Completion of Dr. John Stockton’s Lagrange analysis by Michael Behrend (May 2012)

The case is of three masses A,B,C at the vertices of a triangle with sides a,b,c. It is required to show that the triangle is either collinear or equilateral. The analysis gets as far as an equation labelled (52a):


with equations (52b) and (52c) obtained by cyclic permutation.

To continue: The LH sides are all equal; hence so are the RH sides. From each RH side subtract the cyclically symmetric expression


to get three equal expressions


Negating and rearranging each expression gives


Put A1=Aa(b3c3),B1=Bb(c3a3),C1=Cc(a3b3)  to get


Next put A2=B1C1,B2=C1A1,C2=A1B1 to get


These equations can be solved for A2,B2,C2 up to multiplication by a constant. The result is


where k is arbitrary (we use the fact that a2+b2+c2 etc. are not all 0).

But A2+B2+C2=0, so that


The polynomial in a,b,c is 12 where Δ is the area of the triangle (Heron’s formula). Hence either Δ=0, i.e. the triangle is collinear, or k=0. If k=0 then A2=B2=C2=0, hence A1=B1=C1, i.e. by definition


Here the coefficients Aa,Bb,Cc are all positive; hence the terms b3c3 etc. are either all 0 or all of the same sign. The latter is ruled out since the terms sum to 0. Hence in the case k=0 the triangle is equilateral.