China Mathematical Competition

Let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(t^2, 2t)$. From
$$\begin{cases} x - 2y - 1 = 0, \\ y^2 = 4x, \end{cases}$$
we get $y^2 - 8y - 4 = 0$, which means $y_1 + y_2 = 8$, $y_1 \cdot y_2 = -4$.
Since $x_1 = 2y_1 + 1$, $x_2 = 2y_2 + 1$, we have
$$\begin{aligned} x_1 + x_2 &= 2(y_1 + y_2) + 2 = 18, \\ x_1 \cdot x_2 &= 4y_1 \cdot y_2 + 2(y_1 + y_2) + 1 = 1. \end{aligned}$$
Furthermore, by $\angle ACB = 90^\circ$, we have $\vec{CA} \cdot \vec{CB} = 0$, which means
$$(t^2 - x_1)(t^2 - x_2) + (2t - y_1)(2t - y_2) = 0,$$
that is
$$t^4 - (x_1 + x_2)t^2 + x_1 \cdot x_2 + 4t^2 - 2(y_1 + y_2)t + y_1 \cdot y_2 = 0.$$
Then
$$t^4 - 14t^2 - 16t - 3 = 0,$$
or
$$(t^2 + 4t + 3)(t^2 - 4t - 1) = 0.$$
Obviously, $t^2 - 4t - 1 \neq 0$; otherwise, we have $t^2 - 2 \cdot 2t - 1 = 0$, which means $C$ is on $x - 2y - 1 = 0$, i.e., $C$ coincides with either $A$ or $B$. So $t^2 + 4t + 3 = 0$. Then $t_1 = -1$, $t_2 = -3$.
Therefore, the coordinate of $C$ is $(1, -2)$ or $(9, -6)$.