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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 dimensions <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.  
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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 <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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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 12:35, 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 $y=\pi x$, 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 with length $1$ because $y=4*\pi$, and continuing on we have $9$, $6$, and $3$ for x-values for $1$, $2$, and $3$ in the equation $y=\pi x$. So there are $12+9+6+3 = 30$ squares with length $1$ in the figure. For $2*2$ squares, each square takes up $2$ un left and $2$ un up. Squares can also overlap. For $2*2$ squares, the back row stretches from $(3,0)$ to $(3,3\pi)$, so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next row stretches from $(2,0)$ to $(2,2\pi)$, so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure. Squares with length $3$ in the back row start at $(2,0)$ and end at $(2,2\pi)$, so there are $4$ such squares in the back row. As the front row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$. As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{\textbf{B) 50}}$.

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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