2020 AMC 10A Problems/Problem 16

Revision as of 17:37, 1 February 2020 by Emerald block (talk | contribs) (added alternate solution)

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

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. $\textbf{- Emathmaster}$

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

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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