AMC12 2021 A

AMC12 2021 A · Q10

AMC12 2021 A · Q10. It mainly tests Similarity, 3D geometry (volume).

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

如图所示，两个顶点朝下的右圆锥含有相同量的液体。液体表面的顶面半径分别为 $3$ cm 和 $6$ cm。在每个圆锥中放入一颗半径 $1$ cm 的球形弹珠，该弹珠沉到底部，完全浸没且不溢出液体。求窄锥中液面上升高度与宽锥中液面上升高度的比值。

(A) 1:1 1:1

(B) 47:43 47:43

(D) 40:13 40:13

(E) 4:1 4:1

Answer

Correct choice: (E)

正确答案：（E）

Solution

Initial Scenario Let the heights of the narrow cone and the wide cone be $h_1$ and $h_2,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 & \end{array}\] Equating the volumes gives $3\pi h_1=12\pi h_2,$ which simplifies to $\frac{h_1}{h_2}=4.$ Furthermore, by similar triangles: - For the narrow cone, the ratio of the base radius to the height is $\frac{3}{h_1},$ which always remains constant. - For the wide cone, the ratio of the base radius to the height is $\frac{6}{h_2},$ which always remains constant. Two solutions follow from here: Final Scenario For the narrow cone and the wide cone, let their base radii be $3x$ and $6y$ (for some $x,y>1$), respectively. By the similar triangles discussed above, their heights must be $h_1x$ and $h_2y,$ respectively. We have the following table: \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & & \frac13\pi(3x)^2(h_1x)=3\pi h_1 x^3 & \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.0625mm}\frac13\pi(6y)^2(h_2y)=12\pi h_2 y^3 & \end{array}\] Recall that $\frac{h_1}{h_2}=4.$ Equating the volumes gives $3\pi h_1 x^3=12\pi h_2 y^3,$ which simplifies to $x^3=y^3,$ or $x=y.$ Finally, the requested ratio is \[\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4:1}.\]

初始情景设窄锥和宽锥的高度分别为 $h_1$ 和 $h_2$。有如下表格： \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3 & h_1 & & \frac13\pi(3)^2h_1=3\pi h_1 & \\ [2ex] \textbf{Wide Cone} & 6 & h_2 & & \hspace{2mm}\frac13\pi(6)^2h_2=12\pi h_2 & \end{array}\] 体积相等得 $3\pi h_1=12\pi h_2$，简化得 $\frac{h_1}{h_2}=4$。此外，由相似三角形： - 窄锥，底半径与高度之比为 $\frac{3}{h_1}$，恒定。 - 宽锥，底半径与高度之比为 $\frac{6}{h_2}$，恒定。最终情景设窄锥和宽锥的底半径分别为 $3x$ 和 $6y$（$x,y>1$）。由相似三角形，其高度分别为 $h_1x$ 和 $h_2y$。有如下表格： \[\begin{array}{cccccc} & \textbf{Base Radius} & \textbf{Height} & & \textbf{Volume} & \\ [2ex] \textbf{Narrow Cone} & 3x & h_1x & & \frac13\pi(3x)^2(h_1x)=3\pi h_1 x^3 & \\ [2ex] \textbf{Wide Cone} & 6y & h_2y & & \hspace{2.0625mm}\frac13\pi(6y)^2(h_2y)=12\pi h_2 y^3 & \end{array}\] 忆及 $\frac{h_1}{h_2}=4$。体积相等得 $3\pi h_1 x^3=12\pi h_2 y^3$，简化得 $x^3=y^3$，即 $x=y$。最后，所求比值为 \[\frac{h_1 x - h_1}{h_2 y - h_2}=\frac{h_1 (x-1)}{h_2 (y-1)}=\frac{h_1}{h_2}=\boxed{\textbf{(E) }4:1}\]。

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