2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 16 Hard

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 \frac{1}{2}. (A point (x, y) is a lattice point if x and y are both integers.) What is d to the nearest tenth?

  • A.

    0.3

  • B.

    0.4

  • C.

    0.5

  • D.

    0.6

  • E.

    0.7

Answer:B

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 \Longrightarrow(\text{B}) 0.4

Note: To be more rigorous, note that d<0.5 since if d \geq 0.5_{\text {then }} clearly the probability is greater than \frac{1}{2}. This would make sure the above solution works, as if d \geq 0.5 there is overlap with the quartercircles.

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 \overline{6}. We notice that 0.16 than 0.4^{2}=0.16, so d is roughly (\text{B}) 0.4 .

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