2000 AMC 12 Problems/Problem 8
Problem
Figures ,
,
, and
consist of
,
,
, and
non-overlapping squares. If the pattern continued, how many non-overlapping squares would there be in figure
?
Solution
By counting the squares starting from the center of each figure, the figure 0 has 1 square, the figure 1 has squares, figure 2 has
squares, and so on. Figure 100 would have
.
Alternate solution:
Note that figure 0 has 1 square, figure 1 has 5 squares, figure 2 has 13 squares, and so on. If we let the number of the figure = , note that
represents the number of squares in the figure. For example, figure 4 has
squares. Therefore, the number of squares in figure 100 has
.
alternate solution:
For the
figure, note that it could be constructed by making a
square, and then removing the
triangular number from each of its corners. So, if
represents the amount of squares in figure
,
. Therefore,
, which gives
.
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 |