Difference between revisions of "2016 AMC 12B Problems/Problem 11"
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==Solution== | ==Solution== | ||
+ | Solution by e_power_pi_times_i | ||
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+ | (Note: diagram is needed) | ||
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+ | 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 <math>16 (y=5.1*\pi)</math>, and the limit for the x-value is <math>5</math>. First we count the <math>1*1</math> squares. In the back row, there are <math>12</math> squares <math>y=4*\pi</math>, and continuing on we have <math>9</math>, <math>6</math>, and <math>3</math> for x-values for <math>1</math>, <math>2</math>, and <math>3</math>. So there are <math>12+9+6+3 = 30</math> <math>1*1</math> squares in the figure. For <math>2*2</math> squares, each square takes up <math>2</math> un left and <math>2</math> un up. Squares can also overlap. | ||
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+ | I'm running late so can someone do the calculations? Thanks. | ||
==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 02:40, 27 February 2016
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
(Note: diagram is needed)
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 , and the limit for the x-value is
. First we count the
squares. In the back row, there are
squares
, and continuing on we have
,
, and
for x-values for
,
, and
. So there are
squares in the figure. For
squares, each square takes up
un left and
un up. Squares can also overlap.
I'm running late so can someone do the calculations? Thanks.
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.