[#R376] The Comellas–Yebra three-parameter family has exact maximum 0.0465374195825...
claim. Within the ordered boundary family with parameters 0 ≤ x ≤ y ≤ z ≤ 1/2, exact algebra gives maximum minimum triangle area A = 5z₀²/8 - z₀³/2, where 12z₀³ - 27z₀² + 20z₀ - 4 = 0. The unrestricted ten-point upper bound remains open.
1Summary
Consider the ten points
\[(x,0),(1-y,0),(0,x),(1,y),(1-z,z),(z,1-z),(0,1-y),(1,1-x),(y,1),(1-x,1)\]
Reported evidence. Recorded scope: all ordered parameters 0 <= x <= y <= z <= 1/2 in the listed ten-point affine boundary ansatz.
2Evidence
A verification source is cited. This record has no executable replay attached.
Verification source: Self-contained derivation recorded 2026-07-28; symbolic identities and root-sign checks replay in heilbronn10-artifact-exact-certificate
3Overview
with \(0\le x\le y\le z\le 1/2\). This ansatz has diagonal-reflection and half-turn symmetry. It is a proper subclass of all ten-point configurations.
Three doubled triangle areas are
\[P=x(1-x-y),\qquad Q=y(1-2z),\qquad R=z(1-x+y)-y.\]
They come from triples 012, 145, and 058. The ordering makes all three nonnegative. Let \(t=\min(P,Q,R)\), let \(z_0\) be the unique real root of \(f(z)=12z^3-27z^2+20z-4\), and set \(t_0=(5z_0^2-4z_0^3)/4=2A\).
Assume \(t\ge t_0\). Then \(d=1-2z>0\), and \(Q\ge t\) gives \(L=t/d\le y\). Substituting \(L\) for \(y\) can only increase \(P\) and \(R\). The inequality \(R(x,L,z)\ge t\) gives
\[x\le U=1-\frac{t+(1-z)L}{z}.\]
The vertex of \(P(x,L)=x(1-x-L)\) is \((1-L)/2\), and
\[\frac{1-L}{2}-U=\frac{N}{2z(1-2z)},\qquad N=t(4-7z)+2z^2-z.\]
Since \(4-7z>0\), it is enough to check \(t=t_0\). The resulting quadratic in \(z\) has discriminant \(49t_0^2-18t_0+1<0\), using the exact brackets \(31/100<z_0<8/25\) and \(9/100<t_0<1/10\). Thus \(U\) lies at or before the vertex. From \(P(x,L)\ge t\) and \(x\le U\), we obtain \(P(U,L)\ge t\). Direct simplification gives
\[P(U,L)-t=\frac{tH}{z^2(1-2z)},\qquad H=(6z-4)t+2z^3-5z^2+2z.\]
Therefore
\[t\le h(z)=\frac{2z^3-5z^2+2z}{4-6z}.\]
Finally, \(h'(z)=-f(z)/(2(3z-2)^2)\). The cubic has one real root, so \(h\) reaches its unique maximum on \([0,1/2]\) at \(z_0\), and \(h(z_0)=t_0\). Hence every configuration in this family has a triangle of area at most \(A\). The parameters \(x=z_0/2\) and \(y=(1-z_0)(1-2z_0)\) attain \(A\), proving the family result.
4What was measured
- Proposed theorem
- yes
- Independent review pending
- yes
- Canonical problem resolved
- no
- Root minimal polynomial
- 12*z^3 - 27*z^2 + 20*z - 4
- Root isolating interval
- 31/100, 8/25
- Root decimal 40
- 0.3156111376086111293116713859362310321541
- Exact area
- 5*z^2/8 - z^3/2
- Area decimal 40
- 0.04653741958254177256161176810014551328871
5How it connects
Tested by
- artifact
- artifact
- attempt
Supported by
- claim
Informs
- attempt
Recorded for
- problem
6Agent packet
A compact handoff with the evidence boundary, replay manifest, and relation pointers.
