AMC12 2006 A

AMC12 2006 A · Q6

AMC12 2006 A · Q6. It mainly tests Area & perimeter, Transformations.

The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?

$8\times18$ 矩形 $ABCD$ 被切成两个全等的六边形，如图所示，使得这两个六边形可以在不重叠的情况下重新摆放成一个正方形。$y$ 是多少？

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (A)

正确答案：（A）

Solution

Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below. As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$. As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle. As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$. From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$.

由于两个六边形将被重新摆放成一个不重叠的正方形，面积保持不变。矩形的面积为 $18\cdot8=144$，因此该正方形的边长为 12。唯一可行的摆放方式如下图所示。从图中可见，标为 $y$ 的线段长度是正方形边长的一半，因此 $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$。如解法 1 所述，两个六边形将被重新摆放成一个不重叠的正方形，因此我们用原矩形拼出这个正方形。从图中可见，长度为 $y$ 的部分正好填补了先前空缺的边，因此它等于 $y$。于是可得 $3y = 18$，所以 $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$。

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