=0
(c-a)x^2+2bx+(a+c-2)=0
when the equation of ax^2+bx+c=0 have 2 ame soloution, it have b^2-4ac=0
in this case, a=c-a b=2b c=a+c-2,so
(2b)^2-4(c-a)(a+c-2)=0
4b^2-4(ac+c^2-2c-a^2-ac+2a)=0
b^2=c^2-2c-a^2+2a
b^2=c^2-2c+1-a^2+2a-1
B^2=(c-1)^2-（a^2-2a+1)
B^2=(c-1)^2-(a-1)^2
so if a=c then (c-1)^2=(a-1)^2, (c-1)^2-(a-1)^2=0， b=0
a side of a triangle is always bigger than 0
so a not equal to c
i also alologize for the question should be a(1-x^2)+2bx+(1+x^2)c=0
i am really sorry about that. I typed it in a wrong way
Good job melody and heureka!
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