Question

一辆汽车在一条又长又直的道路上向正东行驶，速度为每分钟 $\frac{2}{3}$ 英里。同时，一个半径为 $51$ 英里的圆形风暴以每分钟 $\frac{1}{2}\sqrt{2}$ 英里的速度向东南方向移动。在 $t = 0$ 时，风暴中心在汽车正北方 $110$ 英里处。在 $t = t_1$ 分钟时，汽车进入风暴圆；在 $t = t_2$ 分钟时，汽车离开风暴圆。求 $\frac{1}{2}(t_1 + t_2)$ 。

A car travels due east at $\frac{2}{3}$ mile per minute on a long, straight road. At the same time, a circular storm, whose radius is $51$ miles, moves southeast at $\frac{1}{2}\sqrt{2}$ mile per minute. At time $t = 0,$ the center of the storm is $110$ miles due north of the car. At time $t = t_1$ minutes, the car enters the storm circle, and at time $t = t_2$ minutes, the car leaves the storm circle. Find $\frac{1}{2}(t_1 + t_2).$

答案

答案：198

知识点：坐标几何距离公式韦达定理

难度评级：2400

解答：

令汽车在 $t = 0$ 时位于原点，正东为 $x$ 轴正方向，正北为 $y$ 轴正方向。时刻 $t$ 时，汽车位置为 $\left(\frac{2t}{3}, 0\right)$ ，风暴中心向东南移动，速度 $\frac{\sqrt{2}}{2}$ 的分量为向东 $\frac{1}{2}$ 、向南 $\frac{1}{2}$ ，所以位置为 $\left(\frac{t}{2}, 110 - \frac{t}{2}\right)$ 。

汽车在风暴边界上时，两者距离平方为 $51^2$ ：即 $\frac{t^2}{36} + \frac{t^2}{4}$ $- 110t + 12100 - 2601 = 0$ ，也就是 $\frac{5}{18}t^2 - 110t + 9499 = 0$ 。 $\begin{aligned} &\left(\frac{2t}{3} - \frac{t}{2}\right)^2 \\ &\quad {}+ \left(110 - \frac{t}{2}\right)^2 = 51^2, \end{aligned}$

两个根为 $t_1$ 和 $t_2$ ，所以由韦达定理 $t_1 + t_2 = \frac{110 \cdot 18}{5} = 396$ ，从而 $\frac{1}{2}(t_1 + t_2) = 198$ 。

Put the car at the origin at $t = 0,$ with east as the positive $x$ -direction and north as the positive $y$ -direction. At time $t$ the car is at $\left(\frac{2t}{3}, 0\right),$ and the storm center, moving southeast at speed $\frac{\sqrt{2}}{2}$ (components $\frac{1}{2}$ east and $\frac{1}{2}$ south), is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right).$

The car is on the storm boundary when the squared distance is $51^2:$ $\begin{aligned} &\left(\frac{2t}{3} - \frac{t}{2}\right)^2 \\ &\quad {}+ \left(110 - \frac{t}{2}\right)^2 = 51^2, \end{aligned}$ that is $\frac{t^2}{36} + \frac{t^2}{4}$ $- 110t + 12100 - 2601 = 0,$ or $\frac{5}{18}t^2 - 110t + 9499 = 0.$

The roots are $t_1$ and $t_2,$ so by Vieta's formulas $t_1 + t_2 = \frac{110 \cdot 18}{5} = 396,$ and $\frac{1}{2}(t_1 + t_2) = 198.$

1997 AIME 第 7 题

答案：198

难度评级：2400

解答：

其他年份的第 7 题