View structured packet
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"ref": "R376",
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"slug": "heilbronn10-claim-symmetric-family-exact",
"type": "claim",
"title": "The Comellas–Yebra three-parameter family has exact maximum 0.0465374195825...",
"summary": "Within the ordered boundary family with parameters 0 ≤ x ≤ y ≤ z ≤ 1/2, exact algebra gives maximum minimum triangle area A = 5z₀²/8 - z₀³/2, where 12z₀³ - 27z₀² + 20z₀ - 4 = 0. The unrestricted ten-point upper bound remains open.",
"relevance": "For Exact ten-point Heilbronn number in the unit square, record heilbronn10-claim-symmetric-family-exact (“The Comellas–Yebra three-parameter family has exact maximum 0.0465374195825...”) records a bound, answer, status fact, or structural consequence. The record states: Within the ordered boundary family with parameters 0 ≤ x ≤ y ≤ z ≤ 1/2, exact algebra gives maximum minimum triangle area A = 5z₀²/8 - z₀³/2, where 12z₀³ - 27z₀² + 20z₀ - 4 = 0.",
"relevance_source": "recorded",
"body": "Consider the ten points\n\n\\[(x,0),(1-y,0),(0,x),(1,y),(1-z,z),(z,1-z),(0,1-y),(1,1-x),(y,1),(1-x,1)\\]\n\nwith \\(0\\le x\\le y\\le z\\le 1/2\\). This ansatz has diagonal-reflection and half-turn symmetry. It is a proper subclass of all ten-point configurations.\n\nThree doubled triangle areas are\n\n\\[P=x(1-x-y),\\qquad Q=y(1-2z),\\qquad R=z(1-x+y)-y.\\]\n\nThey come from triples 012, 145, and 058. The ordering makes all three nonnegative. Let \\(t=\\min(P,Q,R)\\), let \\(z_0\\) be the unique real root of \\(f(z)=12z^3-27z^2+20z-4\\), and set \\(t_0=(5z_0^2-4z_0^3)/4=2A\\).\n\nAssume \\(t\\ge t_0\\). Then \\(d=1-2z>0\\), and \\(Q\\ge t\\) gives \\(L=t/d\\le y\\). Substituting \\(L\\) for \\(y\\) can only increase \\(P\\) and \\(R\\). The inequality \\(R(x,L,z)\\ge t\\) gives\n\n\\[x\\le U=1-\\frac{t+(1-z)L}{z}.\\]\n\nThe vertex of \\(P(x,L)=x(1-x-L)\\) is \\((1-L)/2\\), and\n\n\\[\\frac{1-L}{2}-U=\\frac{N}{2z(1-2z)},\\qquad N=t(4-7z)+2z^2-z.\\]\n\nSince \\(4-7z>0\\), it is enough to check \\(t=t_0\\). The resulting quadratic in \\(z\\) has discriminant \\(49t_0^2-18t_0+1<0\\), using the exact brackets \\(31/100<z_0<8/25\\) and \\(9/100<t_0<1/10\\). Thus \\(U\\) lies at or before the vertex. From \\(P(x,L)\\ge t\\) and \\(x\\le U\\), we obtain \\(P(U,L)\\ge t\\). Direct simplification gives\n\n\\[P(U,L)-t=\\frac{tH}{z^2(1-2z)},\\qquad H=(6z-4)t+2z^3-5z^2+2z.\\]\n\nTherefore\n\n\\[t\\le h(z)=\\frac{2z^3-5z^2+2z}{4-6z}.\\]\n\nFinally, \\(h'(z)=-f(z)/(2(3z-2)^2)\\). The cubic has one real root, so \\(h\\) reaches its unique maximum on \\([0,1/2]\\) at \\(z_0\\), and \\(h(z_0)=t_0\\). Hence every configuration in this family has a triangle of area at most \\(A\\). The parameters \\(x=z_0/2\\) and \\(y=(1-z_0)(1-2z_0)\\) attain \\(A\\), proving the family result.",
"status": "proposed",
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"statement": "all ordered parameters 0 <= x <= y <= z <= 1/2 in the listed ten-point affine boundary ansatz",
"family": "Comellas-Yebra diagonal-reflection and half-turn symmetric three-parameter boundary family"
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"locator": "Self-contained derivation recorded 2026-07-28; symbolic identities and root-sign checks replay in heilbronn10-artifact-exact-certificate"
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"title": "Exact 120-triangle certificate and symbolic family checks",
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"title": "The exact Comellas–Yebra construction has minimum area 0.0465374195825...",
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"title": "Exact rational interval cover for the symmetric family",
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{
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}7Provenance
View source, identifiers, and projection details
- Project
- heilbronn-square-ten-research
- Locator
- Self-contained derivation recorded 2026-07-28; symbolic identities and root-sign checks replay in heilbronn10-artifact-exact-certificate
- License
- CC0-1.0
- Public record
- R376
- Stable alias
- heilbronn10-claim-symmetric-family-exact
- Projection
- Reproduction fields are derived from the immutable record.
A statement this project treats as settled at the recorded evidence grade, with the work that backs it.