China Mathematical Competition

Denote $P$, $B$, $C$ by $P(x_0, y_0)$, $B(0, b)$, $C(0, c)$, and assume that $b > c$. The equation for the line $PB$ is
$$y - b = \frac{y_0 - b}{x_0}x.$$
It can be rewritten as
$$(y_0 - b)x - x_0y + x_0b = 0.$$
Since the distance between the circle center $(1, 0)$ and the line $PB$ is $1$, we have
$$\frac{|y_0 - b + x_0b|}{\sqrt{(y_0 - b)^2 + x_0^2}} = 1.$$
That is to say,
$$(y_0 - b)^2 + x_0^2 = (y_0 - b)^2 + 2x_0b(y_0 - b) + x_0^2b^2.$$
It is easy to see that $x_0 > 2$. Then the last equation can be simplified as
$$(x_0 - 2)b^2 + 2y_0b - x_0 = 0.$$
In a similar way,
$$(x_0 - 2)c^2 + 2y_0c - x_0 = 0.$$
Therefore,
$$b + c = \frac{-2y_0}{x_0 - 2}, \quad bc = \frac{-x_0}{x_0 - 2}.$$
Then we get
$$(b-c)^2 = \frac{4x_0^2 + 4y_0^2 - 8x_0}{(x_0-2)^2}.$$
As $P(x_0, y_0)$ is on the parabola, $y_0^2 = 2x_0$. So we have
$$(b-c)^2 = \frac{4x_0^2}{(x_0-2)^2},$$
or $b-c = \frac{2x_0}{x_0-2}$. Then we have
$$\begin{aligned} S_{\triangle PBC} &= \frac{1}{2}(b-c) \times x_0 \\ &= \frac{x_0}{x_0-2} \times x_0 \\ &= (x_0-2) + \frac{4}{x_0-2} + 4 \\ &\ge 4 + 4 = 8. \end{aligned}$$
The equality holds when $x_0 - 2 = 2$; this means that $x_0 = 4$ and $y_0 = \pm 2\sqrt{2}$. So the minimum of $S_{\triangle PBC}$ is $8$.