Given an equilateral $\triangle ABC$, $D$ is any point on side $BC$. The circumcenter and incenter of $\triangle ABD$ are $O_{1}, I_{1}$, and the circumcenter and incenter of $\triangle ADC$ are $O_{2}, I_{2}$. The line $O_{1} I_{1}$ intersects $O_{2} I_{2}$ at $P$. Try to find: When point $D$ moves on side $BC$, the locus of point $P$.
(Li Shenghong, Liu Yuxin provided)
$$
\begin{array}{l}
\text { Solution 1: As shown in Figure 2, draw auxiliary } \\
\text { lines. From } \\
\angle \mathrm{AO}_{2} \mathrm{D}=2 \angle \mathrm{C}= \\
A D^{2}-O D^{2}=A O^{2}, \\
\text { i.e., }(\sqrt{3} y)^{2}-x^{2}=(\sqrt{3})^{2} \text {, } \\
y^{2}-\frac{x^{2}}{3}=1,-1<x<1, y<0 . \\
\left(2+1-a+\sqrt{a^{2}+3}\right) r_{2}=(1-a) \sqrt{3}, \\
C G=r_{2} \cot 30^{\circ}=\sqrt{3} r_{2} . \\
1-C G=1-\frac{1}{2}\left(3-a-\sqrt{a^{2}+3}\right) \\
=\frac{1}{2}\left(a-1+\sqrt{a^{2}+3}\right) . \\
\left(\frac{1}{2}\left(a-1+\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3-a-\sqrt{a^{2}+3}\right)\right) \text {. } \\
\left(\frac{1}{2}\left(a+1-\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3+a-\sqrt{a^{2}+3}\right)\right) \\
\text { Point } O_{1} \text { has coordinates }\left(\frac{a-1}{2}, \frac{3-a}{2} \cdot \frac{\sqrt{3}}{3}\right) \text {. } \\
\frac{y-\frac{a+3}{2} \cdot \frac{\sqrt{3}}{3}}{x-\frac{a+1}{2}}=\frac{\sqrt{3}}{3} \cdot \frac{2 a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}, \\
=\frac{\sqrt{3}}{3} \cdot \frac{2 a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(x-\frac{a+1}{2}\right) \text {. } \\
\frac{y-\frac{a-3}{2} \cdot \frac{\sqrt{3}}{3}}{x-\frac{a-1}{2}}=\frac{\sqrt{3}}{3} \cdot \frac{2 a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}, \\
y-\frac{\sqrt{3}(3-a)}{6} \\
=\frac{\sqrt{3}}{3} \cdot \frac{2 a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(x-\frac{a-1}{2}\right) . \\
\text { From (1) and (2), we get } x=a \text {. Substituting } x=a \text { into (2) gives } \\
y=\frac{\sqrt{3}(3-a)}{6}+\frac{\sqrt{3}}{3} \cdot \frac{2 a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(a-\frac{a-1}{2}\right) \\
=-\frac{\sqrt{3}}{3} \sqrt{a^{2}+3} \text {. } \\
\end{array}
$$
```
120
    \angle A I _ { 2 } D = 9 0 ^ { \circ } + \frac { 1 } { 2 } \angle C
= 1 2 0 ^ { \circ } \text { ,}
        \angle B =
        \text { knowing }\mp@subsup{\textrm{O}}{2}{}、\mp@subsup{\textrm{I}}{2}{}\text{ are on }\odot\textrm{O
.}
        Similarly, }\mp@subsup{O}{1}{}、\mp@subsup{I}{1}{}\text{ are on }\odot\mp@subsup{O}{2}{
.}
\because \angle A A _ { 2 } O _ { 2 } = 3 0 ^ { \circ } = \angle I _ { 1 } A I _ { 2 } \text { ,}
\therefore A I _ { 1 } / / O _ { 2 } P , \angle O _ { 1 } P O _ { 2 } = \angle O , ~ I ~ I ~ A = 3 0 ^ { \circ }
Thus, D is the circumcenter of }\triangle\mp@subsup{O}{1}{}P\mp@subsup{O}{2}{}\text{.
In }\triangle\mp@subsup{O}{2}{}\mp@subsup{D\mp@code{I}}{2}{}\text{, by
\because \angle O O _ { 2 } D C = \angle I ⿱ 上 丨 ' D C + \angle O _ { 2 } D I _ { 2 }
=30}+2\angle\mp@subsup{O}{2}{}\mp@subsup{\textrm{DI}}{2}{}\text{,
Thus, by }\angleD\mp@subsup{O}{2}{}P=\angleDP\mp@subsup{O}{2}{}\text{, we have
\anglePDC = 180 - \angleCDO
\anglePDC=90^{\circ}\text{,
i.e., }PD\perpBC\text{, and }AD=\sqrt{3}D\mp@subsup{O}{1}{}=\sqrt{3}DP\text{.
Now, establish a Cartesian coordinate system with the line of }BC\text{ as the }x\text{-axis and the midpoint of }BC\text{ as the origin }O
```
as the origin, and assume the side length of the equilateral $\triangle A B C$ is 2, and the coordinates of point $P$ are $(x, y)$. Then, in the right $\triangle A O D$, we have

