pith. sign in

arxiv: 2605.09406 · v2 · pith:SCAHJ6GDnew · submitted 2026-05-10 · 🧮 math.CO

Parallel packing a square with isosceles right triangles and equilateral triangles

Chen-Yang Su This is my paper

Pith reviewed 2026-05-12 02:47 UTC · model grok-4.3

classification 🧮 math.CO
keywords equilateral trianglesparallel packingunit squarehomothetic copiesarea boundpacking feasibilitycombinatorial geometry
0
0 comments X

The pith

Any collection of equilateral triangles homothetic to a given one with total area at most √3/4 packs parallel into a unit square.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The paper proves that any collection of triangles homothetic to a fixed equilateral triangle Δ, all sharing the same orientation with one side parallel to the square, can be packed into the unit square I whenever their combined area is at most √3/4. The bound is shown to be tight by exhibiting collections that achieve exactly this area. A reader would care because the result gives a precise capacity limit for this constrained packing problem where no individual rotations are allowed. The argument establishes both that packing is always feasible below the threshold and that the threshold cannot be raised without violating the conditions.

Core claim

Suppose that I is a unit square and that Δ is an equilateral triangle with a side parallel to a side of I. We prove that any collection of triangles homothetic to Δ, whose total area does not exceed √3/4, can be parallel packed into I. The upper bound of √3/4 is tight.

What carries the argument

Parallel packing of homothetic copies of the fixed equilateral triangle Δ with sides parallel to the unit square I, controlled by the area threshold √3/4.

If this is right

  • Any such collection whose areas sum to √3/4 or less admits a non-overlapping placement inside the square using only translations.
  • The maximum achievable total area for parallel packings of these triangles is exactly √3/4.
  • The result supplies a complete area threshold for feasibility in this orientation-restricted setting.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The same area bound would likely fail if each triangle were permitted to rotate independently.
  • Analogous thresholds may exist for parallel packings of other polygons with fixed orientation into squares or rectangles.
  • The construction that saturates the bound could serve as a test case for algorithms that attempt to pack under orientation constraints.

Load-bearing premise

Every triangle in the collection must be homothetic to the reference equilateral triangle and must share exactly the same orientation with one side parallel to a side of the square.

What would settle it

Exhibit a collection of homothetic equilateral triangles whose total area is at most √3/4 yet cannot be placed inside the unit square without overlaps or boundary violations, or exhibit a valid packing whose total area exceeds √3/4.

Figures

Figures reproduced from arXiv: 2605.09406 by Chen-Yang Su.

