> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# WSFC Round 23
## Strict Monotonicity: Tangent-Gap Refinement and the Remaining Half of the 2007 Conjecture (5)

**English Title:** *Strict Monotonicity: Tangent-Gap Refinement and the Remaining Half of the 2007 Breadth Conjecture*  
**Document ID:** `WSFC-R23-v0.1`  
**Version:** v0.1  
**Date:** 2026-09-23  
**Author:** Neo.K  
**AI Research Collaborator:** Aletheia / OpenAI GPT-5.6 Sol  
**Series:** Worm Sequence Freedom Closure  
**Prerequisites:** `WSFC-R21-v0.1`; `WSFC-R22-v0.1`  
**Status:** `STRICT MONOTONICITY RESEARCH PROOF / CONDITIONAL ON G0-LITE / FORMAL AUTHORITY OPEN`  
**Canonical Source:** UTF-8 Markdown

---

# Abstract

The 2007 conjecture (5) of Panraksa–Wetzel–Wichiramala is:

$$
\boxed{
t_n
\text{ strictly increases to }
t_0.
}
$$

Round 22 has already proved, independently of G0:

$$
\boxed{
t_n\rightarrow t_0
}
$$

and gives:

$$
0\le t_0-t_n
\le
\frac1{\lfloor n/2\rfloor}.
$$

Thus conjecture (5) requires only strictness:

$$
\boxed{
t_n<t_{n+1}.
}
$$

Round 23 compresses strictness into a local **Tangent-Gap Refinement Lemma**.

For two adjacent tangent supports in a breadth-one support-tangent chain, let the outward-normal angular gap be:

$$
0<\delta<\pi.
$$

The original local contribution is:

$$
\ell_{\rm old}
=
2\tan\frac{\delta}{2}.
$$

Insert the bisecting tangent between the two tangents; the gap becomes:

$$
\frac{\delta}{2}+\frac{\delta}{2}.
$$

The new contribution is:

$$
\ell_{\rm new}
=
4\tan\frac{\delta}{4}.
$$

Let:

$$
u=\tan\frac{\delta}{4}.
$$

Then:

$$
0<u<1,
$$

with:

$$
\tan\frac{\delta}{2}
=
\frac{2u}{1-u^2}.
$$

Hence the exact strict gain is:

$$
\boxed{
\ell_{\rm old}-\ell_{\rm new}
=
\frac{4u^3}{1-u^2}
>0.
}
$$

Thus each genuinely active tangent cell that is born strictly reduces the breadth-one chain length.

More importantly, this strictness proof does not require the full G0 reduced formula. It requires only **G0-lite**:

1. after lower-stage normalization, the true stage-$n$ optimizer is a nondegenerate convex width-active tangent chain;
2. lower-stage degeneration admits sound normalization;
3. a genuine stage-$n$ optimizer has at least one positive tangent gap;
4. bisecting-tangent insertion is legal and increases the segment count by exactly one.

This round also gives an exact separation for the induction base:

$$
\boxed{
t_3<\frac7{16}<t_4.
}
$$

The 2007 theorem gives:

$$
t_3(\theta)
=
\frac{\sin2\theta+\sin\theta}{2\cos\theta+3},
\qquad
\theta\in[\pi/3,\pi/2].
$$

Let:

$$
c=\cos\theta\in[0,1/2].
$$

To prove:

$$
t_3<\frac7{16},
$$

squaring reduces the problem to:

$$
\boxed{
P(c)
=
1024c^4+1024c^3-572c^2-436c+185
>0.
}
$$

Exact Sturm count:

```text
roots in [0,1/2] = 0
P(0) = 185
P(1/2) = 16
```

Hence:

$$
\boxed{
t_3<7/16.
}
$$

On the four-segment side, take an existing legal support-tangent candidate:

$$
d=\frac{53}{50}.
$$

Its breadth-one length is fully algebraized:

$$
\boxed{
L_4
=
\frac{932959}{400750}
-
\frac{1573\sqrt{309}}{601125}.
}
$$

And:

$$
\frac{16}{7}-L_4
=
\frac{3146\sqrt{309}-50877}{1202250}.
$$

Because:

$$
3146^2\cdot309-50877^2
=
\boxed{
469801515
}
>0,
$$

we have:

$$
L_4<16/7,
$$

Hence the candidate breadth is:

$$
1/L_4>7/16.
$$

Therefore:

$$
\boxed{
t_3<7/16<t_4.
}
$$

The 2007 source also gives:

$$
t_2<t_3.
$$

Now perform induction.

