Question

在一片平坦结冰的湖面上，两名滑冰者艾莉和比莉分别位于点 $A$ 与点 $B$ 。 $A$ 与 $B$ 之间的距离为 $100$ 米。艾莉从 $A$ 出发，以每秒 $8$ 米的速度沿一条与 $AB$ 成 $60^\circ$ 角的直线滑行。在艾莉离开 $A$ 的同时，比莉从 $B$ 出发，以每秒 $7$ 米的速度沿一条直线滑行；在速度给定的条件下，这条路径使两人尽早相遇。相遇前艾莉滑了多少米？

Two skaters, Allie and Billie, are at points $A$ and $B,$ respectively, on a flat, frozen lake. The distance between $A$ and $B$ is $100$ meters. Allie leaves $A$ and skates at a speed of $8$ meters per second on a straight line that makes a $60^\circ$ angle with $AB.$ At the same time Allie leaves $A,$ Billie leaves $B$ at a speed of $7$ meters per second and follows the straight path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?

答案

答案：160

知识点：路程、速度与时间余弦定理二次方程

难度评级：2270

小提示：

$t$ 秒后，艾莉距 $A$ 为 $8t$ 米，而比莉能到达距 $B$ 为 $7t$ 米的位置

After $t$ seconds, Allie is $8t$ meters from $A$ and Billie can be $7t$ meters from $B$

大提示：

对三角形应用余弦定理，并选择较小的正时间

Apply the Law of Cosines to the triangle and select the smaller positive time

解答：

设两人在 $t$ 秒后相遇。他们距 $A$ 和 $B$ 的距离分别为 $8t$ 与 $7t$ ，且 $A$ 处的夹角为 $60^\circ$ 。由余弦定理， $\begin{aligned}(7t)^2&=(8t)^2+100^2\\&\quad-2(8t)(100)\cos60^\circ\end{aligned}\text{。}$ 因此 $3t^2-160t+2000=0$ ，其根为 $20$ 和 $\frac{100}{3}$ 。最早相遇发生在 $t=20$ 时，所以艾莉滑行了 $8(20)=160$ 米。

Suppose the skaters meet after $t$ seconds. Their distances from $A$ and $B$ are $8t$ and $7t,$ and the included angle at $A$ is $60^\circ.$ The Law of Cosines gives $\begin{aligned}(7t)^2&=(8t)^2+100^2\\&\quad-2(8t)(100)\cos60^\circ.\end{aligned}$ Hence $3t^2-160t+2000=0,$ whose roots are $20$ and $\frac{100}{3}.$ The earliest meeting occurs at $t=20,$ so Allie skates $8(20)=160$ meters.

1989 AIME 第 6 题

答案：160

难度评级：2270

小提示：

大提示：

解答：

其他年份的第 6 题