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 | + | 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 23:00, 10 August 2020
Problem
A cube is composed of twenty-seven cubes. The big cube is ‘tunneled’ as follows: First, the six cubes which make up the center of each face as well as the center cube are removed. Second, each of the twenty remaining 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:
Solution
Solution 1
Each cube has eight faces on each side, for a surface area of 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 . Thus the total surface area for each is .
There are of these cubes, for an area of . However, many of the cubes share a common face; each corner cube has three hidden faces and each edge cube has two hidden faces, for a total of hidden faces. Each hidden face has a surface area of , so the surface area of the final figure is .
Solution 2
After the first step, twenty cubes remain, with corner cubes and edge cubes. Each one of these corner cubes contributes square units of area, and each edge cube contributes square units of area.
The second stage takes away square units of area ( for each exposed face) from each of the eight corner cubes, and adds an additional more units from the center facial cubes removed. Similarly, the twelve edge cubes each lose square units but gain units. Thus, the total surface area is
See also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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