2012 AMC 12A Problem 23

Attempt Problem 23 of the 2012 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12A solutions, or check the answer key.

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23.

Let SS be the square one of whose diagonals has endpoints (0.1,0.7)(0.1, 0.7) and (−0.1,−0.7).(-0.1, -0.7). A point v=(x,y)v = (x, y) is chosen uniformly at random over all pairs of real numbers xx and yy such that 0≤x≤20120 \le x \le 2012 and 0≤y≤2012.0 \le y \le 2012. Let T(v)T(v) be a translated copy of SS centered at v.v. What is the probability that the square region determined by T(v)T(v) contains exactly two points with integer coordinates in its interior?
0.1250.125
0.140.14
0.160.16
0.250.25
0.320.32
Answer: C
Concepts:geometric probabilitylattice pointarea
Small hint:
The square SS has diagonals of length 2\sqrt2 and area 1;1; by periodicity, reduce to one unit cell of the lattice
Big hint:
T(v)T(v) contains a lattice point exactly when vv lies in the copy of SS centered there; containing two requires vv in the overlap of copies centered at adjacent lattice points
Solution:
The diagonal from (0.1,0.7)(0.1, 0.7) to (−0.1,−0.7)(-0.1, -0.7) has length 0.22+1.42=2,\sqrt{0.2^2 + 1.4^2} = \sqrt2, so SS is a square of area 1.1. The translate T(v)T(v) contains a lattice point exactly when vv lies inside the copy of SS centered at that point.
Containing exactly two interior lattice points requires vv to lie in the overlap of two copies centered at adjacent lattice points. By periodicity the answer is the total such overlap area within one unit cell.
Consider copies centered at (0,0)(0,0) and (1,0).(1,0). Their overlap is a rectangle. A quarter-turn of the listed half-diagonal gives the relevant vertices (0.7,−0.1)(0.7,-0.1) and (0.3,0.1).(0.3,0.1). Following the two pairs of parallel sides through their intersections gives side lengths 0.40.4 and 0.2,0.2, so the overlap area is 0.08.0.08. Diagonally centered copies do not overlap, and horizontal and vertical adjacencies contribute equally. Accounting for the cell boundaries gives probability 2⋅0.08=0.16.2\cdot0.08=0.16.
Thus, the correct answer is C.
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