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Statement. If and , there is exactly one orientation-preserving Euclidean isometry with and .
Source. Mathialagan, published 2021 PDF, p. 10, Proposition 20. The printed statement omits , although its proof uses it. The nonzero condition is part of the application to the paper's energy.
Proof. A proper Euclidean isometry has form , where is a planar rotation matrix. Indeed, after subtracting , preservation of squared distances and the polarization identity preserve inner products; the images of the standard orthonormal basis then determine an orthogonal linear map. Preservation of orientation selects determinant .
The two endpoint requirements force . Two equal-length nonzero vectors determine a unique rotation: their normalized directions specify its sine and cosine. This fixes , and then is forced. Conversely these choices meet both requirements. If , the motion is a translation, including the identity. Otherwise is invertible, since for its angle . Its unique fixed point is , so it is a nonidentity rotation about that point.
If the common length were zero, arbitrary rotations followed by the forced translation would satisfy the requirements, explaining the qualification.
Application. Each positive-energy quadruple has one proper motion with and , since the source segment and target segment have the same positive length. Partition these motions into translations and nonidentity rotations.
Verification scope. Verified within the independently reviewed Theorem 3 chain, retained in the final review; the corrected hypothesis and the motion classification are components of the living Theorem 3 record.
Bears on. Problem 661.