AMC10 2001 A

AMC10 2001 A · Q21

AMC10 2001 A · Q21. It mainly tests Similarity, Circle theorems.

A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter 10 and altitude 12, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.

一个直圆柱，其直径等于其高度，内接于一个直圆锥中。圆锥的直径为10，高为12，圆柱与圆锥的轴线重合。求圆柱的半径。

(A) $\frac{8}{3}$ $\frac{8}{3}$

(B) $\frac{30}{11}$ $\frac{30}{11}$

(D) $\frac{25}{8}$ $\frac{25}{8}$

(E) $\frac{7}{2}$ $\frac{7}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Let the cylinder have radius $r$ and height $2r$. Since $\triangle APQ$ is similar to $\triangle AOB$, we have $$\frac{12-2r}{r}=\frac{12}{5},\ \text{so } r=\frac{30}{11}.$$

（B）设圆柱的半径为 $r$，高为 $2r$。由于 $\triangle APQ$ 与 $\triangle AOB$ 相似，我们有 $$\frac{12-2r}{r}=\frac{12}{5}，\ \text{所以 } r=\frac{30}{11}。$$

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