Polygon diagonals calculator

Definition

The number of diagonals of a convex (or simple) polygon with n vertices, n ≥ 3, is the number of segments joining two non-adjacent vertices. The combinatorial formula is D = n(n − 3)/2. Each vertex connects by diagonals to n − 3 others (all except itself and its two neighbours); summing n(n − 3) counts each diagonal twice, hence the division by 2.

For a triangle (n = 3) one obtains D = 0 — no diagonals. For a quadrilateral D = 2, for a pentagon D = 5, for a hexagon D = 9. MathWorld (Polygon Diagonal) gives the same formula. In Euclidean geometry, diagonals of a convex polygon lie inside the figure; with concavity some may go outside, but the vertex-pair counting formula remains the same. The calculator reports that combinatorial count for integer n ≥ 3.

Historical and conceptual context

Counting diagonals belongs to elementary geometric combinatorics and appears in classical school exercises and in Euclid’s Elements in the context of dissecting polygons into triangles. The number of ways to choose 2 vertices among n is C(n, 2) = n(n − 1)/2; among them n are sides, so the number of diagonals is C(n, 2) − n = n(n − 3)/2.

Conceptually D is a nonnegative integer. The calculator takes n as an integer ≥ 3; non-integers or n < 3 do not describe a polygon in this sense. The formula does not depend on side lengths or on regularity — only on the number of vertices and adjacency on the cycle. Side lengths and angles affect geometry, not this discrete count.

Formal properties

Domain: n ∈ ℤ, n ≥ 3. Result: D = n(n − 3)/2 ∈ {0, 1, 2, …}. Sequence: 0, 2, 5, 9, 14, 20, … (for n = 3, 4, 5, 6, 7, 8, …). Increment: D(n+1) − D(n) = n − 1 — adding a vertex adds n − 1 new diagonals after the cycle is updated.

Equivalent forms: D = C(n, 2) − n; D = (n² − 3n)/2. For a regular polygon the same D diagonals can be drawn in the circumscribed circle; interior intersections are a separate problem (number of crossing points), which this formula does not solve.

Triangulation: a convex n-gon splits into n − 2 triangles via n − 3 non-crossing diagonals from one vertex — a different count from D (the total number of all diagonals). Full D also includes diagonals that cross inside.

Full derivation of the formula D = n(n − 3)/2

Step 1. A polygon has n vertices joined in a cycle: each has exactly two neighbours along the boundary.

Step 2. From a fixed vertex one can draw a segment to n − 1 other vertices. Two of them are sides; so the number of diagonals from one vertex is n − 3.

Step 3. Summing over all vertices: n(n − 3). Each diagonal has two endpoints, so it was counted twice. Hence D = n(n − 3)/2.

Step 4 (combinatorial variant). The number of all segments among vertices is C(n, 2). Subtract n sides: C(n, 2) − n = n(n − 1)/2 − n = n(n − 3)/2. Both derivations are equivalent and standard in MathWorld and discrete-geometry texts.

Special cases

n = 3: D = 0. n = 4: D = 2 (quadrilateral diagonals). n = 5: D = 5. n = 6: D = 9. n = 10: D = 35. n = 100: D = 4850. Parity check: n(n − 3) is always even for integer n ≥ 3, so D is an integer.

As n grows, D grows quadratically (~ n²/2). For large n drawing all diagonals becomes dense; the formula still gives the correct count, independent of visualization. For n < 3 the formula may formally give a negative or fractional value — outside the model domain.

Numerical and computational notes

Computation: D = n * (n - 3) / 2 for integer n ≥ 3. In a computer program prefer integer arithmetic or check that n is an integer. Division by 2 is exact when n(n−3) is even — always for integer n.

Non-integer input should be rejected or rounded only if the page clearly says so; this model assumes a vertex count. Comma/period do not change the meaning if the value must be integral. For very large n the result may exceed the largest exactly representable integer (2⁵³ − 1) — then big-integer arithmetic is needed.

Related concepts and authorities

Related tools: sum of interior angles ((n−2)·180°), number of triangles in a triangulation (n−2), perimeter and area of a regular polygon. The diagonal count does not depend on the circumradius — unlike side lengths.

Authorities: Euclid, Elements (polygons and dissections); MathWorld — Polygon Diagonal; combinatorics: C(n, 2) − n; standard school geometry texts. The calculator does not replace an exam proof — it evaluates D = n(n−3)/2. A useful cross-check is to compute C(n, 2) and subtract n sides: the difference must equal the displayed diagonal count for every integer n ≥ 3.