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 $$(y_0-b)x-x_{0}y+x_{0}b=0.$$ Since the distance between the circle center $(1,0)$ and the line $PB$ is $1$, we have $$\frac{|y_0-b+x_{0}b|}{\sqrt{(y_0-b{)}^2+x_0^2}}=1.$$ That is to say, $$(y_0-b{)}^2+x_0^2=(y_0-b{)}^2+2x_{0}b(y_0-b)+x_0^2{b}^2.$$ It is easy to see that $x_0>2$. Then the last equation can be simplified as $$(x_0-2)b^2+2y_{0}b-x_0=0.$$ In a similar way, $$(x_0-2)c^2+2y_{0}c-x_0=0.$$ Therefore, $$b+c=\frac{-2y_0}{x_0-2},bc=\frac{-x_0}{x_0-2}.$$ 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 $$\begin{array}{rl}{S}_{△PBC}& =\frac{1}{2}(b-c)\times x_0\\ & =\frac{x_0}{x_0-2}\times x_0\\ & =(x_0-2)+\frac{4}{x_0-2}+4\\ & \ge 4+4=8.\end{array}$$ 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 $S_{△PBC}$ is $8$.