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

Revision as of 23:50, 2 February 2020

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

$[asy] size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.4, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.4, 90, 180)); filldraw((arc((1,0), 0.4, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.4, 180, 270)); filldraw((arc((1,1), 0.4, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.4, 270, 360)); filldraw(arc((0,1), 0.4, 270, 360)--(0,1)--cycle, gray); [/asy]$

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

$4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}$

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. $\textbf{- Emathmaster}$

Diagram by Shurong.ge Using Asymptote

Solution 2

As in the previous solution, we obtain the equation $4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}$ , 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

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

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.

Revision as of 23:49, 2 February 2020 (view source) Shurong.ge (talk \| contribs) ← Older edit		Revision as of 23:50, 2 February 2020 (view source) Shurong.ge (talk \| contribs) (→‎Solution 1) Newer edit →
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−	Diagram by [[Shurong.ge]] Using Asymptote	+	Diagram by [[User:Shurong.ge\|Shurong.ge]] Using Asymptote

	== Solution 2 ==		== Solution 2 ==

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