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

Pick a random point inside the 2020 × 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

\frac{1}{2}

. Find d rounded 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.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): the 2020 × 2020 square is just 2020² copies of a single 1 × 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 π d². Tool #3 (Eliminate): the resulting equation π d² = $\frac{1}{2}$ gives d² = $\frac{1}{2π}$ ≈ 0.159 — square each answer choice and pick the one closest to 0.159.

1STEP 1

The lattice rule looks identical in every one of the 2020² unit tiles, so the big square's probability equals one unit square's.

P_big = P_unit

2STEP 2

In one unit square only the four corners matter; their four quarter-disks merge into one full disk of area π d² (needs d ≤ 0.5).

A_good = 4 ·

\frac{1}{4}

π d² = π d²

3STEP 3

The unit square has area 1, so P = π d²; setting it to the given $\frac{1}{2}$ gives d² = $\frac{1}{2π}$ .

P =

\frac{π d²}{1}

\frac{1}{2}

→ d² =

\frac{1}{2π}

4STEP 4

Since $\frac{1}{2π}$ ≈ 0.159, squaring the choices makes d² = 0.16 the closest, so d ≈ 0.4 — choice (B).

d² ≈ 0.159 → d ≈ 0.4 → (B)

Answer

0.4

Sanity-check the magnitude: if d=0.5, the four quarter-disks would just touch and have total area π(0.5)² ≈ 0.785 — way too much. If d=0.3, they cover π(0.3)² ≈ 0.283 — too little. Half-coverage sits between, closer to d=0.4 giving area ≈ 0.503, almost exactly

\frac{1}{2}

. The assumption d ≤ 0.5 also checks out, so the four quarter-disks really don't overlap.

💡Key takeaway

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 $\frac{1}{2}$ , and check that d ≈ 0.4 does the trick. The answer is (B).

More like this