AMC10 2015 B

AMC10 2015 B · Q24

AMC10 2015 B · Q24. It mainly tests Sequences & recursion (algebra), Coordinate geometry.

Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin $p_0=(0,0)$ facing to the east and walks one unit, arriving at $p_1=(1,0)$. For $n=1,2,3,\ldots$, right after arriving at the point $p_n$, if Aaron can turn $90^\circ$ left and walk one unit to an unvisited point $p_{n+1}$, he does that. Otherwise, he walks one unit straight ahead to reach $p_{n+1}$. Thus the sequence of points continues $p_2=(1,1)$, $p_3=(0,1)$, $p_4=(-1,1)$, $p_5=(-1,0)$, and so on in a counterclockwise spiral pattern. What is $p_{2015}$?

蚂蚁 Aaron 按照如下规则在坐标平面上行走。他从原点 $p_0=(0,0)$ 出发，面朝东走 1 个单位，到达 $p_1=(1,0)$。对 $n=1,2,3,\ldots$，当 Aaron 刚到达点 $p_n$ 后，如果他可以向左转 $90^\circ$ 并走 1 个单位到一个未到访过的点 $p_{n+1}$，他就这样做；否则，他就沿着当前方向直走 1 个单位到达 $p_{n+1}$。因此点列继续为 $p_2=(1,1)$，$p_3=(0,1)$，$p_4=(-1,1)$，$p_5=(-1,0)$，以此类推，形成一个逆时针的螺旋路径。求 $p_{2015}$。

(A) (-22, -13) (-22, -13)

(B) (-13, -22) (-13, -22)

(D) (13, -22) (13, -22)

(E) (22, -13) (22, -13)

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Note that for any natural number $k$, when Aaron reaches point $(k,-k)$, he will have just completed visiting all of the grid points within the square with vertices at $(k,-k)$, $(k,k)$, $(-k,k)$, and $(-k,-k)$. Thus the point $(k,-k)$ is equal to $p_{(2k+1)^2-1}$. It follows that $p_{2024}=p_{(2\cdot22+1)^2-1}=(22,-22)$. Because $2024-2015=9$, the point $p_{2015}=(22-9,-22)=(13,-22)$.

答案（D）：注意对任意自然数 $k$，当 Aaron 到达点 $(k,-k)$ 时，他恰好完成了访问以 $(k,-k)$、$(k,k)$、$(-k,k)$、$(-k,-k)$ 为顶点的正方形内的所有格点。因此点 $(k,-k)$ 等于 $p_{(2k+1)^2-1}$。于是 $p_{2024}=p_{(2\cdot22+1)^2-1}=(22,-22)$。因为 $2024-2015=9$，所以 $p_{2015}=(22-9,-22)=(13,-22)$。

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