17th BMO 1981 Problem 1 ABC is a triangle. Three lines divide the triangle into four triangles and three pentagons. One of the triangle has its three sides along the new lines, the others each have just two sides along the new lines. If all four triangles are congruent, find the area of each in terms of the area of ABC.
The obvious configuration has ∠EDF = ∠EJI, so KH is parallel to AC and similarly LI parallel to BC, and GJ parallel to AB. Then to get equal area we need E,F to trisect LI and similarly for the other two lines. Suppose GH = kBC. Then GJ = 3kAB, so CG = 3kBC. Hence CH = 2kBC and similarly BG = 2kBC, so k = 1/5 and the area of each small triangle is 1/25 area ABC. However, there are other configurations. For example, suppose we take GJ parallel to AB and KH parallel to AC, but LI not parallel to BC. Take GH = a, DH = b, GD = c. Then GDH is evidently similar to BAC because of the parallel lines. We find DE = b, EJ = b, so GJ = 2b+c. GJC is similar to GDH, so GC = a(2b+c)/c. Similarly, DF = KF = c, so KH = b+2c. Hence BH = a(b+2c)/b. Hence BC = a(bc+2b 2 +2c 2 )/bc. So BC/GH = (bc+2b 2 +2c 2 )/bc and hence the area of a small triangle is 1/(1 + 2b/c + 2c/b) 2 times the area of ABC. Here the ratio can take any value less than 1/25 depending on the ratio b/c. 17th BMO 1981 © John Scholes jscholes@kalva.demon.co.uk 10 February 2004 Last corrected/updated 10 Feb 04