2025 AIME II Problem 3

Attempt Problem 3 of the 2025 AIME II 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 2025 AIME II solutions, or check the answer key.

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3.

Four unit squares form a 2×22 \times 2 grid. Each of the 1212 unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has 22 red sides and 22 blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.

Answer: 82
Concepts:arrangements with restrictionscaseworkmultiplication principle
Difficulty rating: 2440
Solution:

The 1212 segments split into the 44 interior segments forming the central cross and 88 boundary segments, and each unit square has exactly two interior sides (its two cross arms) and two boundary sides. Color the cross first. A square that already has jj red interior sides needs 2j2 - j red boundary sides, which can be chosen in (22j)\binom{2}{2-j} ways: 11 way if j=0j = 0 or j=2,j = 2, and 22 ways if j=1.j = 1.

Group the 24=162^4 = 16 cross colorings by the set of red arms. If all four arms have the same color (22 colorings), every square has j=0j = 0 or j=2,j = 2, contributing 11 each: total 2.2. If exactly one arm is red or exactly one is blue (88 colorings), the two squares touching the odd arm have j=1j = 1 and the others do not, contributing 22=42 \cdot 2 = 4 each: total 32.32. If two adjacent arms are red (44 colorings), the squares have j=2,1,1,0,j = 2, 1, 1, 0, contributing 44 each: total 16.16. If two opposite arms are red (22 colorings), all four squares have j=1,j = 1, contributing 24=162^4 = 16 each: total 32.32.

The number of colorings is 2+32+16+32=82.2 + 32 + 16 + 32 = 82.

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