AMC12 2008 B

AMC12 2008 B · Q11

AMC12 2008 B · Q11. It mainly tests Quadratic equations, 3D geometry (volume).

A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top $\frac{1}{8}$ of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?

一座锥形山以其底面位于海底，高度为8000英尺。山体体积的顶端 $\frac{1}{8}$ 在水上。该山的底面处海洋深度是多少英尺？

(A) 4000 4000

(B) 2000(4 - $\sqrt{2}$) 2000(4 - $\sqrt{2}$)

(D) 6400 6400

(E) 7000 7000

Answer

Correct choice: (A)

正确答案：（A）

Solution

In a cone, radius and height each vary inversely with increasing height (i.e. the radius of the cone formed by cutting off the mountain at $4,000$ feet is half that of the original mountain). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone): $V_I\times \text{Height}^3 = V_N$ Plugging in our given condition, $\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}$. $8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}$.

在圆锥中，半径和高度都随高度的增加按比例缩小（例如在 $4,000$ 英尺处截去山体形成的圆锥，其半径是原山体的一半）。因此体积随高度（占圆锥总高度的百分比）的立方成比例变化： $V_I\times \text{Height}^3 = V_N$ 代入已知条件，$\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}$。 $8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}$.

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