IMO 2001 SL G3

Origin: GBR | Category: Geometry

Problem

Let ABC be a triangle with centroid G. Determine, with proof, the position of the point P in the plane of ABC such that AP \cdot AG + BP \cdot BG + CP \cdot CG is a minimum, and express this minimum value in terms of the side lengths of ABC.

Solution

Let us investigate a more general problem, in which G is any point of the plane such that AG, BG, CG are sides of a triangle. Let F be the point in the plane such that BC : CF : FB = AG : BG : CG and F, A lie on different sides of BC. Then by Ptolemy’s inequality, on BPCF we have AG \cdot AP + BG \cdot BP + CG \cdot CP = AG \cdot AP + AG BC (CF \cdot BP + BF \cdot CP) \geqAG \cdot AP + AG BC BC \cdot PF. Hence AG \cdot AP + BG \cdot BP + CG \cdot CP \geqAG \cdot AF, (1) where equality holds if and only if P lies on the segment AF and on the circle BCF. Now we return to the case of G the centroid of \triangleABC. We claim that F is then the point <G in which the line AG meets again the circumcircle of \triangleBGC. Indeed, if M is the midpoint of AB, by the law of sines we have BC C <G

sin \angleB <GC sin \angleCB <G = sin \angleBGM sin \angleAGM = AG BG, and similarly BC B <G = AG CG. Thus (1) im- plies A B C P G F AG \cdot AP + BG \cdot BP + CG \cdot CP \geqAG \cdot A <G. It is easily seen from the above considerations that equality holds if and only if P \equivG, and then the (minimum) value of AG \cdot AP + BG \cdot BP + CG \cdot CP equals

AG2 + BG2 + CG2 = a2 + b2 + c2 . Second solution. Notice that AG \cdot AP \geq−\to AG \cdot −\to AP = −\to AG \cdot (−\to AG + −−\to PG). Summing this inequality with analogous inequalities for BG \cdot BP and CG \cdot CP gives us AG \cdot AP + BG \cdot BP + CG \cdot CP \geqAG2 + BG2 + CG2 + (−\to AG + −−\to BG + −−\to CG) \cdot −−\to PG = AG2 + BG2 + CG2 = a2+b2+c2 . Equality holds if and only if P \equivQ.