October 20, 2017

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In planar Euclidean geometry, Thales’ theorem states that any triangle pqr circumscribing a circle with one pair (p,q) of antipodal points is necessarily a right triangle. A pair (p,q) of antipodal points of a smooth convex object is such that the tangent lines at p and q are parallel to each other. See Figure 1 for an illustration, and [3] for a historical account (Thales of Miletus, 624–546 BC).

Theorem 1 (Thales’ circle theorem) Any triangle circumscribed by a circle with one side being a diameter is right-angle.

Let D_{A}(p,q) denote the Mahalanobis distance between two points p and q, for a positive definite matrix
A ≻ 0:

A Mahalanobis circle [2] C_{A}(c,r) of center c and radius r is defined as follows:

Let us generalize Thales’ theorem as follows:

Theorem 2 (Thales’ Mahalanobis circle theorem) Any triangle circumscribed by a Mahalanobis circle with one pair of points being antipodal is right-angle.

Proof:
Consider the Cholesky decomposition of A: A = LL^{⊤} = U^{⊤}U with L (U = L^{⊤}) a lower triangular matrix (an
upper triangular matrix, respectively) with positive diagonal elements. The Mahalanobis distance
amounts to calculate an ordinary Euclidean distance on affinely transformed points x′ = L^{⊤}x = Ux:

Thus a Mahalanobis circles C_{A} transforms affinely to a Euclidean circle C_{E} = C_{I}, and antipodal pairs of points
on C_{A} remain antipodal in C_{E}.

Two vectors u and v are perpendicular in the Mahalanobis geometry if and only if u^{⊤}Av = 0. That is, if
u^{⊤}Av = u^{⊤}LL^{⊤}v = (L^{⊤}u)^{⊤}L^{⊤}v = u′^{⊤}v′ = 0.

A triangle pqr circumscribing the Mahalanobis circle C_{A} with (p,q) an antipodal pair in Mahalanobis geometry
transforms into a triangle p′q′r′ circumscribing the Euclidean circle C′ = {L^{⊤}x : x ∈ C_{A}(c,r)} with (p′,q′) an
antipodal pair. Therefore p′q′r′ is a right-angle triangle in Euclidean geometry, and:

Note that Mahalanobis geometry is not conformal when A ⁄= λI (for λ > 0), the scaled identity matrix. Therefore angles are not preserved in Mahalanobis geometry: That is, a Mahalanobis right-angle cannot be visualized as a Euclidean right-angle in general.

Squared Mahalanobis distances are the only symmetric Bregman divergences [1]. But Thales’ theorem do not extend to other (asymmetric) Bregman divergences.

[1] Jean-Daniel Boissonnat, Frank Nielsen, and Richard Nock. Bregman Voronoi diagrams. Discrete & Computational Geometry, 44(2):281–307, 2010.

[2] Frank Nielsen and Richard Nock. On the smallest enclosing information disk. Information Processing Letters, 105(3):93–97, 2008.

[3] Christoph J Scriba and Peter Schreiber. Geometry in the Greek-Hellenistic era and late antiquity. In 5000 Years of Geometry, pages 27–116. Springer, 2015.