Active pairs of a min–max problem on zero-sum planar configurations: the house pentagon, the 3-4 rectangle, an optimum that is not active-maximal, and how many pairs a configuration can activate

the house pentagon of arXiv:2608.15444 Conjecture 4.3 is an admissible strictly convex zero-sum lattice configuration with perimeter 112 and maximal triangle perimeter 80, so delta(house) = 5/7 exactly and pi_(5,2)ᵐax <= 5/7

exactly 7 of the C(5,2) = 10 pairs of the house are active; the three inactive pair values are 41 + sqrt(241) (twice) and 32, all strictly below 80

for every even n = 2m with m >= 2 the 3-4 rectangle configuration of arXiv:2608.15444 Conjecture 4.4 has mass 6m+2 and maximal triangle perimeter 12, so delta = 4/(n + 2/3) exactly and pi_(n,2)ᵐax <= 4/(n + 2/3)

exactly m² + m - 2 of the C(2m,2) = 2m² - m pairs of the 3-4 rectangle are active, for every m >= 2, proved uniformly in m

4/(x + 2/3) < (4/x) cos²(pi/2x) for every real x >= 5, hence pi_(n,2)ᵐax is strictly below the regular n-gon's value for every even n >= 6; and delta(regular pentagon) = (5+sqrt5)/10 exactly, with the house improving on it by exactly (5 sqrt5 - 11)/14, between 0.0128 and 0.0129

the collapsed pair (1, -1, 0,..., 0) has exactly 2n - 3 active pairs at every n >= 2; the square has 4 of its 6 with delta = (2+sqrt2)/4, so pi_(4,2)ᵐax <= (2+sqrt2)/4

at n = 4 an optimal configuration for pi_(4,2)ᵐax does NOT achieve the maximal number of simultaneously active pairs: the square, the unique optimum, activates 4 pairs while the non-trivial zero-sum configuration (1,-1,0,0) activates 5 – a negative answer to the second half of the source's closing open question at n = 4

a uniform lower bound delta(z) >= 2/n for every configuration of n >= 2 nonzero zero-sum vectors, giving the non-empty brackets 2/5 <= pi_(5,2)ᵐax <= 5/7 and 1/m <= pi_(2m,2)ᵐax <= 4/(2m + 2/3)

negative controls: delta(house) is neither <= 0.714 nor >= 0.715 and the house's active count is neither 6, 8 nor 10; the 3-4 rectangle's maximum at m = 3 is neither <= 11 nor >= 13 and its count is neither 9 nor 11; every side condition bites – at m = 1 the 3-4 rectangle degenerates to (4i,-4i) with maximum 8 not 12 and count 1 not 0, at x = 4 the regular-polygon inequality is FALSE, the collapsed pair satisfies WeakConfig but not Config, and (1,1) fails the zero-sum clause

at n = 4 the maximal number of simultaneously active pairs is exactly 5, attained only by (u,-u,0,0) up to permutation; among configurations with every zₖ nonzero it is exactly 4, attained by the square and by the 3-4 rectangle

numerical search for the maximal number of simultaneously active pairs: 5 at n = 4 (only with vanishing entries), 7 at n = 5 (attained by the house), and at n = 6 nothing with 11 or more active pairs, so the 3-4 rectangle's 10 is the numerical maximum there

at n = 4 the maximal number of simultaneously active pairs over the source's own non-trivial class is exactly 5, and the maximisers are exactly the configurations (u, -u, 0, 0) up to permutation

at n = 4 the maximal number of simultaneously active pairs among configurations with every zₖ nonzero is exactly 4, attained by the square and by the 3-4 rectangle

at n = 3 every zero-sum triple has all C(3,2) = 3 pairs active, with pmax = mass and delta = 1

negative controls for the n = 4 maxima: (1,1,1,1) has all six pairs active so the zero-sum constraint is what forbids six; (0,0,0,0) has all six active so the non-triviality hypothesis bites; 4 is not an upper bound in the non-trivial class and 3 is not one in the nondegenerate class; no WeakConfig 4 reaches 6; and (1,1,i) shows the n = 3 statement needs its zero-sum hypothesis

