AMC10 2005 A

AMC10 2005 A · Q19

AMC10 2005 A · Q19. It mainly tests Pythagorean theorem, Coordinate geometry.

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45°, 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?

三个一英寸正方形放置，其底边在一根直线上。中间的正方形被取出并旋转45°，如图所示。然后将其置中并降回原位置，直到它接触到相邻的两个正方形。点B离原始正方形底边所在直线的距离有多少英寸？

(A) 1 1

(B) $\sqrt{2}$ $\sqrt{2}$

(D) $\sqrt{2} + \frac{1}{2}$ $\sqrt{2} + \frac{1}{2}$

(E) 2 2

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Consider the rotated middle square shown in the figure. It will drop until length $DE$ is 1 inch. Thus $FC = DF = FE = \frac{1}{2}$ and $BC = \sqrt{2}$. Hence $BF = \sqrt{2} - 1/2$. This is added to the 1 inch height of the supporting squares, so the overall height of point $B$ above the line is $1 + BF = \sqrt{2} + \frac{1}{2}$ inches.

（D）考虑图中旋转的中间正方形。它会下落直到线段 $DE$ 的长度为 1 英寸。因此 $FC = DF = FE = \frac{1}{2}$，且 $BC = \sqrt{2}$。因此 $BF = \sqrt{2} - 1/2$。把它加到支撑正方形的 1 英寸高度上，所以点 $B$ 相对于该直线的总高度为 $1 + BF = \sqrt{2} + \frac{1}{2}$ 英寸。

Topics