B.IS Given an equilateral $\triangle A B C, D$ is any point on side $B C$, the circumcenter and incenter of $\triangle A B D$ are $O_{1}$ and $I_{1}$, respectively, and the circumcenter and incenter of $\triangle A D C$ are $O_{2}$ and $I_{2}$, respectively. The line $O_{1} I_{1}$ intersects $O_{2} I_{2}$ at point $P$. Try to find: When point $D$ moves on side $B C$, what is the locus of point $P$?
Let's assume the side length of the equilateral triangle $\triangle ABC$ is 2. As shown in the figure, we establish a coordinate system, $BE \perp AC, CF \perp AB$. Let the coordinates of point $D$ be $(a, 0)$, and assume $0 \leqslant a<1$. Then $O_{2}, I_{1}$ lie on $BE$ or its extension, and $O_{1}, I_{2}$ lie on $CF$ or its extension. Since $O_{2}$ is on the perpendicular bisector of $DC$, the x-coordinate of $O_{2}$ is $\frac{a+1}{2}$. Given $\angle CBO_{2}=30^{\circ}$, the y-coordinate of $O_{2}$ 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 $\triangle ADC$. By $AD=\sqrt{a^{2}+3}$ and the area relationship, we have $\left(2+1-a+\sqrt{a^{2}+3}\right) r_{2}-(1-a) \sqrt{3}$, which simplifies to $r_{2}=\frac{\sqrt{3}}{6}\left(3-a-\sqrt{a^{2}+3}\right)$. Draw $I_{2}G \perp BC$ at $G$. Then, by $\angle GCI_{2}=30^{\circ}$, we know $CG=r_{2} \cot 30^{\circ}=\sqrt{3} r_{2}$. The x-coordinate of point $I_{2}$ is
$$
1-CG=1-\frac{1}{2}\left(3-a-\sqrt{a^{2}+3}\right)=\frac{1}{2}\left(a-1+\sqrt{a^{2}+3}\right).
$$

Therefore, the coordinates of point $I_{2}$ are $\left(\frac{1}{2}\left(a-1+\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3-a-\sqrt{a^{2}+3}\right)\right)$. Similarly, the coordinates of point $I_{1}$ are $\left(\frac{1}{2}\left(a+1-\sqrt{a^{2}+3}\right), \frac{\sqrt{3}}{6}\left(3+a-\sqrt{a^{2}+3}\right)\right)$. The coordinates of point $O_{1}$ are $\left(\frac{a-1}{2}, \frac{3-a}{2} \cdot \frac{\sqrt{3}}{3}\right)$. The equation of $O_{2}I_{2}$ is
$$
\frac{y-\frac{a+3}{2} \cdot \frac{\sqrt{3}}{3}}{x-\frac{a+1}{2}}=\frac{\sqrt{3}}{3} \cdot \frac{2a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}},
$$

which simplifies to
$$
y-\frac{\sqrt{3}(a+3)}{6}=\frac{\sqrt{3}}{3} \cdot \frac{2a+\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(x-\frac{a+1}{2}\right).
$$

Similarly, the equation of $O_{1}I_{1}$ is
$$
\frac{y-\frac{3-a}{2} \cdot \frac{\sqrt{3}}{3}}{x-\frac{a-1}{2}}=\frac{\sqrt{3}}{3} \cdot \frac{2a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}},
$$

which simplifies to
$$
y-\frac{\sqrt{3}(3-a)}{6}=\frac{\sqrt{3}}{3} \cdot \frac{2a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(x-\frac{a-1}{2}\right).
$$

By solving (1) and (2), we get $x=a$. Substituting $x=a$ into (2) gives
$$
y=\frac{\sqrt{3}(3-a)}{6}+\frac{\sqrt{3}}{3} \cdot \frac{2a-\sqrt{a^{2}+3}}{2-\sqrt{a^{2}+3}}\left(a-\frac{a-1}{2}\right)=-\frac{\sqrt{3}}{3} \sqrt{a^{2}+3}.
$$

Therefore, $y=-\frac{\sqrt{3}}{3} \sqrt{x^{2}+3}$, which simplifies to $y^{2}-\frac{x^{2}}{3}=1, -1<x<1, y<0$. Thus, the trajectory of point $P$ is part of a hyperbola.