Pick's Theorem and Lattice-Point Geometry — Mathematical Foundations¶

Table of Contents¶

1. Formal Definition
2. Proof via Additivity over Triangulation
3. Proof of the Primitive-Triangle Lemma (Euler's Formula)
4. The gcd Boundary Count — Proof
5. Worked Triangulation Trace
6. Ehrhart Theory — Pick Generalized
7. Worked Ehrhart Computation
8. Higher Dimensions — The Reeve Tetrahedron
9. Polygons with Holes — Euler-Characteristic Form
10. Minkowski, Determinants, and the Primitive Lattice
11. Complexity and Exactness
12. Comparison with Alternatives
13. Open Threads and Connections
14. Summary

1. Formal Definition¶

Let Λ = ℤ² be the integer lattice in the plane. A lattice polygon P is a simple (non-self-intersecting) polygon whose vertices all lie in Λ. Define:

A  = area(P)                         (Lebesgue area of the closed region)
I  = | interior(P) ∩ Λ |             (lattice points strictly inside P)
B  = | ∂P ∩ Λ |                      (lattice points on the boundary of P)

Theorem (Pick, 1899). For every lattice polygon P,

A = I + B/2 − 1.

Equivalently, in the integer-only form used in code:

2A = 2I + B − 2,   so   I = (2A − B + 2) / 2,

where 2A is the absolute shoelace sum, a non-negative integer.

The logical structure of the whole topic — from the atomic primitive triangle up to the higher-dimensional generalizations and the 3-D impossibility — is shown below.

We prove the theorem by reducing P to a triangulation into primitive lattice triangles and showing two facts: (i) each primitive triangle has area exactly 1/2, and (ii) Pick's formula is additive under gluing two lattice polygons along a shared lattice edge. Together these force the formula for every lattice polygon.

Definition 1.1 (Primitive / unimodular triangle). A lattice triangle T is primitive if its only lattice points are its three vertices: I(T) = 0, B(T) = 3. Equivalently, the two edge vectors from any vertex form a basis of Λ (determinant ±1).

2. Proof via Additivity over Triangulation¶

We establish Pick's theorem in three steps.

Step 1 — Define the Pick functional and prove additivity¶

For any lattice polygon Q, define

Pick(Q) := I(Q) + B(Q)/2 − 1.

The claim is Pick(Q) = area(Q). We first show Pick is additive: if Q = Q₁ ∪ Q₂ where Q₁ and Q₂ are lattice polygons meeting along a single shared lattice edge e (and disjoint interiors), then

Pick(Q) = Pick(Q₁) + Pick(Q₂).

Proof of additivity. Let the shared edge e contain k ≥ 2 lattice points (its two endpoints plus k − 2 interior-to-the-edge points). Count contributions:

• Interior points. Every lattice point interior to Q is either interior to Q₁, interior to Q₂, or one of the k − 2 points strictly inside edge e (these become interior to Q after gluing). So

  I(Q) = I(Q₁) + I(Q₂) + (k − 2).
  

• Boundary points. The k points of e are boundary points of both Q₁ and Q₂, but after gluing, the k − 2 interior-to-e points are no longer on ∂Q; only the 2 endpoints of e remain on ∂Q. Every other boundary point of Q₁ or Q₂ stays on ∂Q. Hence

  B(Q) = B(Q₁) + B(Q₂) − 2(k − 2) − 2.
  

  (We subtract the k − 2 shared interior-edge points twice — once from each piece — and subtract 2 because the 2 shared endpoints were counted in both pieces but appear once in Q.)

Now compute:

Pick(Q₁) + Pick(Q₂)
  = I(Q₁) + I(Q₂) + (B(Q₁) + B(Q₂))/2 − 2
  = [I(Q) − (k − 2)] + [B(Q) + 2(k − 2) + 2]/2 − 2
  = I(Q) − (k − 2) + B(Q)/2 + (k − 2) + 1 − 2
  = I(Q) + B(Q)/2 − 1
  = Pick(Q).                                              ∎ (additivity)

Since area is trivially additive (area(Q) = area(Q₁) + area(Q₂)), proving Pick = area for the building blocks (triangles, then primitive triangles) extends to all lattice polygons by induction on the triangulation.