Assume:

$$
t_{n-1}<t_n,
\qquad
n\ge4.
$$

Take a true stage-$n$ optimizer:

$$
\Gamma_n.
$$

If it can be normalized to at most $n-1$ segments, its trace can be represented as an $(n-1)$-segment unit arc, so:

$$
t_n\le t_{n-1},
$$

a contradiction.

Thus $\Gamma_n$ is genuinely stage-$n$.

By G0-lite, it has at least one positive tangent gap:

$$
0<\delta<\pi.
$$

Insert the bisecting tangent to obtain a legal breadth-one $(n+1)$-segment chain, with:

$$
L'<L_n=\frac1{t_n}.
$$

After scaling back to unit length, its breadth is:

$$
\frac1{L'}>t_n.
$$

Therefore:

$$
\boxed{
t_{n+1}>t_n.
}
$$

Induction yields:

$$
\boxed{
t_2<t_3<t_4<t_5<\cdots.
}
$$

Together with Round 22:

$$
t_n\rightarrow t_0,
$$

we obtain:

$$
\boxed{
t_n
\text{ strictly increases to }
t_0.
}
$$

Thus the 2007 conjecture (5) can be upgraded to:

```text
RESEARCH-PROOF-LEVEL PROVISIONALLY CLOSED
FORMAL AUTHORITY OPEN
INDEPENDENT REVIEW OPEN
```

---

# 1. Source Status

The 2007 paper proves only:

$$
t_n\le t_{n+1}<t_0,
$$

and explicitly lists strict increase to $t_0$ as conjecture (5).

Its exact theorem gives the unique three-segment trine and explicitly states:

$$
t_0>t_3>0.435>t_2.
$$

Thus:

$$
t_2<t_3
$$

is a classical base.

The targeted 2026 search found no subsequent work explicitly claiming that general-$n$ strictness had been solved.

This is not an exhaustive novelty proof.

---

# 2. Breadth-One Dual

The unit-length optimizer satisfies:

$$
b(\Gamma_n)=t_n.
$$

Scale by:

$$
1/t_n.
$$

to obtain a breadth-one chain:

$$
\widetilde\Gamma_n
$$

of length:

$$
L_n=\frac1{t_n}.
$$

Strict breadth increase is equivalent to:

$$
\boxed{
L_{n+1}<L_n.
}
$$

---

# 3. Local Tangent Split

A positive support gap:

$$
\delta.
$$

Old:

$$
2\tan\frac{\delta}{2}.
$$

Insert the bisector tangent:

$$
4\tan\frac{\delta}{4}.
$$

Difference:

$$
\boxed{
\frac{4\tan^3(\delta/4)}
{1-\tan^2(\delta/4)}
>0.
}
$$

This is the actual local engine of strictness.

---

# 4. Perspective View

Round 04's function:

$$
g(q,s)
=
2q\tan\frac{s}{2q}
$$

for fixed:

$$
s>0
$$

strictly decreases with $q$.

Let:

$$
z=\frac{s}{2q}.
$$

Then:

$$
\frac{\partial g}{\partial q}
=
2(\tan z-z\sec^2z).
$$

And:

$$
f(z)=z\sec^2z-\tan z
$$

satisfies:

$$
f'(z)=2z\sec^2z\tan z>0,
$$

$$
f(0)=0.
$$

Hence:

$$
\boxed{
\partial_q g<0.
}
$$

---

# 5. G0-lite

Strictness does not use:

```text
balanced allocation
even/odd symmetry
exact d optimizer
exact x optimizer
uniqueness
```

It uses only:

```text
support-width saturation
lower-stage normalization
positive tangent gap
legal tangent insertion
```

Thus the proof dependency is smaller than that of the full finite-stage formula.

