2011 AMC 12B 第 11 题

先试着解答 2011 AMC 12B 第 11 题,然后核对你的答案与精心整理的解答,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 2011 AMC 12B 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。

11.

一只青蛙位于 (x,y)(x, y),其中 xx 和 yy 都是整数。它连续跳跃,每次跳跃长度为 55,并且总是落在整数坐标点上。若青蛙从 (0,0)(0, 0) 出发,最后到达 (1,0)(1, 0),它至少需要跳多少次?
A frog located at (x,y),(x, y), with both xx and yy integers, makes successive jumps of length 55 and always lands on points with integer coordinates. Suppose that the frog starts at (0,0)(0, 0) and ends at (1,0).(1, 0). What is the smallest possible number of jumps the frog makes?
22
33
44
55
66
答案:B
知识点:格点距离公式
小提示:
一次跳不可能,因为两点距离是 11,不是 55。
One jump is impossible since the distance is 1,1, not 55
大提示:
两次跳的中间点必须在 x=12x=\tfrac12 上,而那里没有格点;试着构造三次跳。
A midpoint of two jumps would lie on x=12,x=\tfrac12, which has no lattice points; try three jumps
解答:
一次跳不行,因为 (0,0)(0,0) 与 (1,0)(1,0) 的距离只有 11。两次跳也不行:中间点必须同时距两端 55,只能在垂直平分线 x=12x=\dfrac12 上,而这条线上没有格点。
三次跳可以做到,例如 (0,0)→(3,4)→(6,0)→(1,0), (0,0)\to(3,4)\to(6,0)\to(1,0)\text{,} 其中每一步长度都是 55。
所以正确答案是 B。
One jump cannot work, since (0,0)(0,0) and (1,0)(1,0) are only 11 apart. Two jumps also fail: the intermediate point would be at distance 55 from both, forcing it onto the perpendicular bisector x=12,x=\dfrac12, which contains no lattice points.
Three jumps suffice, for example (0,0)→(3,4)→(6,0)→(1,0), (0,0)\to(3,4)\to(6,0)\to(1,0), where each step has length 5.5.
Thus, the correct answer is B.
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