1. Let $0<a<b<c$, real numbers $x, y$ satisfy $2 x+$ $2 y=a+b+c, 2 x y=a c$. Then the range of $x, y$ is ( ).
(A) $a<x<b, 0<y<a$
(B) $b<x<c, 0<y<a$
(C) $0<x<a, a<y<c$
(D) $a<x<c, b<y<c$
,$- 1 . C$.
It can be known that $x, y$ are the roots of the equation $f(t)=t^{2}-\frac{1}{2}(a+b+c) t+$ $\frac{1}{2} a c=0$, and $f(0)=\frac{1}{2} a c>0, f(a)=\frac{1}{2} a(a$ $-b)0$, so the two roots $x, y$ of the equation $f(t)=0$ satisfy $0<x<a, a<y<c$.