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Difference between revisions of "2016 AMC 12B Problems/Problem 11"

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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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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.  
  
 
I'm running late so can someone do the calculations? Thanks.
 
I'm running late so can someone do the calculations? Thanks.

Revision as of 01:41, 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 $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

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 $16 (y=5.1*\pi)$, and the limit for the x-value is $5$. First we count the $1*1$ squares. In the back row, there are $12$ squares $y=4*\pi$, and continuing on we have $9$, $6$, and $3$ for x-values for $1$, $2$, and $3$. So there are $12+9+6+3 = 30$ $1*1$ squares in the figure. For $2*2$ squares, each square takes up $2$ un left and $2$ 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

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