2016 AMC 12B Problems/Problem 11

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 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 (ProblemsAnswer KeyResources)
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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