42nd IMO 2001 shortlist Problem G3 ABC is a triangle with centroid G and side lengths a = BC, b = CA, c = AB. Find the point P in the plane which minimises AP.AG + BP.BG + CP.CG and find the minimum in terms of a, b, c.
Applying the sine rule to triangle BGL, we have BG/sin BLG = BL/sin BGK. Similarly applying the sine rule to triangle CGL, we have CG/sin CLG = CL/sin CGK. But L is the midpoint of BC and sin BLG = sin CLG, so BG/CG = sin CGK/sin BGK. We have BK = 2 R sin BGK, where R is the radius of the circle (because if O is its center then ∠BOK = 2 ∠BGK). Similarly, CK = 2 R sin CGK, so CK/BK = BG/CG, and hence BG/CK = CG/BK. Similarly, AG/BG = sin BGN/sin AGN = sin BGN/sin CGK (opposite angles). Also BC = 2 R sin BKC = 2 R sin BGN (since BGCK is cyclic, so angle BKC = angle BGN). Hence BC/CK = sin BGN/sin CGK = AG/BG, so BG/CK = AG/BC and hence BC : CK : BK = AG : BG : CG. We now use Ptolemy's theorem on the quadrilateral PBKC: PK.BC <= BP.CK + CP.BK. Hence PK.AG <= BP.BG + CP.CG. Hence (AP + PK)AG <= AP.AG + BP.BG + CP.CG and so finally AK.AG <= AP.AG + BP.BG + CP.CG with equality iff (1) P lies on the circle between B and C (for equality in Ptolemy) and (2) P lies on AK (for equality in the triangle inequality). Thus the unique minimum is when P = G. Using the cosine formula, we have AB 2 = AL 2 + BL 2 - 2 BL.AL cos ALB, AC 2 = AL 2 + CL 2 - 2 CL.AL cos ALC. Adding, AB 2 + AC 2 = 2 AL 2 + BC 2 /2, so AG 2 =(4/9) AL 2 = (2 AB 2 + 2 AC 2 - BC 2 )/9. Hence AG 2 + BG 2 + CG 2 = (AB 2 + BC 2 + CA 2 )/3. 42nd IMO shortlist 2001 © John Scholes jscholes@kalva.demon.co.uk 14 Oct 2002