AMC10 2010 B

AMC10 2010 B · Q20

AMC10 2010 B · Q20. It mainly tests Area & perimeter, Polygons.

Two circles lie outside regular hexagon $ABCDEF$. The first is tangent to $AB$, and the second is tangent to $DE$. Both are tangent to lines $BC$ and $FA$. What is the ratio of the area of the second circle to that of the first circle?

两个圆位于正六边形 $ABCDEF$ 外。第一圆切 $AB$，第二圆切 $DE$。两者均切直线 $BC$ 和 $FA$。第二圆与第一圆面积之比是多少？

(A) 18 18

(B) 27 27

(D) 81 81

(E) 108 108

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): It may be assumed that hexagon $ABCDEF$ has side length $1$. Let lines $BC$ and $FA$ intersect at $G$, let $H$ and $J$ be the midpoints of $AB$ and $DE$, respectively, let $K$ be the center of the second circle, and let that circle be tangent to line $BC$ at $L$. Equilateral $\triangle ABG$ has side length $1$, so the first circle, which is the inscribed circle of $\triangle ABG$, has radius $\frac{\sqrt{3}}{6}$. Let $r$ be the radius of the second circle. Then $\triangle GLK$ is a $30$–$60$–$90^\circ$ right triangle with $LK=r$ and $2r=GK=GH+HJ+JK=\frac{\sqrt{3}}{2}+\sqrt{3}+r$. Therefore $r=\frac{3\sqrt{3}}{2}=9\left(\frac{\sqrt{3}}{6}\right)$. The ratio of the radii of the two circles is $9$, and the ratio of their areas is $9^2=81$.

答案（D）：不妨设六边形 $ABCDEF$ 的边长为 $1$。令直线 $BC$ 与 $FA$ 交于 $G$，令 $H$ 与 $J$ 分别为 $AB$ 与 $DE$ 的中点，令 $K$ 为第二个圆的圆心，并且该圆与直线 $BC$ 在 $L$ 处相切。等边三角形 $\triangle ABG$ 的边长为 $1$，因此第一个圆（即 $\triangle ABG$ 的内切圆）的半径为 $\frac{\sqrt{3}}{6}$。设第二个圆的半径为 $r$。则 $\triangle GLK$ 为 $30$–$60$–$90^\circ$ 直角三角形，且 $LK=r$，并有 $2r=GK=GH+HJ+JK=\frac{\sqrt{3}}{2}+\sqrt{3}+r$。因此 $r=\frac{3\sqrt{3}}{2}=9\left(\frac{\sqrt{3}}{6}\right)$。两圆半径之比为 $9$，面积之比为 $9^2=81$。

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