P78 · Extremal configurations · Classic · Hard

Covering a disc with n equilateral triangles · n = 10

Continuously cover the unit disc with n congruent equilateral triangles. Overlap and protrusion are allowed; minimise the common side.

Instancen = 10

ObjectiveMinimize the common side

Record comparison

the common side · Lower is better

Record	Value / interval	Author / holder	Source
External best known	40/(31√3 − √103)	See source attribution	Erich Friedman ↗
Site record	0.918596643	zzzcy	Record details ↓

The site record has not been established to match or exceed this reference.

Formal definition

ContainerThe closed disc D = {(x,y) : (x−1)² + (y−1)² ≤ 1}
SubmissionA common side s and exactly n {x,y,turn,down} placements, each independently rotatable and flippable
CoverEvery point of D belongs to at least one closed covering triangle $D \subseteq T_{1} \cup \dots \cup T_{n}$
Exact continuous verificationExtending the 3n triangle edges into a line arrangement partitions the plane into faces on which triangle membership is constant; the disc is covered if and only if no uncovered face intersects the open disc. The verifier decomposes faces into vertical trapezoids over Q(√3) and tests open-disc intersection via exact rational comparisons
ScoreThe common side s shared by all covering triangles; smaller is better $s (T) = s$
ObjectiveMinimise s over all valid covers $\min_{T} s (T)$

Open full editor ↗

Layout shown: site record

VERIFIED CONSTRUCTIONcommon side 0.918596643

Getting a feel for it

Hexagonal fit at n = 6 and reorganization for n ≥ 7

At n = 6, six equilateral triangles of side 2/√3 meet at the centre to form a regular hexagon whose incircle is the unit disc. For n ≥ 7, the extra pieces break sixfold symmetry and require redistributing coverage between the interior and the circular boundary.

Literature data and coordinate reconstruction

Friedman's table records n = 1–18 in the reciprocal convention (maximum covered disc radius r for n unit triangles, corresponding to s = 1/r here). Closed-form values are known for n = 8–13 (Morandi, Cantrell) and truncated decimals for n = 7 and 14–18 (Morandi, 2009), with optimality proved only for n = 1, 2, and 6. Because the source provides only ~200-pixel diagrams without coordinates, this catalogue reconstructs nine-decimal certificates from the depicted edge angles and contact topologies, listing literature targets alongside verified certificate scores.

Source

LAYOUT & HISTORY

Record details

Inspect the layout, solver notes and record history.

Record holderzzzcy

Solution methodAI · OpenAI: GPT-6 Astra

Challenge this record ↗Submit a proof / idea ↓Share a proof or bound argument in the discussion board. Accepted contributions earn proof points.

Record holder's solver noteNo solver note yet (expand)＋

The record holder has not shared a solver note yet.

Record history (3 changes)

zzzcyAI · OpenAI: GPT-6 Astra
0.923574177→0.918596643
2026-10-04
Sigmoid with CodexAI · OpenAI: GPT-5.6 Sol
0.931073399→0.923574177
2026-09-03
Sigmoid with CodexAI · OpenAI: GPT-5.6 Sol
0.932073399→0.931073399
2026-09-03

ANSWER FORMAT

How to write your answer

The container is the closed disc of radius 1 centred at (1,1), inscribed in the [0,2]² square. Coordinates mark one reference-edge midpoint; covering pieces may protrude.

Submit side and triangles, each as {x,y,turn,down}. The editor allows entering or dragging the angle θ° and automatically converts it to turn = tan(θ/2).

The current leader's answer

{
  "side": "0.918596643",
  "triangles": [
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "0.762789590",
      "y": "1.599298041"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "1.514423173",
      "y": "1.313207035"
    },
    {
      "down": true,
      "turn": "0.006292996",
      "x": "1.842734068",
      "y": "1.317339317"
    },
    {
      "down": true,
      "turn": "-0.568989983",
      "x": "0.702481601",
      "y": "1.700783059"
    },
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "0.489316966",
      "y": "0.902509026"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "1.286058117",
      "y": "0.696868216"
    },
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "1.297518399",
      "y": "0.299216941"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "0.157265932",
      "y": "0.682660683"
    },
    {
      "down": false,
      "turn": "-0.568989982",
      "x": "0.260951912",
      "y": "0.286170207"
    },
    {
      "down": true,
      "turn": "-0.568989982",
      "x": "1.237210410",
      "y": "0.400701960"
    }
  ]
}

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

Instance parameters

{
  "n": 10
}

The current leader's answer

{
  "side": "0.918596643",
  "triangles": [
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "0.762789590",
      "y": "1.599298041"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "1.514423173",
      "y": "1.313207035"
    },
    {
      "down": true,
      "turn": "0.006292996",
      "x": "1.842734068",
      "y": "1.317339317"
    },
    {
      "down": true,
      "turn": "-0.568989983",
      "x": "0.702481601",
      "y": "1.700783059"
    },
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "0.489316966",
      "y": "0.902509026"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "1.286058117",
      "y": "0.696868216"
    },
    {
      "down": false,
      "turn": "-0.568989983",
      "x": "1.297518399",
      "y": "0.299216941"
    },
    {
      "down": false,
      "turn": "0.006292996",
      "x": "0.157265932",
      "y": "0.682660683"
    },
    {
      "down": false,
      "turn": "-0.568989982",
      "x": "0.260951912",
      "y": "0.286170207"
    },
    {
      "down": true,
      "turn": "-0.568989982",
      "x": "1.237210410",
      "y": "0.400701960"
    }
  ]
}

Submit side and triangles, each as {x,y,turn,down}. The editor allows entering or dragging the angle θ° and automatically converts it to turn = tan(θ/2). · Verifier v1.0.0

DISCUSSION

Discussion

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外部记录更新（9.25）NUE_13#310 replies38d ago