P77 · Extremal configurations · Classic · Hard

Covering an equilateral triangle with n equilateral triangles · n = 6

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

Instancen = 6

ObjectiveMinimize the common side

Best known, unproven3/7published value in this metric 0.428571428571429exhibited site certificate 0.428571429Friedman's table covers an equilateral container with unit triangles; normalisation gives common side 3/7 ≈ 0.428571428571429. The site's nine-decimal reconstruction verifies at 0.428571429. Optimality is open.Submissions 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 equilateral triangle K = conv{(0,0),(1,0),(0.5,√3/2)}
SubmissionA common side s and exactly n {x,y,turn,down} placements, each independently rotatable and flippable
CoverEvery point of K belongs to at least one closed covering triangle $K \subseteq T_{1} \cup \dots \cup T_{n}$
Exact continuous verificationThe verifier enumerates container edges, piece edges and every intersection in Q(√3), then performs an exact vertical sweep rather than pixel sampling
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 0.428571429

Getting a feel for it

A similar container is still not a scaling exercise

Only square values of n can fill the similar container without area waste on a regular triangular grid. Other sizes force overlap or protrusion, and one extra piece may reorganise the whole layered arrangement.

The published nontrivial range starts at n=6

Friedman's table lists frontier constructions for n=6–9 and n=11–15. At n=9 the 3×3 triangular grid meets the area bound and is proved optimal. The n=10 row starts here from that grid plus one redundant triangle and is not presented as an external record. The site reconstructs the other published figures and separates their source values from the nine-decimal certificates.

Source

Current leader

0.428571429

the common side

Answer sourceDavid Cantrell

Solution methodPublished reference construction

Challenge this record ↗

ANSWER FORMAT

How to write your answer

The container has vertices (0,0), (1,0), and (0.5,√3/2). 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": "0.428571429",
  "triangles": [
    {
      "down": false,
      "turn": "0",
      "x": "0.5",
      "y": "0.494871659"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.5",
      "y": "0"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.785714286",
      "y": "0"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.214285714",
      "y": "0"
    },
    {
      "down": true,
      "turn": "0",
      "x": "0.642857143",
      "y": "0.494871659"
    },
    {
      "down": true,
      "turn": "0",
      "x": "0.357142857",
      "y": "0.494871659"
    }
  ]
}

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

Instance parameters

{
  "n": 6
}

The current leader's answer

{
  "side": "0.428571429",
  "triangles": [
    {
      "down": false,
      "turn": "0",
      "x": "0.5",
      "y": "0.494871659"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.5",
      "y": "0"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.785714286",
      "y": "0"
    },
    {
      "down": false,
      "turn": "0",
      "x": "0.214285714",
      "y": "0"
    },
    {
      "down": true,
      "turn": "0",
      "x": "0.642857143",
      "y": "0.494871659"
    },
    {
      "down": true,
      "turn": "0",
      "x": "0.357142857",
      "y": "0.494871659"
    }
  ]
}

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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