Step 2 — Every lattice polygon triangulates into lattice triangles, each into primitive triangles¶

A simple polygon admits a triangulation using only its vertices (ear-clipping); for a lattice polygon every triangle is a lattice triangle. Any lattice triangle that contains an extra lattice point (interior or on an edge) can be subdivided at that point into smaller lattice triangles. This terminates because area strictly decreases and is bounded below by 1/2. The result is a triangulation of P into primitive lattice triangles. By additivity (Step 1), it suffices to prove Pick for one primitive triangle.

Step 3 — A primitive triangle has area 1/2, satisfying Pick¶

For a primitive triangle T: I(T) = 0, B(T) = 3, so

Pick(T) = 0 + 3/2 − 1 = 1/2.

We must show area(T) = 1/2. The two edge vectors u, v from one vertex form a basis of ℤ² (Definition 1.1 / proved in Section 3), so |det(u, v)| = 1, and area(T) = ½|det(u, v)| = 1/2. Thus Pick(T) = area(T) for primitive triangles, and by Steps 1–2, for all lattice polygons. ∎ (Pick's theorem)

The one remaining gap is why a primitive triangle's edge vectors form a unimodular basis — i.e. why "no interior or edge lattice points" forces |det| = 1. That is the primitive-triangle lemma, which we prove next via Euler's formula (an alternative route that also re-derives Pick globally).

3. Proof of the Primitive-Triangle Lemma (Euler's Formula)¶

Lemma 3.1. A lattice triangle with no lattice points other than its 3 vertices has area exactly 1/2.

Proof (via a unit-cell / determinant argument). Let the triangle have vertices 0, u, v (translate so one vertex is the origin). The lattice points are exactly {0, u, v}, so the parallelogram {αu + βv : 0 ≤ α, β < 1} contains no lattice point except 0 (any such point would lie in the triangle or its reflected copy, contradicting primitivity). A fundamental domain of the sublattice ℤu + ℤv containing no other point of ℤ² means ℤu + ℤv = ℤ², i.e. {u, v} is a basis of ℤ², hence |det(u, v)| = 1 and area = ½|det| = 1/2. ∎

Global re-derivation of Pick via Euler's formula¶

Triangulate P into T primitive triangles. View this triangulation as a planar graph with V vertices, E edges, F faces. Euler's formula for a connected planar graph:

V − E + F = 2.

Here F = T + 1 (the T triangles plus the unbounded outer face), and V = I + B (all lattice points are triangulation vertices). Count edge incidences: each triangle has 3 edges; interior edges are shared by 2 triangles, boundary edges by 1 triangle and the outer face. Let E_in and E_bd be interior and boundary edges. Then E_bd = B (the polygon boundary has B lattice points, hence B unit boundary edges) and counting 3T = 2E_in + E_bd gives 2E = 3T + B (since E = E_in + E_bd). Substitute into Euler:

(I + B) − E + (T + 1) = 2
⟹  (I + B) − (3T + B)/2 + T + 1 = 2
⟹  I + B − (3T)/2 − B/2 + T + 1 = 2
⟹  I + B/2 − T/2 − 1 = 0
⟹  T = 2I + B − 2.

Each primitive triangle has area 1/2, so A = T/2 = I + B/2 − 1. ∎ (Pick via Euler)

This Euler-formula route is the most-cited proof: it shows Pick's −1 is precisely the Euler characteristic χ = 1 of a disk, foreshadowing the hole-corrected form (Section 8) and the Ehrhart constant term (Section 6).

4. The gcd Boundary Count — Proof¶

Proposition 4.1. The number of lattice points on the closed segment from p = (x₁, y₁) to q = (x₂, y₂), with p, q ∈ ℤ², is gcd(|x₂ − x₁|, |y₂ − y₁|) + 1. Excluding one endpoint, the segment contributes exactly g = gcd(|dx|, |dy|) to the polygon's boundary count.

