AIME 2025 II · 第 3 题

AIME 2025 II — Problem 3

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Four unit squares form a $2 \times 2$ grid. Each of the $12$ unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has $2$ red sides and $2$ blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.

解析

Solution

Let the red edges be "used" edges. In the digarams below, dashed lines are uncolored lines yet to be decided. Since all four center edges are common to both squares, we consider five distinct cases:

$\textbf{Case 1:}$ All center edges are used. There is only one way to do this.

$AIME diagram$

$\textbf{Case 2:}$ Three center edges are used, meaning two squares are missing an edge. For each square, there are 2 ways to choose an edge, resulting in $2 \times 2 = 4$ ways. Additionally, considering the rotational symmetry of the arrangement, there are 4 possible rotations, giving a total of $4 \times 4 = 16$ configurations.

$AIME diagram$

$\textbf{Case 3:}$ Two center edges are used. There are two sub-cases:

$\textbf{Scenario 1:}$ The two selected sides are perpendicular to each other. The square diagonally opposite its adjacent square has only one choice, while the other two squares each have two choices. This gives a total of $1 \times 2 \times 2 = 4$ choices. Considering the 4 possible rotations, the total number of configurations is $4 \times 4 = 16$ .

$AIME diagram$

$\textbf{Scenario 2:}$ The two selected sides are aligned along the same straight line. Each of the four squares has 2 choices, yielding $2^4 = 16$ possible choices. Taking into account the 2 possible rotations, the total number of configurations is $16 \times 2 = 32$ .

$AIME diagram$

$\textbf{Case 4:}$ Only one center edge is used. This case is similar to Case 2, yielding 16 possible configurations.

$AIME diagram$

$\textbf{Case 5:}$ No center edge is used. This is similar to Case 1, with only 1 possible configuration.

$AIME diagram$

In conclusion, the total number of configurations is:

$1 + 16 + 16 + 32 + 16 + 1 = \boxed{\textbf{082}}$ ~ Athmyx

~ LaTeX by eevee9406

~ Additional edits by aoum

Video Solutions 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=98hzWL-cUvk&t=6s