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

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Solving for <math>r</math>, we obtain <math>r = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>r = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>r \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes
 
Solving for <math>r</math>, we obtain <math>r = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>r = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>r \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes
  
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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>. Then would make sure the above solution works. <math>\textbf{- Emathmaster}</math>
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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. <math>\textbf{- Emathmaster}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:52, 1 February 2020

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

We consider an individual one by one block.

If we draw a quarter of a circle from each corner, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write

\[4 * \frac{1}{4} * \pi r^2 = \frac{1}{2}\]

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

Video Solution

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

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