Proof. Let dx = x₂ − x₁, dy = y₂ − y₁, g = gcd(|dx|, |dy|). A point on the segment is p + λ(dx, dy) for λ ∈ [0,1]. It is a lattice point iff λ·dx ∈ ℤ and λ·dy ∈ ℤ. Write dx = g·a, dy = g·b with gcd(a, b) = 1. Then λ·g·a ∈ ℤ and λ·g·b ∈ ℤ. Since gcd(a,b)=1, there exist integers with ma + nb = 1, so λg = λg(ma+nb) = m(λga) + n(λgb) ∈ ℤ. Thus λg ∈ ℤ, i.e. λ ∈ {0, 1/g, 2/g, …, 1}. Conversely each such λ = j/g gives an integer point (x₁ + j·a, y₁ + j·b). There are g + 1 values j = 0,…,g, giving g + 1 lattice points including both endpoints, hence g excluding one. ∎

Corollary 4.2 (Total boundary). Summing g over the n edges of a simple polygon counts each vertex exactly once (each is the excluded endpoint of one edge and the included endpoint of the next), so B = Σ_e gcd(|dx_e|, |dy_e|). ∎

5. Worked Triangulation Trace¶

Triangulate the quadrilateral P = (0,0), (4,0), (4,3), (0,3) (a 4×3 rectangle) and verify additivity and the primitive count.

(0,3) ---- (4,3)
  |  \        |
  |    \      |     diagonal (0,0)-(4,3) splits P into T1, T2
  |      \    |
(0,0) ---- (4,0)

Rectangle totals (direct):

A = 12,  B = 2*(4) + 2*(3) = 14,  I = A − B/2 + 1 = 12 − 7 + 1 = 6.

Split by diagonal (0,0)–(4,3) into T1 = (0,0),(4,0),(4,3) and T2 = (0,0),(4,3),(0,3).

T1:  2A = |4*3 − 0*0 + ...| = 12,  A = 6
     B = gcd(4,0)+gcd(0,3)+gcd(4,3) = 4 + 3 + 1 = 8
     I = 6 − 8/2 + 1 = 3
T2:  by symmetry  A = 6,  B = 8,  I = 3

Additivity check. Shared edge (0,0)–(4,3) has gcd(4,3)+1 = 2 lattice points (just the endpoints, k = 2):

I(P) = I(T1) + I(T2) + (k − 2) = 3 + 3 + 0 = 6        ✓
B(P) = B(T1) + B(T2) − 2(k−2) − 2 = 8 + 8 − 0 − 2 = 14 ✓
A(P) = 6 + 6 = 12                                      ✓

Now triangulate T1 fully into primitive triangles. T1 has area 6, so it must consist of 2A = 12 primitive triangles (each area 1/2). Indeed T = 2I + B − 2 = 2·3 + 8 − 2 = 12. The Euler-formula relation T = 2I + B − 2 is confirmed numerically, closing the loop with Section 3.

6. Ehrhart Theory — Pick Generalized¶

Pick's theorem is the 2-D shadow of a far more general phenomenon discovered by Eugène Ehrhart (1962).

Definition 6.1 (Ehrhart counting function). For a d-dimensional lattice polytope P and a positive integer t, let tP = {t·x : x ∈ P} be the t-fold dilation, and define

L_P(t) = | tP ∩ ℤ^d |   (lattice points in the scaled copy).

Theorem 6.2 (Ehrhart). L_P(t) agrees with a polynomial of degree d in t for all integers t ≥ 1:

L_P(t) = vol(P)·t^d + … + 1.

The leading coefficient is the volume; the constant term is the Euler characteristic (= 1 for a convex polytope). For d = 2:

L_P(t) = A·t² + (B/2)·t + 1.

Connecting to Pick. Evaluate at t = 1:

L_P(1) = A + B/2 + 1.

But L_P(1) counts all lattice points in P (interior + boundary) = I + B. Therefore

I + B = A + B/2 + 1  ⟹  A = I + B/2 − 1,

which is exactly Pick's theorem. So Pick is the t = 1 instance of the 2-D Ehrhart polynomial, and the polynomial's three coefficients are precisely (A, B/2, 1) — area, half-boundary, Euler characteristic.

