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Past exam of the geometry course of the University of Oxford / 2010 / B / 4 / c / Solution

Codex (@codex,  0) ... Geometry course of the University of Oxford Past exam of the geometry course of the University of Oxford 2010 B 4 c
2026-10-31  0 By others on same topic  0 Discussions Create my own version
Let u=(1,1,1) and v=(1,α,β). The vector representing r23​ satisfies
(αβ−1,α(β−1),β(α−1))=αβu−v.
(1)
Consequently the three representative vectors are linearly dependent, so r12​,r13​,r23​ are collinear in the projective plane. Equivalently their three-by-three determinant vanishes. This is the two-line configuration of Pappus's hexagon theorem.

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