Putnam 40/A8 solution

3rd Putnam 1940

Problem A8

Show that the area of the triangle bounded by the lines a_ix + b_iy + c_i = 0 (i = 1, 2, 3) is Δ²/|2(a₂b₃ - a₃b₂)(a₃b₁ - a₁b₃)(a₁b₂ - a₂b₁)|, where D is the 3 x 3 determinant with columns a_i, b_i, c_i.

Solution

Fairly easy if you remember some formulae for determinants (which people did in those days). Of course, you can just slog through the expression (*) below in terms of a_i, b_i, c_i. That is doable, but completely mindless, the only required skill is doing elementary algebra fast without mistakes. I suppose I am fairly out of sympathy with the rather common Putnam style of problem where the basic idea is obvious, but you have to be skilful at evaluating integrals, determinants etc, often using tricks.

Take A_i be the cofactor of a_i in Δ. Similarly, B_i and C_i. [So, for example, A₁ = b₂c₃ - b₃c₂, A₂ = b₃c₁ - b₁c₃, A₃ = b₁c₂ - b₂c₁.]

The lines a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 intersect at (A₁/C₁, B₁/C₁). Similarly, the other two points of intersection are (A₂/C₂, B₂/C₂) and (A₃/C₃, B₃/C₃).

The area of the triangle is therefore (the absolute value of) the determinant K with rows A₁/C₁, B₁/C₁, 1; A₂/C₂, B₂/C₂, 1; A₃/C₃, B₃/C₃, 1. (*) [For example, take a z-coordinate perpendicular to the plane, and take the cross product of the vectors along two sides.] But K |C₁C₂C₃| is the determinant whose elements are the cofactors of the original determinant. This has value equal to Δ². For example, on multiplying it by Δ, we get Δ down the diagonal and zeros elsewhere, and hence Δ³.

3rd Putnam 1940