Theorem 6.3 (Ehrhart–Macdonald reciprocity). Interior lattice points satisfy

L_P(−t) = (−1)^d · L_{P°}(t),

where P° is the open interior. For d = 2, L_P(−1) = L_{P°}(1) = I. Check: L_P(−1) = A − B/2 + 1 = I. Reciprocity is the rearranged Pick formula. This duality between boundary-inclusive and interior counts is the deep structural reason Pick has the symmetric I ↔ A form.

Ehrhart theory thus subsumes Pick, scales to all dimensions, and — via Barvinok's algorithm — gives polynomial-time lattice-point counting in fixed dimension. The naive-but-important caveat: the coefficient interpretation (area, half-boundary, 1) that makes Pick so clean does not survive into dimension 3, as the next section shows.

7. Worked Ehrhart Computation¶

Make the Ehrhart polynomial concrete on the unit square S = [0,1]² (vertices (0,0),(1,0),(1,1),(0,1)), then on a non-trivial triangle, and verify the coefficient interpretation L(t) = A t² + (B/2) t + 1.

7.1 The unit square¶

A = 1, B = 4, so the predicted Ehrhart polynomial is

L_S(t) = 1·t² + (4/2)·t + 1 = t² + 2t + 1 = (t + 1)².

Check by direct counting. The dilation tS = [0,t]² is a t×t axis-aligned square; its lattice points are {0,1,…,t}², of which there are (t+1)². ✓

At t = 1: L_S(1) = 4 = I + B = 0 + 4. ✓ — this is Pick for the unit square.

Reciprocity: L_S(−1) = (−1+1)² = 0 = I. The square has no interior lattice points, confirming L_S(−1) = I = 0. ✓

7.2 A triangle with interior points¶

Take T = (0,0), (4,0), (0,3) (from junior.md): A = 6, B = 8, I = 3.

L_T(t) = 6 t² + 4 t + 1.

The doubled triangle 2T = (0,0),(8,0),(0,6) has A' = 24, B' = gcd(8,0)+gcd(8,6)+gcd(0,6) = 8 + 2 + 6 = 16, I' = 24 − 8 + 1 = 17, total I' + B' = 33. ✓ matches L_T(2).

This is the payoff of Ehrhart: the same coefficients (A, B/2, 1) predict the lattice count of every integer dilation, and Pick's theorem is just the t = 1 (and reciprocity the t = −1) evaluation of that one quadratic.

8. Higher Dimensions — The Reeve Tetrahedron¶

A natural hope: in 3-D, volume should be some clean function of interior and boundary lattice points, mirroring Pick. It is provably false, and the counterexample is the Reeve tetrahedron (Reeve, 1957).

Definition 7.1 (Reeve tetrahedron). For an integer r ≥ 1, let

T_r = conv{ (0,0,0), (1,0,0), (0,1,0), (1,1,r) }.

Lattice-point count. For every r ≥ 1:

Interior lattice points  I(T_r) = 0,
Boundary lattice points  B(T_r) = 4   (just the four vertices).

That the four vertices are the only lattice points (no edge, face, or interior points) can be checked directly: every edge vector is primitive, and the slanted face (1,0,0),(0,1,0),(1,1,r) contains no interior lattice point for any r.

Volume.

vol(T_r) = (1/6) | det[ (1,0,0), (0,1,0), (1,1,r) ] | = r/6.

The contradiction. I and B are constant (0 and 4) for all r, yet the volume r/6 takes infinitely many distinct values. No formula vol = f(I, B) can output infinitely many different numbers from a single fixed input (I, B) = (0, 4). Therefore no Pick-type theorem exists in 3 dimensions. ∎

Why it fails — the missing data. In 2-D, "no interior points, three boundary points" pins down a primitive triangle of area exactly 1/2. In 3-D, the analogous "empty" tetrahedron (the empty / unimodular-vs-non-unimodular distinction) can have arbitrarily large volume because lattice emptiness does not bound determinant. The fix that does work is Ehrhart's: count lattice points in dilations and read volume off the leading coefficient of the Ehrhart polynomial — but you need the whole polynomial (multiple t values), not just (I, B). Reeve himself introduced these tetrahedra to motivate the Ehrhart approach.

