AMC 10 · 2020 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

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 $(0, 0), (2020, 0), (2020, 2020),$ 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 $?$

Pick an answer.

(A)

0.3

(B)

0.4

(C)

0.5

(D)

0.6

(E)

0.7

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

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

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 $\textbf{(B)}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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