AMC12 2016 A

AMC12 2016 A · Q9

AMC12 2016 A · Q9. It mainly tests Quadratic equations, Pythagorean theorem.

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 $\frac{a-\sqrt{2}}{b}$, where $a$ and $b$ are positive integers. What is $a+b$?

在这个单位正方形内有五个阴影小正方形，它们全等且内部互不重叠。如图所示，中间那个正方形的每一条边的中点都与另外四个小正方形的某个顶点重合。它们的公共边长为 $\frac{a-\sqrt{2}}{b}$，其中 $a$ 和 $b$ 为正整数。求 $a+b$。

(A) 7 7

(B) 8 8

(D) 10 10

(E) 11 11

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $x$ be the common side length. Draw a diagonal between opposite corners of the unit square. The length of this diagonal is $\sqrt{2}$. The diagonal consists of two small-square diagonals and one small-square side length. Combining the previous two observations yields $2x\sqrt{2} + x = \sqrt{2}.$ Solving this equation for $x$ gives $x = \frac{4-\sqrt{2}}{7}$. The requested sum is $4 + 7 = 11$.

答案（E）：设 $x$ 为公共边长。在单位正方形的相对顶点之间画一条对角线。这条对角线的长度为 $\sqrt{2}$。该对角线由两条小正方形的对角线和一条小正方形的边长组成。结合以上两点可得 $2x\sqrt{2} + x = \sqrt{2}.$ 解此方程得 $x = \frac{4-\sqrt{2}}{7}$。所求的和为 $4 + 7 = 11$。

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