2016 AMC 12B Problems/Problem 11
Problem
The grid below contains the points whose - and -coordinates are in the set :A square with all four of its vertices among these points has area . What is the sum of all possible values of ?
Solution
Solution by e_power_pi_times_i Revised by Kinglogic
(red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is squares , and the limit for the -value is squares. First we count the squares. In the back row, there are squares with length because generates squares from to , and continuing on we have , , and for -values for , , and in the equation . So there are squares with length in the figure. For squares, each square takes up units left and units up. Squares can also overlap. For squares, the back row stretches from to , so there are squares with length in a by box. Repeating the process, the next row stretches from to , so there are squares. Continuing and adding up in the end, there are squares with length in the figure. Squares with length in the back row start at and end at , so there are such squares in the back row. As the front row starts at and ends at there are squares with length . As squares with length would not fit in the triangle, the answer is which is .
See Also
2016 AMC 12B (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 |
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