Pancake cutting with exotic knives

DefinitionNumber of regions cut out of the plane#✓fully proved

For a set \(K \subseteq \mathbb{R}^2\) of cuts, the number of regions into which \(K\) divides the plane is the number of connected components of the complement \(\mathbb{R}^2 \setminus K\), or \(0\) if there are infinitely many components.

public noncomputable def planeRegionCount (K : Set (ℝ × ℝ)) : ℕ :=
  Set.ncard {C : Set (ℝ × ℝ) | ∃ x ∈ Kᶜ, C = connectedComponentIn Kᶜ x}
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Section 1.
Used by 4
Used byMaximum region count for long-legged A's · Maximum region count for three-armed V's · Maximum region count for 3-chains · No cuts, one region

TheoremNo cuts, one region#✓fully proved

The empty set of cuts divides the plane into exactly one region.

public theorem planeRegionCount_empty : planeRegionCount ∅ = 1
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Section 1 (n = 0).
Uses 1
Statement usesNumber of regions cut out of the plane

TheoremMaximum region count for three-armed V's#●ready to prove

A long-legged three-armed V is the union of three rays \(\{c + s u_j : s \ge 0\}\), \(j = 1, 2, 3\), from a common tip \(c \in \mathbb{R}^2\), with nonzero direction vectors \(u_1, u_2, u_3 \in \mathbb{R}^2\) in arbitrary directions. For every natural number \(n\), the largest number of regions into which the union of \(n\) long-legged three-armed V's can divide the plane is \((9n^2 - 5n + 2)/2\).

public theorem_wanted isGreatest_planeRegionCount_threeArmedV (n : ℕ) :
    IsGreatest {N : ℕ | ∃ (c : Fin n → ℝ × ℝ) (u : Fin n → Fin 3 → ℝ × ℝ),
      (∀ i j, u i j ≠ 0) ∧
        N = planeRegionCount (⋃ i, ⋃ j, (fun s : ℝ => c i + s • u i j) '' Set.Ici 0)}
      ((9 * n ^ 2 - 5 * n + 2) / 2)
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Theorem 1 (jis-2026-26.4.5-thm-1), with the definition in the section The k-armed V.
Uses 1
Statement usesNumber of regions cut out of the plane

TheoremMaximum region count for 3-chains#●ready to prove

For points \(p, q \in \mathbb{R}^2\) and nonzero vectors \(u, v \in \mathbb{R}^2\), the 3-chain is the union of the ray \(\{p + s u : s \ge 0\}\), the segment \([p, q]\) and the ray \(\{q + s v : s \ge 0\}\); the directions are arbitrary and the 3-chain may intersect itself. For every natural number \(n\), the largest number of regions into which the union of \(n\) 3-chains can divide the plane is \((9n^2 - 5n + 2)/2\).

public theorem_wanted isGreatest_planeRegionCount_threeChain (n : ℕ) :
    IsGreatest {N : ℕ | ∃ (p q u v : Fin n → ℝ × ℝ), (∀ i, u i ≠ 0 ∧ v i ≠ 0) ∧
        N = planeRegionCount (⋃ i, ((fun s : ℝ => p i + s • u i) '' Set.Ici 0 ∪
          segment ℝ (p i) (q i) ∪ (fun s : ℝ => q i + s • v i) '' Set.Ici 0))}
      ((9 * n ^ 2 - 5 * n + 2) / 2)
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Theorem 1 (jis-2026-26.4.5-thm-1), with the definition in the section The k-chain.
Uses 1
Statement usesNumber of regions cut out of the plane

TheoremMaximum region count for long-legged A's#●ready to prove

For a tip \(c \in \mathbb{R}^2\), nonzero vectors \(u_1, u_2 \in \mathbb{R}^2\) and reals \(a, b > 0\), the long-legged A is the union of the rays \(\{c + s u_1 : s \ge 0\}\) and \(\{c + s u_2 : s \ge 0\}\) and the crossbar segment \([c + a u_1, c + b u_2]\). For every natural number \(n\), the largest number of regions into which the union of \(n\) long-legged A's can divide the plane is \((9n^2 - 5n + 2)/2\).