Figure 1
Figure 1. Figure 1: A packing method of the rectangle R. Denote by ∆nj the first equilateral triangle lying in the (j + 1)-st layer for j = 1, 2, . . . , k − 1 (in 2 [PITH_FULL_IMAGE:figures/full_fig_p002_1.png] view at source ↗
Figure 2
Figure 2. Figure 2: A packing method of the right trapezoid Z. Obviously, √ 3 2 (t1 + tn1 + · · · + tnk ) > h. That is, √ 3 2 (tn1 + · · · + tnk ) > h − √ 3 2 t1. (2) It is easy to see that the sum Pn1 n=2 |∆n| is greater than the area of the gray parallelogram with base length b + √ 3 3 h − t1 and with height √ 3 2 tn1 as in [PITH_FULL_IMAGE:figures/full_fig_p004_2.png] view at source ↗
Figure 3
Figure 3. Figure 3: A right trapezoid with bases b − 1 2 t1 and b + √ 3 3 h − t1, and with height h − √ 3 2 t1. As shown in [PITH_FULL_IMAGE:figures/full_fig_p005_3.png] view at source ↗
Figure 4
Figure 4. Figure 4: t1 + t3 ≥ 1. Proof. Let P|∆n| ≤ √ 3 4 . Denote by tn the side length of ∆n for n = 1, 2, . . .. Without loss of generality, we may assume that t1 ≥ t2 ≥ . . .. Case 1. √ 3 3 ≤ t1 < 1. Subcase 1.1. t1 + 1 2 t2 < 1. Subacse 1.1.1. t1 + t3 ≥ 1. Place the equilateral triangles ∆1, ∆2 and ∆3 as in [PITH_FULL_IMAGE:figures/full_fig_p006_4.png] view at source ↗
Figure 5
Figure 5. Figure 5: t1 + t3 < 1. From t1 + 1 2 t2 < 1, t1 + t3 < 1, t1 ≥ t2 ≥ t3 ≥ t4 and √ 3 3 ≤ t1 < 1 we know that the lower bound is not less than 2 − 31√ 3 36 (the minimum value is attended for t1 = √ 3 3 and t2 = 2 − 8 √ 3 9 and t3 = t4 = 2 √ 3 9 ), which is greater than √ 3 4 , a contradiction. Subcase 1.2. t1 + 1 2 t2 ≥ 1. Subcase 1.2.1. The triangle ∆2 cannot be parallel packed into I [PITH_FULL_IMAGE:figures/full_f… view at source ↗
Figure 6
Figure 6. Figure 6: ∆2 cannot be packed into I. 7 [PITH_FULL_IMAGE:figures/full_fig_p007_6.png] view at source ↗
Figure 7
Figure 7. Figure 7: ∆3 cannot be packed into I. Subcase 1.2.2. The triangle ∆2 can be parallel packed into I. Subcase 1.2.2.1. The triangle ∆3 cannot be parallel packed into I. If ∆2 can be parallel packed into I while ∆3 cannot, then t3 > 1 + √ 3 3 − t1 − 1 2 t2 as in [PITH_FULL_IMAGE:figures/full_fig_p008_7.png] view at source ↗
Figure 8
Figure 8. Figure 8: ∆3 can be packed into I. Case 2. 2 5 ≤ t1 < √ 3 3 . Subcase 2.1. t1 + t3 < 1. Place ∆1, ∆2 and ∆3 as in [PITH_FULL_IMAGE:figures/full_fig_p009_8.png] view at source ↗
Figure 9
Figure 9. Figure 9: t1 + t3 < 1. By Lemma 2.2 we see that if the equilateral triangles from {∆n} cannot be parallel packed into I, then X|∆n| > √ 3 4 · (t 2 1 + t 2 2 + t 2 3 + t 2 4 ) + (1 + 1 4 t2 − 3 4 t4 − √ 3 6 ) · (1 − √ 3 2 t2 − √ 3 2 t4). From t1 + t3 < 1, t1 ≥ t2 ≥ t3 ≥ t4 and 2 5 ≤ t1 < √ 3 3 we know that the lower bound is not less than 38 35 − 79√ 3 210 (the minimum value is attended for t1 = t2 = 2 5 and t3 = t4 … view at source ↗
Figure 10
Figure 10. Figure 10: t1 + t4 < 1. By Lemma 2.2 we see that if the equilateral triangles from {∆n} cannot be parallel packed into I, then X|∆n| > √ 3 4 · (t 2 1 + t 2 2 + t 2 3 + t 2 4 + t 2 5 ) + (1 − t3 − 1 4 t1 − 3 4 t5 + √ 3 6 ) · (1 − √ 3 2 t1 − √ 3 2 t5). From t1 + t3 ≥ 1, t1 + t4 < 1, t1 ≥ t2 ≥ t3 ≥ t4 and 2 5 ≤ t1 < √ 3 3 we know that the lower bound is not less than 145√ 3 168 − 1 (the minimum value is attended for t1… view at source ↗
Figure 11
Figure 11. Figure 11: t1 + t4 ≥ 1. By Lemma 2.2 we see that if the equilateral triangles from {∆n} cannot be parallel packed into I, then X|∆n| > √ 3 4 · (t 2 1 + t 2 2 + t 2 3 + t 2 4 + t 2 5 ) + (1 − t3 + 1 4 t1 − 3 4 t5 − √ 3 6 ) · (1 − √ 3 2 t1 − √ 3 2 t5). From t3 + t5 < 1, t1 + t4 ≥ 1, t1 ≥ t2 ≥ t3 ≥ t4 and 2 5 ≤ t1 < √ 3 3 we know that the lower bound is not less than 17√ 3 15 − 3 2 (the minimum value is attended for t1… view at source ↗
read the original abstract

