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

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==Solution 1==
 
==Solution 1==
 
The region is a right triangle which contains the following lattice points:
 
The region is a right triangle which contains the following lattice points:
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<math>(0,0); (1,0)-(1,3); (2,0)-(2,6); (3,0)-(3,9); (4,0)-(4,12); (5,0)-(5,15)</math>
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<math>(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)</math>
  
 
<asy>size(10cm);
 
<asy>size(10cm);
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Squares <math>3\times 3</math>:
 
Squares <math>3\times 3</math>:
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Similarly this produces <math>5</math> squares.
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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>.
 
No other squares will fit in the region. Therefore the answer is <math>\boxed{\textbf{(D) }50}</math>.
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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 1: Square has length <math>\bf1</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 2: Square has length <math>\bf2</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 3: Square has length <math>\bf3</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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Therefore, <math>30+15+5 = \boxed{\textbf{(D)}\ 50}</math>.
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==Video Solution==
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[https://www.youtube.com/watch?v=0CgqeYEARgA&ab_channel=RyanYang Video Solution by Ryan Yang]
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~pixelpyguy
  
 
==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}}

Latest revision as of 07:50, 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

Video Solution by Ryan Yang ~pixelpyguy

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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