China Western Mathematical Olympiad

Denote the lengths of three sides of $△ABC$ by $BC=a$, $CA=b$ and $AB=c$ respectively. No loss of generality, we can suppose $b\ne c$. We choose a rectangular coordinate system (see the figure), then we have $A(m,n)$, $B(0,0)$, $C(a,0)$ and $P(x,y)$.



Since $A{F}^2-B{F}^2=A{P}^2-B{P}^2$, it follows that $$A{F}^2-(c-AF{)}^2=A{P}^2-B{P}^2.$$ Therefore, $$2c\cdot AF-c^2=(x-m{)}^2+(y-n{)}^2-x^2-y^2,$$ and $$AF=\frac{m^2+n^2-2mx-2ny}{2c}+\frac{c}{2}.$$

Similarly, we can compute $BD$ and $CE$ in terms of $x$ and $y$.

Since $AF+BD+CE=\frac{l}{2}$, we get $$\frac{m^2+n^2-2mx-2ny}{2c}+\frac{c}{2}+x+\frac{2mx+2ny-m^2-n^2-2ax+a^2}{2b}+\frac{b}{2}=\frac{l}{2},$$ that is, $$\left(\frac{m}{b}-\frac{a}{b}-\frac{m}{c}+1\right)x+\left(\frac{n}{b}-\frac{n}{c}\right)y+\frac{a^2}{2b}+\frac{b+c}{2}+\frac{m^2+n^2}{2}\left(\frac{1}{c}-\frac{1}{b}\right)-\frac{l}{2}=0.$$ Since $b\ne c$ and $n\ne 0$, point $P$ is on a fixed straight line. Since the condition $2(AF+BD+CE)=l$ is satisfied for both incenter and circumcenter, we complete the proof.