This is the single most important caveat a professional must carry: Pick is intrinsically 2-D. Counting lattice points in higher dimensions requires Ehrhart polynomials / Barvinok's algorithm, not a (I, B)-style identity.

9. Polygons with Holes — Euler-Characteristic Form¶

Theorem 8.1. For a polygon P with h holes (a region homotopy-equivalent to a disk with h punctures),

A = I + B/2 − 1 + h,

where B counts lattice points on all boundary components (outer + the h inner loops), and I counts lattice points in the material region (excluding hole interiors).

Proof sketch. Re-run the Euler-formula derivation of Section 3 with the region's Euler characteristic χ = 1 − h instead of 1. The −1 in Pick was exactly −χ for the disk; replacing χ = 1 by χ = 1 − h turns it into −(1 − h) = −1 + h. ∎

This makes precise the senior-level rule that "each hole adds +1," and exposes the unifying theme: the additive constant in every Pick/Ehrhart statement is an Euler characteristic.

10. Minkowski, Determinants, and the Primitive Lattice¶

The primitive-triangle area 1/2 is one face of a broader fact about lattices that is worth stating precisely, because it is what fails in 3-D.

Theorem 10.1 (Determinant of a sublattice). Let u, v ∈ ℤ² be linearly independent. The sublattice L = ℤu + ℤv has index [ℤ² : L] = |det(u, v)| in ℤ². Equivalently, the fundamental parallelogram spanned by u, v has area |det(u, v)| and contains exactly |det(u, v)| lattice points in a half-open fundamental domain.

Corollary 10.2. A lattice triangle with edge vectors u, v has area ½|det(u, v)|, and it is primitive (empty of other lattice points) iff |det(u, v)| = 1, i.e. iff {u, v} is a unimodular basis of ℤ². This is the rigorous version of Lemma 3.1.

Connection to Minkowski's theorem. Minkowski's convex-body theorem states that a centrally symmetric convex set of area > 4 contains a nonzero lattice point. A primitive triangle's "doubled" symmetric hull has area 2·(1/2)·2 = 2 < 4, consistent with containing no nonzero lattice point — the boundary case that makes 1/2 the minimal positive lattice-triangle area. The deep statement is:

In 2-D, lattice emptiness pins down the determinant (= ±1), hence the area (= 1/2). In 3-D, it does not — there exist empty tetrahedra of arbitrarily large volume (Section 8). This single dimensional difference is exactly why Pick exists in 2-D and dies in 3-D.

Unimodular invariance (restated). Any M ∈ GL₂(ℤ) (det = ±1) maps ℤ² bijectively to itself, so it preserves I, B, and A of every lattice polygon. Combined with Theorem 10.1, a general integer matrix M scales area by |det M| and maps ℤ² onto the sublattice of index |det M|; counting lattice points then requires dividing by that index — the bridge to counting points in arbitrary affine-image polygons.

11. Complexity and Exactness¶

Shoelace 2A:        O(n) integer mults/adds; values up to n·C²
Boundary B:         O(n · log C) (one Euclidean gcd per edge)
Pick I:             O(1) after the above
Total:              O(n · log C),  independent of polygon AREA.

Exactness theorem. For a lattice polygon, 2A, B, and 2A − B + 2 are integers, and 2A − B + 2 is even, so I = (2A − B + 2)/2 ∈ ℤ is computed with no rounding. Proof of evenness: Pick gives 2A = 2I + B − 2, so 2A − B + 2 = 2I, manifestly even. ∎ The parity check (2A − B + 2) mod 2 = 0 is therefore a necessary condition for valid lattice input and a cheap corruption detector.

Overflow bound. |2A| ≤ Σ|x_i y_{i+1} − x_{i+1} y_i| ≤ 2n·C². Choose 64-bit when 2n·C² < 2^{63}; otherwise use big integers (see senior.md). Vertex-shifting by v₀ reduces effective C to the polygon's diameter, often re-enabling 64-bit.

