Frank Morgan

A new ingredient in Hales’s 2001 proof of the Hexagonal Honeycomb Conjecture* is his Chordal Isoperimetric Inequality (see Chapter 15 of my Geometric Measure Theory book). It provides a lower bound on the perimeter of a bulging curvilinear polygon in terms of the extra area X enclosed by the bulges. With some help from my new colleague Steven Miller, I’ve realized that a minor modification of the proof improves the constant π/8 to the optimal π/4 as the requisite upper bound on X.

            To modify the proof, add to the auxiliary, less constrained problem the hypothesis that the longest chord is no longer than the sum of the lengths of the others. In the delicate case 1 < L₀ ≤ 2, by the hypothesis the length of the longest chord is at most L₀/2 and a minimizer consists of arcs of equal curvature over two chords of length L₁ = L₂ = L₀/2. If one of the arcs were more than a semicircle, X would exceed the area of a circle of unit diameter, π/4, a contradiction. Hence we are in the convex range of the arc function and the old convexity argument kicks in.

            Similarly if L₀ > 2, a minimizer conists of m ≥ 1 arcs over chords of length L_i = 1 and perhaps one lower arc over a chord of length 0 ≤ L_m+1 < 1. If any of the first m arcs were more than a semicircle, again the area under two would exceed π/4, a contradiction, putting us back in the convex range of the arc function and the old proof.

*The Hexagonal Honeycomb Conjecture says that congruent regular hexagons provide a least-perimeter way to partition the plane into unit areas.