Given $a, \sqrt{2}<a<2$. A convex quadrilateral $ABCD$ is inscribed in the unit circle $\Gamma$ and satisfies the following conditions:
(1) The center of the circle is inside this convex quadrilateral;
(2) The longest side is $a$, and the shortest side is $\sqrt{4-a^{2}}$. Tangents to the circle $\Gamma$ at points $A, B, C, D$ are denoted as $L_{A}, L_{B}, L_{C}, L_{D}$, respectively. It is known that $L_{A}$ intersects $L_{B}$, $L_{B}$ intersects $L_{C}$, $L_{C}$ intersects $L_{D}$, and $L_{D}$ intersects $L_{A}$ at points $A^{\prime}, B^{\prime}, \dot{C}^{\prime}$, respectively. 

（Supplied by Xuan-Guo Huang）
Let the center of $\odot \Gamma$ be $O$, and denote $\angle A O B=2 \theta_{1}$, $\angle B O C=2 \theta_{2}$, $\angle C O D=2 \theta_{3}$, $\angle D O A=2 \theta_{4}$. Thus, $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}$ are all acute angles and $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=\pi$. It is not difficult to find:
$$
\begin{array}{l}
S_{\text {quadrilateral } A^{\prime} B^{\prime} C^{\prime}}=\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3}+\tan \theta_{4}, \\
S_{\text {quadrilateral } A B C D}=\frac{1}{2}\left(\sin 2 \theta_{1}+\sin 2 \theta_{2}+\sin 2 \theta_{3}+\sin 2 \theta_{4}\right) \text {. } \\
\end{array}
$$

Therefore, we have
$S_{\text {quadrilateral } A^{\prime} B C D^{\prime}}$
$S_{\text {quadrilateral } B \text { BC }}$
$$
=\frac{2\left(\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3}+\tan \theta_{4}\right)}{\sin 2 \theta_{1}+\sin 2 \theta_{2}+\sin 2 \theta_{3}+\sin 2 \theta_{4}} .
$$

Since the right side of equation (1) is symmetric with respect to $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}$, we can assume $A D=a, C D=\sqrt{4-a^{2}}$. Thus,
$$
\sin \theta_{4}=\frac{a}{2}, \sin \theta_{3}=\frac{1}{2} \sqrt{4-a^{2}} .
$$

Furthermore,
$$
\begin{array}{l}
\cos \left(\theta_{1}+\theta_{2}\right)=\cos \theta_{3} \cdot \cos \theta_{4}-\sin \theta_{3} \cdot \sin \theta_{4} \\
=\sqrt{1-\left(\frac{a}{2}\right)^{2}} \sqrt{1-\frac{1}{4}\left(4-a^{2}\right)} \\
\quad-\frac{a}{2} \cdot \frac{1}{2} \sqrt{4-a^{2}}=0,
\end{array}
$$

Therefore, $\theta_{3}+\theta_{4}=\frac{\pi}{2}$. Thus, $\theta_{1}+\theta_{2}=\frac{\pi}{2}$.
By trigonometric formulas, we have
$$
\begin{array}{l}
\tan \theta_{3}+\tan \theta_{4}=\frac{\sin \left(\theta_{3}+\theta_{4}\right)}{\cos \theta_{3} \cdot \cos \theta_{4}} \\
=\frac{1}{\sin \theta_{4} \cdot \sin \theta_{3}}=\frac{4}{a \sqrt{4-a^{2}}}, \\
\tan \theta_{1}+\tan \theta_{2}=\frac{\sin \left(\theta_{1}+\theta_{2}\right)}{\cos \theta_{1} \cdot \cos \theta_{2}} \\
=\frac{1}{\cos \theta_{1} \cdot \cos \theta_{2}}=\frac{2}{\sin 2 \theta_{1}}, \\
\sin 2 \theta_{3}+\sin 2 \theta_{4}=2 \sin 2 \theta_{3}=4 \sin \theta_{3} \cdot \cos \theta_{3} \\
=4 \sin \theta_{3} \cdot \sin \theta_{4}=a \sqrt{4-a^{2}}, \\
\sin 2 \theta_{1}+\sin 2 \theta_{2}=2 \sin 2 \theta_{1} .
\end{array}
$$

Let $\alpha=\tan \theta_{3}+\tan \theta_{4}, \beta=\sin 2 \theta_{3}+\sin 2 \theta_{4}, t=$ $\sin 2 \theta_{1}$, then $\alpha, \beta$ are constants while $t$ is a variable. Substituting (2) to (5) into (1), we get
$\frac{S_{\text {quadrilateral } A^{\prime} A^{\prime} B C^{\prime} D^{\prime}}}{S_{\text {quadrilateral } 5 A B C D}}=\frac{2 \alpha+\frac{4}{t}}{\beta+2 t}$.
It is easy to see that the right side of equation (6) is a strictly decreasing function, and the maximum value of $t$ is 1, hence we get
$$
\min \frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} C D^{\prime}}}{S_{\text {quadrilateral } A B C D}}=\frac{4}{a \sqrt{4-a^{2}}} .
$$

On the other hand, since $\theta_{1}, \theta_{2}, \theta_{3}, \theta_{4}$ are all acute angles and $a$ is the longest side length, $\sqrt{4-a^{2}}$ is the shortest side length, we have $\theta_{3} \leqslant \theta_{1}, \theta_{2} \leqslant$ $\theta_{4}$.
Since $\theta_{1}+\theta_{2}=\frac{\pi}{2}=\theta_{3}+\theta_{4}$, we know that the minimum value of $t$ is $t_{0}=2 \sin \theta_{3} \cdot \sin \theta_{4}=\frac{1}{2} a \sqrt{4-a^{2}}$.
Thus, by the strict monotonicity of the right side of equation (6), we get
$$
\max \frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} B^{\prime} C^{\prime}}}{S_{\text {quadrilateral } A B A C D}}=\frac{8}{a^{2}\left(4-a^{2}\right)} .
$$