The triangle ABC has area 1. Take X on AB and Y on AC so that the centroid G is on the opposite of XY to B and C. Show that area BXGY + area CYGX ≥ 4/9. When do we have equality?
Solution
Solution by Nizameddin Ordulu and Ali Adalý:
Answer: equality iff XY is parallel to BC and passes through G.
Area BXGY + area CXGY = 2 area XGY + area BXY + area CXY. Let M be the midpoint of BC. Then the distance of M from the line XY is the average of the distances of B and C, so area BXY + area CXY = 2 area MXY. Hence area BXGY + area CXGY = 2 area XGY + 2 area MXY = 2 area GXM + 2 area GYM = area AXG + area AYG (since AG = 2 GM).
Take B' on the side AB and C' on the side AC so that B'C' is parallel to BC and passes through G. AB'C' is similar to ABC with sides 2/3 as long, so area AB'C' = 4/9. Thus we wish to show that area AXG + area AYG ≥ area AB'C'. If X lies between B and B' and Y lies between C and C', then the distance of X from AG is greater than the distance of B', so area AXG > AB'G, and, similarly, area AYG > AC'G, so we are done. So assume X lies between A and B'. Let the ray XG meet AC at Y'. Then since G lies on the opposite side of XY to B and C, Y must lie between Y' and C. So area GYC' ≥ GY'C' and it is sufficient to show that area AXY' ≥ AB'C', or equivalently that area XGB' ≤ area Y'GC'.
By Menelaus' theorem applied to the triangle AB'C', we have (XB'/XA) (GC'/GB') (Y'C'/Y'A) = 1, or (XB'/XA) (Y'A/Y'C') = 1, since G is the midpoint of B'C'. So AB'/AC' ≥ AX/AY' = XB'/Y'C'. Let hB be the distance of G from AB and hC the distance of G from AC. Then (hBAB')/(hCAC') = area AGB'/area AGC' = 1, so hC/hB = AB'/AC'. Hence hC/hB ≥ XB'/Y'C', so area Y'GC' ≥ XGB' as required.
For equality we must have X = B' and Y = C'.
© John Scholes
jscholes@kalva.demon.co.uk
12 Aug 2002