China Mathematical Competition

Let $P(x_0,x_0+\frac{2}{x_0})$. The expression for line $PA$ is then $$y-\left(x_0+\frac{2}{x_0}\right)=-(x-x_0),$$ or $$y=-x+2x_0+\frac{2}{x_0}.$$ From $$\{\begin{array}{l}y=x,\\ y=-x+2x_0+\frac{2}{x_0},\end{array}$$ we get $A(x_0+\frac{1}{x_0},x_0+\frac{1}{x_0})$. On the other hand, we have $B(0,x_0+\frac{2}{x_0})$. Then $\overrightarrow{PA}=(\frac{1}{x_0},-\frac{1}{x_0})$ and $\overrightarrow{PB}=(-x_0,0)$. Therefore, $$\overrightarrow{PA}\cdot \overrightarrow{PB}=\frac{1}{x_0}\cdot (-x_0)=-1.$$ The answer is $-1$.

---

Alternative solution.

As seen in Fig. 1.1, the distances from $P(x_0,x_0+\frac{2}{x_0})$ to lines $y=x$ and $y$-axis, respectively, are

Fig. 1.1 $$|PA|=\frac{|x_0-\left(x_0+\frac{2}{x_0}\right)|}{\sqrt{2}}=\frac{\sqrt{2}}{x_0}$$ and $$|PB|=x_0.$$ Since $O$, $A$, $P$ and $B$ are concyclic points, then $$\angle APB=\pi -\angle AOB=\frac{3\pi }{4}.$$ Therefore, $\overrightarrow{PA}\cdot \overrightarrow{PB}=|\overrightarrow{PA}|\cdot |\overrightarrow{PB}|\cdot \cos \frac{3\pi }{4}=-1$. $\square$