the square-root-free activity criterion: (M - |zᵢ| - |zⱼ|)² - |zᵢ + zⱼ|² = (M - 2|zᵢ|)(M - 2|zⱼ|) - 2(|zᵢ||zⱼ| + <zᵢ,zⱼ>) for all zᵢ, zⱼ and all real M, so under 2|zᵢ| <= M and 2|zⱼ| <= M one has pv(i,j) <= M iff 2(|zᵢ||zⱼ| + <zᵢ,zⱼ>) <= (M - 2|zᵢ|)(M - 2|zⱼ|), with equality iff the pair is active; and 2|zₖ| <= pmax(z) for every entry of every configuration of length >= 2, so the side condition is free

the deficit identity sum_(i<j) [(M - |zᵢ| - |zⱼ|)² - |zᵢ+zⱼ|²] = C(n,2) M² - 2(n-1) M s + s² - |sumₖ zₖ|² with s = mass(z), valid for every list and every real M; hence the closure bound mass(z) <= cₙ pmax(z) with cₙ = (n-1) - sqrt((n-1)(n-2)/2) for every list of length n >= 2, i.e. delta(z) >= 1/cₙ (2+sqrt2)/n – strictly stronger than AP8's 2/n for every n >= 3 and strictly weaker than the source's 4/(n+1) for every n >= 4 – with equality iff all C(n,2) pairs are simultaneously active, so at n = 4 the bound is strict: mass(z) < (3 - sqrt3) pmax(z)

at every odd n = 2g+1 with g >= 2 the configuration T(g) – g copies of (c,s), g copies of (c,-s) and one (-2gc,0), where c in (0,1) solves g c² - 3gc + 2 = 0 and s = sqrt(1-c²) – is an admissible all-nonzero zero-sum configuration with pmax = 4 and exactly g² + g simultaneously active pairs; since g² + g > 2n - 3 for every g >= 3, the collapsed pair (1,-1,0,...,0) is NOT the odd-n maximiser, and at n = 7 the maximal number of simultaneously active pairs is at least 12 > 11

negative controls for AP16-AP18 and AP20: the side condition 2|zᵢ| <= M of the activity criterion bites, the factor 2 in the square-root-free form is not decoration, Lemma 1 needs length >= 2; the closure bound is attained at n = 3 and fails if strengthened by 1/10, and needs length >= 2; T(g) needs its defining quadratic, does not activate all C(2g+1,2) pairs, and at g = 2 gives 6 < 2n-3 = 7; the uniform g >= 4 argument for the regular-polygon comparison is FALSE at g = 3; and delta(T(2)) is exactly the regular pentagon's (5+sqrt5)/10

delta(T(g)) = 4/(2g + 2gc), so pi_(n,2)ᵐax <= delta(T(g)) at every odd n = 2g+1 and, for every odd n >= 7, pi_(n,2)ᵐax < (4/n) cos²(pi/2n): the regular n-gon is never optimal at odd n >= 7, certified by an explicit configuration; at n = 7 exactly, pi_(7,2)ᵐax <= 4/(15 - sqrt57) = (15 + sqrt57)/42 = 0.5369008... against the regular heptagon's (4/7)cos²(pi/14) = 0.5431339...

Via the shared exact SOS/PSD checker: for every z in C⁵ with sum zₖ = 0 and sum |zₖ| = 1 some pair has |zᵢ| + |zⱼ| + |zᵢ + zⱼ| > 9/14 (no zₖ!= 0 hypothesis, so it bounds the source's pi_(5,2)ᵐax from below); the modulus relaxation's exact ceiling is (4 + sqrt 6)/10, so the source's 2/3 is a route limit

