2022 AIME I 第 5 题

Question

一条笔直的河宽

264

米，水流以每分钟

14

米的速度自西向东流。Melanie 和 Sherry 坐在河的南岸，Melanie 位于 Sherry 下游

D

米处。相对于水，Melanie 的游泳速度为每分钟

80

米，Sherry 的游泳速度为每分钟

60

米。两人同时开始沿直线游向北岸上的同一点，该点到她们的起点距离相等。两人同时到达该点。求

D

。

A straight river that is

264

meters wide flows from west to east at a rate of

14

meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of

D

meters downstream from Sherry. Relative to the water, Melanie swims at

80

meters per minute, and Sherry swims at

60

meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find

D.

答案

答案：550

知识点：路程、速度与时间向量平方差

难度评级：2390

小提示：

交汇点位于两人中点的正北方。每位游泳者相对于水的速度等于其对地速度减去水流速度

(14, 0)

。

The meeting point is due north of the midpoint between them. Each swimmer’s velocity relative to the water is her ground velocity minus the current

(14, 0).

大提示：

令

u = \frac{D}{2t}

，将两个速度方程相减，得到

(u + 14)^2 - (u - 14)^2

= 6400 - 3600

With

u = \frac{D}{2t},

subtracting the two speed equations gives

(u + 14)^2 - (u - 14)^2

= 6400 - 3600

解答：

把 Sherry 放在原点，Melanie 放在南岸的

(D, 0)

。北岸上到两人等距的点是

\left(\frac{D}{2}, 264\right)

。若两人都在时间

t

到达，则每位游泳者相对于水的速度等于她的对地速度减去水流速度

(14, 0)

，所以

\begin{aligned} &\left(\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 60^2 \end{aligned}\text{，}

\begin{aligned} &\left(-\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 80^2 \end{aligned}\text{。}

令

u = \frac{D}{2t}

。将两式相减，得到

(u + 14)^2 - (u - 14)^2

= 56u

= 6400 - 3600

= 2800

，所以

u = 50

。代回可得

(50 - 14)^2 + \left(\frac{264}{t}\right)^2 = 3600

，因此

\frac{264}{t} = 48

，

t = \frac{11}{2}

。

因此

D = 2ut = 100t = 550

。

Put Sherry at the origin and Melanie at

(D, 0)

on the south bank. A point on the north bank equidistant from both is

\left(\frac{D}{2}, 264\right).

If both arrive at time

t,

then each swimmer’s velocity relative to the water is her ground velocity minus the current

(14, 0),

so

\begin{aligned} &\left(\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 60^2, \end{aligned}

\begin{aligned} &\left(-\frac{D}{2t} - 14\right)^2 \\ &\quad {}+ \left(\frac{264}{t}\right)^2 = 80^2. \end{aligned}

Subtracting, with

u = \frac{D}{2t}:

(u + 14)^2 - (u - 14)^2

= 56u

= 6400 - 3600

= 2800,

so

u = 50.

Substituting back,

(50 - 14)^2 + \left(\frac{264}{t}\right)^2 = 3600

gives

\frac{264}{t} = 48,

so

t = \frac{11}{2}.

Therefore

D = 2ut = 100t = 550.

2022 AIME I 第 5 题

答案：550

难度评级：2390

小提示：

大提示：

解答：

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