AMC8 2009

AMC8 2009 · Q25

AMC8 2009 · Q25. It mainly tests Fractions, Area & perimeter.

A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C and D. The pieces are then glued together end to end in the order C, B, A, D to make a long solid as shown below. What is the total surface area of this solid in square feet?

一个 1 立方英尺的立方体被三刀平行于立方体顶面切割成四块。第一刀距顶面 $rac{1}{2}$ 英尺。第二刀在第一刀下方 $rac{1}{3}$ 英尺，第三刀在第二刀下方 $rac{1}{17}$ 英尺。从上到下，块依次标记为 A、B、C 和 D。然后将这些块按 C、B、A、D 的顺序端对端粘合，形成一个长固体，如下图所示。这个固体的总表面积是多少平方英尺？

(A) 6 6

(B) 7 7

(D) \frac{158}{17} \frac{158}{17}

(E) 11 11

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Looking from either end, the visible area totals $\frac{1}{2}$ square foot because piece $A$ measures $\frac{1}{2}\times 1=\frac{1}{2}\ \text{ft}^2$, and the other pieces decrease in height from that piece. The two side views each show four blocks that can stack to a unit cube. So the area as seen from each side is $1\ \text{ft}^2$. Finally, the top and bottom views each show four unit squares. So the top and bottom view each contribute $4\ \text{ft}^2$ to the area. Summing, the total surface area is $\frac{1}{2}+\frac{1}{2}+1+1+4+4=11$ square feet. CHALLENGE: Suppose the cuts are $\frac{1}{2}$, $\frac{1}{4}$ and $\frac{1}{8}$. Does this change the solution?

答案（E）：从任一端看，可见面积总计为$\frac{1}{2}$平方英尺，因为块$A$的面积为$\frac{1}{2}\times 1=\frac{1}{2}\ \text{ft}^2$，其余块的高度从该块开始逐渐降低。两个侧视图各显示四个小方块，它们可以堆叠成一个单位立方体。因此从每个侧面看到的面积是$1\ \text{ft}^2$。最后，俯视图和仰视图各显示四个单位正方形，所以顶面和底面视图各贡献$4\ \text{ft}^2$的面积。相加可得总表面积为 $\frac{1}{2}+\frac{1}{2}+1+1+4+4=11$平方英尺。挑战：假设切割比例是$\frac{1}{2}$、$\frac{1}{4}$和$\frac{1}{8}$。这会改变解法吗？

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