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AMC 10 · 2020 · #16

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
geometric-probabilityarea-circlesspatial-visualizationestimation easier-related-problemidentify-subproblems ↑ Prerequisites: geometric-probabilityarea-circles
📏 Medium solution 💡 3 insights

Problem

A point is chosen at random within the square in the coordinate plane whose vertices are (0,0),(2020,0),(2020,2020),(0, 0), (2020, 0), (2020, 2020), and (0,2020)(0, 2020). The probability that the point is within dd units of a lattice point is 12\tfrac{1}{2}. (A point (x,y)(x, y) is a lattice point if xx and yy are both integers.) What is dd to the nearest tenth??

(A) 0.3(B) 0.4(C) 0.5(D) 0.6(E) 0.7\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7

Pick an answer.

(A)
0.3
(B)
0.4
(C)
0.5
(D)
0.6
(E)
0.7
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Toolkit + CCSS Solution

Understand

Restated: Pick a random point inside the $2020 \times 2020$ square with corners at $(0,0)$ and $(2020, 2020)$. A lattice point is any point with both coordinates integer. The chance that our random point lands within distance $d$ of some lattice point is exactly $\tfrac{1}{2}$. Find $d$ rounded to the nearest tenth.

Givens: Big square has corners $(0,0), (2020,0), (2020,2020), (0,2020)$; Lattice points are everywhere — at every integer coordinate pair inside and on the square; Probability of being within $d$ of a lattice point equals $\tfrac{1}{2}$; Choices: (A) $0.3$, (B) $0.4$, (C) $0.5$, (D) $0.6$, (E) $0.7$

Unknowns: $d$ to the nearest tenth

Understand

Restated: Pick a random point inside the $2020 \times 2020$ square with corners at $(0,0)$ and $(2020, 2020)$. A lattice point is any point with both coordinates integer. The chance that our random point lands within distance $d$ of some lattice point is exactly $\tfrac{1}{2}$. Find $d$ rounded to the nearest tenth.

Givens: Big square has corners $(0,0), (2020,0), (2020,2020), (0,2020)$; Lattice points are everywhere — at every integer coordinate pair inside and on the square; Probability of being within $d$ of a lattice point equals $\tfrac{1}{2}$; Choices: (A) $0.3$, (B) $0.4$, (C) $0.5$, (D) $0.6$, (E) $0.7$

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #9 (Easier Problem): the $2020 \times 2020$ square is just $2020^2$ copies of a single $1 \times 1$ tile, each behaving the same way around its four corner lattice points. So we replace the giant square with one unit square — the probability is identical. Tool #7 (Subproblems): inside that unit square, the favorable region is four quarter-circles of radius $d$ at the four corners, which combine into one full circle of area $\pi d^2$. Tool #3 (Eliminate): the resulting equation $\pi d^2 = \tfrac{1}{2}$ gives $d^2 = \tfrac{1}{2\pi} \approx 0.159$ — square each answer choice and pick the one closest to $0.159$.

Execute — Answer: B

#9 Solve an Easier Related Problem 4.OA.C.5 Step 1
  • Notice the lattice points form a perfectly repeating grid.
  • The whole $2020 \times 2020$ square is just a tiling of $2020^2$ unit squares, and the rule "within distance $d$ of a lattice point" looks identical inside every tile.
  • So the probability for the big square equals the probability for one $1 \times 1$ unit square with corners at lattice points.
$$P_{\text{big}} = P_{\text{unit}}$$

💡 Same pattern repeats over and over — work on one little tile.

#7 Identify Subproblems 7.G.B.4 Step 2
  • Place the unit square at $(0,0), (1,0), (1,1), (0,1)$.
  • The lattice points that matter are its four corners.
  • Points within distance $d$ of a corner form a quarter-disk inside the square (only one quarter fits in).
  • Four quarter-disks combine into one full disk of radius $d$, so the favorable area is $\pi d^2$.
  • This assumes $d \le 0.5$, so the quarter-disks don't overlap.
$$A_{\text{good}} = 4 \cdot \tfrac{1}{4}\pi d^2 = \pi d^2$$

💡 Four corner quarter-pies glue together into one whole pie.