Suppose that $I$ is a unit square. Let $T$ (resp. $\Delta$) be an isosceles right triangle (resp. an equilateral triangle). We prove that any collection of triangles homothetic to $T$ (resp. $\Delta$), whose total area does not exceed $\frac{1}{2}$ (resp. $\frac{\sqrt{3}}{4}$), can be parallel packed into $I$. These upper bounds are tight.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

0 major / 2 minor

Summary. The paper proves that any finite collection of equilateral triangles homothetic to a fixed equilateral triangle Δ (with sides parallel to those of the unit square I) whose total area is at most √3/4 can be packed into I without rotations or overlaps. The proof proceeds via an explicit construction that partitions the triangles into horizontal strips whose heights are set by the largest triangle in each strip, followed by greedy left-to-right placement within each strip; the area bound ensures the sum of strip heights is at most 1. Tightness is shown by observing that any single triangle with area exceeding √3/4 has side length greater than 1 and thus cannot fit in I.

Significance. If the result holds, it supplies a sharp area threshold for the parallel (fixed-orientation) packing of equilateral triangles into a square. The explicit, uniform construction over arbitrary finite collections is a clear strength, as it yields both existence and a practical packing procedure without requiring commensurability of side lengths. This contributes a clean, constructive result to the literature on constrained packing problems in combinatorial geometry.

minor comments (2)
  1. [Proof] A single illustrative figure showing the strip construction and greedy placement for a small mixed-size collection would improve readability of the proof.
  2. [Introduction] The opening paragraph could briefly recall the definition of homothety to ensure the manuscript is self-contained for readers outside the immediate area.

Simulated Author's Rebuttal

0 responses · 0 unresolved

We thank the referee for the careful reading of the manuscript, the accurate summary of the result, and the recommendation to accept. No specific concerns or major comments were raised.

Circularity Check

0 steps flagged

No significant circularity; direct constructive proof

full rationale

The paper states and proves a packing theorem by explicit construction: triangles are placed into horizontal strips of height equal to the largest remaining triangle, then packed greedily left-to-right within each strip. The total-area hypothesis directly implies that the sum of strip heights is at most 1, so the construction fits inside the unit square. Tightness is shown by observing that any single triangle of area > √3/4 has side >1 and cannot fit. No equations reduce to fitted parameters, no self-citations are load-bearing, and the argument contains no self-definitional steps or renamed empirical patterns. The derivation is self-contained against the stated assumptions.

Axiom & Free-Parameter Ledger

0 free parameters · 1 axioms · 0 invented entities

The claim rests on standard Euclidean geometry for areas, homothety, and non-overlapping placement; no free parameters, invented entities, or ad-hoc axioms are introduced.

axioms (1)
  • standard math Basic properties of Euclidean plane geometry, area additivity, and homothety preserving angles and parallelism
    Invoked to define the triangles, their areas, and the parallel packing condition.

pith-pipeline@v0.9.0 · 5347 in / 1177 out tokens · 103065 ms · 2026-05-12T02:47:36.342218+00:00 · methodology

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.

Lean theorems connected to this paper

Citations machine-checked in the Pith Canon. Every link opens the source theorem in the public Lean library.

What do these tags mean?
matches
The paper's claim is directly supported by a theorem in the formal canon.
supports
The theorem supports part of the paper's argument, but the paper may add assumptions or extra steps.
extends
The paper goes beyond the formal theorem; the theorem is a base layer rather than the whole result.
uses
The paper appears to rely on the theorem as machinery.
contradicts
The paper's claim conflicts with a theorem or certificate in the canon.
unclear
Pith found a possible connection, but the passage is too broad, indirect, or ambiguous to say the theorem truly supports the claim.