AMC10 2013 A

AMC10 2013 A · Q22

AMC10 2013 A · Q22. It mainly tests Coordinate geometry, Geometry misc.

Six spheres of radius 1 are positioned so that their centers are at the vertices of a regular hexagon of side length 2. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

六个半径为1的球体，其中心位于边长为2的正六边形的顶点上。这六个球体都与一个较大的球体内切，较大球的中心是六边形的中心。还有一个第八个球体，与六个小球体外切，并与较大球体内切。这个第八个球体的半径是多少？

(A) $\sqrt{2}$ $\sqrt{2}$

(B) $\frac{3}{2}$ $\frac{3}{2}$

(D) $\sqrt{3}$ $\sqrt{3}$

(E) 2 2

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let the vertices of the regular hexagon be labeled in order $A, B, C, D, E,$ and $F$. Let $O$ be the center of the hexagon, which is also the center of the largest sphere. Let the eighth sphere have center $G$ and radius $r$. Because the centers of the six small spheres are each a distance $2$ from $O$ and the small spheres have radius $1$, the radius of the largest sphere is $3$. Because $G$ is equidistant from $A$ and $D$, the segments $GO$ and $AO$ are perpendicular. Let $x$ be the distance from $G$ to $O$. Then $x + r = 3$. The Pythagorean Theorem applied to $\triangle AOG$ gives $(r + 1)^2 = 2^2 + x^2 = 4 + (3 - r)^2$, which simplifies to $2r + 1 = 13 - 6r$, so $r = \frac{3}{2}$. Note that this shows that the eighth sphere is tangent to $AD$ at $O$.

答案（B）：设正六边形的顶点按顺序标为 $A, B, C, D, E, F$。设 $O$ 为六边形的中心，它也是最大球的球心。第八个球的球心为 $G$，半径为 $r$。由于六个小球的球心到 $O$ 的距离均为 $2$，且小球半径为 $1$，所以最大球的半径为 $3$。由于 $G$ 到 $A$ 与 $D$ 的距离相等，线段 $GO$ 与 $AO$ 垂直。设 $x$ 为 $G$ 到 $O$ 的距离，则 $x + r = 3$。对 $\triangle AOG$ 应用勾股定理得 $(r + 1)^2 = 2^2 + x^2 = 4 + (3 - r)^2$，化简得到 $2r + 1 = 13 - 6r$，因此 $r = \frac{3}{2}$。注意这表明第八个球在 $O$ 点与 $AD$ 相切。

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