Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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A triangle has zero Jordan content if and only if its vertices are collinear

Statement

A triangle has zero Jordan content if and only if its vertices are collinear.

Here collinear means that the displacement list (B−A,C−A) is linearly dependent.

Facts & Assumptions

Given: Vertices A,B,C∈R2.

[L1]

Every triangle T(A,B,C) has content 12∣det⁡[B−A C−A]∣ (A triangle has content 12∣det⁡[B−A C−A]∣, equal to half base times height when the chosen side is nonzero).

[L2]

For a real square matrix, ∣det⁡M∣ is the ordinary absolute value of its real determinant (For n≥1, the determinant over a commutative ring by the Leibniz formula, and ∣det⁡A∣ for a real matrix).

Proof

technique · direct
1.1L1L2L3algebra

For the forward implication from collinearity to zero content, dependence in [L3] makes one of the two displacement vectors a scalar multiple of the other, including when either is zero; the two columns then have determinant zero, so [L1] and [L2] give content zero.

2.1L1L2L3algebra∎

For the converse implication, suppose the content is zero. By [L1] and [L2], det⁡[B−A C−A]=0. If B−A=0 the list is dependent by [L3]; otherwise one coordinate of B−A=(u1,u2) is nonzero, and the equation u1v2−u2v1=0 for C−A=(v1,v2) shows by division in that nonzero coordinate that C−A is a scalar multiple of B−A. Thus [L3] gives collinearity.

Depends on

Used by

Dependency tree · two levels

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Sources