Thales circle theorem extended to Mahalanobis geometry

Frank Nielsen
Frank.Nielsen@acm.org

October 20, 2017

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1 Thales’ theorem in Euclidean geometry

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.

Figure 1: Thales’ circle theorem: Triangle pqr is right-angle at r where [pq] is a diameter. (p,q) is a pair of antipodal points.

2 Thales’ theorem in Mahanalobis geometry

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

∘ --------------- DA(p,q) = (p- q)⊤A(p - q) = ∥p- q∥A.

When A = I is the 2 × 2 identity matrix, the Mahalanobis distance amounts to the Euclidean distance D_E(p,q) = ∥p - q∥ = ∘(p-- q)⊤(p-- q)

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

CA (c,r) = {x : DA(c,x) = r}.

A Mahalanobis circle has an ellipsoid (Euclidean) shape.

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:

∘ ----------------- DA (p,q) = (p- q)⊤LL ⊤(p- q), = DE (L⊤p,L⊤q ).

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:

′ ′ ⊤ ′ ′ (q- r ) (p - r ) = 0, (1) (L⊤ (q - r))⊤L ⊤(p- r) = 0, (2) (q- r)⊤LL ⊤(p- r) = 0, (3) (q- r)A (p- r) = 0. (4)

Therefore, pqr is a right-angle triangle in Mahalanobis geometry. □

Figure 2: Thales’ circle theorem in Mahalanobis geometry: Triangle pqr is right-angle at r where [pq] is an antipodal pair of points. Notice that Mahalanobis geometry is generally not conformal so that a Mahalanobis right-angle does not visualize as a Euclidean right-angle.

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.

References

[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.