1. Notation
If p = q then I(x) = 2/|x − p|², and some corner is at distance at least √2/2 from p, where the intensity is at most 4 < 24/5. From now on p ≠ q, and
m = (p + q)/2, v = (q − p)/2, s = |v| > 0, t = s², e = v/s
and e⊥ is a unit vector perpendicular to e. The midpoint m lies in the square, which is convex. Let α₊ and α₋ be the distances from m to the boundary along +e⊥ and −e⊥, and d = max(α₊, α₋).
2. First witness: a boundary point on the bisector
For y = m + λe⊥ we have y − p = λe⊥ + v and y − q = λe⊥ − v, with perpendicular terms, so |y − p|² = |y − q|² = λ² + t and I(y) = 2/(λ² + t). The farther exit point (λ = d, replacing e⊥ by −e⊥ if needed) lies in the square, and
I(y) = 2/(d² + t)
So L(p, q) ≤ 24/5 whenever d² + t ≥ 5/12. It remains to treat t < 5/12 − d².
3. Intensity at a corner (Lemma C)
For a corner C let R = |C − m|² and w = (C − m)·e. From C − p = (C − m) + v and C − q = (C − m) − v,
|C − p|² = R + t + 2sw, |C − q|² = R + t − 2sw
Their product is D = (R + t)² − 4tw². By Cauchy–Schwarz w² ≤ R, so D = (R − t)² + 4t(R − w²) ≥ 0, and D = 0 exactly when a light sits at C. When D > 0,
I(C) = 2(R + t) / ((R + t)² − 4tw²)
Lemma C: for t > 0 put
Q_C(t) = 24t² + (48R − 96w² − 10)t + R(24R − 10)
If Q_C(t) ≥ 0 then I(C) ≤ 24/5. Proof: expanding gives Q_C(t) = 24D − 10(R + t). If D = 0 then R = t = w², and substituting gives Q_C(t) = −20t < 0, so this case is excluded; hence D > 0, and then I(C) ≤ 24/5 ⇔ 10(R + t) ≤ 24D ⇔ Q_C(t) ≥ 0.
4. Summing over the corners (Lemma S)
Let c₀ = (1/2, 1/2), μ = m − c₀, ρ = |μ|² and γ = (μ·e)². The corners are C = c₀ + κ with κ ∈ {(±1/2, ±1/2)}, Σκ = 0, and Σ(κ·z)² = |z|² for every vector z. With C − m = κ − μ this gives
Σ R_C = 2 + 4ρ, Σ R_C² = 4ρ² + 8ρ + 1, Σ w_C² = 1 + 4γ
(For instance R_C = 1/2 + ρ − 2κ·μ; on squaring and summing, the cross term vanishes because Σκ = 0, and Σ(κ·μ)² = ρ.) Substituting into Q_C and summing:
Σ_C Q_C(t) = 96t² + (192ρ − 384γ − 40)t + 96ρ² + 152ρ + 4
5. Lemma G: a central midpoint exits far away
If |μ| ≤ (√2 − 1)/2 then d ≥ 1/2. Proof: the reflections in x₁ = 1/2 and x₂ = 1/2 preserve the square and the whole problem, as well as |μ| and d, so we may take u = e⊥ with u₁, u₂ ≥ 0. Then
α₊ = min over uᵢ > 0 of (1/2 − μᵢ)/uᵢ, α₋ = min over uᵢ > 0 of (1/2 + μᵢ)/uᵢ
If both were below 1/2, some i would have μᵢ > (1 − uᵢ)/2 and some j would have μⱼ < (uⱼ − 1)/2. Taking i = j would force uᵢ > 1, which is impossible, so i ≠ j. As uᵢ, uⱼ ≤ 1 these give |μᵢ| > (1 − uᵢ)/2 and |μⱼ| > (1 − uⱼ)/2, and since uᵢ + uⱼ ≤ √2,
√2|μ| ≥ |μ₁| + |μ₂| > (2 − uᵢ − uⱼ)/2 ≥ (2 − √2)/2
that is |μ| > (√2 − 1)/2, a contradiction.
6. Upper bound: L(p, q) ≤ 24/5
Let t > 0 and t < 5/12 − d² (otherwise Section 2 already applies). Since γ ≤ ρ and t > 0,
Σ_C Q_C(t) ≥ S(t, ρ) = 96t² − (192ρ + 40)t + 96ρ² + 152ρ + 4
As a function of t, S is an upward parabola with minimum 112ρ − 1/6 at t_S = ρ + 5/24.
- If ρ ≥ 1/672, then S ≥ 112ρ − 1/6 ≥ 0.
- If ρ < 1/672, then |μ| < 1/√672 < (√2 − 1)/2, so Lemma G gives d ≥ 1/2 and t < 5/12 − 1/4 = 1/6 < t_S. S decreases for t ≤ t_S, so S(t, ρ) ≥ S(1/6, ρ) = 96ρ² + 120ρ ≥ 0.
Either way the four numbers Q_C(t) have a nonnegative sum, so one of them is nonnegative, and by Lemma C that corner has I(C) ≤ 24/5. With the case p = q from Section 1, L(p, q) ≤ 24/5 for all p, q.
7. Lower bound: the symmetric layout reaches 24/5
Take p, q = (1/2 ∓ a, 1/2) with a = 1/√6; both lie in the square. For x = (1/2 + ξ, 1/2 + η) put A = ξ² + η² + 1/6. Then |x − p|² + |x − q|² = 2A and |x − p|²·|x − q|² = A² − 4a²ξ², so (away from the lights)
I(x) = 2A / (A² − 4a²ξ²)
With 4a² = 2/3, I(x) ≥ 24/5 ⇔ 24A² − 10A − 16ξ² ≤ 0. With u = ξ² and v = η² (both in [0, 1/4]) and w = u + v, half the left side is exactly
F(u, v) = 12(u + v)² − 9u − v − 1/2 = 12w² − w − 1/2 − 8u
- For 0 ≤ w ≤ 1/4: u ≥ 0, so F ≤ 12w² − w − 1/2, a convex function of w equal to −1/2 at w = 0 and 0 at w = 1/4, hence ≤ 0.
- For 1/4 ≤ w ≤ 1/2: v ≤ 1/4 gives u ≥ w − 1/4, so F ≤ 12w² − 9w + 3/2, convex and equal to 0 at both w = 1/4 and w = 1/2, hence ≤ 0.
So I(x) ≥ 24/5 throughout the square (at a light the intensity is infinite), and L = 24/5. Equality holds at (u, v) = (0, 1/4) and (1/4, 1/4): the four corners and the midpoints (1/2, 0) and (1/2, 1) of the bottom and top sides. With Section 6, the optimum is exactly 24/5. ∎
Credit and precision
The symmetric layout and the value 24/5 were published by Erich Friedman in July 2026 on his Packing Center page, and both are his. NUE_13 #31 posted the optimality proof in the P51 discussion on 18 October, followed by a second part and a corrections reply, stating that DeepSeek v4.1 Flash generated it. This page follows the corrected argument, and MinMax Arena checked every step. The corrections reply withdrew a claim that the optimal layout is unique; this page proves only the value and claims no uniqueness. This adopted contribution earns one permanent +2 proof award.
The attaining coordinates 1/2 ∓ 1/√6 are irrational and cannot be written in nine decimals, so no grid optimum is claimed; the site’s reconstruction of Friedman’s figure verifies at 4.798582. The instance closes through the existing proven-optimal mechanism, with history retained and the record holder unchanged.