
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

$DM$ is a tangent drawn from vertex $D$, $ON$ is a perpendicular from point $O$ to $DM$, $PQ$ is a tangent to the circle parallel to side $DC$, $OK$ is perpendicular to $PQ$, $OD=2$, and $ON=1$. According to the conditions, $\angle ODN=\frac{\pi}{6}$, $\angle NOK=\angle MDC=\frac{\pi}{12}$.

Let $PQ$ be an arbitrary tangent, $ON$ forms an angle $\alpha \in\left[0, \frac{\pi}{12}\right]$. In the coordinate system $XOY$ with center at point $O$, the equation of the line $PQ$ is $x \cos \alpha+y \sin \alpha=1$, and the area of the trapezoid $QPCD$ is given by

$$
S_{1}(\alpha)=\int_{-\sqrt{2}}^{\sqrt{2}}\left(\sqrt{2}-\frac{1-y \sin \alpha}{\cos \alpha}\right) dy=4-\frac{2\sqrt{2}}{\cos \alpha}
$$

The function $S_{1}(\alpha)$ decreases monotonically over the interval $\alpha \in\left[0, \frac{\pi}{12}\right]$ and $S_{1,\text{min}}=4-\frac{2\sqrt{2}}{\cos \pi /12}$. Note that the ratio $S_{1}(\alpha):(8-S_{1}(\alpha))$ is minimized when $S_{1}(\alpha)$ attains its minimum value.

Indeed, $\left(\frac{S_{1}(\alpha)}{8-S_{1}(\alpha)}\right)^{\prime}=\frac{8S_{1}^{\prime}(\alpha)}{\left(8-S_{1}(\alpha)\right)^2}$ and the signs of the derivative of the ratio and $S_{1}^{\prime}(\alpha)$ coincide.

If $\alpha \in\left[\frac{\pi}{12}; \frac{5\pi}{12}\right]$, the line cuts off a triangle from the square.

The equation of the line $PQ$ is $x \cos \alpha + y \sin \alpha=1$, with the coordinates of point $P$ being $\left(\frac{1-\sqrt{2}\sin \alpha}{\cos \alpha}; \sqrt{2}\right)$ and point $Q$ being $\left(\sqrt{2}; \frac{1-\sqrt{2}\cos \alpha}{\sin \alpha}\right)$. The length of the leg $PC$ is $\sqrt{2}-\frac{1-\sqrt{2}\sin \alpha}{\cos \alpha}=\frac{\sqrt{2}(\sin \alpha+\cos \alpha)-1}{\cos \alpha}$, and the length of the leg $QC$ is $\sqrt{2}-\frac{1-\sqrt{2}\cos \alpha}{\sin \alpha}=\frac{\sqrt{2}(\sin \alpha+\cos \alpha)-1}{\sin \alpha}$. Therefore, the area of the triangle $PCQ$ is

$$
S_{2}(\alpha)=\frac{(\sqrt{2}(\sin \alpha+\cos \alpha)-1)^2}{2\sin \alpha \cos \alpha}
$$

Its derivative (after transformation) is given by:

$$
S_{2}^{\prime}(\alpha)=\frac{2(\cos \alpha-\sin \alpha)(\sqrt{2}(\sin \alpha+\cos \alpha)-1)(\sqrt{2}+(\sin \alpha+\cos \alpha))}{\sin^2 2 \alpha}
$$

The function $S_{2}(\alpha)$ on the interval $\alpha \in\left[\frac{\pi}{12}; \frac{5\pi}{12}\right]$ has only one critical point $\alpha=\frac{\pi}{4}$ (minimum).

By symmetry, the behavior of the function $S_{2}(\alpha)$ over the interval $\left[\frac{5\pi}{12}; \frac{\pi}{2}\right]$ is the same as over the interval $\alpha \in\left[0, \frac{\pi}{12}\right]$.

Therefore, the minimum value $S_{1,\text{min}}=S_{2}(\pi/4)=1$ and the minimum ratio $\left(S_{1}:S_{2}\right)_{\text{min}}=1:7$.

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