Place n points in general position in the plane. How many empty convex pentagons do they determine? Astudillo-Marbán and Solé-Pi prove the answer is at least Ω(n^{20/11}) — a substantial improvement toward the conjectured Ω(n²) lower bound.
The result is notable not for what it proves but for how. Previous bounds on empty convex pentagons relied heavily on computational verification — checking cases by machine. This proof is purely human, achieving a better bound through mathematical structure rather than exhaustive enumeration.
The key insight is that pentagons in point configurations are far more abundant than naive counting suggests. The same set of n points contains not just scattered pentagons but structured families of them, and the structure enables counting arguments that outperform case analysis.
The broader pattern: in discrete geometry, the simplest objects (empty triangles, quadrilaterals) have tight bounds. The next step up — empty convex pentagons — has resisted precise counting for decades. The gap between the known lower bound (Ω(n^{20/11})) and the conjecture (Ω(n²)) is the gap between what structure we can currently exploit and what structure we believe exists. Every improvement in the exponent corresponds to finding one more layer of geometric organization in seemingly arbitrary point sets.