AMC10 2007 A

AMC10 2007 A · Q15

AMC10 2007 A · Q15. It mainly tests Pythagorean theorem, Coordinate geometry.

Four circles of radius $1$ are each tangent to two sides of a square and externally tangent to a circle of radius $2$, as shown. What is the area of the square?

四个半径为 $1$ 的圆，每个都与正方形的两条边相切，并与半径为 $2$ 的圆外切，如图所示。正方形的面积是多少？

(A) 32 32

(B) $22 + 12\sqrt{2}$ $22 + 12\sqrt{2}$

(D) 48 48

(E) $36 + 16\sqrt{2}$ $36 + 16\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $s$ be the length of a side of the square. Consider an isosceles right triangle with vertices at the centers of the circle of radius $2$ and two of the circles of radius $1$. This triangle has legs of length $3$, so its hypotenuse has length $3\sqrt{2}$. The length of a side of the square is $2$ more than the length of this hypotenuse, so $s = 2 + 3\sqrt{2}$. Hence the area of the square is $$ s^2 = (2 + 3\sqrt{2})^2 = 22 + 12\sqrt{2}. $$

答案（B）：设正方形的一边长为 $s$。考虑一个等腰直角三角形，其顶点位于半径为 $2$ 的圆的圆心以及两个半径为 $1$ 的圆的圆心处。该三角形的两条直角边长为 $3$，因此其斜边长为 $3\sqrt{2}$。正方形的边长比这条斜边长多 $2$，所以 $s = 2 + 3\sqrt{2}$。因此正方形的面积为 $$ s^2 = (2 + 3\sqrt{2})^2 = 22 + 12\sqrt{2}. $$

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