Sum of Squares of Distances to Vertices

Elsewhere we established the identity

where H is the orthocenter, O the circumcenter and R the circumradius of triangle ΔABC with side lengths a, b, c.

Another way to obtain this result is via another identity that holds for any point M in the plane of the triangle

where G is the centroid of ΔABC.

The derivation is as follows. Let M be O, the circumcenter of the triangle. Then (1) implies that

AG, BG, and CG are two thirds of the medians m_a, m_b, m_c, respectively. The medians are expressed in terms of the side lengths as follows

(*)

2 ma²

= b² + c² - a²/2, etc.,

meaning, for example, that

and similarly for BG and CG. Substituting these into (2) gives

Three centers O, G, and H lie on the Euler line such that OH = 3OG, reducing (3) to the desired identity

Now the task is to prove (1).

Solution

For the proof, let D be the midpoint of AG and A' the midpoint of BC. Then DA' = AG. Apply (*) to the medians in triangles MBC, MDA', MAG:

Multiplying the second of these by 2 and adding, we arrive, after simplification, at

Considering the medians BB' and CC' leads to analogous identities. Adding the three yields

However, from (*),

which, combined with (5), gives the desired result

References

1. N. Altshiller-Court, College Geometry, Dover, 2007, pp. 70-71.

Copyright © 1996-2009 Alexander Bogomolny