AMC12 2012 B

AMC12 2012 B · Q15

AMC12 2012 B · Q15. It mainly tests Circle theorems, 3D geometry (volume).

Jesse cuts a circular disk of radius 12, along 2 radii to form 2 sectors, one with a central angle of 120. He makes two circular cones using each sector to form the lateral surface of each cone. What is the ratio of the volume of the smaller cone to the larger cone?

Jesse 将一个半径为 12 的圆盘沿两条半径切开，得到两个扇形，其中一个的圆心角为 120。他用每个扇形分别制作一个圆锥，使扇形成为圆锥的侧面。较小圆锥的体积与较大圆锥的体积之比是多少？

(A) \frac{1}{8} \frac{1}{8}

(B) \frac{1}{4} \frac{1}{4}

(D) \frac{\sqrt{5}}{6} \frac{\sqrt{5}}{6}

(E) \frac{\sqrt{10}}{5} \frac{\sqrt{10}}{5}

Answer

Correct choice: (C)

正确答案：（C）

Solution

If the original radius is $12$, then the circumference is $24\pi$; since arcs are defined by the central angles, the smaller arc, a $120$ degree angle, is half the size of the larger sector. so the smaller arc is $8\pi$, and the larger is $16\pi$. Those two arc lengths become the two circumferences of the new cones; so the radius of the smaller cone is $4$ and the larger cone is $8$. Using the Pythagorean theorem, the height of the larger cone is $4\cdot\sqrt{5}$ and the smaller cone is $8\cdot\sqrt{2}$, and now for volume just square the radii and multiply by $\tfrac{1}{3}$ of the height to get the volume of each cone: $128\cdot\sqrt{2}$ and $256\cdot\sqrt{5}$ [both multiplied by three as ratio come out the same. now divide the volumes by each other to get the final ratio of $\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}$

若原半径为 $12$，则周长为 $24\pi$；由于弧长由圆心角决定，圆心角为 $120$ 的较小扇形的弧长是较大扇形的一半。因此较小弧长为 $8\pi$，较大弧长为 $16\pi$。这两段弧长分别成为新圆锥的底面周长；因此较小圆锥的底面半径为 $4$，较大圆锥的底面半径为 $8$。由勾股定理，较大圆锥的高为 $4\cdot\sqrt{5}$，较小圆锥的高为 $8\cdot\sqrt{2}$。体积计算中将半径平方并乘以 $\tfrac{1}{3}$ 的高，得到体积分别为 $128\cdot\sqrt{2}$ 和 $256\cdot\sqrt{5}$（两者都乘以 3 时比值不变）。将体积相除得到最终比值为 $\boxed{\textbf{(C) } \frac{\sqrt{10}}{10}}$。

Topics