Difference between revisions of "2016 AMC 10B Problems/Problem 14"

Revision as of 07:48, 9 March 2024

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: $(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$

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

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

Solution 3

The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases.

Case 1: Square has length $\bf1$ .

The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12 = 30$ squares for this case.

Case 2: Square has length $\bf2$ .

The $x$ coordinates must be $(1,3)$ or $(2,4)$ or $(3,15)$ and so now it starts at $y=2.$ It ends up being $2+5+8 = 15.$

Case 3: Square has length $\bf3$ .

The $x$ coordinates must be $(1,4)$ or $(2,5)$ so there is $1+4 = 5$ squares for this case.

Therefore, $30+15+5 = \boxed{\textbf{(D)}\ 50}$ .

Video Solution

<url>https://www.youtube.com/watch?v=yJFQ2lsh6wg&ab_channel=TheBeautyofMath</url> - the Beauty of Math ~pixelpyguy

@@ Line 58: / Line 58: @@
 Therefore, <math>30+15+5 = \boxed{\textbf{(D)}\ 50}</math>.
+==Video Solution==
+<url>https://www.youtube.com/watch?v=yJFQ2lsh6wg&ab_channel=TheBeautyofMath</url> - the Beauty of Math
+~pixelpyguy
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=13|num-a=15}}
 {{MAA Notice}}

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