AMC10 2005 B

AMC10 2005 B · Q7

AMC10 2005 B · Q7. It mainly tests Circle theorems, Area & perimeter.

A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?

一个圆内接于一个正方形，然后一个正方形内接于这个圆，最后一个圆内接于这个正方形。小圆的面积与大正方形的面积之比是多少？

(A) $\frac{\pi}{16}$ $\frac{\pi}{16}$

(B) $\frac{\pi}{8}$ $\frac{\pi}{8}$

(D) $\frac{\pi}{4}$ $\frac{\pi}{4}$

(E) $\frac{\pi}{2}$ $\frac{\pi}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let the radius of the smaller circle be $r$. Then the side length of the smaller square is $2r$. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$. Hence the larger square has sides of length $2\sqrt{2}r$. The ratio of the area of the smaller circle to the area of the larger square is therefore \[ \frac{\pi r^2}{(2\sqrt{2}r)^2}=\frac{\pi}{8}. \]

（B）设较小圆的半径为 $r$。则较小正方形的边长为 $2r$。较大圆的半径等于较小正方形对角线长度的一半，因此为 $\sqrt{2}r$。因此较大正方形的边长为 $2\sqrt{2}r$。所以较小圆面积与较大正方形面积之比为 \[ \frac{\pi r^2}{(2\sqrt{2}r)^2}=\frac{\pi}{8}. \]

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