Difference between revisions of "2016 AMC 12B Problems/Problem 11"
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For <math>2*2</math> squares, each square takes up <math>2</math> units left and <math>2</math> units up. Squares can also overlap. For <math>2*2</math> squares, the rightmost column stretches from <math>(3,0)</math> to <math>(3,3\pi)</math>, so there are <math>8</math> squares with length <math>2</math> in a <math>2</math> by <math>9</math> box. Repeating the process, the next column stretches from <math>(2,0)</math> to <math>(2,2\pi)</math>, so there are <math>5</math> squares. Continuing and adding up in the end, there are <math>8+5+2=15</math> squares with length <math>2</math> in the figure. | For <math>2*2</math> squares, each square takes up <math>2</math> units left and <math>2</math> units up. Squares can also overlap. For <math>2*2</math> squares, the rightmost column stretches from <math>(3,0)</math> to <math>(3,3\pi)</math>, so there are <math>8</math> squares with length <math>2</math> in a <math>2</math> by <math>9</math> box. Repeating the process, the next column stretches from <math>(2,0)</math> to <math>(2,2\pi)</math>, so there are <math>5</math> squares. Continuing and adding up in the end, there are <math>8+5+2=15</math> squares with length <math>2</math> in the figure. | ||
− | Squares with length <math>3</math> in the rightmost column start at <math>(2,0)</math> and end at <math>(2,2\pi)</math>, so there are <math>4</math> such squares in the right column. As the left row starts at <math>(1,0)</math> and ends at <math>(1,\pi)</math> there are <math>4+1=5</math> squares with length <math>3</math>. As squares with length <math>4</math> would not fit in the triangle, the answer is <math>30+15+5</math> which is <math>\boxed{\textbf{D)}\ 50}</math>. | + | Squares with length <math>3</math> in the rightmost column start at <math>(2,0)</math> and end at <math>(2,2\pi)</math>, so there are <math>4</math> such squares in the right column. As the left row starts at <math>(1,0)</math> and ends at <math>(1,\pi)</math> there are <math>4+1=5</math> squares with length <math>3</math>. As squares with length <math>4</math> would not fit in the triangle, the answer is <math>30+15+5</math> which is <math>\boxed{\textbf{(D)}\ 50}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2016|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:05, 20 October 2024
Problem
How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line , the line and the line
Solution
Solution by e_power_pi_times_i Revised by Kinglogic and RJ5303707
(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 rightmost column, 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 rightmost column stretches from to , so there are squares with length in a by box. Repeating the process, the next column 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 rightmost column start at and end at , so there are such squares in the right column. As the left 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.