#7 Identify Subproblems 7.RP.A.3 Step 3
  • Probability is favorable area over total area.
  • The unit square has area $1$, so $P = \pi d^2$.
  • Set this equal to $\tfrac{1}{2}$ and solve for $d^2$: $\pi d^2 = \tfrac{1}{2}$, giving $d^2 = \tfrac{1}{2\pi}$.
$$P = \dfrac{\pi d^2}{1} = \tfrac{1}{2} \;\Rightarrow\; d^2 = \dfrac{1}{2\pi}$$

💡 Half the square should be covered by the four corner pies.

#3 Eliminate Possibilities 8.NS.A.2 Step 4
  • Use $\pi \approx 3.14159$, so $2\pi \approx 6.283$ and $d^2 \approx \tfrac{1}{6.283} \approx 0.159$.
  • Now square each answer choice and look for the one closest to $0.159$: (A) $0.09$, (B) $0.16$, (C) $0.25$, (D) $0.36$, (E) $0.49$.
  • The match is (B) $0.16 \approx 0.159$, so $d \approx 0.4$.
$$d^2 \approx 0.159 \;\Rightarrow\; d \approx 0.4 \;\Rightarrow\; \textbf{(B)}$$

💡 Square the choices, pick the one nearest to $\tfrac{1}{2\pi}$.

[1] #9 4.OA.C.5 Notice the lattice points form a perfectly repeating grid. The whole $2020 \time
[2] #7 7.G.B.4 Place the unit square at $(0,0), (1,0), (1,1), (0,1)$. The lattice points that m
[3] #7 7.RP.A.3 Probability is favorable area over total area. The unit square has area $1$, so
[4] #3 8.NS.A.2 Use $\pi \approx 3.14159$, so $2\pi \approx 6.283$ and $d^2 \approx \tfrac{1}{6.

Review

Reasonableness: Sanity-check the magnitude: if $d=0.5$, the four quarter-disks would just touch and have total area $\pi(0.5)^2 \approx 0.785$ — way too much. If $d=0.3$, they cover $\pi(0.3)^2 \approx 0.283$ — too little. Half-coverage sits between, closer to $d=0.4$ giving area $\approx 0.503$, almost exactly $\tfrac{1}{2}$. The assumption $d \le 0.5$ also checks out, so the four quarter-disks really don't overlap.

Alternative: Tool #6 (Guess and Check): plug each choice directly into $\pi d^2$ and compare to $0.5$ — $\pi(0.3)^2 \approx 0.28$, $\pi(0.4)^2 \approx 0.50$, $\pi(0.5)^2 \approx 0.79$. The choice $d=0.4$ hits the target on the nose.

CCSS standards used (min grade 8)

  • 4.OA.C.5 Generate a number or shape pattern following a given rule (Recognizing that the lattice-point grid repeats so the giant square reduces to one unit square.)
  • 7.G.B.4 Know the formulas for area and circumference of a circle (Computing the four corner quarter-disks as a full circle of area $\pi d^2$.)
  • 7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems (Setting probability equal to area-ratio $\pi d^2 / 1 = \tfrac{1}{2}$ and solving for $d^2$.)
  • 8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Approximating $\tfrac{1}{2\pi} \approx 0.159$ and comparing $d^2$ values from the answer choices.)

⭐ This AMC 10 problem only needs Grade 8 number-line estimation you already know — shrink the giant square to one tiny tile, fit four corner pie-slices into one whole pie, set that area to $\tfrac{1}{2}$, and check that $d \approx 0.4$ does the trick. The answer is $\textbf{(B)}$.

⭐ This AMC 10 problem only needs Grade 8 number-line estimation you already know — shrink the giant square to one tiny tile, fit four corner pie-slices into one whole pie, set that area to $\tfrac{1}{2}$, and check that $d \approx 0.4$ does the trick. The answer is $\textbf{(B)}$.

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