Question $P$ is any point on the incircle $\odot O$ of the equilateral $\triangle A_{1} A_{2} A_{3}$, and the distances from $P$ to $A_{1} A_{2}$, $A_{2} A_{3}$, and $A_{3} A_{1}$ are $d_{1}$, $d_{2}$, and $d_{3}$, respectively. Is $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}$ a constant when the position of point $P$ changes? Is $d_{1}^{4}+d_{2}^{4}+d_{3}^{4}$ a constant? Explain your reasoning ${ }^{[1]}$.
Solution 1: When the position of point $P$ changes, $d_{1}^{2}+d_{2}^{2}+d_{3}^{2}$ is a constant; $d_{1}^{4}+d_{2}^{4}+d_{3}^{4}$ is not a constant.

As shown in Figure 1, $\triangle A_{1} A_{2} A_{3}$ is inscribed in $\odot O$ at points $D$, $E$, and $F$. Suppose $P$ is any point on the minor arc $\overparen{D F}$, and $P P_{1} \perp A_{1} A_{2}$, $P P_{2} \perp A_{2} A_{3}$, $P P_{3} \perp A_{3} A_{1}$, with the feet of the perpendiculars being $P_{1}$, $P_{2}$, and $P_{3}$, respectively. Then
$$
d_{1}=P P_{1}, d_{2}=P P_{2}, d_{3}=P P_{3}.
$$

Let $P P_{2}$ intersect $D F$ at point $Q$, then $P Q \perp D F$.
It is easy to see that $P$, $Q$, $D$, $P_{1}$ and $P$, $Q$, $F$, $P_{3}$ are concyclic.
Therefore, $\angle P P_{3} Q = \angle P F Q = \angle P F D = \angle P D P_{1} = \angle P Q P_{1}$.
Similarly, $\angle P Q P_{3} = \angle P P_{1} Q$.
Thus, $\triangle P Q P_{1} \sim \triangle P P_{3} Q$.
Hence $\frac{P Q}{P P_{3}} = \frac{P P_{1}}{P Q}$, which means $P Q^{2} = P P_{3} \cdot P P_{1}$.
In the equilateral $\triangle A_{1} D F$,
$$
\begin{array}{l}
P Q + P P_{1} + P P_{3} \\
= \frac{\sqrt{3}}{2} A_{1} D = \frac{\sqrt{3}}{4} A_{1} A_{2} = Q P_{2}.
\end{array}
$$

Therefore, $P P_{2} = P Q + Q P_{2} = 2 P Q + P P_{1} + P P_{3}$
$$
= 2 \sqrt{P P_{1} \cdot P P_{3}} + P P_{1} + P P_{3},
$$

which means $d_{2} - d_{1} - d_{3} = 2 \sqrt{d_{1} d_{3}}$.
Squaring both sides, we get
$$
d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 2 d_{1} d_{2} + 2 d_{2} d_{3} + 2 d_{1} d_{3}.
$$

Also, $d_{1} + d_{2} + d_{3} = \frac{\sqrt{3}}{2} a$ (where $a$ is the side length of the equilateral $\triangle A_{1} A_{2} A_{3}$), so
$$
d_{1}^{2} + d_{2}^{2} + d_{3}^{2} + 2 d_{1} d_{2} + 2 d_{2} d_{3} + 2 d_{1} d_{3} = \frac{3}{4} a^{2}.
$$

Therefore, $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = \frac{3}{8} a^{2}$ (a constant).
Let $A_{1} E$ intersect $\odot O$ at point $G$. When point $P$ is at points $D$ and $G$, the values of $d_{1}^{4} + d_{2}^{4} + d_{3}^{4}$ are $\frac{9}{128} a^{4}$ and $\frac{129}{1152} a^{4}$, respectively. Therefore, when the position of point $P$ changes, $d_{1}^{4} + d_{2}^{4} + d_{3}^{4}$ is not a constant.

Solution 2: As shown in Figure 2, with the center $O$ as the origin and a line through $O$ parallel to side $A_{2} A_{3}$ as the $x$-axis, establish a Cartesian coordinate system.

Assume the radius of $\odot O$ is 1, then the equations of lines $A_{1} A_{2}$, $A_{2} A_{3}$, and $A_{3} A_{1}$ are
$$
\begin{array}{l}
\sqrt{3} x - y + 2 = 0, y = -1, \sqrt{3} x + y - 2 = 0. \\
\text{Let } P(\cos \theta, \sin \theta), \text{ then} \\
d_{1}^{2} + d_{2}^{2} + d_{3}^{2} \\
= \left(\frac{|\sqrt{3} \cos \theta - \sin \theta + 2|}{2}\right)^{2} + |\sin \theta + 1|^{2} + \\
\left(\frac{|\sqrt{3} \cos \theta + \sin \theta - 2|}{2}\right)^{2} \\
= \frac{1}{2} \left[3 \cos^{2} \theta + (\sin \theta - 2)^{2}\right] + (\sin \theta + 1)^{2} \\
= \frac{9}{2} \text{ (a constant).} \\
d_{1}^{4} + d_{2}^{4} + d_{3}^{4} \text{ is not a constant, as shown in Solution 1.} \\
\end{array}
$$