The Inevitable Pentagon

Place n points in general position in the plane. An empty convex pentagon — a “5-hole” — is a convex hull of five points containing no other points in its interior. How many must exist?

Until recently, the best bound was Omega(n · (log n)^{4/5}). Astudillo-Marban and Sole-Pi push this to Omega(n^{20/11}).

The jump is enormous. From barely super-linear to a polynomial with exponent 1.818… — substantially closing the gap toward the conjectured Omega(n²). The proof is purely combinatorial, requiring no computational verification.

The result means empty pentagons are geometrically inevitable at near-quadratic density. You cannot arrange a large point set to avoid them. The empty triangles are trivially abundant (Omega(n²)). The empty quadrilaterals are known to be quadratic. Pentagons are harder — the convexity and emptiness constraints interact nontrivially for five vertices — but the new bound shows they’re far more abundant than linear.

The proof works by identifying substructures that force 5-holes through a chain of Ramsey-type arguments. The key insight: certain configurations of four points in convex position, combined with density arguments about the fifth point, generate pentagons faster than local avoidance strategies can prevent them.

Five-sided emptiness is not a rare geometric accident. It’s a structural inevitability, and the question is no longer whether it’s polynomial but whether it’s exactly quadratic.