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

(Problem 16)
(Solution)
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== Solution ==
 
== Solution ==
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We consider an individual one by one block.
  
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If we draw a quarter of a circle from each corner, 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
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<cmath>4 * \frac{1}{4} * \pi r^2 = \frac{1}{2}</cmath>.
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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 {\textbf{(B) } 0.4.}</math>
  
 
==See Also==
 
==See Also==

Revision as of 22:52, 31 January 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 $r \approx 0.4 \implies {\textbf{(B) } 0.4.}$

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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