ABC is a triangle and k > 0. Take A' on BC, B' on CA, C' on AB so that AB' = k B'C, CA' = k A'B, BC' = k C'A. Let the three points of intersection of AA', BB', CC' be P, Q, R. Show that the area PQR (k2 + k + 1) = area ABC (k - 1)2.
Solution
There does not seem to be a geometric solution, so one is faced with a messy algebraic calculation. There are various ways of doing it. The worst is probably to use ordinary Cartesian coordinates.
Let us use points in the Argand diagram. Take A to be the origin and C to be the real point 1+k. Take B to be z(1+k). Then B' is k and A' is zk+1 (check: ( (zk+1) - (1+k) )/( (z(1+k) - (zk+1) ) = (zk-k)/(z-1) = k). We want to find R, the intersection of AA' and BB'. It lies on AA', so it is r(zk+1) for some 0 < r < 1. But it lies on BB', so it is k + s(zk+z-k) for some 0 < s < 1. After some manipulation, we get r = (k2+k)/(k2+k+1).
Now take k < 1. Then area ARB = (AR/AA') area AA'B = r area AA'B (this is where we need k < 1, because if k > 1, then the geometry changes). But area AA'B = (A'B/CB) area ABC, so area ARB = r/(1+k) area ABC = k/(k2+k+1) area ABC. Similarly, taking P as the intersection of BB' and CC' and Q as the intersection of AA' and CC', we have that area AQC = k/(k2+k+1) area ABC, and area CPB = k/(k2+k+1) area ABC. Adding we get (area ABC - area PQR), so area PQR/area ABC = 1 - 3k/(k2+k+1) = (1 - k)2/(k2+k+1).
If k > 1, then we get the result with 1/k replacing k, but multiplying through by k2 gives the same formula.
© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002