A student asks,
- prove for triangle ABC with centroid G, and any point M,
- |MA|^2 + |MB|^2 + |MC|^2 = 3|MG|^2 + (1/3)(|AB|^2 + |BC|^2 + |CA|^2)
- Let M=(m1,m2,m3),
- let A=(a1,a2,a3),
- let B=(b1,b2,b3),
- let C=(c1,c2,c3), and
- let G=(g1,g2,g3).
Notice in the statement to be proved that all lengths (distances) are squared, and gi depends only on ai, bi, and ci, so the statement is true of each dimension independently. Then it's a matter of proving the identity
- (m-a)2 + (m-b)2 + (m-c)2 = 3(m-g)2 + ((a-b)2 + (b-c)2 + (c-a)2)/3
To do this, start by expressing g in terms of the a, b, and c, expand both sides, and collect terms.
- 3(m-g)2 + ((a-b)2 + (b-c)2 + (c-a)2)/3 =
- ((3m-a-b-c)2 + (a-b)2 + (b-c)2 + (c-a)2)/3 =
- (9m2+a2+b2+c2-6ma-6mb-6mc+2bc+2ca+2ab + a2-2ab+b2 + b2-2bc+c2 + c2-2ca+a2)/3 =
- (9m2+3a2+3b2+3c2-6ma-6mb-6mc)/3 =
- (m-a)2 + (m-b)2 + (m-c)2
Brute force, yes, but it works.