AMC10 2004 A

AMC10 2004 A · Q25

AMC10 2004 A · Q25. It mainly tests Triangles (properties), 3D geometry (volume).

Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?

三个半径为1的互切球体放置在水平平面上。一个半径为2的球体放置在它们上面。求该大球顶点到平面的距离。

(A) $3 + \sqrt{30}/2$ $3 + \frac{\sqrt{30}}{2}$

(B) $3 + \sqrt{69}/3$ $3 + \frac{\sqrt{69}}{3}$

(D) $52/9$ $\frac{52}{9}$

(E) $3 + 2\sqrt{2}$ $3 + 2\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let $A$, $B$, $C$ and $E$ be the centers of the three small spheres and the large sphere, respectively. Then $\triangle ABC$ is equilateral with side length $2$. If $D$ is the intersection of the medians of $\triangle ABC$, then $E$ is directly above $D$. Because $AE=3$ and $AD=\frac{2\sqrt{3}}{3}$, it follows that $DE=\sqrt{3^2-\left(\frac{2\sqrt{3}}{3}\right)^2}=\frac{\sqrt{69}}{3}.$ Because $D$ is $1$ unit above the plane and the top of the larger sphere is $2$ units above $E$, the distance from the plane to the top of the larger sphere is $3+\frac{\sqrt{69}}{3}.$

（B）设 $A$、$B$、$C$ 和 $E$ 分别为三个小球和大球的球心。则 $\triangle ABC$ 是边长为 $2$ 的等边三角形。若 $D$ 为 $\triangle ABC$ 的三条中线的交点，则 $E$ 在 $D$ 的正上方。因为 $AE=3$ 且 $AD=\frac{2\sqrt{3}}{3}$，可得 $DE=\sqrt{3^2-\left(\frac{2\sqrt{3}}{3}\right)^2}=\frac{\sqrt{69}}{3}.$ 由于 $D$ 在平面上方 $1$ 个单位，而大球的最高点在 $E$ 上方 $2$ 个单位，所以从该平面到大球最高点的距离为 $3+\frac{\sqrt{69}}{3}.$

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