Number theory: $a\in A\iff \frac{1}{2}-a\in A$.

Solution 1:

Let $$ a = \frac{xy+yz+zx}{x^2+y^2+z^2} $$ as above, note that this is equivalent to $$ 1+2a = \frac{(x+y+z)^2}{x^2+y^2+z^2} $$ and let $a'=\frac12-a$. If $x=y=z$ then $a=1$ and $$ a' = -\frac12 = \frac{-1\cdot 2 -1\cdot 2 + 1}{(-1)^2+(-1)^2+2^2} \in A $$ Otherwise, if $x,y,z$ are not all equal then $$ 1+2a' = 2-2a = \frac{u+v+w}{x^2+y^2+z^2} = \frac{(u+v+w)^2}{u^2+v^2+w^2} $$ where $$ u = x^2+y^2-z(x+y) \\ v = y^2+z^2-x(y+z) \\ w = z^2+x^2-y(z+x) \\ u+v+w = (x-y)^2+(y-z)^2+(z-x)^2>0 $$ clearly $u,v,w$ are not all zero and hence $$ a' = \frac{uv+vw+wu}{u^2+v^2+w^2} \in A $$ Thus $a\in A \implies \frac12-a \in A$, and the other direction follows immediately because $a'\in A \implies \frac12-a' = a \in A$.