ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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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.
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The four midpoint subtriangles of the $0, 1, i$ triangle display all three cancelling interior edges

Example

For $a = 0$ , $b = 1$ , and $c = i$ , the side midpoints are

$p = \frac{1}{2}, q = \frac{1 + i}{2}, r = \frac{i}{2} .$

The midpoint subdivision, with the orientation of Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter, consists of $Δ [0, p, r]$ , $Δ [p, 1, q]$ , $Δ [r, q, i]$ , and $Δ [p, q, r]$ .

Facts & Assumptions

Given: The displayed vertices and midpoints, with every triangle carrying the orientation fixed by its ordered vertices.

[L1]

If $p, q, r$ are the side midpoints of $Δ [a, b, c]$ and $f$ is continuous on that filled triangle, then $I_{f} [a, b, c] = I_{f} [a, p, r] + I_{f} [p, b, q] + I_{f} [r, q, c] + I_{f} [p, q, r]$ (Midpoint subdivision of a triangle cancels every interior edge and preserves its outer boundary integral).

Verification

technique · direct

1.1given

The directed edge lists are $(0 \to p, p \to r, r \to 0)$ , $(p \to 1, 1 \to q, q \to p)$ , $(r \to q, q \to i, i \to r)$ , and $(p \to q, q \to r, r \to p)$ .

2.1step 1.1algebra

The interior segment pairs are $p \to r, r \to p$ , $q \to p, p \to q$ , and $r \to q, q \to r$ ; each pair consists of one directed edge and its reversal, so the formal oriented edges cancel.

3.1step 1.1step 2.1L1∎

The surviving half-edges concatenate as $0 \to p \to 1$ , $1 \to q \to i$ , and $i \to r \to 0$ , which is the positive outer boundary $0 \to 1 \to i \to 0$ ; for every continuous integrand, this is exactly the integral identity in [L1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Theorem 1.1 (standard reference, not scraped)

The four midpoint subtriangles of the 0,1,i0,1,i triangle display all three cancelling interior edges