Difference between revisions of "2016 AMC 12B Problems/Problem 11"
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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> squares with | + | 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 <math>x</math>-value is <math>5</math>. First we count the <math>1*1</math> squares. In the back row, there are <math>12</math> squares with length <math>1</math> because <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> in the equation <math>y=\pi x</math>. So there are <math>12+9+6+3 = 30</math> squares with length <math>1</math> 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. For <math>2*2</math> squares, the back row 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 row 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 back row start at <math>(2,0)</math> and end at <math>(2,2\pi)</math>, so there are <math>4</math> such squares in the back row. As the front 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{B) 50}}</math>. |
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==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 13:35, 28 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
-value is
. First we count the
squares. In the back row, there are
squares with length
because
, and continuing on we have
,
, and
for x-values for
,
, and
in the equation
. So there are
squares with length
in the figure. For
squares, each square takes up
un left and
un 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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