Clothing 125 If the center of the square $A_{1} A_{2} A_{3} A_{4}$ is $O, \odot O$ is inside the square $A_{1} A_{2} A_{3} A_{4}$, $P$ is any point on $\odot O$, and the distances from $P$ to $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{1}$ are $d_{1}, d_{2}, d_{3},$ and $d_{4}$, respectively, then $d_{1}^{3}+d_{2}^{3}+d_{3}^{3}+d_{4}^{3}$ is a constant.
Prove: (1) If $\odot O$ is exactly the incircle of the square $A_{1} A_{2} A_{3} A_{4}$, we prove that $d_{1}+d_{2}+d_{3}+d_{4}, d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}, d_{1}^{3}+d_{2}^{3}+d_{2}^{3}+d_{a}^{3}$ are constants.

As shown in Figure 3, let the incircle touch $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{1}$ at $E, F, G, H$, respectively. Let the projections of $P$ on $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{1}$ be $E^{\prime}, F^{\prime}, G^{\prime}, H^{\prime}$, respectively. Draw the diameter $P M$ of the incircle, and connect $P G$ and $G M$. Let $\angle G P M=\theta$, and the radius of the incircle be $r$, then
$$
\begin{array}{l}
d_{3}=2 r \cos ^{2} \theta, d_{4}=2 r \cos ^{2}\left(\frac{\pi}{4}-\theta\right), \\
d_{2}=2 r \cos ^{2}\left(\frac{\pi}{4}+\theta\right), d_{1}=2 r \cos ^{2}\left(\frac{\pi}{2}-\theta\right) .
\end{array}
$$

Clearly, $d_{1}+d_{2}+d_{3}+d_{4}=4 r$ is a constant.
$$
\begin{array}{l} 
d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \\
= 4 r^{2} \cos ^{4}\left(\frac{\pi}{2}-\theta\right)+4 r^{2} \cos ^{4}\left(\frac{\pi}{4}+\theta\right)+ \\
4 r^{2} \cos ^{4} \theta+4 \cos ^{2}\left(\frac{\pi}{4}-\theta\right) \\
= r^{2}\left\{\left[1+\cos \left(\pi \frac{3}{4}\right)\right]^{2}+\left[1+\cos \left(\frac{\pi}{2}+2 \theta\right)\right]^{2}+\right. \\
{\left.[1+\cos 2 \theta]^{2}+\left[1+\cos \left(\frac{\pi}{2}-2 \theta\right)\right]^{2}\right\} } \\
= r^{2}\left\{4+2\left[\cos (\pi-2 \theta)+\cos \left(\frac{\pi}{2}+2 \theta\right)+\cos 2 \theta+\right.\right. \\
\left.\cos \left(\frac{\pi}{2}-2 \theta\right)\right]+\left[\cos ^{2}(\pi-2 \theta)+\cos ^{2}\left(\frac{\pi}{2}+2 \theta\right)\right. \\
\left.\left.\cos 2 \theta+\cos \left(\frac{\pi}{2}-2 \theta\right)\right]\right\} \\
= 4 r^{2}+r^{2}\left[\frac{1+\cos (2 \pi-4 \theta)}{2}+\frac{1+\cos (\pi+4 \theta)}{2}+\right. \\
\left.\frac{1+\cos 4 \theta}{2}+\frac{1+\cos (\pi-4 \theta)}{2}\right]=6 r^{2}
\end{array}
$$

