AMC 10 · 2012 · #23

Grade 10 geometry-2d
geometric-probabilitycoordinate-geometrysymmetry-argumentarea-rectangles identify-subproblemseasier-related-problemcasework ↑ Prerequisites: geometric-probabilitycoordinate-geometry
📏 Long solution 💡 4 insights
Problem
A tilted square is slid to a random centre and must contain exactly two grid points. Find the probability.

Pick an answer.

(A)
$\frac{1}{8}$
(B)
$\frac{7}{50}$
(C)
$\frac{4}{25}$
(D)
$\frac{1}{4}$
(E)
$\frac{8}{25}$
How to solve
Strategy Change Focus / Count the Complement

A square wandering over a 2012 × 2012 board is hopeless to track directly, so Tool #16 (Change Focus) flips the roles: nail a copy of S to every lattice point and ask how many of those fixed copies cover the random point. The count is then a fixed map of the plane. Two facts have to be nailed down before any area is computed, and both are easy to skip. Tool #14 (Extreme Principle) supplies the first: the largest possible distance between two interior points of a unit square is strictly under √(2), which forces any two captured lattice points to be orthogonal neighbours. Tool #3 (Eliminate Possibilities) supplies the second: three lattice points can never be captured, which is what makes 'exactly two' the same event as 'some neighbouring pair is captured' and — the part usually left unsaid — makes the regions belonging to different pairs disjoint, so their areas may simply be added. Tool #9 (Solve an Easier Related Problem) then shrinks the board to a single unit cell using periodicity, and Tool #7 (Identify Subproblems) reduces everything to one measurable object: the overlap of two neighbouring copies of S. Tool #1 (Draw a Diagram) underlies the whole thing by first identifying S as a tilted unit square.

1STEP 1

Pin down the square

The diagonal fixes the square's size and tilt.

Center (0,0); diagonal √(2); side 1; vertices (1/10,7/10), (-7/10,1/10), (-1/10,-7/10), (7/10,-1/10)
2STEP 2

Let the point pick the squares

Containing a point becomes a condition on the centre.

p ∈ int T(v) ⇔ p - v ∈ int S ⇔ v - p ∈ int S ⇔ v ∈ int S_p
3STEP 3

Two inside means side by side

Two points inside forces them to be neighbours.

|p - q| < √(2) and |p - q| ∈ {1, √(2), 2, √(5), …} → |p - q| = 1
4STEP 4

Three is impossible, so nothing overlaps

Three inside is impossible, so nothing overlaps.

N(v) ≤ 2, so {v : N(v) = 2} = bigsqcup_{{p,q} adjacent} (int S_p ∩ int S_q)
5STEP 5

One unit cell is the whole story

One unit cell is the whole story.

P = (good area in [0,1)²)/1, since 2012² identical cells tile the sampling region
6STEP 6

Measure one overlap

Each overlap is a small rectangle.

S ∩ (S+e) has strip widths 1 - |e · n₁| and 1 - |e · n₂|; for e = (1,0) these are 1 - 3/5 = 2/5 and 1 - 4/5 = 1/5
7STEP 7

Add the two directions

Adding both directions gives 4/25, choice (C).

2/5 · 1/5 = 2/25 per pair direction, and 2/25 + 2/25 = 4/25 → (C)
Answer
4/25
Three checks. First, an identity the numbers must satisfy: every copy S_p has area 1 and there is one copy per unit cell, so the average of N(v) over a cell equals 1. Writing p₀, p₁, p₂ for the fractions of a cell where N equals 0, 1, 2, this gives p₁ + 2p₂ = 1 alongside p₀ + p₁ + p₂ = 1, hence p₀ = p₂. So the chance of capturing no lattice point must equal the chance of capturing two. Computing p₀ independently agrees: the copies cover 1 - 4/25 = 21/25 of a cell by inclusion-exclusion (total area 1, minus the doubly covered 4/25, with no triple overlaps to add back), leaving 4/25 uncovered, and p₁ = 17/25 closes the books. Second, the tilt is load-bearing: an axis-parallel unit square would tile the plane, T(v) would contain exactly one lattice point for almost every v, and the answer would be 0 — so a nonzero answer of moderate size is exactly what the 3-4-5 tilt should produce. Third, the distractors behave as expected: 8/25 is exactly twice the answer, which is what appears if the four neighbour directions are counted as four separate regions instead of two unordered pairs; 1/8 = 0.125 and 7/50 = 0.14 sit just below 4/25 = 0.16 and are what slightly mis-measured strip widths produce; 1/4 is the round-number guess.
💡Key takeaway

Nail a copy of the tilted square to every grid point: it is just wide enough to cover two side-by-side grid points at once but never two diagonal ones, so the answer is simply how much of a cell two neighbouring copies share.

  • Pin down the square
  • Let the point pick the squares
  • Two inside means side by side
  • Three is impossible, so nothing overlaps
  • One unit cell is the whole story
  • Measure one overlap
  • Add the two directions