China Mathematical Competition (Jiangxi)

The slope of the tangent line passing through $A$ is $y^{\prime }=2x{|}_{x=1}=2$. So the equation of the tangent line $AB$ is $y=2x-1$. Hence the coordinates of $B$ and $D$ are $B(0,-1)$, $D(\frac{1}{2},0)$. Thus $D$ is the midpoint of line segment $AB$.

Consider $P(x,y)$, $C(x_0,x_0^2)$, $E(x_1,y_1)$, $F(x_2,y_2)$. Then by $\frac{AE}{EC}={\lambda }_1$, we know $x_1=\frac{1+{\lambda }_1{x}_0}{1+{\lambda }_1}$, $y_1=\frac{1+{\lambda }_1{x}_0^2}{1+{\lambda }_1}$. From $\frac{BF}{FC}={\lambda }_2$, we get $x_2=\frac{{\lambda }_2{x}_0}{1+{\lambda }_2}$, $y_2=\frac{-1+{\lambda }_2{x}_0^2}{1+{\lambda }_2}$.

$$\frac{y-\frac{1+{\lambda }_1{x}_0^2}{1+{\lambda }_1}}{\frac{-1+{\lambda }_2{x}_0^2}{1+{\lambda }_2}-\frac{1+{\lambda }_1{x}_0^2}{1+{\lambda }_1}}=\frac{x-\frac{1+{\lambda }_1{x}_0}{1+{\lambda }_1}}{\frac{{\lambda }_2{x}_0}{1+{\lambda }_2}-\frac{1+{\lambda }_1{x}_0}{1+{\lambda }_1}}$$

Simplifying it, we get $$[({\lambda }_2-{\lambda }_1)x_0-(1+{\lambda }_2)]y=[({\lambda }_2-{\lambda }_1)x_0^2-3]x+1+x_0-{\lambda }_2{x}_0^2.(1)$$

When $x_0\ne \frac{1}{2}$, the equation of line $CD$ is $$y=\frac{2x_0^{2}x-x_0^2}{2x_0-1}.(2)$$

From (1) and (2), we get $$\{\begin{array}{l}x=\frac{x_0+1}{3},\\ y=\frac{x_0}{3}.\end{array}$$

Eliminating $x_0$, we get the equation of the trail of point $P$ as $y=\frac{1}{3}(3x-1{)}^2$.

When $x_0=\frac{1}{2}$, the equation of $EF$ is $-\frac{3}{2}y=(\frac{1}{4}{\lambda }_2-\frac{1}{4}{\lambda }_1-3)x+\frac{3}{2}-\frac{1}{4}{\lambda }_2$, the equation of $CD$ is $x=\frac{1}{2}$. Combining them, we conclude that $(x,y)=(\frac{1}{2},\frac{1}{12})$ is on the trail of $P$. Since $C$ and $A$ cannot be congruent, $x_0\ne 1,x\ne \frac{2}{3}$.

Therefore the equation of the trail is $y=\frac{1}{3}(3x-1{)}^2$, $x\ne \frac{2}{3}$.

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Alternative solution.

From Solution I, the equation of $AB$ is $y=2x-1$, $B(0,-1)$, $D(\frac{1}{2},0)$. Thus $D$ is the midpoint of $AB$.

Set $\gamma =\frac{CD}{CP}$, $t_1=\frac{CA}{CE}=1+{\lambda }_1$, $t_2=\frac{CB}{CF}=1+{\lambda }_2$. Then $t_1+t_2=3$. Since $AD$ is a median of $△ABC$, $S_{△CAB}=2S_{△CAD}=2S_{△CBD}$ where $S_{△}$ denotes the area of $△$. But $$\frac{1}{t_1{t}_2}=\frac{CE\cdot CF}{CA\cdot CB}=\frac{S_{△CEF}}{S_{△CAB}}=\frac{S_{△CEP}}{2S_{△CAD}}+\frac{S_{△CFP}}{2S_{△CED}}$$ $$=\frac{1}{2}\left(\frac{1}{t_1\gamma }+\frac{1}{t_2\gamma }\right)=\frac{t_1+t_2}{2t_1{t}_2\gamma }=\frac{3}{2t_1{t}_2\gamma },$$ so $\gamma =\frac{3}{2}$ and $P$ is the center of gravity for $△ABC$.

Consider $P(x,y)$ and $C(x_0,x_0^2)$. Since $C$ is different from $A$, $x_0\ne 1$. Thus the coordinates of the center of gravity $P$ are $x=\frac{0+1+x_0}{3}=\frac{1+x_0}{3}$, $x\ne \frac{2}{3}$, $y=\frac{-1+1+x_0^2}{3}=\frac{x_0^2}{3}$. Eliminating $x_0$, we get $y=\frac{1}{3}(3x-1{)}^2$. Thus the equation of the trail is $y=\frac{1}{3}(3x-1{)}^2$, $x\ne \frac{2}{3}$.