圆内矩形

设单位圆上点 $P=(\\cos\\alpha,\\sin\\alpha)$，圆内均匀点 $Q$。以 $PQ$ 为对角线且边平行坐标轴时，矩形在圆内当且仅当 $Q$ 落在以 $P$ 与其关于坐标轴对称点构成的矩形内。\n\n该矩形面积为 $4\\sin\\alpha\\cos\\alpha$，圆面积为 $\\pi$，故\n\n$$\n\\mathbb{P}(\\text{满足}\\mid\\alpha)=\\frac{4\\sin\\alpha\\cos\\alpha}{\\pi}.\n$$\n\n对 $\\alpha$ 平均可得\n\n$$\n\\mathbb{P}=\\boxed{\\frac{4}{\\pi^2}}.\n$$

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英文解析

Denote by $E$ the event that the described rectangle is inside the circle. Let $\alpha = \angle POx$ , where $O$ is the center of the circle. Denote by $E_{1}$ the event that the point $P$ is chosen in the first quadrant, i.e., $\alpha \in \left(0, \frac{\pi}{2}\right)$ . Then,

$$\mathbb{P}(E) = 4\mathbb{P}\left(E \cap E_{1}\right).$$

Let $V$ be the intersection of the line $PO$ with the unit circle. Let $U$ and $W$ be the points symmetric to $P$ with respect to $Oy$ and $Ox$ axes. Then, $PUVW$ is a rectangle. The event $E \cap E_1$ occurs if and only if the point $Q$ belongs to the interior of the rectangle $PUVW$ . Since the point $Q$ is uniformly chosen inside the unit circle of area $\pi$ , we obtain that

$$\begin{array}{r}\mathbb{P}(E\cap E_1\mid \angle POx = \alpha) = \frac{\sigma(PUVW)}{\pi}\\ = \frac{4\sin\alpha\cos\alpha}{\pi}. \end{array}$$

From (2.246) and (2.247), we conclude that

$$\begin{array}{l}\mathbb{P}(E) = 4\int_{0}^{\frac{\pi}{2}}\frac{4\sin\alpha\cos\alpha}{\pi}\cdot \frac{1}{2\pi} d\alpha \\ = \frac{4}{\pi^{2}}\int_{0}^{\frac{\pi}{2}}\sin (2\alpha)d\alpha \\ = \frac{4}{\pi^{2}}. \end{array}$$