public theorem_wanted isGreatest_planeRegionCount_longLeggedA (n : ℕ) :
    IsGreatest {N : ℕ | ∃ (c u₁ u₂ : Fin n → ℝ × ℝ) (a b : Fin n → ℝ),
      (∀ i, u₁ i ≠ 0 ∧ u₂ i ≠ 0 ∧ 0 < a i ∧ 0 < b i) ∧
        N = planeRegionCount (⋃ i, ((fun s : ℝ => c i + s • u₁ i) '' Set.Ici 0 ∪
          (fun s : ℝ => c i + s • u₂ i) '' Set.Ici 0 ∪
          segment ℝ (c i + a i • u₁ i) (c i + b i • u₂ i)))}
      ((9 * n ^ 2 - 5 * n + 2) / 2)
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Theorem 1 (jis-2026-26.4.5-thm-1), with the definition in the section The long-legged A.
Uses 1
Statement usesNumber of regions cut out of the plane

TheoremFour-point intersections of constrained T's#●ready to prove

For a point \(c \in \mathbb{R}^2\) and an angle \(\theta \in \mathbb{R}\), the constrained long-legged T \(\bar T(c, \theta)\) is the union of the crossbar, the line \(\{c + s(-\sin\theta, \cos\theta) : s \in \mathbb{R}\}\), and the stem, the ray \(\{c + s(\cos\theta, \sin\theta) : s \ge 0\}\). For all \(c_1, c_2 \in \mathbb{R}^2\) and \(\theta_1, \theta_2 \in \mathbb{R}\), if \(\bar T(c_1, \theta_1) \cap \bar T(c_2, \theta_2)\) has exactly four points, then \(\cos(\theta_1 - \theta_2) < 0\), that is, the two stem directions make an angle greater than \(\pi/2\).

public theorem_wanted cos_sub_neg_of_tbar_inter_ncard_four (c₁ c₂ : ℝ × ℝ) (θ₁ θ₂ : ℝ) :
    let T : ℝ × ℝ → ℝ → Set (ℝ × ℝ) := fun c θ =>
      Set.range (fun s : ℝ => c + s • (-Real.sin θ, Real.cos θ)) ∪
        (fun s : ℝ => c + s • (Real.cos θ, Real.sin θ)) '' Set.Ici 0
    (T c₁ θ₁ ∩ T c₂ θ₂).ncard = 4 → Real.cos (θ₁ - θ₂) < 0
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Lemma 2 (jis-2026-26.4.5-lem-2), with the definition in the subsection The constrained long-legged T̄.

TheoremStem–circle intersections of two lollipops#●ready to prove

For a point \(c = (a, b) \in \mathbb{R}^2\), a radius \(r > 0\) and an angle \(\theta \in \mathbb{R}\), the lollipop with center \(c\), radius \(r\) and stem direction \((\cos\theta, \sin\theta)\) consists of the circle \(\{(x, y) \in \mathbb{R}^2 : (x - a)^2 + (y - b)^2 = r^2\}\) and the stem \(\{c + s(\cos\theta, \sin\theta) : s \ge r\}\), the part of the ray from \(c\) in that direction outside the open disc. For all \(c_1, c_2 \in \mathbb{R}^2\), \(r_1, r_2 > 0\) and \(\theta_1, \theta_2 \in \mathbb{R}\), consider the lollipops with centers \(c_i\), radii \(r_i\) and stem directions \((\cos\theta_i, \sin\theta_i)\). If the number of points where the stem of the first meets the circle of the second plus the number of points where the stem of the second meets the circle of the first is \(3\) or \(4\), then \(\cos(\theta_1 - \theta_2) < 0\).

public theorem_wanted cos_sub_neg_of_lollipop_good_pair (c₁ c₂ : ℝ × ℝ) {r₁ r₂ : ℝ}
    (hr₁ : 0 < r₁) (hr₂ : 0 < r₂) (θ₁ θ₂ : ℝ) :
    let circ : ℝ × ℝ → ℝ → Set (ℝ × ℝ) := fun c r =>
      {x | (x.1 - c.1) ^ 2 + (x.2 - c.2) ^ 2 = r ^ 2}
    let stem : ℝ × ℝ → ℝ → ℝ → Set (ℝ × ℝ) := fun c r θ =>
      (fun s : ℝ => c + s • (Real.cos θ, Real.sin θ)) '' Set.Ici r
    (stem c₁ r₁ θ₁ ∩ circ c₂ r₂).ncard + (stem c₂ r₂ θ₂ ∩ circ c₁ r₁).ncard ∈
        ({3, 4} : Set ℕ) → Real.cos (θ₁ - θ₂) < 0
Sources
  • David O. H. Cutler, Jonas Karlsson and Neil J. A. Sloane, Cutting a Pancake with an Exotic Knife, Journal of Integer Sequences 29 (2026), Article 26.4.5, Lemma 3 (jis-2026-26.4.5-lem-3), with the definition in the subsection The lollipop or qoppa.
2 fully proved5 ready to prove

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