AMC10 2014 B

AMC10 2014 B · Q22

AMC10 2014 B · Q22. It mainly tests Circle theorems, Coordinate geometry.

Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?

如图，八个半圆沿边长为2的正方形内侧排列。求与所有这些半圆相切的圆的半径。

(A) $\frac{1+\sqrt{2}}{4}$ $\frac{1+\sqrt{2}}{4}$

(B) $\frac{\sqrt{5}-1}{2}$ $\frac{\sqrt{5}-1}{2}$

(D) $\frac{2\sqrt{3}}{5}$ $\frac{2\sqrt{3}}{5}$

(E) $\frac{\sqrt{5}}{3}$ $\frac{\sqrt{5}}{3}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $O$ be the center of the circle and choose one of the semicircles to have center point $B$. Label the point of tangency $C$ and point $A$ as in the figure. In $\triangle OAB$, $AB=\frac{1}{2}$ and $OA=1$, so $OB=\frac{\sqrt{5}}{2}$. Because $BC=\frac{1}{2}$, $OC=\frac{\sqrt{5}}{2}-\frac{1}{2}=\frac{\sqrt{5}-1}{2}$.

答案（B）：设 $O$ 为圆心，并选取其中一个半圆，使其圆心为点 $B$。如图所示，标出切点 $C$ 和点 $A$。在 $\triangle OAB$ 中，$AB=\frac{1}{2}$ 且 $OA=1$，所以 $OB=\frac{\sqrt{5}}{2}$。由于 $BC=\frac{1}{2}$，因此 $OC=\frac{\sqrt{5}}{2}-\frac{1}{2}=\frac{\sqrt{5}-1}{2}$。

Topics