2025 AMC 8 第 5 题

先试着解答 2025 AMC 8 第 5 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2025 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

5.

Betty 开卡车在一个街区送包裹,街道地图如下。Betty 从工厂(标为 FF)出发,依次开到地点 AA、BB、CC,然后回到 FF。完成这条路线的最短距离是多少个街区?
Betty drives a truck to deliver packages in a neighborhood whose street map is shown below. Betty starts at the factory (labeled FF) and drives to location A,A, then B,B, then C,C, before returning to F.F. What is the shortest distance, in blocks, she can drive to complete the route?
2020
2222
2424
2626
2828
答案:C
知识点:坐标几何最优化
小提示:
把路线分成四段最短的网格路径
Break the route into four shortest grid trips
大提示:
在街道网格上,最短距离等于水平街区数加竖直街区数
On a street grid, count horizontal blocks plus vertical blocks
视频讲解:
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文字解答:
关键是:若在街道网格中从坐标 (x1,y1)(x_1, y_1) 开到 (x2,y2)(x_2, y_2),最短距离为 ∣x2−x1∣+∣y2−y1∣|x_2 - x_1| + |y_2 - y_1|。这通常称为曼哈顿距离。它也等于两地之间的水平街区数加上竖直街区数。
从 FF 到 AA 的最短距离是 1+2=31 + 2 = 3。
从 AA 到 BB 的最短距离是 7+3=107 + 3 = 10。
从 BB 到 CC 的最短距离是 2+4=62 + 4 = 6。
从 CC 到 FF 的最短距离是 4+1=54 + 1 = 5。
相加得到 3+10+6+5=243 + 10 + 6 + 5 = 24,选 C。
一个小捷径是注意,从 BB 到 CC 再到 FF 时,经过 CC 正好位于从 BB 到 FF 的某条最短路径上,所以甚至可以先忽略必须停在 CC 的要求。
The key idea is that if driving from coordinates (x1,y1)(x_1, y_1) to (x2,y2),(x_2, y_2), then the shortest distance is ∣x2−x1∣+∣y2−y1∣.|x_2 - x_1| + |y_2 - y_1|. This is often called the Manhattan distance. It is also equal to the number of horizontal blocks between the locations, plus the number of vertical blocks between the locations.
The shortest distance from FF to AA is then 1+2=3.1 + 2 = 3.
The shortest distance from AA to BB is 7+3=10.7 + 3 = 10.
The shortest distance from BB to CC is 2+4=6.2 + 4 = 6.
The shortest distance from CC to FF is 4+1=5.4 + 1 = 5.
Adding up all of these numbers, we get 3+10+6+5=24,3 + 10 + 6 + 5 = 24, which is choice C.
One small possible shortcut for the solution is to notice that when going from BB to CC to F,F, the visit to CC is conveniently along a shortest path from BB to FF anyway, so we can even remove the requirement to stop at CC from the problem.
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