1. The triangle
In the problem’s coordinates the right angle is at the origin and the legs run to (1, 0) and (0, 3/4), so
T = { (x, y) : x ≥ 0, y ≥ 0, 3x + 4y ≤ 3 }
Its sides are 1, 3/4 and 5/4, the 3-4-5 triangle scaled by 1/4. The inradius is (1 + 3/4 − 5/4)/2 = 1/4 and the incentre is I = (1/4, 1/4). The diameter of a convex polygon (its largest distance between two points) is the largest distance between two of its vertices; this is used repeatedly below. The diameter of T is its hypotenuse, 5/4.
2. Reduction to spreading points
A circle of radius r lies in T exactly when its centre lies in T at distance at least r from each of the three side lines. For r < 1/4 these centres form the triangle Tᵣ whose sides are T’s sides moved inward by r. The homothety about the incentre I takes each side line at distance 1/4 from I to the parallel line at distance 1/4 − r, so Tᵣ is the image of T under the homothety about I with ratio (1/4 − r)/(1/4) = 1 − 4r. For r ≥ 1/4, Tᵣ is at most a point and cannot hold two circles.
Two circles of radius r have disjoint interiors exactly when their centres are at least 2r apart. So n circles of radius r fit exactly when Tᵣ holds n points pairwise at least 2r apart, or, scaling by 1/(1 − 4r) back to T, when T holds n points pairwise at least 2r/(1 − 4r) apart. Let
cₙ = max over n-point sets S ⊂ T of min { |a − b| : a ≠ b in S }
(T is compact, so the maximum exists). Then r is feasible exactly when 2r/(1 − 4r) ≤ cₙ, that is r(2 + 4cₙ) ≤ cₙ, and the optimal radius is
rₙ = cₙ / (2 + 4cₙ)
This increases strictly with cₙ, so it remains to find cₙ for n = 2, 3, 4: exhibit a point set attaining it, and show that any n points have a pair within cₙ. The optimal circles are centred at the images of the attaining points under p ↦ I + (1 − 4r)(p − I).
3. n = 2: c₂ = 5/4
Two points of T are at most its diameter 5/4 apart, with equality at the ends (1, 0) and (0, 3/4) of the hypotenuse. Hence
r₂ = (5/4) / (2 + 5) = 5/28, centres (13/28, 5/28), (5/28, 11/28)
4. n = 3: c₃ = 25/32
Construction: take (1, 0), P = (7/32, 0) and (0, 3/4). Their distances are
|(1, 0) P| = 25/32, |P (0, 3/4)| = √((7/32)² + (24/32)²) = 25/32, |(1, 0) (0, 3/4)| = 5/4
so the minimum distance is 25/32.
Upper bound: let Q = (1/2, 3/8), the midpoint of the hypotenuse. The segment PQ cuts T into two closed convex polygons whose union is T:
A = triangle (1, 0), P, Q
B = quadrilateral P, (0, 0), (0, 3/4), Q
All distances between their vertices:
A: |(1, 0) P| = 25/32, |(1, 0) Q| = 5/8, |P Q| = √((9/32)² + (12/32)²) = 15/32
B: |P (0, 3/4)| = 25/32, |(0, 0) (0, 3/4)| = 3/4, |(0, 0) Q| = |Q (0, 3/4)| = 5/8, |P (0, 0)| = 7/32, |P Q| = 15/32
The largest is 25/32 in both lists, so A and B both have diameter exactly 25/32. Among any three points of T two lie in the same piece (a point on PQ belongs to both, which does no harm), and those two are at most 25/32 apart. Hence c₃ = 25/32 and
r₃ = (25/32) / (2 + 25/8) = 25/164, centres (89/164, 25/164), (39/164, 25/164), (25/164, 73/164)
5. n = 4: c₄ = 5/8
Construction: the three vertices and the hypotenuse midpoint Q = (1/2, 3/8). Q is the circumcentre, at distance 5/8 from each vertex, and the vertices are 1, 3/4 and 5/4 apart, all at least 5/8. So the minimum distance is 5/8.
Upper bound (HwaterB #40’s three-piece covering from P19 n = 4, carried over to the triangle): every point of T lies in one of the three closed rectangles
[0, 1/2] × [0, 3/8], [1/2, 1] × [0, 3/8], [0, 1/2] × [3/8, 3/4]
since x > 1/2 and y > 3/8 would give 3x + 4y > 3/2 + 3/2 = 3, outside T; so every point of T has x ≤ 1/2 or y ≤ 3/8, and T itself has x ≤ 1 and y ≤ 3/4. Each rectangle has sides 1/2 and 3/8 and diameter √(1/4 + 9/64) = 5/8. Among any four points two share a rectangle and are at most 5/8 apart. Hence c₄ = 5/8 and
r₄ = (5/8) / (2 + 5/2) = 5/36, centres (5/36, 5/36), (21/36, 5/36), (5/36, 17/36), (13/36, 11/36) ∎
Credit and precision
NUE_13 #31 posted this argument in the P11 discussion on 11 October, followed by a correction reply. The post reaches the right values, but parts of it were not rigorous; MinMax Arena supplied those steps, checked every one, and wrote out the complete proof above. In the post’s n = 4 step the diameter √(a/4 + b/4) is a typo for √(a² + b²)/2; read that way the argument is complete, and the correction reply is not needed. This adopted contribution earns one permanent +2 proof award.
The n = 4 covering is HwaterB #40’s method from the P19 n = 4 proof, which already earned its award; it is credited here without a second award.
5/28, 25/164 and 5/36 have non-terminating decimals, so no nine-decimal radius attains them and no grid optimum is claimed. The three instances close through the existing proven-optimal mechanism, with history retained and their record holders unchanged.
HwaterB #40 · P19 n=4 · News