Question

三个一英寸正方形的底边放在同一直线上。中间的正方形被取出并旋转 $45^\circ$ ，如图所示。然后将它居中并放回原位置，直到它接触相邻两个正方形。点 $B$ 距原来正方形底边所在直线多少英寸？

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated $45^\circ,$ as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?

$1$

$\sqrt{2}$

$\dfrac{3}{2}$

$\sqrt{2} + \dfrac{1}{2}$

$2$

答案：D

知识点：变换正方形（几何）特殊直角三角形

难度评级：1760

解答：

旋转后的正方形向下移动时，它的两条下边落在相邻正方形的内侧上角，这些角的高度为 $1.$ 下顶点位于两角正中，所以每个角与它的水平距离为 $\tfrac12$ 。下边斜率的绝对值为 $1$ ，因此从下顶点到角会升高 $\tfrac12$ 。所以下顶点的高度是 $1-\tfrac12=\tfrac12.$

点 $B$ 是对角顶点，比下顶点高一条长度为 $\sqrt2$ 的完整竖直对角线。因此它的高度是 $\tfrac12+\sqrt2.$

所以正确答案是 D。

When lowered, the rotated square's two lower edges rest on the inner top corners of the adjoining squares, which are at height $1.$ The bottom vertex is centered between those corners, so each corner is horizontally $\tfrac12$ unit from it. A lower edge has slope $1$ in magnitude, so it rises $\tfrac12$ unit on the way to a corner. Therefore the bottom vertex is at height $1-\tfrac12=\tfrac12.$

Point $B$ is the opposite vertex, a full vertical diagonal of length $\sqrt2$ higher. Its height is therefore $\tfrac12+\sqrt2.$

Thus, the correct answer is D.

2005 AMC 10A 第 19 题

答案：D

难度评级：1760

解答：

其他年份的第 19 题