AMC12 2011 B

AMC12 2011 B · Q11

AMC12 2011 B · Q11. It mainly tests Coordinate geometry, Distance / midpoint.

A frog located at $(x,y)$, with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$. What is the smallest possible number of jumps the frog makes?

一只青蛙位于 $(x,y)$，其中 $x$ 和 $y$ 都是整数。它连续进行长度为 $5$ 的跳跃，并且总是落在整数坐标点上。假设青蛙从 $(0,0)$ 出发，最终到达 $(1,0)$。青蛙最少需要跳跃多少次？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (B)

正确答案：（B）

Solution

Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle). Because either $1$, $5$, or $7$ is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, the frog cannot reach $(1,0)$ in two moves. However, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$. Thus, the answer is $= \boxed{3 \textbf{}}$.

由于青蛙每次跳跃长度都是 $5$ 且落在格点上，它的坐标和的变化要么是 $5$（沿 $x$ 轴或 $y$ 轴方向跳），要么是 $3$ 或 $4$（构成 $3$-$4$-$5$ 直角三角形）。因为坐标和的变化总是 $1$、$5$ 或 $7$ 之一，所以坐标和每次都会从奇数变为偶数或从偶数变为奇数。因此，它不可能用偶数次跳跃从 $(0,0)$ 到达 $(1,0)$。所以青蛙不可能用两次跳跃到达 $(1,0)$。然而，用 $3$ 次跳跃可以实现：从 $(0,0)$ 到 $(3,4)$，再到 $(6,0)$，再到 $(1,0)$。因此答案是 $= \boxed{3 \textbf{}}$。

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