Let d be the distance between the circumcenter and the centroid of a triangle. Let R be its circumradius and r the radius of its inscribed circle. Show that d2 ≤ R(R - 2r).
Solution
Use vectors. Take the origin as the circumcenter O. Let A be the vector OA, B be the vector OB and C be the vector OC. Then the vector OG = (A + B + C)/3. Hence 9 OG2 = A2 + B2 + C2 + 2(A.B + B.C + C.A). Now A2 = B2 = C2 = R2. We have A.B = 2R2 cos 2C = 2R2 - AB2 = 2R2 - c2 (with the usual notation AB = c, BC = a, CA = b). Hence 9 OG2 = 9 R2 - (a2 + b2 + c2). By the arithmetic geometric mean inequality we have (a2 + b2 + c2) ≥ 3 (abc)2/3. Hence R2 - OG2 ≥ (abc)2/3/3.
Now the area of ABC = area IAB + area IBC + area ICA = rc/2 + ra/2 + rb/2, where I is the incenter. It is also 1/2 ab sin C. But considering the triangle ABO, we have c = 2R sin C and hence 2 area ABC = r(a + b + c) = abc/(2R). So 2rR = abc/(a + b + c) ≤ 1/3 (abc)2/3. Hence R2 - OG2 ≥ 2rR.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002