Solution 2: As shown in Figure 3, establish a coordinate system, $B E \perp$ $A C, C F \perp A B$. Let the coordinates of point $D$ be $(a, 0)$, and assume $0 \leqslant a<1$. Then $O_{2} 、 I_{1}$ are on $B E$ or its extension, and $O_{1} 、 I_{2}$ are on $F$ or its extension. Since $\mathrm{O}_{2}$ is on $L C$:

Thus, the x-coordinate of $\mathrm{O}_{2}$ is $\frac{a+1}{2}$. From $\angle \mathrm{CBO}_{2}=30^{\circ}$, the y-coordinate of $\mathrm{O}_{2}$ is $\frac{3+a}{2} \cdot \frac{\sqrt{3}}{3}$. Therefore, the coordinates of point $O_{2}$ are $\left(\frac{a+1}{2}, \frac{3+a}{2} \cdot \frac{\sqrt{3}}{3}\right)$. The y-coordinate of point $I_{2}$ is the radius $r_{2}$ of the incircle of $\triangle A D C$. From $A D=\sqrt{a^{2}+3}$ and the area relationship, we have

i.e., $r_{2}=\frac{\sqrt{3}}{6}\left(3-a-\sqrt{a^{2}+3}\right)$.
Draw $I_{2} G \perp B C$ intersecting $B C$ at $G$. Then, from $\angle G C I_{2}=$ $30^{\circ}$, we know
the x-coordinate of point $I_{2}$ is
Therefore, the coordinates of point $I_{2}$ are
Similarly, the coordinates of point $I_{1}$ are
$$
\begin{array}{l}
\left(\frac{1}{2}\left(a+1-\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3+a-\sqrt{a^{2}+3}\right)\right) . \\
\text { Point } O_{1} \text { has coordinates }\left(\frac{a-1}{2}, \frac{3-a}{2} \cdot \frac{\sqrt{3}}{3}\right) . \\
O_{2} I_{2} \text { has the equation } \\
\frac{y-\frac{a+3}{2} \cdot \frac{\sqrt{3}}{3}}{x-\frac{a+1}{2}}=\frac{\sqrt{3}}{3} \cdot \frac{2 a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}},
\end{array}
$$
The equation of $O_{2} I_{2}$ is

i.e., $y-\frac{\sqrt{3}(a+3)}{6}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{3} \cdot \frac{2 a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(x-\frac{a+1}{2}\right) . \\
O_{1} I_{1} \text { has the equation }
\end{array}
$$
The equation of $O_{1} I_{1}$ is

i.e.,
$$
y=\sqrt{3}(3-a)+\sqrt{3} \cdot 2 a-\sqrt{a^{2}+3}(a
$$

Therefore, $y=-\frac{\sqrt{3}}{2} \sqrt{x^{2}+3}$,
