OC632 from Crux Mathematicorum Vol. 49, No. 5

OC632 (from Crux Mathematicorum Vol. 49, No. 5)

Triangle \(ABC\) is inscribed in a circle \(C(O, 1)\). Let \(G_1, G_2, G_3\) be the centroids of triangles \(OBC\), \(OAC\) and \(OAB\), respectively. Prove that triangle \(ABC\) is equilateral if and only if \(AG_1 + BG_2 + CG_3 = 4\).

Proof

Part I: Equilateral Triangle implies \(AG_1 + BG_2 + CG_3 = 4\).

Assume that triangle \(ABC\) is equilateral, with \(O\) as its center. In this special case, \(A, O, G_1\) are colinear, and we have \(AG_1 = AO + OG_1 = 1 + \frac{1}{2}\frac{2}{3} = \frac{4}{3}\). By symmetry, \(BG_2 = CG_3 = AG_1 = \frac{4}{3}\). Therefore \(AG_1 + BG_2 + CG_3 = 4\) if \(ABC\) is equilateral.

Part II: \(AG_1 + BG_2 + CG_3 = 4\) implies Equilateral Triangle

Extend line segment \(OG_1\) to intersect \(BC\) at \(P\). Extend line segment \(OG_2\) to intersect \(AC\) at \(Q\). Extend line segment \(OG_3\) to intersect \(AB\) at \(R\). From the given conditions, we have \(OP \perp BC\), \(OQ \perp AC\), and \(OR \perp AB\). We also have \(3OG_1 = 2OP\), \(3OG_2 = 2OQ\), and \(3OG_3 = 2OR\) since \(G_1, G_2, G_3\) are centroids.

Applying the Erdos-Mordell inequality, we have \[OA + OB + OC \ge 2(OP+OQ+OR) = 3(OG_1 + OG_2 + OG_3).\] By the triangle inequality, we have \[AG_1 + BG_2 + CG_3 \le (OA + OG_1) + (OB + OG_2) + (OC + OG_3) \le \frac{4}{3}(OA + OB + OC) = 4\] in our case. Apparently the equality holds when \(ABC\) is equilateral and \(O\) is the center of it.

Thus we have completed the proof that triangle \(ABC\) is equilateral if and only if \(AG_1 + BG_2 + CG_3 = 4\).