Question

对每个满足 $0 \le a \le 1$ 的实数 $a$ ，从区间 $[0, a]$ 和 $[0, 1]$ 中分别独立随机选择数 $x$ 和 $y$ 。令 $P(a)$ 为下列不等式成立的概率： $P(a)$ 的最大值是多少？ $\sin^2(\pi x) + \sin^2(\pi y) \gt 1.$

For each real number $a$ with $0 \le a \le 1,$ let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1],$ respectively, and let $P(a)$ be the probability that $\sin^2(\pi x) + \sin^2(\pi y) \gt 1.$ What is the maximum value of $P(a)?$

$\dfrac{7}{12}$

$2 - \sqrt{2}$

$\dfrac{1 + \sqrt{2}}{4}$

$\dfrac{\sqrt{5} - 1}{2}$

$\dfrac58$

答案：B

知识点：几何概率最优化微积分

难度评级：2540

解答：

由 $\sin^2(\pi y) = 1 - \cos^2(\pi y)$ 条件等价于 $|\sin \pi x| \gt |\cos \pi y|$ 对固定的 $x$ 当 $y \in [0, 1]$ 时，所求概率为 $g(x) = 2x$ （ $0 \le x \le \tfrac12$ ），以及 $g(x) = 2 - 2x$ （ $\tfrac12 \le x \le 1$ ）。

因此 $P(a) = \tfrac1a \int_0^a g(x)\,dx$ 当 $a \le \tfrac12$ 有 $P(a) = a$ 并递增到 $\tfrac12$ 当 $a \ge \tfrac12$ 有 $\begin{gathered} P(a) = \frac{2a - a^2 - \tfrac12}{a} \\ {}= 2 - a - \frac{1}{2a}. \end{gathered}$

将导数令为零得 $2a^2 = 1$ 所以 $a = \tfrac{1}{\sqrt2}$ 且 $P\!\left(\tfrac{1}{\sqrt2}\right) = 2 - \sqrt2$ 。

所以正确答案是 B。

Since $\sin^2(\pi y) = 1 - \cos^2(\pi y),$ the condition is $|\sin \pi x| \gt |\cos \pi y|.$ For fixed $x,$ the probability over $y \in [0, 1]$ is $g(x) = 2x$ for $0 \le x \le \tfrac12$ and $g(x) = 2 - 2x$ for $\tfrac12 \le x \le 1.$

Then $P(a) = \tfrac1a \int_0^a g(x)\,dx.$ For $a \le \tfrac12,$ $P(a) = a,$ increasing to $\tfrac12.$ For $a \ge \tfrac12,$ $\begin{gathered} P(a) = \frac{2a - a^2 - \tfrac12}{a} \\ {}= 2 - a - \frac{1}{2a}. \end{gathered}$

Setting the derivative to zero gives $2a^2 = 1,$ so $a = \tfrac{1}{\sqrt2},$ and $P\!\left(\tfrac{1}{\sqrt2}\right) = 2 - \sqrt2.$

Thus, the correct answer is B.

2020 AMC 12B 第 25 题

答案：B

难度评级：2540

解答：

其他年份的第 25 题