2016 AMC 10B Problems/Problem 14

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 1

The region is a right triangle which contains the following lattice points: $(0,0); (1,0)-(1,3); (2,0)-(2,6); (3,0)-(3,9); (4,0)-(4,12); (5,0)-(5,15)$

$[asy]size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow); draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray); [/asy]$

Squares $1\times 1$ : Suppose that the top-right corner is $(x,y)$ , with $2\le x\le 5$ . Then to include all other corners, we need $1\le y\le 3(x-1)$ . This produces $3+6+9+12=30$ squares.

Squares $2\times 2$ : Here $3\le x\le 5$ . To include all other corners, we need $2\le y\le 3(x-2)$ . This produces $2+5+8=15$ squares.

Squares $3\times 3$ : Similarly this produces $5$ squares.

No other squares will fit in the region. Therefore the answer is $\boxed{\textbf{(D) }50}$ .

Solution 2

The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ . This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of $1\cdot1$ , $2\cdot2$ , and $3\cdot3$ squares and getting $30$ , $15$ , and $5$ respectively, and we end up with $\boxed{\textbf{(D)}\ 50}$ .

Solution by Wwang

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2016 AMC 10B Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See Also