Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 66 2 i Solution Created 2026-10-03 Updated 2026-10-14
Put , , and . A point on each moving segment has the form and , with . Contact is therefore equivalent toThe contact time of translating line segments is the minimum feasible . This reduces the entire continuous-motion problem to three linear equations and linear inequalities.
In three-dimensional space form the matrix . If it is nonsingular, solveusing a stable linear solver. A candidate with in the unit interval and is the only contact event and hence the earliest one; otherwise no contact occurs. Check initial intersection explicitly, so an already touching pair returns time zero.
If is singular, use a rank-revealing factorization to test consistency. Write every solution as , with columns of spanning the null space, and minimize its time coordinate subject to the box and nonnegative-time constraints. This is a small linear programming problem. It handles parallel segments, collinear overlap, a continuum of contact times and segments degenerated to points. Infeasibility means no future contact. When , the relative geometry is stationary: initially disjoint segments never meet. The first contact is the least feasible nonnegative time, not just an intersection of the two supporting lines.