Question

一个圆外接于一个等腰三角形，该三角形两个相等角的度数为 $x$ 。在圆上独立且均匀随机选取两个点，并连接它们作一条弦。该弦与三角形相交的概率为 $\frac{14}{25}$ 。求 $x$ 的最大可能值与最小可能值之差。

A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x.$ Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}.$ Find the difference between the largest and smallest possible values of $x.$

答案

答案：48

知识点：几何概率圆周角二次方程

难度评级：2650

解答：

三角形的每个圆周角截得一段度数为其两倍的弧，所以三个顶点把圆分成度数为 $2x$ 、 $2x$ 和 $360 - 4x$ 的弧。弦不与三角形相交，当且仅当两个随机点落在同一段弧内，其概率为 $\begin{aligned} &\left(\frac{2x}{360}\right)^2 + \left(\frac{2x}{360}\right)^2 \\ &\quad {}+ \left(\frac{360 - 4x}{360}\right)^2 \\ &= 1 - \frac{14}{25} = \frac{11}{25}. \end{aligned}$

令 $y = \frac{x}{180}$ ，得到 $2y^2 + (1 - 2y)^2 = \frac{11}{25}$ ，化简为 $75y^2 - 50y + 7 = 0$ ，根为 $y = \frac{1}{5}$ 和 $y = \frac{7}{15}$ 。它们给出 $x = 36$ 和 $x = 84$ ，两者都可以作为等腰三角形的底角。

所求差为 $84 - 36 = 48$ 。

Each inscribed angle of the triangle subtends an arc of twice its measure, so the vertices split the circle into arcs of $2x,$ $2x,$ and $360 - 4x$ degrees. The chord fails to intersect the triangle exactly when both random points fall in the same arc, which has probability $\begin{aligned} &\left(\frac{2x}{360}\right)^2 + \left(\frac{2x}{360}\right)^2 \\ &\quad {}+ \left(\frac{360 - 4x}{360}\right)^2 \\ &= 1 - \frac{14}{25} = \frac{11}{25}. \end{aligned}$

Setting $y = \frac{x}{180},$ this reads $2y^2 + (1 - 2y)^2 = \frac{11}{25},$ which simplifies to $75y^2 - 50y + 7 = 0,$ with roots $y = \frac{1}{5}$ and $y = \frac{7}{15}.$ These give $x = 36$ and $x = 84,$ both legitimate base angles of an isosceles triangle.

The requested difference is $84 - 36 = 48.$

2017 AIME I 第 6 题

答案：48

难度评级：2650

解答：

其他年份的第 6 题