AMC12 2014 B

AMC12 2014 B · Q19

AMC12 2014 B · Q19. It mainly tests Quadratic equations, 3D geometry (volume).

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the truncated cone. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

如图所示，一个球体内接于一个右圆锥台中。圆锥台的体积是球体体积的两倍。求圆锥台底面半径与顶面半径的比。

(A) \frac{3}{2} \frac{3}{2}

(B) \frac{1+\sqrt{5}}{2} \frac{1+\sqrt{5}}{2}

(D) 2 2

(E) \frac{3+\sqrt{5}}{2} \frac{3+\sqrt{5}}{2}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Assume without loss of generality that the radius of the top base of the truncated cone (frustum) is $1$. Denote the radius of the bottom base by $r$ and the radius of the sphere by $a$. The figure on the left is a side view of the frustum. Applying the Pythagorean Theorem to the triangle on the right yields $r=a^2$. The volume of the frustum is $$\frac{1}{3}\pi(r^2+r\cdot1+1^2)\cdot2a=\frac{1}{3}\pi(a^4+a^2+1)\cdot2a.$$ Setting this equal to twice the volume of the sphere, $\frac{4}{3}\pi a^3$, and simplifying gives $a^4-3a^2+1=0$, or $r^2-3r+1=0$. Therefore $r=\frac{3+\sqrt{5}}{2}$.

答案（E）：不失一般性，设截锥（圆台）上底半径为 $1$。设下底半径为 $r$，球的半径为 $a$。左图是圆台的侧视图。对右侧三角形应用勾股定理得 $r=a^2$。圆台体积为 $$\frac{1}{3}\pi(r^2+r\cdot1+1^2)\cdot2a=\frac{1}{3}\pi(a^4+a^2+1)\cdot2a.$$ 令其等于球体体积的两倍 $\frac{4}{3}\pi a^3$，化简得 $a^4-3a^2+1=0$，或 $r^2-3r+1=0$。因此 $r=\frac{3+\sqrt{5}}{2}$。

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