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

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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

The sum of squares of the distances from a point to the vertices of a triangle is constant for points lying on a circle with center G. In particular, for an equilateral triangle, the sum of squares of the distances to the vertices for points on the incircle is constant and the same holds for points on the circumcircle, the 9-point circle, Adams' circle, and the rest. This remark complements a result by Leo Moser.

References

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

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Copyright © 1996-2012 Alexander Bogomolny