11b. The Three Proof Strategies — Compared¶

We have given (or sketched) three independent proofs of Pick's theorem. A professional should know which to reach for in which context.

All three rely on the same atomic fact — a primitive lattice triangle has area 1/2 — and on additivity over a triangulation. The differences are pedagogical: the Euler route best reveals why the constant is −1 (topology), the additivity route best reveals how it scales (Ehrhart), and the elementary route best reveals that nothing deep is needed (arithmetic). The unifying invariant across all three is unimodularity (det = ±1), which is also what fails in dimension 3.

A note on numerical correctness. Because every quantity (2A, B, T, 2A − B + 2) is an integer, all three proofs hold exactly in integer arithmetic — there is no epsilon, no tolerance, no floating-point subtlety. This is rare in computational geometry, where most identities (orientation, in-circle, segment intersection) require careful exact-arithmetic or interval techniques to be provably correct. Pick's lattice restriction buys total numerical robustness, and the only failure mode is integer overflow (a magnitude bound, |2A| ≤ 2n·C²), never round-off.

12. Comparison with Alternatives¶

The hierarchy: Pick ⊂ 2-D Ehrhart ⊂ general Ehrhart ⊂ Barvinok counting. Grid scan is the brute-force oracle; the Reeve tetrahedron is the negative result that forces the jump from Pick to Ehrhart in dimension ≥ 3.

12b. Elementary Inductive Proof — Rectangles and Right Triangles¶

For completeness, here is a self-contained proof avoiding both Euler's formula and the abstract primitive-triangle machinery. It builds Pick from three explicit cases, each verified by hand, then glued by additivity (Section 2).

Case 1 — Axis-aligned rectangle. A rectangle with integer width w and height h:

A = w·h.
B = 2w + 2h          (perimeter lattice points; corners shared).
I = (w−1)(h−1)       (interior grid points).
Pick:  I + B/2 − 1 = (w−1)(h−1) + (w + h) − 1
                   = wh − w − h + 1 + w + h − 1 = wh = A.   ✓

Case 2 — Right triangle (legs axis-aligned). Cut an w×h rectangle along its diagonal into two congruent right triangles. The diagonal from (0,0) to (w,h) carries gcd(w,h) − 1 interior-to-diagonal lattice points plus 2 endpoints. Let d = gcd(w,h). For one right triangle:

B_tri = w + h + d            (two legs of gcd w and h, plus the hypotenuse's d).
By symmetry the two triangles split the rectangle's interior and the
diagonal's interior points:
   2·I_tri + (d − 1) = I_rect = (w−1)(h−1)
   ⟹  I_tri = [(w−1)(h−1) − (d − 1)] / 2.
Pick:  I_tri + B_tri/2 − 1
     = [(w−1)(h−1) − d + 1]/2 + (w + h + d)/2 − 1
     = [wh − w − h + 1 − d + 1 + w + h + d − 2]/2
     = wh/2 = A_tri.          ✓

Case 3 — Arbitrary triangle. Any lattice triangle embeds in its bounding box; subtract the (at most three) right triangles and (at most one) rectangle that fill the box around it. Each piece satisfies Pick (Cases 1–2), and additivity (Section 2) gives Pick for the arbitrary triangle. Triangulating any simple lattice polygon into such triangles extends Pick to all polygons. ∎

This route is the one usually taught first because every step is a concrete arithmetic identity — no Euler's formula, no determinant theory — at the cost of more case-bookkeeping.

12c. Algorithmic Lattice-Point Counting Beyond Pick¶

Pick answers "how many lattice points" only for simple 2-D lattice polygons. The general computational problem — count |P ∩ ℤ^d| for a rational polytope P in dimension d — has a rich algorithmic theory.

Barvinok's algorithm (1994). For fixed dimension d, the number of lattice points in a rational polytope can be computed in polynomial time, via signed decompositions of cones into unimodular cones and generating functions Σ_{x ∈ P ∩ ℤ^d} z^x. Pick is the trivial d = 2, single-polytope shadow; Barvinok is the general engine (implemented in LattE, barvinok).