is a constant.
$$
\begin{aligned}
& d_{1}^{3}+d_{2}^{3}+d_{3}^{3}+d_{4}^{3} \\
= & 8 r^{3}\left[\cos ^{6}\left(\frac{\pi}{2}-\theta\right)+\cos ^{6}\left(\frac{\pi}{4}+\theta\right)+\cos ^{6} \theta+\right. \\
& \left.\cos ^{6}\left(\frac{\pi}{4}-\theta\right)\right] \\
= & r^{3}\left\{\left[\frac{5}{2}+\frac{15}{4} \cos (\pi-2 \theta)+\frac{3}{2} \cos (2 \pi-4 \theta)+\right.\right. \\
& \left.\frac{1}{4} \cos (3 \pi-6 \theta)\right]+\left[\frac{5}{2}+\frac{15}{4} \cos \left(\frac{\pi}{2}+2 \theta\right)+\right. \\
& \left.\frac{3}{2} \cos (\pi+4 \theta)+\frac{1}{4} \cos \left(\frac{3 \pi}{2}+6 \theta\right)\right]+ \\
& {\left[\frac{5}{2}+\frac{15}{4} \cos 2 \theta+\frac{3}{2} \cos 4 \theta+\frac{1}{4} \cos 6 \theta\right]+} \\
& {\left[\frac{5}{2}+\frac{15}{4} \cos \left(\frac{\pi}{2}-2 \theta\right)+\frac{3}{2} \cos (\pi-4 \theta)+\right.} \\
& \left.\left.\frac{1}{4} \cos \left(\frac{3 \pi}{2}-6 \theta\right)\right]\right\}=10 r^{3}
\end{aligned}
$$

is a constant.
(2) If $\odot O$ is not the incircle of the square $A_{1} A_{2} A_{3} A_{4}$, construct the circumscribed square $A_{1}^{\prime} A_{2}^{\prime} A_{3}^{\prime} A_{4}^{\prime}$ of $\odot O$, such that
$$
\begin{array}{l}
A_{1}^{\prime} A_{2}^{\prime} / / A_{1} A_{2}, A_{2}^{\prime} A_{3}^{\prime} / / A_{2} A_{3}, \\
A_{3}^{\prime} A_{4}^{\prime} / / A_{3} A_{4}, A_{4}^{\prime} A_{\mathrm{t}}^{\prime} / / A_{4} A_{1} .
\end{array}
$$

Let the distances from $P$ to $A_{1}^{\prime} A_{2}^{\prime}, A_{2}^{\prime} A_{3}^{\prime}, A_{3}^{\prime} A_{4}^{\prime}, A_{4}^{\prime} A_{1}^{\prime}$ be $d_{1}^{\prime}, d_{2}^{\prime}, d_{3}^{\prime}, d_{4}^{\prime}$, respectively. Then
$$
d_{1}-d_{1}^{\prime}=d_{2}-d_{2}^{\prime}=d_{3}-d_{3}^{\prime}=d_{4}-d_{4}^{\prime}=a \text { (constant). }
$$

Thus, $d_{1}=d_{1}^{\prime}+a, d_{2}=d_{2}^{\prime}+a, d_{3}=d_{3}^{\prime}+a, d_{4}=d_{4}^{\prime}+a$.
Therefore, $d_{1}^{3}+d_{2}^{3}+d_{3}^{3}+d_{4}^{3}$
$$
\begin{aligned}
= & \left(d_{1}^{\prime}+a\right)^{3}+\left(d_{2}^{\prime}+a\right)^{3}+\left(d_{3}^{\prime}+a\right)^{3}+\left(d_{4}^{\prime}+a\right)^{3} \\
= & \left(d_{1}^{\prime 3}+d_{2}^{\prime 3}+d_{3}^{\prime 3}+d_{4}^{\prime 3}\right)+3 a\left(d_{1}^{\prime 2}+d_{2}^{\prime 2}+d_{3}^{\prime 2}+\right. \\
& \left.d_{4}^{\prime 2}\right)+3 a^{2}\left(d_{1}^{\prime}+d_{2}^{\prime}+d_{3}^{\prime}+d_{4}^{\prime}\right)+4 a^{3} .
\end{aligned}
$$

Since (1) has already proven that $d_{1}^{\prime}+d_{2}^{\prime}+d_{3}^{\prime}+d_{4}^{\prime}, d_{1}^{\prime 2}+d_{2}^{\prime 2}+d_{3}^{\prime 2}+d_{4}^{\prime 2}, d_{1}^{\prime 3}+d_{2}^{\prime 3}+d_{3}^{\prime 3}+d_{4}^{\prime 3}$ are all constants,
thus $d_{1}^{3}+d_{2}^{3}+d_{3}^{3}+d_{4}^{3}$ is also a constant.
(Xu Dao, Jiangsu Ruzhi Teacher Training School, 226500)