---

# 6. Exact $t_3$ Bound

Let:

$$
c=\cos\theta.
$$

$$
t_3(c)
=
\frac{\sqrt{1-c^2}(2c+1)}{2c+3}.
$$

For:

$$
c\in[0,1/2].
$$

compare:

$$
t_3(c)<7/16.
$$

After squaring:

$$
P(c)>0
$$

where:

$$
P(c)
=
1024c^4+1024c^3-572c^2-436c+185.
$$

The Sturm certificate has no roots and both endpoint values are positive, so the inequality holds.

---

# 7. Exact $t_4$ Competitor

When $d=53/50$:

$$
\sqrt{d^2-1}
=
\frac{\sqrt{309}}{50}.
$$

the support-tangent candidate has:

$$
L_4
=
\frac{932959}{400750}
-
\frac{1573\sqrt{309}}{601125}.
$$

with:

$$
16/7-L_4>0.
$$

Hence:

$$
t_4>7/16.
$$

---

# 8. Lower-Stage Induction

Assume:

$$
t_{n-1}<t_n.
$$

If a stage-$n$ optimum has a redundant segment:

$$
\Gamma_n
\rightarrow
\Gamma_k,
\qquad
k<n,
$$

then dormant subdivision can represent $\Gamma_k$ as an $(n-1)$-segment arc.

Hence:

$$
t_n\le t_{n-1},
$$

a contradiction.

Thus the optimizer is genuinely stage-$n$.

---

# 9. Positive Gap

A genuine nondegenerate support-tangent chain cannot have every adjacent normal gap equal to $0$.

Otherwise, support lines could be merged and the stage would decrease.

Hence:

$$
\exists\delta\in(0,\pi).
$$

---

# 10. Strict Refinement

Insert the tangent normal at the angular midpoint.

The existing width-active support structure is preserved; the new support is also tangent to the same opposite support circle.

Thus breadth remains at least $1$.

Segment count:

$$
n\rightarrow n+1.
$$

Length strictly drops.

Thus breadth strictly rises after scaling back.

---

# 11. Final Induction

Base:

$$
t_3<t_4.
$$

Step:

$$
t_{n-1}<t_n
\Rightarrow
t_n<t_{n+1}.
$$

Hence:

$$
\boxed{
t_n<t_{n+1}
\quad
\forall n\ge3.
}
$$

Add:

$$
t_2<t_3.
$$

to complete all $n\ge2$.

---

# 12. Round 22 + 23

Round 22:

$$
t_n\rightarrow t_0.
$$

Round 23:

$$
t_n<t_{n+1}.
$$

Therefore:

$$
\boxed{
t_n\uparrow t_0
\text{ strictly}.
}
$$

---

# 13. Authority Discipline

Currently permitted:

```text
Conjecture (5):
  research-proof-level provisionally closed
```

Not permitted:

```text
formally verified theorem
priority established
peer-reviewed resolution
```

---

# 14. Numerical Audit

Re-optimize the reduced formulas for:

$$
n=3,\ldots,30.
$$

At every stage:

$$
\widehat t_{n+1}-\widehat t_n>0.
$$

This result serves only as an implementation red-team check.

---

# 15. Next Steps

Conjecture (6) now requires only:

$$
\boxed{
b_{n+1}<b_n.
}
$$

because Round 22 has already proved:

$$
b_n\rightarrow b_0.
$$

But endpoint distance is not the objective, and its strictness does not follow automatically from tangent splitting.

Another potentially more fundamental direction is:

$$
\boxed{
\textbf{Finite-Stage Uniqueness}
}
$$

namely the 2007 conjecture (1).

If the reduced scalar optimizer can be proved unique, this will also strengthen:

- the definition of endpoint distance;
- parity/symmetry;
- G0 authority;
- the strictness analysis for conjecture (6).

---

# Conclusion

Round 23 compresses strict monotonicity to:

$$
\boxed{
2\tan\frac{\delta}{2}
>
4\tan\frac{\delta}{4}.
}
$$

The local strict gain is:

$$
\boxed{
\frac{4u^3}{1-u^2}>0.
}
$$

Together with the exact base:

$$
\boxed{
t_3<\frac7{16}<t_4
}
$$

and lower-stage induction, we obtain:

$$
\boxed{
t_n<t_{n+1}
}
$$

at research-proof level conditional on G0-lite.

Combined with Round 22:

$$
\boxed{
t_n
\text{ strictly increases to }
t_0.
}
$$

===END===