AI 中文总结
本文求解三个顶点位于同心圆上时构成钝角三角形的概率，给出闭式解与范围，建立圆与高斯采样约定的联系，模拟验证结果。
AI 中文摘要
刘易斯·卡罗尔的《枕边问题》询问：平面内随机选取的三个点构成钝角三角形的概率。该问题只有在确定采样方案后才有答案，最简洁的确定方式是将点置于圆上，此时概率为3/4。本文求解三个顶点分别位于半径为$r_1$、$r_2$、$r_3$的三个同心圆上的版本，答案是三个项的和，每个项为两个独立反正弦变量的加权和超过阈值的概率，当半径相等时该和退化为3/4。该公式在两个子族中给出闭式解，一个为反正弦函数，另一个为勒让德χ函数；存在一个勾股条件，决定哪个顶点可成为钝角顶点；概率范围为$1/2 \leq P < 1$，最小值仅在一个顶点位于公共圆心且另外两个半径相等时取得。对半径进行条件分析表明，同一公式是所有独立旋转对称采样方案的核，因此该范围适用于所有此类方案；对瑞利半径取平均后，以闭式形式得到已知的高斯值3/4，建立了文献中缺失的圆约定与高斯约定之间的联系。全程通过模拟验证了结果。
英文摘要
Lewis Carroll's Pillow Problem asks for the probability that three points chosen at random in the plane form an obtuse triangle. The question has no answer until the sampling scheme is pinned down, and the cleanest way to pin it down is to put the points on a circle, which gives 3/4. This paper solves the version in which the three vertices lie on three concentric circles of radii $r_1$, $r_2$, and $r_3$. The answer is a sum of three terms, each the probability that a weighted sum of two independent arcsine variables exceeds a threshold, and it collapses to 3/4 when the radii are equal. Reading the formula gives closed forms in two subfamilies, one an arcsine and one the Legendre chi function; a Pythagorean condition deciding which vertex can carry the obtuse angle; and the bound $1/2 \leq P < 1$, with the minimum attained only when one vertex sits at the common center and the other two radii are equal. Conditioning on the radii shows that the same formula is the kernel for every independent rotationally symmetric sampling scheme, so the bound applies to all of them at once and averaging over Rayleigh radii recovers the known Gaussian value of 3/4 in closed form, supplying a link between the circular and Gaussian conventions that the literature records as missing. Simulation confirms the results throughout.
Comments15 pages, 5 figures, 3 tables. Reference implementation, tests, and reproduction materials at https://github.com/bgreenwell/rtcc
