P11 n = 2, 3, 4: equal circles in a right triangle
The centres must lie in the inner triangle shrunk by the factor 1 − 4r, so the problem becomes spreading n points in the triangle, and the optimal radius is r = c/(2 + 4c). For n = 2, c is the hypotenuse 5/4; for n = 3 one segment cuts the triangle into two convex pieces of diameter 25/32; for n = 4 three rectangles of diameter 5/8 cover it. The optimal radii are 5/28, 25/164 and 5/36. The site completed the steps of the post that were not rigorous; the n = 4 covering is HwaterB’s three-piece method from P19 n = 4, credited without a second award. NUE_13 receives +2.
P51 n = 2: two lights in a square
For any two lights, the farther boundary point along the perpendicular bisector or one of the corners has intensity at most 24/5: the four corners’ quadratics have a nonnegative sum, so one corner qualifies. Lights at (1/2 ∓ 1/√6, 1/2) give intensity at least 24/5 everywhere; that layout and value are the ones Friedman published. The proof was posted by NUE_13 (generated by DeepSeek v4.1 Flash, as disclosed); the page follows the corrected argument and claims no uniqueness of the optimal layout. NUE_13 receives +2.
Precision and awards
All four are continuous optima: P11’s 5/28, 25/164 and 5/36 are not nine-decimal numbers, and the attaining P51 layout has irrational coordinates, so no grid optimum is claimed; the instances close through the existing proven-optimal mechanism with their record holders unchanged. NUE_13’s two distinct proofs each earn a permanent +2 proof award. Verifiers, scoring units and historical submissions are unchanged.