41st IMO shortlist 2000/G4 solution

41st IMO 2000 shortlist

Problem G4

Show that a convex n-gon can be inscribed in a circle iff there is a pair of real numbers a_i, b_i associated with each vertex P_i such that the distance between each pair of vertices P_iP_j with i < j is a_ib_j - a_jb_i.

Solution

Suppose the real numbers exist. By Ptolemy's theorem P₁P₂P₃P₄ is cyclic if (P₁P₃)(P₂P₄) = (P₁P₂)(P₃P₄) + (P₂P₃)(P₁P₄). But rhs = (a₁b₂ - a₂b₁)(a₃b₄ - a₄b₃) + (a₂b₃ - a₃b₂)(a₁b₄ - a₄b₁) = a₁a₃ = a₁a₃b₂b₄ - a₁a₄b₂b₃ - a₂a₃b₁b₄ + a₂a₄b₁b₃ + a₁a₂b₃b₄ - a₂a₄b₁b₃ - a₁a₃b₂b₄ + a₃a₄b₁b₂ = a₁a₂b₃b₄ - a₁a₄b₂b₃ - a₂a₃b₁b₄ + a₃a₄b₁b₂ = lhs. So P₁P₂P₃P₄ is cyclic. Similarly, P₁P₂P₃P_i is cyclic for i = 5, 6, ... , n. So all of P₄, P₅, ... , P_n lie on the circle P₁P₂P₃, so the n-gon is cyclic.

Conversely suppose the n-gon is cyclic. Put a₁ = 0, b₁ = - P₁P₂, a₂ = 1, b₂ = 0, and a_i = P₁P_i/P₁P₂, b_i = P₂P_i for i = 3, 4, ... , n. If 2 < i < j, then P₁P₂P_iP_j is cyclic, so (P₁P_i)(P₂P_j) = (P₁P₂)(P_iP_j) + (P₁P_j)(P₂P_i). Dividing by P₁P₂, we get a_ib_j = P_iP_j + a_jb_i, as required. Also, a₁b₂ - a₂b₁ = 0 + P₁P₂, which is correct. Similarly, a₁b_i - a_ib₁ = 0 + (P₁P_i)/(P₁P₂) x (P₁P₂) = P₁P_i, which is correct. Finally, a₂b_i - a_ib₂ = P₂P_i, which is correct.

41st IMO shortlist 2000