1. Given $a, \sqrt{2}<a<2$. A convex quadrilateral $ABCD$ inscribed in the unit circle $\Gamma$ satisfies the following conditions:
(1) The center of the circle is inside the convex quadrilateral;
(2) The longest side is $a$, and the shortest side is $\sqrt{4-a^{2}}$. Tangents $L_{A}, L_{B}, L_{C}, L_{D}$ are drawn to the circle $\Gamma$ through points $A, B, C, D$ 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 $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$ respectively. Find the maximum and minimum values of the ratio $\frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} C^{\prime} D^{\prime}}}{S}$.
$S_{\text {quadrilateral } A B C D}$
Let the center of circle $\Gamma$ be $O$, and let $\angle A O B=2 \theta_{1}, \angle B O C=2 \theta_{2}$, $\angle C O D=2 \theta_{3}, \angle D O A=2 \theta_{4}$. 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 that
$$
\begin{array}{c}
S_{\text {quadrilateral } A^{\prime} B^{\prime} C^{\prime} D^{\prime}} = \tan \theta_{1} + \tan \theta_{2} + \tan \theta_{3} + \tan \theta_{4} \\
S_{\text {quadrilateral } A B C D} = \frac{1}{2} \left( \sin 2 \theta_{1} + \sin 2 \theta_{2} + \sin 2 \theta_{3} + \sin 2 \theta_{4} \right)
\end{array}
$$

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

Since the right-hand 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,
$$
\begin{aligned}
\cos \left( \theta_{1} + \theta_{2} \right) = & \cos \theta_{3} \cos \theta_{4} - \sin \theta_{3} \sin \theta_{4} = \\
& \sqrt{1 - \left( \frac{a}{2} \right)^{2}} \sqrt{1 - \frac{1}{4} \left( 4 - a^{2} \right)} - \\
& \frac{a}{2} \cdot \frac{1}{2} \sqrt{4 - a^{2}} = 0
\end{aligned}
$$

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

Let $\alpha = \tan \theta_{3} + \tan \theta_{4}, \beta = \sin 2 \theta_{3} + \sin 2 \theta_{4}, t = \sin 2 \theta_{1}$. Then $\alpha \beta$ is a constant, and $t$ is a variable. Substituting (2) to (5) into (1), we get
$$
\frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} C D^{\prime}}}{S_{\text {quadrilateral } A B C D}} = \frac{2 \alpha + \frac{4}{t}}{\beta + 2 t}
$$

It is easy to see that the right-hand side of equation (6) is a strictly decreasing function of $t$, and the maximum value of $t$ is 1, hence
$$
\frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} C D^{\prime}}}{S_{\text {quadrilateral } A B C D \text { min }}} = \frac{4}{a \sqrt{4 - a^{2}}}
$$

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 $\theta_{3} \leqslant \theta_{1}, \theta_{2} < \theta_{4}$.
Since $\theta_{1} + \theta_{2} = \frac{\pi}{2} = \theta_{3} + \theta_{4}$, we know that the minimum value of $t$ is
$$
t_{0} = 2 \sin \theta_{3} \sin \theta_{4} = \frac{1}{2} a \sqrt{4 - a^{2}}
$$

Thus, by the strict decreasing nature of the right-hand side function of equation (6), we get
$$
\frac{S_{\text {quadrilateral } A^{\prime} B^{\prime} C D^{\prime}}}{S_{\text {quadrilateral } A B C D \text { max }}} = \frac{8}{a^{2} \left( 4 - a^{2} \right)}
$$