Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 8-13.
Visit Beast Academy ‚
Books for Ages 8-13 Beast Academy Online
AoPS Academy
Nationwide learning centers
for students in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 10A Problems/Problem 16"
m (→‎Diagram)
 
(26 intermediate revisions by 17 users not shown)
Line 2: Line 2:
  
 
== Problem ==
 
== Problem ==
−
 
 
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>
  
−
== Solution 1 ==
+
== Solutions ==
 +
==== 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>
 +
 
 +
The diagram represents each unit square of the given <math>2020 \times 2020</math> square.
 +
 
 +
==== Solution 1 ====
 
We consider an individual one-by-one block.
 
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 <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write
 
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write
  
−
<cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath>
+
<cmath>4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}</cmath>
 +
 
 +
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}</math>, and from here, we see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.</math>  
  
−
Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes
+
~Crypthes
 +
 
 +
~ Minor Edits by BakedPotato66
  
 
<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>
  
−
== Solution 2 ==
+
=== Solution 2 ===
−
As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \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]]
 +
 
 +
=== Solution 3 (Estimating) ===
 +
As above, we find that we need to estimate <math>d = \frac{1}{\sqrt{2\pi}}</math>.
 +
 
 +
Note that we can approximate <math>2\pi \approx 6.28318 \approx 6.25</math> and so <math>\frac{1}{\sqrt{2\pi}}</math> <math>\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4</math>.
 +
 
 +
And so our answer is <math>\boxed{\textbf{(B) } 0.4}</math>.
 +
 
 +
~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 <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.
 +
 
 +
Since the closest is <math>0.4</math> which turns out to be <math>(0.4)^2\times \pi = 0.16 \times \pi</math> which is about <math>0.502</math>, we find that the answer rounded to the nearest tenth is <math>0.4</math> or <math>\boxed{\textbf{(B)}}</math>.
 +
 
 +
~RuiyangWu
 +
 
 +
=== 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>.
 +
 
 +
<math>\sqrt{2} \approx 1.4 = \frac{7}{5}</math>.
 +
 
 +
<math>\sqrt{\pi}</math> is between <math>1.7</math> and <math>1.8</math> <math>((1.7)^2 = 2.89</math> and <math>(1.8)^2 = 3.24)</math>, so we can say <math>\sqrt{\pi} \approx 1.75 = \frac{7}{4}</math>.
  
−
==Video Solution==
+
So <math>d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}</math>. This is slightly above <math>\boxed{\textbf{(B) } 0.4}</math>, since <math>\frac{20}{49} \approx \frac{2}{5}</math>.
 +
 
 +
-Solution by Joeya
 +
 
 +
== Video Solutions ==
 +
=== Video Solution 1 ===
 +
Education, The Study of Everything
 +
 
 +
https://youtu.be/napCkujyrac
 +
 
 +
=== Video Solution 2 ===
 
https://youtu.be/RKlG6oZq9so
 
https://youtu.be/RKlG6oZq9so
  
 
~IceMatrix
 
~IceMatrix
 +
 +
=== Video Solution 3 ===
 +
https://youtu.be/R220vbM_my8?t=238
 +
 +
~ amritvignesh0719062.0
  
 
==See Also==
 
==See Also==
Line 30: Line 87:
 
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2020|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2020|ab=A|num-b=15|num-a=17}}
 +
 +
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:43, 8 November 2022

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

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]

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

Solution 1

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 $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$, and from here, we see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.$

~Crypthes

~ Minor Edits by BakedPotato66

$\textbf{Note:}$ 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 Solutions

Video Solution 1

Education, The Study of Everything

https://youtu.be/napCkujyrac

Video Solution 2

https://youtu.be/RKlG6oZq9so

~IceMatrix

Video Solution 3

https://youtu.be/R220vbM_my8?t=238

~ amritvignesh0719062.0

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_16&oldid=180474"
Invalid username
Login to AoPS