AMC12 2004 A

AMC12 2004 A · Q22

AMC12 2004 A · Q22. 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 $1$ 且两两相切的球放在一个水平平面上。一个半径为 $2$ 的球放在它们上面。从平面到较大球的顶部的距离是多少？

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

(B) $3 + \dfrac{\sqrt{69}}{3}$ $3 + \dfrac{\sqrt{69}}{3}$

(D) $\dfrac{52}{9}$ $\dfrac{52}{9}$

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

Answer

Correct choice: (B)

正确答案：（B）

Solution

The height from the center of the bottom sphere to the plane is $1$, and from the center of the top sphere to the tip is $2$. Error creating thumbnail: Unable to save thumbnail to destination We now need the vertical height of the centers. If we connect centers, we get a rectangular pyramid with an equilateral triangle base. The distance from the vertex of the equilateral triangle to its centroid can be found by $30\text-60\text-90 \triangle$s to be $\frac{2\sqrt{3}}{3}$. Error creating thumbnail: Unable to save thumbnail to destination By the Pythagorean Theorem, we have $\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}$. Adding the heights up, we get $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow 3+\frac{\sqrt{69}}{3}$, or $\boxed{B}$

底部小球的球心到平面的高度为 $1$，顶部大球的球心到球顶的高度为 $2$。现在需要求球心之间的竖直高度差。连接各球心，可得到一个底面为等边三角形的直角锥。等边三角形的顶点到其重心的距离可由 $30\text-60\text-90 \triangle$s 求得为 $\frac{2\sqrt{3}}{3}$。由勾股定理，$\left(\frac{2}{\sqrt{3}}\right)^2 + h^2 = 3^2 \Longrightarrow h = \frac{\sqrt{69}}{3}$。将各段高度相加，得到 $\frac{\sqrt{69}}{3} + 1 + 2 \Rightarrow 3+\frac{\sqrt{69}}{3}$，即 $\boxed{B}$。

Topics