Theorem Z, the degenerate case at every n: if some entry of a zero-sum configuration satisfies 2|zᵢ| = pmax(z) then every other entry is a non-positive real multiple of it (|zₖ| zᵢ + |zᵢ| zₖ = 0), the configuration is collinear with mass = pmax so delta(z) = 1, and activeCount(z) <= 2n-3; the maximum over degenerate non-trivial configurations of size n is EXACTLY 2n-3 for every n >= 3, attained by the collapsed pair – the first upper bound on the active count in this family valid at every n. Contrapositive: delta(z) < 1, or activeCount(z) > 2n-3, forces 2|zₖ| < pmax(z) for every entry

negative controls for AP22: the zero-sum hypothesis is what makes delta = 1 (the saturated list (1,0) has pmax = 2 and delta = 2); the degeneracy hypothesis is what makes 2n-3 a bound (T(3) at n = 7 has 12 > 11 = 2n-3 active pairs, row AP18); the collapsed pair is saturated and attains 2n-3 at every n >= 2; the saturated entry need not be unique – both 1 and -1 are saturated in the collapsed pair – and the parallelism relation is genuinely signed; and the pair-mass bound is about distinct positions, since (1,1) is not a pair of [1] and |1| + |1| = 2 > 1 = mass [1]

the separation model: for every entry with 2|z| < M there is W in C = R² with (M - 2|z|) W² = 2z, unique up to sign, and pgap(M,a,b) <Wₐ,W_b>² = qv(a,b), so pv(a,b) <= M iff <Wₐ,W_b>² <= 1 and the pair is active iff <Wₐ,W_b>² = 1, while 4|z| < M iff |W| < 1; hence an active pair has |Wᵢ||Wⱼ| >= 1 (the heavy vertices form an independent set in the active graph, and every partner of a heavy vertex has |W| > 1) and two vertices at one position (cross Wᵢ Wⱼ = 0) have |Wᵢ||Wⱼ| <= 1, so a position carries at most one vertex with |W| > 1 and its unit vertices are pairwise active

a HEAVY entry has active degree at most 2: in any configuration all of whose entries satisfy 2|zₖ| < pmax, an entry w with 4|w| < pmax is active with at most two entries – at every n, with no ordering, convexity or zero-sum hypothesis. The proof is order-free: the plane Gram identity |W|² <U,V> = <W,U><W,V> + (W^U)(W^V) makes (<W,U>(W^U))(<W,V>(W^V)) = s |W|² <U,V> - 1 <= |W|² - 1 < 0 for any two active partners of a heavy vertex, and three reals cannot have all three pairwise products negative

negative controls for AP24-AP25: heaviness is what bounds the degree – the saturated entry 1 of the collapsed pair at n = 5 has active degree 4, not <= 2 – and the bound 2 is attained, the house's heavy edge (0,16) having 4|u| = 64 < 80 = pmax and active degree exactly 2

the interior lemma: if (zᵢ,zⱼ) is an active pair of a non-degenerate configuration, zᵢ and zⱼ are not parallel, and an entry lies strictly inside the cone they span (zᵤ = lam zᵢ + mu zⱼ with lam, mu > 0), then zᵤ is HEAVY: 4|zᵤ| < pmax. Cone membership in the z-plane is arc-betweenness on the half-argument circle R/pi, so this is the strategist's Lemma 5 with every angle removed; the bridge is the cross-product dictionary 2 (a ^ b) = (M-2|a|)(M-2|b|) <Wₐ,W_b> (Wₐ ^ W_b), Cramer's rule in the plane, and a real sign inequality whose only input is |Wᵢ||Wⱼ| > 1

controls for AP27: a positive control on the house – u₂ = (0,16) = (2/5) u₁ + (4/5) u₃ lies strictly inside the cone of the active non-parallel pair (u₁,u₃), and is heavy, 4*16 = 64 < 80 – plus two negative ones: dropping either feasibility hypothesis breaks the conclusion (Wᵢ = 2, Wⱼ = 1/2 + i at lam = mu = 1 give |Wᵤ|² = 29/4 > 1 with <Wᵢ,Wᵤ>² = 25 > 1), and dropping non-parallelism breaks it by an equality (Wᵢ = Wⱼ = 1 at lam = mu = 1/2 give |Wᵤ|² = 1 exactly, and the interior is then empty)