Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The arc length of a unit semicircle is pi

Statement

Every once-traversed semicircle of radius 1 has arc length π. In particular, the upper unit semicircle γ(t)=(cos⁡t,sin⁡t), 0≤t≤π, has length π.

Facts & Assumptions

Given: The path γ(t)=(cos⁡t,sin⁡t) on [0,π].

[L2]

The functions sine and cosine are differentiable, with (sin⁡t)′=cos⁡t and (cos⁡t)′=−sin⁡t (The derivatives of sine and cosine are cosine and minus sine).

[L3]

For every real t, sin⁡2t+cos⁡2t=1 (Parity and the Pythagorean identity for sine and cosine).

[L6]

Path length is invariant under every continuous surjective monotone reparametrization (Arc length is invariant under every continuous surjective monotone reparametrization, including pauses and reversal).

Proof

technique · direct
1.1givenL1L2

By [L1] and [L2], γ′(t)=(−sin⁡t,cos⁡t) on [0,π].

2.1step 1.1L3algebra

By [L3], ∥γ′(t)∥2=sin⁡2t+cos⁡2t=1.

3.1step 2.1L4L5

By [L4] and [L5], L(γ)=∫0π1 dt=π.

4.1step 3.1L6algebra∎

Translating or rotating the displayed path does not change the differences between its points, and reversing or monotonically reparametrizing it does not change its length by [L6]. Thus every once-traversed unit semicircle has length π.

Depends on

Dependency tree · 2 levels

Sources