Question

在这个单位正方形内有五个全等的小阴影正方形，它们内部互不重叠。如图，中间正方形每条边的中点分别与另外四个小正方形的一个顶点重合。它们的公共边长为 $\dfrac{a-\sqrt{2}}{b}$ ，其中 $a$ 、 $b$ 为正整数。 $a+b$ 是多少？

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\dfrac{a-\sqrt{2}}{b},$ where $a$ and $b$ are positive integers. What is $a+b?$

$7$

$8$

$9$

$10$

$11$

答案：E

知识点：对角线分母有理化

难度评级：1510

解答：

设公共边长为 $x$ 。单位正方形的对角线长为 $\sqrt{2}$ ，由两条各长 $x\sqrt2$ 的小正方形对角线和一条长为 $x$ 的小正方形边组成，所以 $2x\sqrt2+x=\sqrt2.$

解得因此 $a=4$ 、 $b=7$ ，所以 $a+b=11$ 。 $\begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered}$

所以正确答案是 E。

Let $x$ be the common side length. The diagonal of the unit square has length $\sqrt{2}$ and consists of two small-square diagonals (each $x\sqrt2$ ) plus one small-square side length $x,$ so $2x\sqrt2+x=\sqrt2.$

Solving, $\begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered}$ Thus $a=4,$ $b=7,$ and $a+b=11.$

Thus, the correct answer is E.

2016 AMC 12A 第 9 题

答案：E

难度评级：1510

解答：

其他年份的第 9 题