AMC10 2009 B

AMC10 2009 B · Q13

AMC10 2009 B · Q13. It mainly tests Rounding & estimation, Geometry misc.

As shown below, convex pentagon ABCDE has sides AB = 3, BC = 4, CD = 6, DE = 3, and EA = 7. The pentagon is originally positioned in the plane with vertex A at the origin and vertex B on the positive x-axis. The pentagon is then rolled clockwise to the right along the x-axis. Which side will touch the point x = 2009 on the x-axis?

如图所示，凸五边形ABCDE的边长AB = 3, BC = 4, CD = 6, DE = 3, EA = 7。五边形最初位于平面中，顶点A在原点，顶点B在正x轴上。然后五边形沿x轴向右顺时针滚动。哪条边会接触x轴上的点x = 2009？

(A) AB AB

(B) BC BC

(D) DE DE

(E) EA EA

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Define a rotation of the pentagon to be a sequence that starts with $AB$ on the $x$-axis and ends when $AB$ is on the $x$-axis the first time thereafter. Because the pentagon has perimeter $23$ and $2009 = 23 \cdot 87 + 8$, it follows that after $87$ rotations, point $A$ will be at $x = 23 \cdot 87 = 2001$ and point $B$ will be at $x = 2001 + 3 = 2004$. Points $C$ and $D$ will next touch the $x$-axis at $x = 2004 + 4 = 2008$ and $x = 2008 + 6 = 2014$, respectively. Therefore a point on $CD$ will touch $x = 2009$.

答案（C）：将五边形的一次“旋转”定义为：从线段 $AB$ 位于 $x$ 轴开始，到之后第一次再次出现 $AB$ 位于 $x$ 轴时结束的一段过程。由于五边形的周长为 $23$，且 $2009 = 23 \cdot 87 + 8$，可知经过 $87$ 次旋转后，点 $A$ 将位于 $x = 23 \cdot 87 = 2001$，点 $B$ 将位于 $x = 2001 + 3 = 2004$。接着，点 $C$ 和点 $D$ 将分别在 $x = 2004 + 4 = 2008$ 与 $x = 2008 + 6 = 2014$ 处再次接触 $x$ 轴。因此，线段 $CD$ 上会有一点接触到 $x = 2009$。

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