Show that the area of the triangle bounded by the lines a_{i}x + b_{i}y + c_{i} = 0 (i = 1, 2, 3) is Δ^{2}/|2(a_{2}b_{3} - a_{3}b_{2})(a_{3}b_{1} - a_{1}b_{3})(a_{1}b_{2} - a_{2}b_{1})|, 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, A_{1} = b_{2}c_{3} - b_{3}c_{2}, A_{2} = b_{3}c_{1} - b_{1}c_{3}, A_{3} = b_{1}c_{2} - b_{2}c_{1}.]

The lines a_{2}x + b_{2}y + c_{2} = 0 and a_{3}x + b_{3}y + c_{3} = 0 intersect at (A_{1}/C_{1}, B_{1}/C_{1}). Similarly, the other two points of intersection are (A_{2}/C_{2}, B_{2}/C_{2}) and (A_{3}/C_{3}, B_{3}/C_{3}).

The area of the triangle is therefore (the absolute value of) the determinant K with rows A_{1}/C_{1}, B_{1}/C_{1}, 1; A_{2}/C_{2}, B_{2}/C_{2}, 1; A_{3}/C_{3}, B_{3}/C_{3}, 1. (*) [For example, take a z-coordinate perpendicular to the plane, and take the cross product of the vectors along two sides.] But K |C_{1}C_{2}C_{3}| is the determinant whose elements are the cofactors of the original determinant. This has value equal to Δ^{2}. For example, on multiplying it by Δ, we get Δ down the diagonal and zeros elsewhere, and hence Δ^{3}.

© John Scholes

jscholes@kalva.demon.co.uk

15 Sep 1999