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Difference between revisions of "1998 AHSME Problems/Problem 27"

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After the first step, twenty <math>3 \times 3 \times 3</math> cubes remain, with <math>8</math> corner cubes and <math>12</math> edge cubes. Each one of these <math>3 \times 3 \times 3</math> corner cubes contributes <math>27</math> square units of area, and each edge cube contributes <math>36</math> square units of area.  
 
After the first step, twenty <math>3 \times 3 \times 3</math> cubes remain, with <math>8</math> corner cubes and <math>12</math> edge cubes. Each one of these <math>3 \times 3 \times 3</math> corner cubes contributes <math>27</math> square units of area, and each edge cube contributes <math>36</math> square units of area.  
  
The second stage takes away <math>3</math> square units of area (<math>1</math> for each exposed face) from each of the eight <math>3 \times 3 \times 3</math> corner cubes, and adds an additional <math>24</math> more units from the center facial cubes removed. Similarly, the twelve <math>3\times 3\times 3</math> edge cubes each lose <math>4</math> square nits but gain <math>24</math> units. Thus, the total surface area is  
+
The second stage takes away <math>3</math> square units of area (<math>1</math> for each exposed face) from each of the eight <math>3 \times 3 \times 3</math> corner cubes, and adds an additional <math>24</math> more units from the center facial cubes removed. Similarly, the twelve <math>3\times 3\times 3</math> edge cubes each lose <math>4</math> square units but gain <math>24</math> units. Thus, the total surface area is  
 
<cmath>8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056</cmath>
 
<cmath>8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056</cmath>
  

Latest revision as of 22:00, 10 August 2020

Problem

A $9 \times 9 \times 9$ cube is composed of twenty-seven $3 \times 3 \times 3$ cubes. The big cube is ‘tunneled’ as follows: First, the six $3 \times 3 \times 3$ cubes which make up the center of each face as well as the center $3 \times 3 \times 3$ cube are removed. Second, each of the twenty remaining $3 \times 3 \times 3$ cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The surface area of the final figure is:

1998 AHSME num. 27.png

$\mathrm{(A)}\ 384 \qquad\mathrm{(B)}\ 729 \qquad\mathrm{(C)}\ 864 \qquad\mathrm{(D)}\ 1024 \qquad\mathrm{(E)}\ 1056$

Solution

Solution 1

Each $3 \times 3 \times 3$ cube has eight faces on each side, for a surface area of $6 \cdot 8 \cdot (1 \cdot 1) = 48$ on the outside. Each face also has to count the surface area in the inside area of the removed cube, for an additional surface area of $6 \cdot 4 \cdot (1 \cdot 1) = 24$. Thus the total surface area for each $3 \times 3 \times 3$ is $72$.

There are $20$ of these cubes, for an area of $72 \cdot 20 = 1440$. However, many of the cubes share a common face; each corner $3\times 3\times 3$ cube has three hidden faces and each edge cube has two hidden faces, for a total of $8\cdot 3 + 12\cdot 2 = 48$ hidden faces. Each hidden face has a surface area of $8$, so the surface area of the final figure is $1440 - 48 \cdot 8 = 1056 \Rightarrow \mathrm{(E)}$.

Solution 2

After the first step, twenty $3 \times 3 \times 3$ cubes remain, with $8$ corner cubes and $12$ edge cubes. Each one of these $3 \times 3 \times 3$ corner cubes contributes $27$ square units of area, and each edge cube contributes $36$ square units of area.

The second stage takes away $3$ square units of area ($1$ for each exposed face) from each of the eight $3 \times 3 \times 3$ corner cubes, and adds an additional $24$ more units from the center facial cubes removed. Similarly, the twelve $3\times 3\times 3$ edge cubes each lose $4$ square units but gain $24$ units. Thus, the total surface area is \[8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056\]

See also

Menger Sponge

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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