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Difference between revisions of "2020 AMC 10A Problems/Problem 16"

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== Problem ==
 
== Problem ==
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A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math>
 
A point is chosen at random within the square in the coordinate plane whose vertices are <math>(0, 0), (2020, 0), (2020, 2020),</math> and <math>(0, 2020)</math>. The probability that the point is within <math>d</math> units of a lattice point is <math>\tfrac{1}{2}</math>. (A point <math>(x, y)</math> is a lattice point if <math>x</math> and <math>y</math> are both integers.) What is <math>d</math> to the nearest tenth<math>?</math>
  
 
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math>
 
<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math>
  
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== Solution 1 ==
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== Solutions ==
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=== Diagram ===
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=== Solution 1 ===
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==== Diagram ====
 
<asy>
 
<asy>
 
size(10cm);
 
size(10cm);
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Note: The diagram represents each unit square of the given <math>2020 \times 2020</math> square.
 
Note: The diagram represents each unit square of the given <math>2020 \times 2020</math> square.
  
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===Solution===
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==== Solution ====
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We consider an individual one-by-one block.
 
We consider an individual one-by-one block.
  
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<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math>
 
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math>
  
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=== Solution 2 ===
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== Solution 2 ==
 
 
As in the previous solution, we obtain the equation <math>4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]
 
As in the previous solution, we obtain the equation <math>4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]]
  
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== Solution 3 (Estimating) ==
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=== Solution 3 (Estimating) ===
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As above, we find that we need to estimate <math>d = \frac{1}{\sqrt{2\pi}}</math>.  
 
As above, we find that we need to estimate <math>d = \frac{1}{\sqrt{2\pi}}</math>.  
  
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~Silverdragon
 
~Silverdragon
  
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==Solution 4 (Estimating but a bit different)==
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=== Solution 4 (Estimating but a bit different) ===
 
We only need to figure out the probability for a unit square, as it will scale up to the <math>2020\times 2020</math> square. Since we want to find the probability that a point inside a unit square that is <math>d</math> units away from a lattice point (a corner of the square) is <math>\frac{1}{2}</math>, we can find which answer will come the closest to covering <math>\frac{1}{2}</math> of the area.  
 
We only need to figure out the probability for a unit square, as it will scale up to the <math>2020\times 2020</math> square. Since we want to find the probability that a point inside a unit square that is <math>d</math> units away from a lattice point (a corner of the square) is <math>\frac{1}{2}</math>, we can find which answer will come the closest to covering <math>\frac{1}{2}</math> of the area.  
  
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~RuiyangWu
 
~RuiyangWu
  
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==Solution 5 (Estimating but differently again)==
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=== Solution 5 (Estimating but differently again) ===
 
As per the above diagram, realize that <math>\pi d^2 = \frac{1}{2}</math>, so <math>d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}</math>.
 
As per the above diagram, realize that <math>\pi d^2 = \frac{1}{2}</math>, so <math>d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}</math>.
  
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-Solution by Joeya
 
-Solution by Joeya
  
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==Video Solution==
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== Video Solution ==
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=== Video Solution 1 ===
 
Education, The Study of Everything
 
Education, The Study of Everything
  
 
https://youtu.be/napCkujyrac
 
https://youtu.be/napCkujyrac
  
 +
=== Video Solution 2 ===
 
https://youtu.be/RKlG6oZq9so
 
https://youtu.be/RKlG6oZq9so
  

Revision as of 13:32, 19 January 2021

The following problem is from both the 2020 AMC 12A #16 and 2020 AMC 10A #16, so both problems redirect to this page.

Problem

A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth$?$

$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

Solutions

Solution 1

Diagram

[asy] size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); [/asy]

Diagram by MathandSki Using Asymptote

Note: The diagram represents each unit square of the given $2020 \times 2020$ square.

Solution

We consider an individual one-by-one block.

If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write

\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]

Solving for $d$, we obtain $d = \frac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d = \frac{1}{\sqrt{6}}$, and from here, we simplify and see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}$ ~Crypthes

$\textbf{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\geq0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\geq0.5$ there is overlap with the quartercircles. $\textbf{- Emathmaster}$

Solution 2

As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$, which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$. Since $\pi$ is slightly more than $3$, $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$. We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$, so $d$ is roughly $\boxed{\textbf{(B) } 0.4}.$ ~emerald_block

Solution 3 (Estimating)

As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$.

Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$.

And so our answer is $\boxed{\textbf{(B) } 0.4}$.

~Silverdragon

Solution 4 (Estimating but a bit different)

We only need to figure out the probability for a unit square, as it will scale up to the $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$, we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.

Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$, we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{\textbf{(B)}}$.

~RuiyangWu

Solution 5 (Estimating but differently again)

As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$, so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$.

$\sqrt{2} \approx 1.4 = \frac{7}{5}$.

$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$, so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$.

So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$. This is slightly above $\boxed{\textbf{(B) } 0.4}$, since $\frac{20}{49} \approx \frac{2}{5}$.

-Solution by Joeya

Video Solution

Video Solution 1

Education, The Study of Everything

https://youtu.be/napCkujyrac

Video Solution 2

https://youtu.be/RKlG6oZq9so

~IceMatrix

See Also

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
2020 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 12 Problems and Solutions

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