Difference between revisions of "2020 AMC 10A Problems/Problem 10"
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It can be seen that the side lengths of the cubes using cube roots are all integers from <math>1</math> to <math>7</math>, inclusive. | It can be seen that the side lengths of the cubes using cube roots are all integers from <math>1</math> to <math>7</math>, inclusive. | ||
− | Only the cubes with side length <math>1</math> and <math>7</math> have <math>5</math> faces in the surface area and the rest have <math> | + | Only the cubes with side length <math>1</math> and <math>7</math> have <math>5</math> faces in the surface area and the rest have <math>4</math>. Also, since the |
cubes are stacked, we have to find the difference between each <math>n^2</math> and <math>(n-1)^2</math> side length as <math>n</math> ranges from <math>7</math> to | cubes are stacked, we have to find the difference between each <math>n^2</math> and <math>(n-1)^2</math> side length as <math>n</math> ranges from <math>7</math> to |
Revision as of 23:53, 7 July 2020
- The following problem is from both the 2020 AMC 12A #7 and 2020 AMC 10A #10, so both problems redirect to this page.
Contents
Problem
Seven cubes, whose volumes are , , , , , , and 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?
Solution 1
The volume of each cube follows the pattern of ascending, for is between and .
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the cube (which is just ). The sides areas can be measured as the sum , giving us . Structurally, if we examine the tower from the top, we see that it really just forms a square of area . Therefore, we can say that the total surface area is . Alternatively, for the area of the tops, we could have found the sum , giving us as well.
~ciceronii
Solution 2
It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7, inclusive.
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is . Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to . Subtracting the overlapped surface area from the total surface area, we get . ~emerald_block
Solution 3 (a bit more tedious than other solutions)
It can be seen that the side lengths of the cubes using cube roots are all integers from to , inclusive.
Only the cubes with side length and have faces in the surface area and the rest have . Also, since the
cubes are stacked, we have to find the difference between each and side length as ranges from to
.
We then come up with this: .
We then add all of this and get .
~aryam
Video Solution
~IceMatrix
Solution 5
Notice that the surface area of the top cube is and the others are . Then we can directly compute. The edge length for the first cube is and has a surface area of . The surface area of the next cube is . The surface area of the next cube . The surface area of the next cube is . The surface area of the next cube is . The surface area of the next cube is . The surface area of the next cube is . The surface area of the next cube is . We then sum up to get . -smartatmath
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 | ||
All AMC 10 Problems and Solutions |
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.