The generating-function viewpoint. Encode lattice points as a rational generating function. For an interval [0, n] ⊂ ℤ, Σ z^k = (1 − z^{n+1})/(1 − z). Polytope generating functions multiply/combine these; evaluating at z → 1 recovers the count, and the Ehrhart polynomial is the lattice-point count of the dilation, read from the same function. Pick's (A, B/2, 1) are the first three Taylor-like data of the 2-D case.

The last row is the boundary: counting lattice points is #P-hard when the dimension is not fixed, so no fully general efficient formula exists. Pick's elegance is a low-dimensional luxury.

12d. Verifying the Reeve Tetrahedron Computationally¶

To make the impossibility concrete, here is the explicit check that T_r = conv{(0,0,0),(1,0,0),(0,1,0),(1,1,r)} is lattice-empty (only its 4 vertices) for any r, while its volume grows. A point p = (x,y,z) is in T_r iff its barycentric coordinates are all in [0,1] and sum to ≤ 1.

Express p = α(1,0,0) + β(0,1,0) + γ(1,1,r),  with α,β,γ ≥ 0, α+β+γ ≤ 1.
Then:  x = α + γ,  y = β + γ,  z = r·γ.
So:    γ = z/r,  α = x − z/r,  β = y − z/r.

For an INTEGER interior/boundary point with 0 ≤ z ≤ r:
   γ = z/r ∈ [0,1].  For α,β ≥ 0 we need x,y ≥ z/r.
   The constraint α+β+γ ≤ 1 forces x + y − z/r ≤ 1.
   With x,y,z integers and 0 < z < r, z/r is a non-integer in (0,1),
   so x,y ≥ z/r ⟹ x,y ≥ 1, giving x + y ≥ 2 > 1 + z/r — contradiction.
   Hence no integer point exists with 0 < z < r.
   For z = 0: the triangle (0,0,0),(1,0,0),(0,1,0) — only its 3 vertices are lattice.
   For z = r: the apex (1,1,r) only.

So I(T_r) = 0, B(T_r) = 4 for every r ≥ 1, confirmed symbolically. The volume vol(T_r) = r/6 is computed from the scalar triple product of the three edge vectors from the origin: det[(1,0,0),(0,1,0),(1,1,r)] = r. Constant counts, unbounded volume — no (I,B) formula can reproduce this, which is the whole point. (A brute-force scan of the bounding box [0,1]×[0,1]×[0,r] reaches the same conclusion numerically and is a good test-case for any lattice-counting code.)

12e. The gcd Trick, Continued Fractions, and the Three-Distance Theorem¶

The boundary-count gcd is the visible tip of a deep number-theoretic structure.

Visibility and Farey fractions. A lattice point (p, q) is visible from the origin (no lattice point strictly between it and 0) iff gcd(p, q) = 1. These are exactly the points the gcd trick "skips over." Ordered by slope in [0,1], the visible points are the Farey sequence F_Q = { p/q : 0 ≤ p ≤ q ≤ Q, gcd(p,q)=1 }.

Stern–Brocot mediants. Between two adjacent Farey fractions p/q < p'/q' (with p'q − pq' = 1, i.e. unimodular), the next fraction to appear is the mediant (p+p')/(q+q'). This is precisely the unimodular-basis condition (Section 10) — adjacent Farey fractions correspond to primitive triangles (0,0),(q,p),(q',p') of area 1/2. So the Stern–Brocot tree is a tree of primitive lattice triangles, and Pick's atom (area 1/2) is its node.

Continued fractions and the Euclidean recursion. Counting lattice points under the line y = (a/b)x reduces, via the floor sum Σ floor(ax/b), to a Euclidean-style recursion that mirrors the continued-fraction expansion of a/b. The same gcd(a,b) that counts boundary points drives the recursion depth O(log b).

Three-distance (Steinhaus) theorem. Placing points {kα mod 1} on a circle leaves at most three distinct gaps; the connection to lattice geometry is that these gaps are governed by the continued-fraction convergents of α — the same convergents that appear when counting lattice points near a line of slope α. This ties Pick-adjacent lattice counting to equidistribution theory.

