P78 · Extremal configurations · Classic · Hard

Covering a disc with n equilateral triangles · n = 8

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

Instancen = 8

ObjectiveMinimize the common side

Best known, unproven752/(453√3 − 3√347)published value in this metric 1.031924891171631exhibited site certificate 1.047235397As printed in the source (the radius r of the largest disc that n unit triangles cover (s = 1/r)): (453√3 − 3√347)/752 ≈ 0.969+Maurizio Morandi's arrangement (June 2009), listed in Friedman's table as the largest covered-disc radius r = (453√3 − 3√347)/752 ≈ 0.969+ for n unit triangles; here it is converted in closed form by s = 1/r and shown rounded upward to fifteen places. The site reconstructed the same contact topology from the public raster, and its nine-decimal certificate verifies at 1.047235397; the gap from the continuous target remains explicit. Optimality is openSubmissions are decimal, so a record chases the writable value. If you believe you have the exact closed form, send it in. minmaxarena.com@gmail.com

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 verificationThe 3n edges are extended to lines; on each face of that arrangement every triangle's membership is constant, so the disc is covered exactly when no face meeting the open disc is empty. Faces are sampled as trapezoids in vertical strips, all arithmetic in Q(√3); the circle enters only as whether a trapezoid meets the open disc, which is a rational comparison. No pixel sampling and no epsilon anywhere
ScoreThe side s shared by every covering triangle; smaller is better $s (T) = s$
ObjectiveMinimise s over all legal covers $\min_{T} s (T)$

Open the full editor ↗

VERIFIED CONSTRUCTIONcommon side 1.047235397

Getting a feel for it

Six fit exactly; the problem starts at seven

Six equilateral triangles around the centre make a regular hexagon whose incircle is exactly the unit disc, side 2/√3, with nothing to spare. From the seventh piece on there is no ready answer to where the extra one goes and how the others make room.

Where the frontier is

Friedman's table records n = 1–18 in the reciprocal normalisation: the largest covered-disc radius r for n unit triangles, so the side here is s = 1/r. n = 8–13 have closed forms (Morandi, Cantrell); n = 7 and 14–18 are truncated decimals (Morandi, 2009), and nothing beyond n = 1, 2 and 6 is proved optimal. The source gives only approximately 200-pixel figures and no coordinates. This site reconstructed each pictured set of horizontal and ±60° edges and its contact topology, then used the exact continuous verifier to remove raster-rounding slivers. The literature target and the nine-decimal reconstruction remain separate, so reconstruction error is never presented as the authors' data.

Source

Current leader

1.047235397

the common side

Answer sourceMaurizio Morandi

Solution methodPublished reference construction

Challenge this record ↗

ANSWER FORMAT

How to write your answer

The container is the closed disc of radius 1 about (1,1), which fills the [0,2]² board. Coordinates mark one reference-edge midpoint; covering pieces may protrude.

Submit side and triangles, each as {x,y,turn,down}. The editor accepts or scrubs θ° and writes turn = tan(θ/2).

The current leader's answer

{
  "side": "1.047235397",
  "triangles": [
    {
      "down": true,
      "turn": "0.000000000",
      "x": "1.350507773",
      "y": "2.005602406"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "0.644023375",
      "y": "1.989890331"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "1.723424464",
      "y": "0.840059616"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "0.998320008",
      "y": "0.824347542"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "1.542161448",
      "y": "0.843957127"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "0.269670322",
      "y": "0.840059616"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "0.457943884",
      "y": "0.843957127"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "1.004785357",
      "y": "-0.014460356"
    }
  ]
}

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

Instance parameters

{
  "n": 8
}

The current leader's answer

{
  "side": "1.047235397",
  "triangles": [
    {
      "down": true,
      "turn": "0.000000000",
      "x": "1.350507773",
      "y": "2.005602406"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "0.644023375",
      "y": "1.989890331"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "1.723424464",
      "y": "0.840059616"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "0.998320008",
      "y": "0.824347542"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "1.542161448",
      "y": "0.843957127"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "0.269670322",
      "y": "0.840059616"
    },
    {
      "down": true,
      "turn": "0.000000000",
      "x": "0.457943884",
      "y": "0.843957127"
    },
    {
      "down": false,
      "turn": "0.000000000",
      "x": "1.004785357",
      "y": "-0.014460356"
    }
  ]
}

Submit side and triangles, each as {x,y,turn,down}. The editor accepts or scrubs θ° and writes turn = tan(θ/2). · Verifier v1.0.0

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