Solution: Riddle of Rings

Part 1: The Vertical Line

For a vertical line to exist, the three gaps must have a common intersection. If we represent the cylinder's circumference as a unit interval [0, 1), each gap is an arc of length L = 1/8. Let X₁, X₂, X₃ be the centers of these gaps, chosen uniformly and independently from [0, 1).

A common intersection exists if and only if all three gap centers X_i lie within some arc of length L. The probability that n points on a circle of circumference 1 lie within an arc of length L (for L < 1/2) is given by the formula:

Substituting n = 3 and L = 1/8:

Part 2: The Helix (Extra Credit)

If we unroll the cylinder, a helix becomes a straight line θ(z) = αz + β. Let the vertical heights of the rings be z = 0, h, 2h (evenly spaced). A helix passes through all three gaps if there exist positions θ₁, θ₂, θ₃ within the respective gaps and integers k₁, k₂, k₃ such that the points (0, θ₁ + k₁), (h, θ₂ + k₂), (2h, θ₃ + k₃) are collinear in the unrolled plane.

Collinearity for evenly spaced z-coordinates implies:

(θ₂ + k₂) - (θ₁ + k₁) = (θ₃ + k₃) - (θ₂ + k₂)

2θ₂ - θ₁ - θ₃ = k₁ + k₃ - 2k₂

Since k_i can be any integers, the right-hand side can be any integer K. Thus, a helix exists if there exist θ_i in the gaps such that 2θ₂ - θ₁ - θ₃ is an integer.

Let θ_i = X_i + ε_i, where X_i is the center of gap i and ε_i ∈ [-L/2, L/2]. The condition becomes:

(2X₂ - X₁ - X₃) + (2ε₂ - ε₁ - ε₃) = K

The term δ = 2ε₂ - ε₁ - ε₃ can take any value in the range [-2L, 2L], which has a total length of 4L. For a solution to exist, the value (2X₂ - X₁ - X₃) mod 1 must fall within an interval of length 4L. Since X_i are uniform and independent, the expression (2X₂ - X₁ - X₃) mod 1 is also uniformly distributed on [0, 1).