These connections explain why an elementary dot-counting formula sits at the crossroads of geometry, number theory, and the theory of continued fractions: the common thread is unimodularity — determinant ±1, coprimality, primitive triangles, Farey neighbors — all the same condition wearing different clothes.

13. Open Threads and Connections¶

1. Lattice-width and flatness. The flatness theorem bounds the lattice width of empty convex bodies; it is the higher-dimensional structure that Reeve tetrahedra exploit and that Pick (trivially) satisfies in 2-D.
2. Coefficient positivity of Ehrhart polynomials. Whether all coefficients are non-negative is false in general (dimension ≥ 3) but holds for special polytopes; the 2-D coefficients (A, B/2, 1) are always positive — another way 2-D is special.
3. Counting lattice points under a line / Stern–Brocot. The gcd boundary count is the same coprimality that generates the Farey sequence and the Stern–Brocot tree; lattice-point counting under y = (a/b)x connects to the three-distance theorem and continued fractions.

4. Toric geometry. Lattice polygons correspond to toric surfaces; Pick's A and B become the self-intersection and degree data of divisors — a bridge from this elementary identity to algebraic geometry. Specifically, a lattice polygon P defines a polarized toric surface (X_P, L) with L² = 2A (twice the area) and arithmetic genus tied to I (the interior points are the dimension of the space of sections of the adjoint bundle). The Riemann–Roch theorem on X_P is Pick's theorem dressed in cohomological language — the most striking instance of an elementary count reappearing as a deep theorem.

5. Scissors congruence. Two lattice polygons of equal area are scissors-congruent via lattice-preserving cuts; Pick's additivity (Section 2) is the discrete shadow of the Dehn-invariant-free 2-D scissors theory, contrasting with dimension 3 where the Dehn invariant (and the Reeve phenomenon) obstruct any such clean equivalence.

6. Empty simplices and the width spectrum. Classifying empty lattice simplices (lattice-point-free, like the Reeve tetrahedron) in dimension d ≥ 3 is an active research area; their volumes are unbounded in d = 3 but the normalized volume is constrained by the lattice width, and a full classification is known only in low dimensions. This is the precise mathematical statement of "why no 3-D Pick," refined.

7. Coefficient interpretation in higher dimensions. While the 2-D Ehrhart coefficients are exactly (A, B/2, 1), in dimension d ≥ 3 only the leading (volume) and constant (= 1) coefficients have a clean universal meaning; the intermediate coefficients are subtle invariants (related to the polytope's faces and the Todd class in the toric dictionary). Pick's tidy three-term reading is, once more, a 2-D luxury.

14. Summary¶

• Statement. A = I + B/2 − 1 for any simple lattice polygon; integer form I = (2A − B + 2)/2.
• Proof. Additive Pick functional + triangulation into primitive triangles (area 1/2), or equivalently Euler's formula V − E + F = 2, which yields T = 2I + B − 2 and exposes the −1 as the disk's Euler characteristic χ = 1.
• Boundary count. B = Σ gcd(|dx_e|, |dy_e|), proven via coprimality of (dx/g, dy/g).
• Generalization up. Ehrhart's L_P(t) = A t² + (B/2) t + 1 makes Pick the t = 1 slice; Ehrhart–Macdonald reciprocity L_P(−1) = I is the rearranged formula; this scales to all dimensions and to Barvinok's polynomial-time counting.
• Generalization sideways. Holes give A = I + B/2 − 1 + h, the constant being −χ.
• Hard limit. The Reeve tetrahedron conv{0, e₁, e₂, (1,1,r)} has I = 0, B = 4 for all r but volume r/6 — proving no (I, B)-style Pick formula can exist in 3-D. Higher dimensions require Ehrhart, not Pick.

Pick (1899) gave the identity; Reeve (1957) showed it cannot extend to 3-D; Ehrhart (1962) found the polynomial that does. A 125-year-old dot-counting formula that turns out to be the first term of one of the deepest series in discrete geometry.

Next step: Pick's Theorem — Interview Preparation