2009 AMC 12A Problems/Problem 11
- The following problem is from both the 2009 AMC 12A #11 and 2009 AMC 10A #15, so both problems redirect to this page.
Problem
The figures , , , and shown are the first in a sequence of figures. For , is constructed from by surrounding it with a square and placing one more diamond on each side of the new square than had on each side of its outside square. For example, figure has diamonds. How many diamonds are there in figure ?
Solution 1
Split into congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to times the th triangular number (i.e. the diamonds within the triangles or between the diagonals) and (the diamonds on sides of the triangles or on the diagonals). The th triangular number is . Putting this together for this gives:
Solution 2
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure , the blue diamonds form a square, and the red diamonds form a square. Hence the total number of diamonds in is .
Solution 3
When constructing from , we add new diamonds. Let be the number of diamonds in . We now know that and .
Hence we get:
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 |
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
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