1. As shown in Figure 1, in the Cartesian coordinate system, the graph of the quadratic function $y=a x^{2}+m c(a \neq$ $0)$ passes through three vertices $A$, $B$, and $C$ of the square $A B O C$, and $a c=-2$. Then the value of $m$ is ( ).
(A) 1
(B) -1
(C) 2
(D) -2
-1.A.
From the problem, we know $A(0, m c), O A=m c=B C$.
Also, $B$ and $C$ are symmetric with respect to the $y$-axis, so $C\left(\frac{1}{2} m c, \frac{1}{2} m c\right)$.
Thus, $\frac{1}{2} m c=a\left(\frac{1}{2} m c\right)^{2}+m c$, which gives $a m c=-2$.
Also, $a c=-2$, so $m=1$.