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

Revision as of 22:20, 20 October 2020

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

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 $1.$

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 $2.$

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 $3.$

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

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

@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 57</math>
+==Solution 1==
+The region is a right triangle which contains the following lattice points:
+<math>(0,0); (1,0)-(1,3); (2,0)-(2,6); (3,0)-(3,9); (4,0)-(4,12); (5,0)-(5,15)</math>
-==Solution==
+<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 <math>1\times 1</math>:
+Suppose that the top-right corner is <math>(x,y)</math>, with <math>2\le x\le 5</math>. Then to include all other corners, we need <math>1\le y\le 3(x-1)</math>.
+This produces <math>3+6+9+12=30</math> squares.
+Squares <math>2\times 2</math>:
+Here <math>3\le x\le 5</math>. To include all other corners, we need <math>2\le y\le 3(x-2)</math>.
+This produces <math>2+5+8=15</math> squares.
+Squares <math>3\times 3</math>:
+Similarly this produces <math>5</math> squares.
+No other squares will fit in the region. Therefore the answer is <math>\boxed{\textbf{(D) }50}</math>.
+==Solution 2==
+The vertical line is just to the right of <math>x = 5</math>, the horizontal line is just under <math>y = 0</math>, and the sloped line will always be above the <math>y</math> value of <math>3x</math>.
+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 <math>1\cdot1</math>, <math>2\cdot2</math>, and <math>3\cdot3</math> squares and getting <math>30</math>, <math>15</math>, and <math>5</math> respectively, and we end up with <math>\boxed{\textbf{(D)}\ 50}</math>.
+Solution by Wwang
+== Solution 3==
+The endpoint lattice points are <math>(1,3), (2,6), (3,9), (4,12), (5,15).</math> Now we split this problem into cases.
+Case <math>1:</math> Square has length <math>1.</math>
+The <math>x</math> coordinates must be <math>(1,2)</math> or <math>(2,3)</math> and so on to <math>(4,5).</math> The idea is that you start at <math>y=1</math> and add at the endpoint, namely <math>y=3.</math> The number ends up being <math>3+6+9+12 = 30</math> squares for this case.
+Case <math>2:</math> Square has length <math>2.</math>
+The <math>x</math> coordinates must be <math>(1,3)</math> or <math>(2,4)</math> or <math>(3,15)</math> and so now it starts at <math>y=2.</math> It ends up being <math>2+5+8 = 15.</math>
+Case <math>3:</math> Square has length <math>3.</math>
+The <math>x</math> coordinates must be <math>(1,4)</math> or <math>(2,5)</math> so there is <math>1+4 = 5</math> squares for this case.
+Answer is <math>30+15+5 = \boxed{\textbf{(D)}\ 50}</math>.
+==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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