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

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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>.
 
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>.
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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 1x1, 2x2, 3x3, squares and getting 30, 15, and 5 respectively, and we end up with <math>\boxed{\textbf{(D)}\ 50}</math>.
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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 by Wwang
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== Solution 3==
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The endpoint lattice points are <math>(1,3), (2,6), (3,9), (4,12), (5,15).</math> Now we split this problem into cases.
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Case <math>1:</math> Square has length <math>1.</math>
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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.
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Case <math>2:</math> Square has length <math>2.</math>
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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>
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Case <math>3:</math> Square has length <math>3.</math>
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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.
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Answer is <math>30+15+5 = \boxed{\textbf{(D)}\ 50}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2016|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

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

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