2020 AMC 10A Problems/Problem 16

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 $?$

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

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

Education, The Study of Everything

https://youtu.be/napCkujyrac

https://youtu.be/RKlG6oZq9so

~IceMatrix

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