# 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):

2Ka3b3c3/G=2(C+B)c3b3+Aa[c3(a2+b2c2)+b3(c2+a2b2)]

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

Aa3b3+Aa3c3+Bb3c3+Bb3a3+Cc3a3+Cc3b3

to get three equal expressions

(B+C)b3c3+Aac3(b2c2)+Aab3(c2b2)Bb3a3Cc3a3=...

Negating and rearranging each expression gives

Aa(b2c2)(b3c3)+Bb3(a3c3)+Cc3(a3b3)=...

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

A1(b2c2)B1b2+C1c2=B1(c2a2)C1c2+A1a2=C1(a2b2)A1a2+B1b2.

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

b2C2+c2B2=c2A2+a2C2=a2B2+b2A2.

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

A2=ka2(a2+b2+c2),B2=kb2(a2b2+c2),C2=kc2(a2+b2c2)

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

But A2+B2+C2=0, so that

k(2b2c2+2c2a2+2a2b2a4b4c4)=0.

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

Aa(b3c3)=Bb(c3a3)=Cc(a3b3).

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.