AMC12 2020 A

AMC12 2020 A · Q7

AMC12 2020 A · Q7. It mainly tests Arithmetic sequences basics, 3D geometry (surface area).

Seven cubes, whose volumes are 1, 8, 27, 64, 125, 216, and 343 cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

有七个立方体，它们的体积分别是 1、8、27、64、125、216 和 343 立方单位。这些立方体垂直堆叠成一座塔，体积从底部到顶部递减。除了最底部的立方体，每个立方体的底面完全位于下方立方体的顶面上。塔的总表面积（包括底部）有多少平方单位？

(A) 644 644

(B) 658 658

(D) 720 720

(E) 749 749

Answer

Correct choice: (B)

正确答案：（B）

Solution

Because the volumes of the cubes are given to be $1^3, 2^3, \dots, 7^3$, the edge lengths are 1, 2, $\dots$, 7. If the cubes were not stacked, the total surface area would be $6 \cdot (1^2 + 2^2 + \dots + 7^2)$. When two cubes meet along a surface, two of the smaller face areas are lost, for a total area of $2 \cdot (1^2 + 2^2 + \dots + 6^2)$ that needs to be subtracted from the surface area of the unstacked cubes to get the total surface area of the tower. The result is $$4 \cdot (1^2 + 2^2 + \dots + 6^2) + 6 \cdot 7^2 = 4 \cdot 91 + 6 \cdot 49 = 658.$$ OR The horizontal surface area is $2 \cdot 7^2 = 98$, and the vertical surface area is $4 \cdot (1^2 + 2^2 + \dots + 7^2) = 560$, for a total of 658.

因为立方体的体积分别是 $1^3, 2^3, \dots, 7^3$，所以棱长分别是 1、2、$\dots$、7。如果不堆叠，这些立方体的总表面积是 $6 \cdot (1^2 + 2^2 + \dots + 7^2)$。当两个立方体沿一个面接触时，丢失了两个较小面的面积，总共需要从未堆叠立方体的表面积中减去 $2 \cdot (1^2 + 2^2 + \dots + 6^2)$ 以得到塔的总表面积。结果是 $$4 \cdot (1^2 + 2^2 + \dots + 6^2) + 6 \cdot 7^2 = 4 \cdot 91 + 6 \cdot 49 = 658。$$ 或者水平表面积是 $2 \cdot 7^2 = 98$，垂直表面积是 $4 \cdot (1^2 + 2^2 + \dots + 7^2) = 560$，总计 658。

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