China Mathematical Competition (Hainan)

(1) The equations of lines $AB$, $AC$ and $BC$ are $y=\frac{4}{3}(x+1)$, $y=-\frac{4}{3}(x-1)$ and $y=0$ respectively. The distances from point $P$ to $AB$, $AC$ and $BC$ are respectively $$d_1=\frac{1}{5}|4x-3y+4|,$$ $$d_2=\frac{1}{5}|4x+3y-4|,d_3=|y|.$$ According to the assumption, $d_1{d}_2=d_3^2$, we have $|16x^2-(3y-4{)}^2|=25y^2$. That is $$16x^2-(3y-4{)}^2+25y^2=0,\text{ or }16x^2-(3y-4{)}^2-25y^2=0.$$ By simplifying the above equations, we obtain that the locus equations of point $P$ consist of $$\text{circle S: }2x^2+2y^2+3y-2=0$$ and $$\text{hyperbola T: }8x^2-17y^2+12y-8=0.$$

(2) The incenter of $△ABC$ is also a point satisfying the assumption. In view of $d_1=d_2=d_3$, solving the equations we have $D(0,\frac{1}{2})$. Line $L$ passes through $D$, and has three common points with the locus of point $P$. So the slope of $L$ is defined. Suppose that the equation of $L$ is $$y=kx+\frac{1}{2}.$$ (i) If $k=0$, then $L$ is tangent to the circle $S$, which means there is a unique common point $D$. In this case line $L$ is parallel to the $x$-axis, which implies that $L$ and hyperbola $T$ have two other common points different from point $D$. Hence, there are just three common points for $L$ and the locus of point $P$.

Case 1: Line $L$ passes through point $B$ or point $C$, which means that the slope of $L$ is $k=\pm \frac{1}{2}$, and the equation of $L$ is $x=\pm (2y-1)$. Substitute it into equation $\stackrel{◯}{2}$ we get $$y(3y-4)=0.$$ Solving it we have $E(\frac{5}{3},\frac{4}{3})$ or $F(-\frac{5}{3},\frac{4}{3})$, which means that line $BD$ and curve $T$ have 2 intersection points $B$ and $E$, and line $CD$ and curve $T$ have 2 intersection points $C$ and $F$.

Case 2: Line $L$ does not pass through point $B$ and point $C$ (i.e. $k\ne \pm \frac{1}{2}$). Since for $L$ and $S$ there are two different intersection points, there exists a unique common point for $L$ and hyperbola $T$. Thus for the following system of equations $$\{\begin{array}{l}8x^2-17y^2+12y-8=0,\\ y=kx+\frac{1}{2},\end{array}$$ there is one and only one real solution. After eliminating $y$ and simplifying we have $$(8-17k^2)x^2-5kx-\frac{25}{4}=0.$$ The above equation has a unique real solution if and only if $$8-17k^2=0$$ or $$(-5k{)}^2+4(8-17k^2)\frac{25}{4}=0.$$ Solving $8-17k^2=0$ we get $k=\pm \frac{2\sqrt{34}}{17}$. And solving $(-5k{)}^2+4(8-17k^2)\frac{25}{4}=0$ we obtain $k=\pm \frac{\sqrt{2}}{2}$. Consequently, the set of all possible values of the slope $k$ of line $L$ is the following finite set: $$\{0,\pm \frac{1}{2},\pm \frac{2\sqrt{34}}{17},\pm \frac{\sqrt{2}}{2}\}$$