AMC 8 2026 Problem 19
AMC8 2026 年第 19 题
行程问题★★★☆☆
Miguel is walking with his dog, Luna. When they reach the entrance to a park, Miguel throws a ball straight ahead and continues to walk at a steady pace. Luna sprints toward the ball, which stops by a tree. As soon as the dog reaches the ball, she brings it back to Miguel. Luna runs 5 times faster than Miguel walks. What fraction of the distance from the entrance to the tree will Miguel have walked when Luna returns the ball to him?
Miguel 带着他的狗 Luna 散步。当他们走到公园入口时，Miguel 向前扔出一个球并继续匀速行走。Luna 冲向球，球停在一棵树旁。Luna 一拿到球就立刻跑回 Miguel 身边。Luna 的奔跑速度是 Miguel 步行速度的 5 倍。当 Luna 把球送回给 Miguel 时，Miguel 已经走了从入口到树这段距离的几分之几？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
D
解析 · Solution
解法1
设总距离为1。往返总路程为2，Miguel走了总路程的1/(1+5)，则2×1/6=1/3。（解法4为此方法的代数展开版本。）
取具体数值简化计算：设入口到树的距离为 30 个单位，Luna 速度 5 单位/秒，Miguel 速度 1 单位/秒。Luna 跑到树需 30÷5=6 秒，此时 Miguel 已走 6 个单位。此后 Miguel 继续向前（1 单位/秒），Luna 折返（5 单位/秒），两人相向而行，相对速度 6 单位/秒，剩余 24 单位需 4 秒相遇。Miguel 总用时 10 秒，走 10 个单位。 。
Luna 到树时，Miguel 走了全程的 （速度比 5:1）。剩余 路程两人相向而行，Miguel 的速度占两人合速的 ，所以 Miguel 再走 。Miguel 总共走： 。
设总距离为 d，Miguel 速度 v，Luna 速度 5v。相遇时 Luna 跑的距离是 Miguel 的 5 倍。两人合计走了 2d（Luna 去程+返程一部分，Miguel 前进步）。Miguel 走 y，Luna 走 5y，y+5y=2d，6y=2d，y=d/3。占比 。
选择题策略：Luna 是 Miguel 速度的 5 倍，Luna 到树时 Miguel 走了 ，排除小于 的选项。取剩余中间项 验证：Miguel 总共走 ，则返程多走 − 。Luna 返程走 5× 。她总跑 1+ 程，Miguel 走 ，两人在 处相遇，验证通过。
Solution 1
Let the distance from entrance to tree be 1. The total distance traveled by both is 2 (Luna runs to the tree and back). Since Luna is 5× faster, Miguel covers 1/(1+5) = 1/6 of this combined distance. Miguel's distance: 2 × 1/6 = 1/3.
Pick convenient numbers: let the distance from entrance to tree be 30 units. Let Luna's speed be 5 units/sec, Miguel's 1 unit/sec. Luna reaches the tree in 30÷5 = 6 seconds. By then, Miguel has walked 6 units. From that point, Miguel continues forward at 1 unit/sec while Luna runs back at 5 units/sec (closing speed 6 units/sec). They meet after 24÷6 = 4 more seconds. Total time for Miguel = 10 seconds, so he walks 10 units. Fraction = .
When Luna reaches the tree, Miguel has gone of the way (since Luna is 5× faster). The remaining of the distance is closed by both: Miguel walks of that (his speed is of their combined speed), adding . Total fraction Miguel walks: .
Let d be the distance, Miguel's speed = v, Luna's = 5v. When they meet, Luna has run 5 times as far as Miguel walked. Together they cover 2d (Luna to the tree and back partway, Miguel forward). If Miguel walks distance y, Luna runs 5y. Total: y+5y = 2d, so 6y = 2d, y = d/3. Fraction = .
Multiple-choice shortcut: since Luna is 5× faster, by the time Luna reaches the tree, Miguel has gone of the way. This eliminates options less than . Test the middle remaining option ( ): if Miguel walks , the extra beyond the first is . Luna covers 5 × on her return. She runs 1 + of the distance, Miguel walks , they meet. Answer: .
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