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

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==Problem==
 
==Problem==
  
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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 <math>y=\pi x</math>, the line <math>y=-0.1</math> and the line <math>x=5.1?</math>
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The grid below contains the <math>16</math> points whose <math>x</math>- and <math>y</math>-coordinates are in the set <math>\{0,1,2,3\}</math>:A square with all four of its vertices among these <math>16</math> points has area <math>A</math>. What is the sum of all possible values of <math>A</math>?
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<math>\textbf{(A)}\ 30 \qquad
 
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\textbf{(B)}\ 41 \qquad
 
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\textbf{(C)}\ 45 \qquad
 
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\textbf{(D)}\ 50 \qquad
 
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\textbf{(E)}\ 57</math>
 
  
 
==Solution==
 
==Solution==

Revision as of 15:45, 1 May 2022

Problem

The grid below contains the $16$ points whose $x$- and $y$-coordinates are in the set $\{0,1,2,3\}$:A square with all four of its vertices among these $16$ points has area $A$. What is the sum of all possible values of $A$?

Solution

Solution by e_power_pi_times_i Revised by Kinglogic

[asy] Label l; l.p=fontsize(8); xaxis(-1,8,Ticks(l, 1.0)); yaxis(-1,16,Ticks(l, 1.0)); real f(real x) { return x * pi; } D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { for(int y = 0; y < 16; ++y) { if(x * pi > y) { D((x,y)); } } } [/asy] (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 $16$ squares $(y=5.1*\pi)$, and the limit for the $x$-value is $5$ squares. First we count the $1*1$ squares. In the back row, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$, 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 $2$ units left and $2$ units 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 $\boxed{\textbf{D)}\ 50}$.

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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