AMC12 2004 B

AMC12 2004 B · Q19

AMC12 2004 B · Q19. It mainly tests 3D geometry (volume), Geometry misc.

A truncated cone has horizontal bases with radii $18$ and $2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?

一个截锥的上下底面为水平圆，半径分别为 $18$ 和 $2$。一个球与该截锥的上底面、下底面以及侧面都相切。该球的半径是多少？

(A) 6 6

(B) 4\sqrt{5} 4\sqrt{5}

(D) 10 10

(E) 6\sqrt{3} 6\sqrt{3}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases: Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$, so $BC = 20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$: By the Pythagorean Theorem, \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\] Therefore, the answer is $\boxed{{A (6)}}.$

考虑沿底面直径截取该截锥的截面梯形（按如下标记为 $ABCD$）：上图中，$E,F,$ 和 $G$ 为切点。由两切线定理，$BF = BE = 18$ 且 $CF = CG = 2$，所以 $BC = 20$。作 $H$ 为从 $D$ 向 $\overline{AB}$ 作高 $\overline{HD}$ 的垂足：由勾股定理， \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\] 因此答案为 $\boxed{{A (6)}}.$

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