AIME 2025 II · 第 11 题

Solution

The segments we draw must be of equal length, corresponding to a specific step size $k$ (number of steps between vertices).

For each step size $k$, we need to determine if it is possible to form a perfect matching (non-overlapping segments covering all vertices). The number of such perfect matchings depends on the greatest common divisor (gcd) of $k$ and 24.

When choosing a step size $k$, the 24-gon is decomposed into $\gcd(k, 24)$ cycles, each of length $\frac{24}{\gcd(k, 24)}$. For a perfect matching to exist, each cycle must be of even length (since each line segment consists of 2 points, the length of the cycle must be divisible by 2).

For each valid step size ($k$):

If the cycle length is 2 (diameters), there is exactly 1 way to match the vertices.

For other even cycle lengths, each cycle contributes a factor of 2 to the number of perfect matchings.

($k = 1$): $\gcd(1, 24) = 1$, cycle length 24, 2 matchings.

($k = 2$): $\gcd(2, 24) = 2$, cycle length 12, $(2^2 = 4)$ matchings.

($k = 3$): $\gcd(3, 24) = 3$, cycle length 8, $(2^3 = 8)$ matchings.

($k = 4$): $\gcd(4, 24) = 4$, cycle length 6, $(2^4 = 16)$ matchings.

($k = 5$): $\gcd(5, 24) = 1$, cycle length 24, 2 matchings.

($k = 6$): $\gcd(6, 24) = 6$, cycle length 4, $(2^6 = 64)$ matchings.

($k = 7$): $\gcd(7, 24) = 1$, cycle length 24, 2 matchings.

($k = 8$): $\gcd(8, 24) = 8$, cycle length 3 (invalid, no matchings).

($k = 9$): $\gcd(9, 24) = 3$, cycle length 8, $(2^3 = 8)$ matchings.

($k = 10$): $\gcd(10, 24) = 2$, cycle length 12, $(2^2 = 4)$ matchings.

($k = 11$): $\gcd(11, 24) = 1$, cycle length 24, 2 matchings.

($k = 12$): $\gcd(12, 24) = 12$, cycle length 2, 1 matching.

Summing these values: $2 + 4 + 8 + 16 + 2 + 64 + 2 + 0 + 8 + 4 + 2 + 1 = \boxed{113}$.

~ Athmyx

~ Minor edits by Christian

Video Solution

2025 AIME II #11

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