3rd Mexico 1989 Problem A1 The triangle ABC has AB = 5, the medians from A and B are perpendicular and the area is 18. Find the lengths of the other two sides.
2√13, √73 Solution Let the medians be AD and BE, meeting at F. Let G be the foot of the perpendicular from F to AB. F is the centroid, so it is one third of the way along towards C on the median from C, and hence FG is one third of the height of the triangle. Hence area AFB = 6, so FG = 2.4. Let AG = x. Then since AFG, FGB are similar, we have FG/AG = FG/FB or x(5-x) - 5.76. Solving the quadratic, x = 1.8 or 3.2. wlog we can take AG = 3.2, so FGA is a 3,4,5 triangle and hence AF = 4. Similarly BF = 3. So EF = BF/2 = 1.5. Now by Pythagoras AE 2 = 4 2 + (3/2) 2 = 73/4. So AC = 2AE = √73. Similarly, BD2 = 3 2 + 2 2 = 13, so BC = 2√13 3rd Mexican 1989 © John Scholes jscholes@kalva.demon.co.uk 25 February 2004 Last corrected/updated 25 Feb 04