P16 · Extremal configurations · Classic

Spreading points in a right triangle · n = 4

Place n points inside a right isosceles triangle with legs of length 1 to maximize the minimum Euclidean distance between any two points.

Instancen = 4

ObjectiveMaximize the smallest distance between two points

Record comparison

the smallest distance between two points · Higher is better

Record	Value / interval	Author / holder	Source
Proven optimum	√2/2	See source attribution	MinMax Arena · Proof ↗
Site record	No player record yet	—	—

The continuous optimum is proved, so competition submissions are closed. Historical records and contribution points are retained. Because finite-decimal coordinates cannot attain this value exactly, the current certificate is not claimed optimal on the nine-decimal grid.

Formal definition

ContainerA right isosceles triangle with legs of length 1: the right angle is at the origin (0, 0), and the other vertices are (1, 0) and (0, 1)
SubmissionExactly n distinct points
ConstraintsEvery point lies inside the triangle or on its boundary
Scoreδ(P) is the minimum Euclidean distance over all point pairs (the verifier compares δ(P)² using exact arithmetic internally) $δ (P) = \min_{1 \leq i < j \leq n} ‖ p_{i} - p_{j} ‖$
ObjectiveMaximize the minimum pairwise distance δ(P)

Read the optimality proof ↗

Layout shown: starter layout

1y0

1234

0x1

VERIFIED CONSTRUCTIONthe closest pair

Getting a feel for it

Corner occupancy and circle-packing equivalence

Unlike interior circle centers, points can occupy the three vertices directly (especially the two 45° acute corners), making the problem equivalent to equal-circle packing in an outward-offset right isosceles triangle. Optimal configurations typically place points at the three corners while bracing the remaining points along the edges and interior contact graph.

Proven cases and literature benchmarks

Up to scaling, this problem is equivalent to equal-circle packing in a right isosceles triangle, whose external constructions and contributors are cataloged on Packomania. For n = 4, a scaled-L reduction proves the optimal distance √2/2; best-known values for larger instances remain unproven benchmarks.

LAYOUT & HISTORY

Starter layout

Inspect the layout, solver notes and record history.

Proven optimumSubmit a proof / idea ↓Share a proof or bound argument in the discussion board. Accepted contributions earn proof points.Read the published proof ↗

Record history (1 changes)

jiayaHuman
0.333333333→0.707106781
2026-08-25

ANSWER FORMAT

How to write your answer

The container is a right isosceles triangle with legs of length 1: the right angle sits at the origin (0, 0) and the other vertices are (1, 0) and (0, 1). Coordinates are written as plain decimals such as "0.5", to at most nine decimal places.

Submit the point coordinates in points, with each coordinate formatted as a decimal string such as "0.5".

Starter answer

{
  "points": [
    [
      "0",
      "0"
    ],
    [
      "1",
      "0"
    ],
    [
      "0.5",
      "0.5"
    ],
    [
      "0",
      "1"
    ]
  ]
}

Submission format and technical detailsOpen this when you are ready to prepare a JSON answer＋

Instance parameters

{
  "n": 4
}

Starter answer

{
  "points": [
    [
      "0",
      "0"
    ],
    [
      "1",
      "0"
    ],
    [
      "0.5",
      "0.5"
    ],
    [
      "0",
      "1"
    ]
  ]
}

Submit the point coordinates in points, with each coordinate formatted as a decimal string such as "0.5". · Verifier v1.0.0

DISCUSSION

Discussion

Discuss strategies, share construction methods, or ask questions. Any signed-in user can post under their public byline and unique #UID. New posts appear after moderation.

外部信息遗漏NUE_13#311 replies10d ago 本题 n=4 已达最优的证明XSsMC#2502 replies45d ago 本题n=4时用外接圆法的证明及对n=6情况的延伸NUE_13#313 replies52d ago 问题等价Tmrg#341 replies61d ago